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1
NH
NHNH
NH
cyclam
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (31th Mar 2012)
Theoretical tasks and solutions
TASK 1
Complexes with cyclam
14811-tetraazacyclotetradecane (cyclam) is a macrocyclic ligand with four
coordination sites at nitrogen atoms
This ligand can form complexes with heavy metal cations (with 11 stoichiometry)
These complexes are applicable as catalysts they can be also applied in medicine as
they affect important processes of biological significance
These complexes are stable in pH range limited by disadvantageous effect of
nitrogen atoms protonation (only neutral molecule exhibits complexing properties) and possibility
of heavy metal hydroxides deposition
This task concerns general conditions of stability of complexes with cyclam in solutions of
constant pH = 7 Basing on obtained data it would be possible to predict stability of complexes of
specific metal cations By solving this task please assume predomination of some protonated forms
of the ligand
Problems
a Calculate concentration of the neutral ligand in solution with total concentration of cyclam
(protonated and non-protonated forms) equal to 001 M (pH = 7)
b Determine the ratio of stability constant and so-called bdquoconditionalrdquo stability constant rsquo
where rsquo = [ML]([M]cL) [ML] is concentration of the complex [M] is concentration of free
metal ions (in symbols M ML the charge is omitted) cL is total concentration of cyclam
(protonated and non-protonated forms) not bound with metal ions
c Derive equation and calculate the minimal value of stability constant for which in solution of pH =
7 at least 999 of metal ions is complexed for cL = 001 M
d Heavy metal ions of charge 2+ can form hydroxide deposits What is the minimal value of
stability constant to avoid hydroxide deposition in solution containing cyclam and metal ions
(at pH = 7) Express the stability constant as a function of solubility product of the hydroxide
Ks0 and other selected equilibrium constants given in the task and concentrations ML and cL
e Are the conditions from sections (c) and (d) fulfilled for cations Cu2+
i Ni2+
for [ML] = 001 M
and cL = 001 M For the case of the complex with Ni(II) ions calculate the concentration of non-
complexed nickel ions How does this concentration change after 2-fold dilution of the solution
Dissociation constants of acids Ka1 (for H4L4+
) = 410
3 Ka2 (for H3L
3+) = 3
10
2
Ka3 (for H2L2+
) = 310
11 Ka4 (dla HL
+) = 3
10
12 (L is the form presented in the figure)
Stability constants of complexes with cyclam for Cu2+
210
27 for Ni
2+ 2
10
22
Solubility products Ks0 for Cu(OH)2 310
19 for Ni(OH)2 2
10
15
wwwShimiPediair
2
TASK 2
Phosphorus oxide ndash acidic or alkaline
Compound A is formed in the reaction of white phosphorus with oxygen carried out at lowered
pressure and at a temperature below 50 degC Reactive compound A can be isolated via vacuum
distillation from the products mixture in the form of white crystalline substance The compound
melts at ca 24 degC and can be easily dissolved in a series of organic solvents eg in benzene Only
one signal was found in the 31
P NMR spectrum of its benzene solution It was also observed that the
solution obtained by dissolution of 0185 g of compound A in 1000 g of benzene freezes at a
temperature that is by 043deg lower than melting point of pure benzene The cryoscopic constant Et
for benzene equals 512 K middot kgmol
Compound A reacts very easily with water to form a solution that contains acid B which forms a
precipitate of an anhydrous barium salt upon addition of the excess of Ba(OH)2 solution
A sample of compound A with a mass m1 = 126 g was reacted with a substantial excess of nickel
carbonyl Ni(CO)4 The reaction was proceeding at room temperature and a colourless gas was being
evolved It was found that after compound A had been consumed the mass of the reaction mixture
decreased by 0642 g 45 g of pure compound C was obtained after removal of the unreacted
carbonyl The 31
P NMR spectrum of this compound comprised only one signal as well
Compound A in the solid state (308 g) was also reacted with gaseous diborane (B2H6) at room
temperature After 24 hours products of the reaction were dissolved in n-pentane and left for
crystallisation 267 g of pure crystalline compound D were collected
It was found that this compound reacts violently with water and in the course of reaction colourless
gas is liberated and the resulting solution contains acids B and E In a carefully controlled reaction
0147 g of compound D were used and ca 80 mL (equivalent for 0 degC and atmospheric pressure) of
gas were given off Compound D reacts easily with nickel carbonyl too and using an excess of
carbonyl leads to the previously described compound C
Problems
a Determine compound A formula and confirm it with appropriate calculations Write the
equation of white phosphorus reaction with oxygen
b Draw Lewis electron structure of compound A molecule taking into account all of the lone
valence electron pairs Justify your answer
c Write the chemical equation in molecular form (reagent form) of compound A reaction with
water
d Write the chemical equation in molecular form of acid B reaction with barium hydroxide Draw
the molecular structure of anion present in the structure of barium salt Justify your answer
e Determine the formula of compound C and write the equation of its formation reaction Justify
your answer and confirm it with calculations
f Draw and describe the structure of nickel coordination sphere in compound C Justify your answer
g Determine the formulae of compounds D and E Write the equation of compound D reaction
with water Justify your answer and confirm it with calculations
h Basing on the appropriate acids and bases definition determine the chemical character of
reagents in the formation reaction of compound D and calculate the reaction yield
i Draw and describe the boron coordination sphere in compound D Justify your answer
Use the following values of molar masses in your calculations (gmol)
B ndash 1081 C ndash 1201 H ndash 1008 Ni ndash 5869 O ndash 1600 P ndash 3097
and the molar volume of gases at 0 degC and under atmospheric pressure Vm = 2241middot10minus3
m3mol wwwShimiPediair
3
Fig 1 Molecular
formula of a 12C4
crown ether
TASK 3
Determination of stoichiometry and stability of complexes with an NMR method
Application of approximations in limiting conditions
One of the problems in the 2nd
stage of the 58 Chemistry Olympiad was
related to the crown ethers ability to complex metal ions Now we
present the results of 1H NMR study on formation reaction of a
12-crown-4 ether (labelled as 12C4) with Na+ cation introduced into the
system in a form of sodium thiocyanate NaSCN
Three independent experiments have been carried out the first and the
second at temperature 23degC and the third one at temperature ndash50degC In
the two first experiments the spectra were recorded at varying NaSCN
concentration and a single averaged signal coming from both the
complexed molecules and the free ether was monitored The signal chemical shift δobs was
dependent on the ratio of sodium ions [Na+]0 and crown ether [E]0 concentrations The appearance
of the averaged signal proves that the complexation reaction is a so called fast exchange reaction
Let us remind that for such reactions the observed chemical shift δobs is an average of characteristic
chemical shifts of the ligand δE and the complexes δEnM multiplied by corresponding
concentration ratios of ligands unbound and bound in complexes to the total concentration of
ligands in the system By writing the general complex formula as EnM (where E means ether and
M means metal) the relationship can be presented as follows
E3M
0
3E2M
0
2EM
0
E
0
obs]E[
]ME[3
]E[
]ME[2
]E[
]EM[
]E[
]E[
(1)
The total concentration of ligands in the system [E]0 = [E] + [EM] + 2[E2M] + 3[E3M]+
Experiment 1 The characteristic chemical shift δE = 2761 ppm has been determined from the
NMR spectrum of the pure crown ether Then small amounts of solid sodium thiocyanate were
added to a 12C4 solution (in deuterated methanol CD3OD) with concentration [E]0 =
0219 moldm3 and the chemical shift δobs was recorded as a function of total concentrations
[NaSCN][12C4] The measurement results are shown in Fig 2 and additionally the data for points
within the linearity range ie at [NaSCN][12C4] le 03 are given in Table 1
The experimental points form a
curve showing two characteristic
bends which prove that two
complexes k1 and k2 with different
stoichiometry and different overall
complex formation constants for
metal ion and for ether βn are
present in the system Based on the
plot given in Fig 2 one can
determine the stoichiometry of the
two complexes being formed and
remaining in equilibrium
Fig 2
complex 1 complex 2
wwwShimiPediair
4
Table 1 Results of experiment 1 [E]0 = 0219 moldm3
[NaSCN][12C4] δobs ppm
0
0045
0096
0150
0240
0300
2761
2846
2937
3025
3190
3282
Experiment 2 In all measurements the initial concentration of the crown ether was kept constant at
[E]0 = 0050 moldm3 while the thiocyanate concentration varied but was much higher than that of
the emerging complex The sodium ion concentrations c and the corresponding resultant chemical
shifts δobs are given in Table 2
Table 2 Results of experiment 2 [E]0 = 0050 moldm3
Total Na+ ion concentration
c (moldm3)
Observed chemical shift
δobs ppm
0204
0271
0359
0495
0646
0768
4437
4558
4660
4757
4821
4856
Experiment 3 The spectrum of the pure 12C4 ether was recorded again at temperature ndash50degC then
a small amount of NaSCN was added and the spectrum was recorded again The spectrum showed
distinctly resolved ether and complex peaks (the fast exchange reaction does not occur here) It
turned out that the chemical shift of the pure ether virtually did not change with decreasing
temperature One can therefore assume that the chemical shift of the k2 complex that is stable in the
presence of a large excess of ether is the same at both 23degC and at ndash50degC and is δk2 = 4575 ppm
Using the results of the three experiments described above determine the stoichiometry of complexes
that are formed in the system calculate their stability constants and characteristic chemical shift of the
k1 complex by following the directions given below
In the solution one should assume the following symbols for the major quantities occurring in the problem
[E]0 ndash total concentration of the crown ether constant in each experiment
[Na+] = c ndash total concentration of sodium ions varying during titration
δE ndash chemical shift of the pure 12C4 crown ether
δk1 δk2 ndash chemical shifts of two complexes
ck1 ck2 ndash concentrations of EnM complexes respectively in equilibrium
To simplify the notation of equations the following symbols can be also introduced
Eobsobs Ek1k1 and Ek2k2
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5
Problems
a Based on the plot in Fig 2 determine the stoichiometric coefficients of the resulting complexes Justify
the answer with a short comment
b Give chemical equations for reactions taking place in the system and the formulae for stability
constants of the complexes by writing them down in a form containing the initial ether concentration
[E]0 and the concentration of sodium ions c
c Using appropriate equations determine complex concentrations ck1 and ck2 by introducing the
simplifications proposed in Note 1
d Give an equation describing the dependence of the averaged chemical shift δobs on corresponding
chemical shifts of the ether δE (data) and the two complexes δk1 (this should be treated as an unknown)
and δk2 (data) Present the equation as a dependence of corresponding differences of chemical shifts that
is obs k1 and k2 Write the two equations in two forms corresponding to the limiting cases by
introducing corresponding complex concentrations determined in Direction c
e Calculate β2 from the simplified equation from Direction d for the limiting case [E]0 gtgt c by
determining the slope of the straight line (δobsndashδE) = f(c) from two selected points from Table 1 and using
known value of δk2 (see Note 2)
f Solve the equation from Direction d in the high Na+ concentration limit (c) by converting the equation
so as to get a linear dependence of 1(δobs ndash δE) on 1c Calculate β1 and δk1 constants (see Note 3)
Notes
1 The equations for equilibrium constants can be simplified if the experiment conditions can
be considered as the limiting ones
In experiment 1 [E]0 gtgt c so the equilibrium is strongly shifted towards the k2 complex and
we can assume that [E]0 gtgt ck2 gtgt ck1 asymp 0
In experiment 2 [E]0 ltlt c so the equilibrium is strongly shifted towards the k1 complex and
we assume that c gtgt ck1 gtgt ck2 asymp 0
Corresponding approximations are best introduced by determining ck1 and ck2 as functions of
complex stability constants β1 and β2
2 Fig 3 shows linear dependence of (δobs ndash δE) on c in the limit [E]0 gtgt c The plot should facilitate the
selection of points of which the coordinates allow to calculate the slope of the straight line
Fig 3
Experiment 1
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6
3 To solve the problem in the high Na+ concentration limit ([E]0 ltlt c) it is necessary to
ldquolineariserdquo the dependence of the corresponding difference of chemical shifts (δobsndashδE) on
varying [Na+] = c concentration The dependence of 1(δobsndashδE) on 1c should be linear
(1(δobsndashδE) = a(1c) + b) in the limit of large values of the ratio c[E]0 ie in the conditions of
experiment 2 Fig 4 shows the relationship plotted using the data from Table 2
Fig 4
TASK 4
Biologically active indole derivatives
Tryptamines are derivatives of indole (a heterocyclic compound consisting of two condensed rings
a benzene ring and a pyrrole ring) abundantly present in living organisms To tryptamines belong
inter alia neurotransmiters as important as melatonin or serotonin Due to that also other
tryptamine derivatives are not neutral to human organism and may have medicinal use or exert
hallucinogenic effects
The phosphorylated tryptamine derivative F initially isolated from plant sources has been obtained
through the reaction sequence depicted in Scheme 1 The starting material for the synthesis was 4-
hydroxyindole It is known that the most reactive position of indole is the C-3 within the pyrrole
ring which easily undergoes aromatic electrophilic substitution but may also be involved in
reactions typical for enamines
First the phenol group of 4-hydroxyindole was protected by reacting it with benzyl chloride in the
presence of sodium methanolate This led to compound A which was then subjected to Friedel-
Crafts acylation with oxalyl chloride leading to compound B Subsequently B without isolation
was reacted with dimethylamine which resulted in compound C In the next step C was reduced to
D (molecular formula C19H22N2O) by means of LiAlH4 and then catalytically hydrogenated to yield
compound E and toluene as a by-product In the last step E was converted into F by applying
following reagents 1 n-buthyllithium 2 tetrabenzyl pyrophosphate (TBPP) 3 catalytic
hydrogenation (PdC H2)
A homologue of compound E known as 4-hydroxygramine and having molecular formula of
C11H14N2O may be obtained in a multi-component reaction (Mannich reaction) by reacting
compound A with formaldehyde and dimethylamine followed by a catalytic hydrogenation
Exploiting a similar multi-component reaction but from different starting materials one can obtain
compound X as a mixture of several stereoisomers (Scheme 2 depicts only one of them)
Experiment 2
wwwShimiPediair
7
O
ClO
Cl NH
A
NH
OH
Cl
OP
OP
O
OO
OO
C19H22N2O
TBPP =Bn
Bn
Bn
Bn
Scheme 1
MeOHMeO-
F
H2 Pd-C
1 n-BuLi2TBPP3 H2 Pd-C
[ B ]C
DE
LiAlH4
CH3
O
CH3
N O
O
NH
OCH
3Chiral
Scheme 2
342
1
X
Problems
a Draw structural formulas for compounds A-F and 4-hydroxygramine
b Draw structural formulas of the starting materials for the synthesis of compound X
c Compound X could be obtained from the same starting materials but in a stereochemically pure
form (as depicted in Scheme 2) Choose which conditions should be applied to achieve this
I the reaction temperature should be increased
II the reagents should be mixed in a certain order
III the reaction should be carried out in the presence of L-proline
IV the reaction mixture should be stirred always in the same direction
d Determine the absolute configuration of all stereogenic centers in compound X
e Explain the reactivity of the C-3 position of indole by means of appropriate mesomeric structures
TASK 5
The Diels-Alder Reaction
Cycloaddition reactions including Diels-Alder reaction in most cases are stereospecific what
means that geometry of substrates (eg dienophile or diene) determine the structure of the formed
product
X
Y Y
X+
General Scheme of the Diels-Alder Reaction
I The carboxylic acid with molecular formula C4H4O4 as well as its dimethyl esters (A B and C)
exist as two geometric isomers and one regioisomer and all of them undergo the Diels-Alder
reaction The reaction of geometric isomers A and B mixture with diene D composed entirely of
carbon and hydrogen atoms provides diastereomeric products E1 E2 and F Ozonolysis of E and F
mixture with zinc dust form product G (diastereomeric mixture) and subsequent reduction with
LiAlH4 excess leads to the mixture of diastereomers of polihydroxylic alcohol H
wwwShimiPediair
8
OHOH
OHOHLiAlH
4A i B
D E1 + E2 i F(diastereomers)
1) O3
2) ZnG
Hmixture of three diastereomers
1) O3
2) Zn
ID
J
mixture of three diastereomers
Compounds A and B after ozonolysis with zinc dust form the same product I which also undergo
cycloaddition reaction with diene D The reaction leads to the cyclic ether J The molecular mass of
J is equal to the sum of mass of reactants D and I
II Stereochemistry in Diels-Alder reaction Isomer A in the reaction with diene D forms the
mixture of two achiral diastereomers E1 and E2 The reduction of diastereomeric mixture of E1 and
E2 with hydrogen on palladium catalyst leads to only one product K
The Diels-Alder reaction of isomer B with diene D provides exclusively racemic mixture of
cycloadduct F The product F after ozonolysis with zinc dust and subsequent reduction with LiAlH4
excess leads to one diastereomer of polihydroxylic alcohol H (as a racemic mixture)
Problems
I
a Draw the structure of geometric isomer A or B and structure of regioisomer C
b Draw the structure of diene D
c Determine general structure of Diels-Alder product EF (no stereochemistry required)
d Draw the general structure of ozonolysis product G (no stereochemistry required)
e Draw the structure of compound I and general structure of J (no stereochemistry required)
II
a Determine unambiguously geometry of isomers A and B
b Draw stereochemical structures of compounds E (E1 and E2) and their reduction product K
c Draw the stereochemical structure of compound F and stereochemical structure of
corresponding diastereomer of polihydroxylic alcohol H
d Draw all possible stereoisomers of the product J Determine enantiomers and diastereomers
SOLUTIONS
SOLUTION OF TASK 1
a The total concentration of cyclam cL = [L] + [HL+] + [H2L
2+] + [H3L
3+] + [H4L
4+] Using acidic
dissociation constants after rearrangement one can write
cL = [L]1 + [H+]Ka4 + [H
+]2(Ka4 Ka3) + [H
+]3(Ka4 Ka3 Ka2) + [H
+]
4(Ka4 Ka3 Ka2 Ka1)
Taking into account the values of dissociation constants it can be assumed that at pH = 7 the
form H2L2+
predominates in the solution ie the above equation can be simplified to the form
cL = [L][H+]
2(Ka3 Ka4) thus [L] = cL Ka3 Ka4 [H
+]
2 After introducing numerical results for
cL = 001 M and [H+] = 10
7 M the obtained [L] = 9
10
11 M
wwwShimiPediair
9
OP
O
PO
P
O O
O
P
b Because stability constant = [ML]([M][L]) the ratio rsquo = cL[L] Basing
on previous data cL[L] = 001 910
11 = 11
10
8
c Because [ML][M] = 999 01 rsquo = [ML]([M]cL) = 999 (01 001) = 1
10
5
Thus = rsquo11
10
8 = 11
10
13
d = [ML]([M][L]) = [ML][H+]2([M]cL Ka3 Ka4) Assuming that a deposit of
hydroxide does not precipitate Ks0 = [M][OH]
2 the above equation can be written in the form
= [ML][H+]
2[OH
]
2(cL Ks0 Ka3 Ka4) = KW
2 ([ML]cL) ( Ks0 Ka3 Ka4)
e For [ML] = cL = 001 M using appropriate constants the minimal values of fulfilling the
conditions from (d) are 410
12 for the Cu
2+ complex and 6
10
8 for the Ni
2+ complex Comparison
of these results with experimental data shows that Cu2+
and Ni2+
complexes obey the conditions
from (c) and (d)
Because for Ni2+
- cyclam complex rsquo = [ML]([M]cL) for [ML] = cL [M] = 1rsquo is obtained
For the Ni2+
complex rsquo = 210
22 11
10
8 = 2
10
14 Thus [M] = [Ni
2+] = 1 (2
10
14) ie
[Ni2+
] = 5
10
15 M
2-fold dilution results in the same value of the [ML]cL ratio therefore Ni2+
ions concentration does
not change as well
SOLUTION OF TASK 2
a White phosphorus is built of P4 molecules Depending on the reaction conditions it may form
P4On (n = 6 7 8 9 or 10) oxides in the reaction with oxygen whose structures stem from P4
molecules structure Freezing point depression of benzene allows one to calculate the molality of
oxide A solution (cmA) and consequently its molar mass
0840125
430A
t
t
m
E
Tc molkg czyli 220
08400100
1850A
mAb
A
cm
mM gmol
This value corresponds to P4O6 molar mass (21988 gmol) Thus compound A is
phosphorus(III) oxide P4O6 Phosphorus reaction with oxygen proceeds according to the
following equation
P4 + 3O2 P4O6
b The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule There is
one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs
This structure agrees with 31
P NMR spectrum indicating all phosphorus nuclei are chemically
equivalent
c P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation
P4O6 + 6H2O 4H3PO3
d Phosphoric(III) acid contains PndashH bond and is a diprotic acid of an average strength It forms
therefore two series of salts and sparsely soluble barium salt containing HPO32minus
anion will be
formed with the excess of barium hydroxide
H3PO3 + Ba(OH)2 3 + 2H2O
HPO32minus
anion exhibits tetrahedral structure (sp3 hybridisation of phosphorus) There are three
oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron wwwShimiPediair
10
Nickel coordination sphere
O
H
O
PO
O
PO
H
O
or
2 2
e Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base
in chemical reactions Therefore Lewis adduct is formed in the reaction with nickel carbonyl in
which phosphorus forms a chemical bond with nickel replacing one of the CO ligands Assuming
that the only gaseous product of the reaction is CO the molar ratio of the evolved CO to the used
P4O6 equals
0642 g 126 g
002292 000573 4 12801 gmol 21988 gmol
which indicates that four nickel atoms
were bound by a P4O6 molecule The reaction proceeds according to the following equation
4Ni(CO)4 + P4O6 P4O6middot4Ni(CO)3 + 4CO
The formula of compound C is P4O6middot4Ni(CO)3 (or P4O6[Ni(CO)3]4) The molar mass of
compound C equals 79076 gmol Such a reaction course is confirmed by the approximate molar
mass of compound C calculated from the reaction product mass
4 6
C
P O
45785 gmol
000573
CmM
n
f It follows from the fact that all of phosphorus nuclei are chemically
equivalent in the molecule of compound C that every phosphorus atom is
bound to one nickel atom and three oxygen atoms The coordination
centres in the form of nickel atoms satisfy the 18 electron rule (10 valence
electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons
coming from phosphorus) Therefore the structure of nickel coordination
sphere is tetrahedral and ligands bound to nickel atom are localised in
vertices of a slightly deformed tetrahedron
g Similarly as in the reaction with nickel carbonyl P4O6 phosphorus oxide serves as a Lewis base
in the reaction with B2H6 forming an adduct with acid (BH3) As a result of the reaction with
water the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and
H3BO3 boric acid (acid E) The liberated gas is molecular hydrogen and its quantity depends on
the amount of reacted diborane
2
2 6
H 3
B H 3
800595 10
6 2241 10 6m
Vn
V
mole which correspond to B2H6 mass equal to
2 6 2 6 2 6
3
B H B H B H 0595 10 2767 00165m n M g The molar ratio of P4O6 to B2H6 in the Lewis
adduct equals 3(0147 00165) g 0595 10 0594 0595 11
21988 gmol
Only two phosphorus atoms are involved in bond formation with BH3 and the formula of
compound D is P4O6 middot 2BH3 (or P4O6 middot
B2H6) The hydrolysis reaction proceeds according to the
following equation
P4O6 middot 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2
P Ni
C
C
CO
OO
O
O
O
wwwShimiPediair
11
h The reaction between P4O6 and B2H6 leading to the formation of compound D
(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition
Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid
The reaction product compound D is a Lewis adduct The yield of compound D formation
reaction equals
4 6 4 6 4 6
D D
P O P O P O
267 24755100 100 100 770
308 21988
Dn m Mw
n m M
i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron
B P
O
O
OH
HH
or
B P
O
O
OH
HH
SOLUTION OF TASK 3
a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows
clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means
that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the
range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex
and in the range c[E]0 ltlt 05 towards the E2M complex
b The complex formation reactions occurring at the equilibrium are as follows
(k1) E + M EM and (k2) 2E + M E2M
The complex formation equilibrium constants k1 and k2 also referred to as the complex
stability constants can be written as
)()2]E([]T[]E[
]ET[β
k21kk21k0
k11
ccccc
c
(1)
)()2E]([]T[]E[
]TE[β
k2k1
2
k21k0
k2
2
22
ccccc
c
(2)
c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be
determined by introducing the simplifications suggested in Note 1 In equation (1) we assume
that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2
in the first term of the denominator and get
cc
c
)]E([β
k10
1k1
(3)
In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0
and in the second term of the denominator we can neglect subtraction of ck1 and we get
)(]E[
βk2
2
0
k22
cc
c
(4)
Then from equations (3) and (4) we determine
1
101k
β1
β]E[
c
cc (5)
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
wwwShimiPediair
14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
wwwShimiPediair
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
wwwShimiPediair
6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
2
TASK 2
Phosphorus oxide ndash acidic or alkaline
Compound A is formed in the reaction of white phosphorus with oxygen carried out at lowered
pressure and at a temperature below 50 degC Reactive compound A can be isolated via vacuum
distillation from the products mixture in the form of white crystalline substance The compound
melts at ca 24 degC and can be easily dissolved in a series of organic solvents eg in benzene Only
one signal was found in the 31
P NMR spectrum of its benzene solution It was also observed that the
solution obtained by dissolution of 0185 g of compound A in 1000 g of benzene freezes at a
temperature that is by 043deg lower than melting point of pure benzene The cryoscopic constant Et
for benzene equals 512 K middot kgmol
Compound A reacts very easily with water to form a solution that contains acid B which forms a
precipitate of an anhydrous barium salt upon addition of the excess of Ba(OH)2 solution
A sample of compound A with a mass m1 = 126 g was reacted with a substantial excess of nickel
carbonyl Ni(CO)4 The reaction was proceeding at room temperature and a colourless gas was being
evolved It was found that after compound A had been consumed the mass of the reaction mixture
decreased by 0642 g 45 g of pure compound C was obtained after removal of the unreacted
carbonyl The 31
P NMR spectrum of this compound comprised only one signal as well
Compound A in the solid state (308 g) was also reacted with gaseous diborane (B2H6) at room
temperature After 24 hours products of the reaction were dissolved in n-pentane and left for
crystallisation 267 g of pure crystalline compound D were collected
It was found that this compound reacts violently with water and in the course of reaction colourless
gas is liberated and the resulting solution contains acids B and E In a carefully controlled reaction
0147 g of compound D were used and ca 80 mL (equivalent for 0 degC and atmospheric pressure) of
gas were given off Compound D reacts easily with nickel carbonyl too and using an excess of
carbonyl leads to the previously described compound C
Problems
a Determine compound A formula and confirm it with appropriate calculations Write the
equation of white phosphorus reaction with oxygen
b Draw Lewis electron structure of compound A molecule taking into account all of the lone
valence electron pairs Justify your answer
c Write the chemical equation in molecular form (reagent form) of compound A reaction with
water
d Write the chemical equation in molecular form of acid B reaction with barium hydroxide Draw
the molecular structure of anion present in the structure of barium salt Justify your answer
e Determine the formula of compound C and write the equation of its formation reaction Justify
your answer and confirm it with calculations
f Draw and describe the structure of nickel coordination sphere in compound C Justify your answer
g Determine the formulae of compounds D and E Write the equation of compound D reaction
with water Justify your answer and confirm it with calculations
h Basing on the appropriate acids and bases definition determine the chemical character of
reagents in the formation reaction of compound D and calculate the reaction yield
i Draw and describe the boron coordination sphere in compound D Justify your answer
Use the following values of molar masses in your calculations (gmol)
B ndash 1081 C ndash 1201 H ndash 1008 Ni ndash 5869 O ndash 1600 P ndash 3097
and the molar volume of gases at 0 degC and under atmospheric pressure Vm = 2241middot10minus3
m3mol wwwShimiPediair
3
Fig 1 Molecular
formula of a 12C4
crown ether
TASK 3
Determination of stoichiometry and stability of complexes with an NMR method
Application of approximations in limiting conditions
One of the problems in the 2nd
stage of the 58 Chemistry Olympiad was
related to the crown ethers ability to complex metal ions Now we
present the results of 1H NMR study on formation reaction of a
12-crown-4 ether (labelled as 12C4) with Na+ cation introduced into the
system in a form of sodium thiocyanate NaSCN
Three independent experiments have been carried out the first and the
second at temperature 23degC and the third one at temperature ndash50degC In
the two first experiments the spectra were recorded at varying NaSCN
concentration and a single averaged signal coming from both the
complexed molecules and the free ether was monitored The signal chemical shift δobs was
dependent on the ratio of sodium ions [Na+]0 and crown ether [E]0 concentrations The appearance
of the averaged signal proves that the complexation reaction is a so called fast exchange reaction
Let us remind that for such reactions the observed chemical shift δobs is an average of characteristic
chemical shifts of the ligand δE and the complexes δEnM multiplied by corresponding
concentration ratios of ligands unbound and bound in complexes to the total concentration of
ligands in the system By writing the general complex formula as EnM (where E means ether and
M means metal) the relationship can be presented as follows
E3M
0
3E2M
0
2EM
0
E
0
obs]E[
]ME[3
]E[
]ME[2
]E[
]EM[
]E[
]E[
(1)
The total concentration of ligands in the system [E]0 = [E] + [EM] + 2[E2M] + 3[E3M]+
Experiment 1 The characteristic chemical shift δE = 2761 ppm has been determined from the
NMR spectrum of the pure crown ether Then small amounts of solid sodium thiocyanate were
added to a 12C4 solution (in deuterated methanol CD3OD) with concentration [E]0 =
0219 moldm3 and the chemical shift δobs was recorded as a function of total concentrations
[NaSCN][12C4] The measurement results are shown in Fig 2 and additionally the data for points
within the linearity range ie at [NaSCN][12C4] le 03 are given in Table 1
The experimental points form a
curve showing two characteristic
bends which prove that two
complexes k1 and k2 with different
stoichiometry and different overall
complex formation constants for
metal ion and for ether βn are
present in the system Based on the
plot given in Fig 2 one can
determine the stoichiometry of the
two complexes being formed and
remaining in equilibrium
Fig 2
complex 1 complex 2
wwwShimiPediair
4
Table 1 Results of experiment 1 [E]0 = 0219 moldm3
[NaSCN][12C4] δobs ppm
0
0045
0096
0150
0240
0300
2761
2846
2937
3025
3190
3282
Experiment 2 In all measurements the initial concentration of the crown ether was kept constant at
[E]0 = 0050 moldm3 while the thiocyanate concentration varied but was much higher than that of
the emerging complex The sodium ion concentrations c and the corresponding resultant chemical
shifts δobs are given in Table 2
Table 2 Results of experiment 2 [E]0 = 0050 moldm3
Total Na+ ion concentration
c (moldm3)
Observed chemical shift
δobs ppm
0204
0271
0359
0495
0646
0768
4437
4558
4660
4757
4821
4856
Experiment 3 The spectrum of the pure 12C4 ether was recorded again at temperature ndash50degC then
a small amount of NaSCN was added and the spectrum was recorded again The spectrum showed
distinctly resolved ether and complex peaks (the fast exchange reaction does not occur here) It
turned out that the chemical shift of the pure ether virtually did not change with decreasing
temperature One can therefore assume that the chemical shift of the k2 complex that is stable in the
presence of a large excess of ether is the same at both 23degC and at ndash50degC and is δk2 = 4575 ppm
Using the results of the three experiments described above determine the stoichiometry of complexes
that are formed in the system calculate their stability constants and characteristic chemical shift of the
k1 complex by following the directions given below
In the solution one should assume the following symbols for the major quantities occurring in the problem
[E]0 ndash total concentration of the crown ether constant in each experiment
[Na+] = c ndash total concentration of sodium ions varying during titration
δE ndash chemical shift of the pure 12C4 crown ether
δk1 δk2 ndash chemical shifts of two complexes
ck1 ck2 ndash concentrations of EnM complexes respectively in equilibrium
To simplify the notation of equations the following symbols can be also introduced
Eobsobs Ek1k1 and Ek2k2
wwwShimiPediair
5
Problems
a Based on the plot in Fig 2 determine the stoichiometric coefficients of the resulting complexes Justify
the answer with a short comment
b Give chemical equations for reactions taking place in the system and the formulae for stability
constants of the complexes by writing them down in a form containing the initial ether concentration
[E]0 and the concentration of sodium ions c
c Using appropriate equations determine complex concentrations ck1 and ck2 by introducing the
simplifications proposed in Note 1
d Give an equation describing the dependence of the averaged chemical shift δobs on corresponding
chemical shifts of the ether δE (data) and the two complexes δk1 (this should be treated as an unknown)
and δk2 (data) Present the equation as a dependence of corresponding differences of chemical shifts that
is obs k1 and k2 Write the two equations in two forms corresponding to the limiting cases by
introducing corresponding complex concentrations determined in Direction c
e Calculate β2 from the simplified equation from Direction d for the limiting case [E]0 gtgt c by
determining the slope of the straight line (δobsndashδE) = f(c) from two selected points from Table 1 and using
known value of δk2 (see Note 2)
f Solve the equation from Direction d in the high Na+ concentration limit (c) by converting the equation
so as to get a linear dependence of 1(δobs ndash δE) on 1c Calculate β1 and δk1 constants (see Note 3)
Notes
1 The equations for equilibrium constants can be simplified if the experiment conditions can
be considered as the limiting ones
In experiment 1 [E]0 gtgt c so the equilibrium is strongly shifted towards the k2 complex and
we can assume that [E]0 gtgt ck2 gtgt ck1 asymp 0
In experiment 2 [E]0 ltlt c so the equilibrium is strongly shifted towards the k1 complex and
we assume that c gtgt ck1 gtgt ck2 asymp 0
Corresponding approximations are best introduced by determining ck1 and ck2 as functions of
complex stability constants β1 and β2
2 Fig 3 shows linear dependence of (δobs ndash δE) on c in the limit [E]0 gtgt c The plot should facilitate the
selection of points of which the coordinates allow to calculate the slope of the straight line
Fig 3
Experiment 1
wwwShimiPediair
6
3 To solve the problem in the high Na+ concentration limit ([E]0 ltlt c) it is necessary to
ldquolineariserdquo the dependence of the corresponding difference of chemical shifts (δobsndashδE) on
varying [Na+] = c concentration The dependence of 1(δobsndashδE) on 1c should be linear
(1(δobsndashδE) = a(1c) + b) in the limit of large values of the ratio c[E]0 ie in the conditions of
experiment 2 Fig 4 shows the relationship plotted using the data from Table 2
Fig 4
TASK 4
Biologically active indole derivatives
Tryptamines are derivatives of indole (a heterocyclic compound consisting of two condensed rings
a benzene ring and a pyrrole ring) abundantly present in living organisms To tryptamines belong
inter alia neurotransmiters as important as melatonin or serotonin Due to that also other
tryptamine derivatives are not neutral to human organism and may have medicinal use or exert
hallucinogenic effects
The phosphorylated tryptamine derivative F initially isolated from plant sources has been obtained
through the reaction sequence depicted in Scheme 1 The starting material for the synthesis was 4-
hydroxyindole It is known that the most reactive position of indole is the C-3 within the pyrrole
ring which easily undergoes aromatic electrophilic substitution but may also be involved in
reactions typical for enamines
First the phenol group of 4-hydroxyindole was protected by reacting it with benzyl chloride in the
presence of sodium methanolate This led to compound A which was then subjected to Friedel-
Crafts acylation with oxalyl chloride leading to compound B Subsequently B without isolation
was reacted with dimethylamine which resulted in compound C In the next step C was reduced to
D (molecular formula C19H22N2O) by means of LiAlH4 and then catalytically hydrogenated to yield
compound E and toluene as a by-product In the last step E was converted into F by applying
following reagents 1 n-buthyllithium 2 tetrabenzyl pyrophosphate (TBPP) 3 catalytic
hydrogenation (PdC H2)
A homologue of compound E known as 4-hydroxygramine and having molecular formula of
C11H14N2O may be obtained in a multi-component reaction (Mannich reaction) by reacting
compound A with formaldehyde and dimethylamine followed by a catalytic hydrogenation
Exploiting a similar multi-component reaction but from different starting materials one can obtain
compound X as a mixture of several stereoisomers (Scheme 2 depicts only one of them)
Experiment 2
wwwShimiPediair
7
O
ClO
Cl NH
A
NH
OH
Cl
OP
OP
O
OO
OO
C19H22N2O
TBPP =Bn
Bn
Bn
Bn
Scheme 1
MeOHMeO-
F
H2 Pd-C
1 n-BuLi2TBPP3 H2 Pd-C
[ B ]C
DE
LiAlH4
CH3
O
CH3
N O
O
NH
OCH
3Chiral
Scheme 2
342
1
X
Problems
a Draw structural formulas for compounds A-F and 4-hydroxygramine
b Draw structural formulas of the starting materials for the synthesis of compound X
c Compound X could be obtained from the same starting materials but in a stereochemically pure
form (as depicted in Scheme 2) Choose which conditions should be applied to achieve this
I the reaction temperature should be increased
II the reagents should be mixed in a certain order
III the reaction should be carried out in the presence of L-proline
IV the reaction mixture should be stirred always in the same direction
d Determine the absolute configuration of all stereogenic centers in compound X
e Explain the reactivity of the C-3 position of indole by means of appropriate mesomeric structures
TASK 5
The Diels-Alder Reaction
Cycloaddition reactions including Diels-Alder reaction in most cases are stereospecific what
means that geometry of substrates (eg dienophile or diene) determine the structure of the formed
product
X
Y Y
X+
General Scheme of the Diels-Alder Reaction
I The carboxylic acid with molecular formula C4H4O4 as well as its dimethyl esters (A B and C)
exist as two geometric isomers and one regioisomer and all of them undergo the Diels-Alder
reaction The reaction of geometric isomers A and B mixture with diene D composed entirely of
carbon and hydrogen atoms provides diastereomeric products E1 E2 and F Ozonolysis of E and F
mixture with zinc dust form product G (diastereomeric mixture) and subsequent reduction with
LiAlH4 excess leads to the mixture of diastereomers of polihydroxylic alcohol H
wwwShimiPediair
8
OHOH
OHOHLiAlH
4A i B
D E1 + E2 i F(diastereomers)
1) O3
2) ZnG
Hmixture of three diastereomers
1) O3
2) Zn
ID
J
mixture of three diastereomers
Compounds A and B after ozonolysis with zinc dust form the same product I which also undergo
cycloaddition reaction with diene D The reaction leads to the cyclic ether J The molecular mass of
J is equal to the sum of mass of reactants D and I
II Stereochemistry in Diels-Alder reaction Isomer A in the reaction with diene D forms the
mixture of two achiral diastereomers E1 and E2 The reduction of diastereomeric mixture of E1 and
E2 with hydrogen on palladium catalyst leads to only one product K
The Diels-Alder reaction of isomer B with diene D provides exclusively racemic mixture of
cycloadduct F The product F after ozonolysis with zinc dust and subsequent reduction with LiAlH4
excess leads to one diastereomer of polihydroxylic alcohol H (as a racemic mixture)
Problems
I
a Draw the structure of geometric isomer A or B and structure of regioisomer C
b Draw the structure of diene D
c Determine general structure of Diels-Alder product EF (no stereochemistry required)
d Draw the general structure of ozonolysis product G (no stereochemistry required)
e Draw the structure of compound I and general structure of J (no stereochemistry required)
II
a Determine unambiguously geometry of isomers A and B
b Draw stereochemical structures of compounds E (E1 and E2) and their reduction product K
c Draw the stereochemical structure of compound F and stereochemical structure of
corresponding diastereomer of polihydroxylic alcohol H
d Draw all possible stereoisomers of the product J Determine enantiomers and diastereomers
SOLUTIONS
SOLUTION OF TASK 1
a The total concentration of cyclam cL = [L] + [HL+] + [H2L
2+] + [H3L
3+] + [H4L
4+] Using acidic
dissociation constants after rearrangement one can write
cL = [L]1 + [H+]Ka4 + [H
+]2(Ka4 Ka3) + [H
+]3(Ka4 Ka3 Ka2) + [H
+]
4(Ka4 Ka3 Ka2 Ka1)
Taking into account the values of dissociation constants it can be assumed that at pH = 7 the
form H2L2+
predominates in the solution ie the above equation can be simplified to the form
cL = [L][H+]
2(Ka3 Ka4) thus [L] = cL Ka3 Ka4 [H
+]
2 After introducing numerical results for
cL = 001 M and [H+] = 10
7 M the obtained [L] = 9
10
11 M
wwwShimiPediair
9
OP
O
PO
P
O O
O
P
b Because stability constant = [ML]([M][L]) the ratio rsquo = cL[L] Basing
on previous data cL[L] = 001 910
11 = 11
10
8
c Because [ML][M] = 999 01 rsquo = [ML]([M]cL) = 999 (01 001) = 1
10
5
Thus = rsquo11
10
8 = 11
10
13
d = [ML]([M][L]) = [ML][H+]2([M]cL Ka3 Ka4) Assuming that a deposit of
hydroxide does not precipitate Ks0 = [M][OH]
2 the above equation can be written in the form
= [ML][H+]
2[OH
]
2(cL Ks0 Ka3 Ka4) = KW
2 ([ML]cL) ( Ks0 Ka3 Ka4)
e For [ML] = cL = 001 M using appropriate constants the minimal values of fulfilling the
conditions from (d) are 410
12 for the Cu
2+ complex and 6
10
8 for the Ni
2+ complex Comparison
of these results with experimental data shows that Cu2+
and Ni2+
complexes obey the conditions
from (c) and (d)
Because for Ni2+
- cyclam complex rsquo = [ML]([M]cL) for [ML] = cL [M] = 1rsquo is obtained
For the Ni2+
complex rsquo = 210
22 11
10
8 = 2
10
14 Thus [M] = [Ni
2+] = 1 (2
10
14) ie
[Ni2+
] = 5
10
15 M
2-fold dilution results in the same value of the [ML]cL ratio therefore Ni2+
ions concentration does
not change as well
SOLUTION OF TASK 2
a White phosphorus is built of P4 molecules Depending on the reaction conditions it may form
P4On (n = 6 7 8 9 or 10) oxides in the reaction with oxygen whose structures stem from P4
molecules structure Freezing point depression of benzene allows one to calculate the molality of
oxide A solution (cmA) and consequently its molar mass
0840125
430A
t
t
m
E
Tc molkg czyli 220
08400100
1850A
mAb
A
cm
mM gmol
This value corresponds to P4O6 molar mass (21988 gmol) Thus compound A is
phosphorus(III) oxide P4O6 Phosphorus reaction with oxygen proceeds according to the
following equation
P4 + 3O2 P4O6
b The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule There is
one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs
This structure agrees with 31
P NMR spectrum indicating all phosphorus nuclei are chemically
equivalent
c P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation
P4O6 + 6H2O 4H3PO3
d Phosphoric(III) acid contains PndashH bond and is a diprotic acid of an average strength It forms
therefore two series of salts and sparsely soluble barium salt containing HPO32minus
anion will be
formed with the excess of barium hydroxide
H3PO3 + Ba(OH)2 3 + 2H2O
HPO32minus
anion exhibits tetrahedral structure (sp3 hybridisation of phosphorus) There are three
oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron wwwShimiPediair
10
Nickel coordination sphere
O
H
O
PO
O
PO
H
O
or
2 2
e Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base
in chemical reactions Therefore Lewis adduct is formed in the reaction with nickel carbonyl in
which phosphorus forms a chemical bond with nickel replacing one of the CO ligands Assuming
that the only gaseous product of the reaction is CO the molar ratio of the evolved CO to the used
P4O6 equals
0642 g 126 g
002292 000573 4 12801 gmol 21988 gmol
which indicates that four nickel atoms
were bound by a P4O6 molecule The reaction proceeds according to the following equation
4Ni(CO)4 + P4O6 P4O6middot4Ni(CO)3 + 4CO
The formula of compound C is P4O6middot4Ni(CO)3 (or P4O6[Ni(CO)3]4) The molar mass of
compound C equals 79076 gmol Such a reaction course is confirmed by the approximate molar
mass of compound C calculated from the reaction product mass
4 6
C
P O
45785 gmol
000573
CmM
n
f It follows from the fact that all of phosphorus nuclei are chemically
equivalent in the molecule of compound C that every phosphorus atom is
bound to one nickel atom and three oxygen atoms The coordination
centres in the form of nickel atoms satisfy the 18 electron rule (10 valence
electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons
coming from phosphorus) Therefore the structure of nickel coordination
sphere is tetrahedral and ligands bound to nickel atom are localised in
vertices of a slightly deformed tetrahedron
g Similarly as in the reaction with nickel carbonyl P4O6 phosphorus oxide serves as a Lewis base
in the reaction with B2H6 forming an adduct with acid (BH3) As a result of the reaction with
water the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and
H3BO3 boric acid (acid E) The liberated gas is molecular hydrogen and its quantity depends on
the amount of reacted diborane
2
2 6
H 3
B H 3
800595 10
6 2241 10 6m
Vn
V
mole which correspond to B2H6 mass equal to
2 6 2 6 2 6
3
B H B H B H 0595 10 2767 00165m n M g The molar ratio of P4O6 to B2H6 in the Lewis
adduct equals 3(0147 00165) g 0595 10 0594 0595 11
21988 gmol
Only two phosphorus atoms are involved in bond formation with BH3 and the formula of
compound D is P4O6 middot 2BH3 (or P4O6 middot
B2H6) The hydrolysis reaction proceeds according to the
following equation
P4O6 middot 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2
P Ni
C
C
CO
OO
O
O
O
wwwShimiPediair
11
h The reaction between P4O6 and B2H6 leading to the formation of compound D
(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition
Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid
The reaction product compound D is a Lewis adduct The yield of compound D formation
reaction equals
4 6 4 6 4 6
D D
P O P O P O
267 24755100 100 100 770
308 21988
Dn m Mw
n m M
i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron
B P
O
O
OH
HH
or
B P
O
O
OH
HH
SOLUTION OF TASK 3
a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows
clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means
that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the
range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex
and in the range c[E]0 ltlt 05 towards the E2M complex
b The complex formation reactions occurring at the equilibrium are as follows
(k1) E + M EM and (k2) 2E + M E2M
The complex formation equilibrium constants k1 and k2 also referred to as the complex
stability constants can be written as
)()2]E([]T[]E[
]ET[β
k21kk21k0
k11
ccccc
c
(1)
)()2E]([]T[]E[
]TE[β
k2k1
2
k21k0
k2
2
22
ccccc
c
(2)
c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be
determined by introducing the simplifications suggested in Note 1 In equation (1) we assume
that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2
in the first term of the denominator and get
cc
c
)]E([β
k10
1k1
(3)
In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0
and in the second term of the denominator we can neglect subtraction of ck1 and we get
)(]E[
βk2
2
0
k22
cc
c
(4)
Then from equations (3) and (4) we determine
1
101k
β1
β]E[
c
cc (5)
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
wwwShimiPediair
14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
wwwShimiPediair
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
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6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
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8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
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10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
3
Fig 1 Molecular
formula of a 12C4
crown ether
TASK 3
Determination of stoichiometry and stability of complexes with an NMR method
Application of approximations in limiting conditions
One of the problems in the 2nd
stage of the 58 Chemistry Olympiad was
related to the crown ethers ability to complex metal ions Now we
present the results of 1H NMR study on formation reaction of a
12-crown-4 ether (labelled as 12C4) with Na+ cation introduced into the
system in a form of sodium thiocyanate NaSCN
Three independent experiments have been carried out the first and the
second at temperature 23degC and the third one at temperature ndash50degC In
the two first experiments the spectra were recorded at varying NaSCN
concentration and a single averaged signal coming from both the
complexed molecules and the free ether was monitored The signal chemical shift δobs was
dependent on the ratio of sodium ions [Na+]0 and crown ether [E]0 concentrations The appearance
of the averaged signal proves that the complexation reaction is a so called fast exchange reaction
Let us remind that for such reactions the observed chemical shift δobs is an average of characteristic
chemical shifts of the ligand δE and the complexes δEnM multiplied by corresponding
concentration ratios of ligands unbound and bound in complexes to the total concentration of
ligands in the system By writing the general complex formula as EnM (where E means ether and
M means metal) the relationship can be presented as follows
E3M
0
3E2M
0
2EM
0
E
0
obs]E[
]ME[3
]E[
]ME[2
]E[
]EM[
]E[
]E[
(1)
The total concentration of ligands in the system [E]0 = [E] + [EM] + 2[E2M] + 3[E3M]+
Experiment 1 The characteristic chemical shift δE = 2761 ppm has been determined from the
NMR spectrum of the pure crown ether Then small amounts of solid sodium thiocyanate were
added to a 12C4 solution (in deuterated methanol CD3OD) with concentration [E]0 =
0219 moldm3 and the chemical shift δobs was recorded as a function of total concentrations
[NaSCN][12C4] The measurement results are shown in Fig 2 and additionally the data for points
within the linearity range ie at [NaSCN][12C4] le 03 are given in Table 1
The experimental points form a
curve showing two characteristic
bends which prove that two
complexes k1 and k2 with different
stoichiometry and different overall
complex formation constants for
metal ion and for ether βn are
present in the system Based on the
plot given in Fig 2 one can
determine the stoichiometry of the
two complexes being formed and
remaining in equilibrium
Fig 2
complex 1 complex 2
wwwShimiPediair
4
Table 1 Results of experiment 1 [E]0 = 0219 moldm3
[NaSCN][12C4] δobs ppm
0
0045
0096
0150
0240
0300
2761
2846
2937
3025
3190
3282
Experiment 2 In all measurements the initial concentration of the crown ether was kept constant at
[E]0 = 0050 moldm3 while the thiocyanate concentration varied but was much higher than that of
the emerging complex The sodium ion concentrations c and the corresponding resultant chemical
shifts δobs are given in Table 2
Table 2 Results of experiment 2 [E]0 = 0050 moldm3
Total Na+ ion concentration
c (moldm3)
Observed chemical shift
δobs ppm
0204
0271
0359
0495
0646
0768
4437
4558
4660
4757
4821
4856
Experiment 3 The spectrum of the pure 12C4 ether was recorded again at temperature ndash50degC then
a small amount of NaSCN was added and the spectrum was recorded again The spectrum showed
distinctly resolved ether and complex peaks (the fast exchange reaction does not occur here) It
turned out that the chemical shift of the pure ether virtually did not change with decreasing
temperature One can therefore assume that the chemical shift of the k2 complex that is stable in the
presence of a large excess of ether is the same at both 23degC and at ndash50degC and is δk2 = 4575 ppm
Using the results of the three experiments described above determine the stoichiometry of complexes
that are formed in the system calculate their stability constants and characteristic chemical shift of the
k1 complex by following the directions given below
In the solution one should assume the following symbols for the major quantities occurring in the problem
[E]0 ndash total concentration of the crown ether constant in each experiment
[Na+] = c ndash total concentration of sodium ions varying during titration
δE ndash chemical shift of the pure 12C4 crown ether
δk1 δk2 ndash chemical shifts of two complexes
ck1 ck2 ndash concentrations of EnM complexes respectively in equilibrium
To simplify the notation of equations the following symbols can be also introduced
Eobsobs Ek1k1 and Ek2k2
wwwShimiPediair
5
Problems
a Based on the plot in Fig 2 determine the stoichiometric coefficients of the resulting complexes Justify
the answer with a short comment
b Give chemical equations for reactions taking place in the system and the formulae for stability
constants of the complexes by writing them down in a form containing the initial ether concentration
[E]0 and the concentration of sodium ions c
c Using appropriate equations determine complex concentrations ck1 and ck2 by introducing the
simplifications proposed in Note 1
d Give an equation describing the dependence of the averaged chemical shift δobs on corresponding
chemical shifts of the ether δE (data) and the two complexes δk1 (this should be treated as an unknown)
and δk2 (data) Present the equation as a dependence of corresponding differences of chemical shifts that
is obs k1 and k2 Write the two equations in two forms corresponding to the limiting cases by
introducing corresponding complex concentrations determined in Direction c
e Calculate β2 from the simplified equation from Direction d for the limiting case [E]0 gtgt c by
determining the slope of the straight line (δobsndashδE) = f(c) from two selected points from Table 1 and using
known value of δk2 (see Note 2)
f Solve the equation from Direction d in the high Na+ concentration limit (c) by converting the equation
so as to get a linear dependence of 1(δobs ndash δE) on 1c Calculate β1 and δk1 constants (see Note 3)
Notes
1 The equations for equilibrium constants can be simplified if the experiment conditions can
be considered as the limiting ones
In experiment 1 [E]0 gtgt c so the equilibrium is strongly shifted towards the k2 complex and
we can assume that [E]0 gtgt ck2 gtgt ck1 asymp 0
In experiment 2 [E]0 ltlt c so the equilibrium is strongly shifted towards the k1 complex and
we assume that c gtgt ck1 gtgt ck2 asymp 0
Corresponding approximations are best introduced by determining ck1 and ck2 as functions of
complex stability constants β1 and β2
2 Fig 3 shows linear dependence of (δobs ndash δE) on c in the limit [E]0 gtgt c The plot should facilitate the
selection of points of which the coordinates allow to calculate the slope of the straight line
Fig 3
Experiment 1
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6
3 To solve the problem in the high Na+ concentration limit ([E]0 ltlt c) it is necessary to
ldquolineariserdquo the dependence of the corresponding difference of chemical shifts (δobsndashδE) on
varying [Na+] = c concentration The dependence of 1(δobsndashδE) on 1c should be linear
(1(δobsndashδE) = a(1c) + b) in the limit of large values of the ratio c[E]0 ie in the conditions of
experiment 2 Fig 4 shows the relationship plotted using the data from Table 2
Fig 4
TASK 4
Biologically active indole derivatives
Tryptamines are derivatives of indole (a heterocyclic compound consisting of two condensed rings
a benzene ring and a pyrrole ring) abundantly present in living organisms To tryptamines belong
inter alia neurotransmiters as important as melatonin or serotonin Due to that also other
tryptamine derivatives are not neutral to human organism and may have medicinal use or exert
hallucinogenic effects
The phosphorylated tryptamine derivative F initially isolated from plant sources has been obtained
through the reaction sequence depicted in Scheme 1 The starting material for the synthesis was 4-
hydroxyindole It is known that the most reactive position of indole is the C-3 within the pyrrole
ring which easily undergoes aromatic electrophilic substitution but may also be involved in
reactions typical for enamines
First the phenol group of 4-hydroxyindole was protected by reacting it with benzyl chloride in the
presence of sodium methanolate This led to compound A which was then subjected to Friedel-
Crafts acylation with oxalyl chloride leading to compound B Subsequently B without isolation
was reacted with dimethylamine which resulted in compound C In the next step C was reduced to
D (molecular formula C19H22N2O) by means of LiAlH4 and then catalytically hydrogenated to yield
compound E and toluene as a by-product In the last step E was converted into F by applying
following reagents 1 n-buthyllithium 2 tetrabenzyl pyrophosphate (TBPP) 3 catalytic
hydrogenation (PdC H2)
A homologue of compound E known as 4-hydroxygramine and having molecular formula of
C11H14N2O may be obtained in a multi-component reaction (Mannich reaction) by reacting
compound A with formaldehyde and dimethylamine followed by a catalytic hydrogenation
Exploiting a similar multi-component reaction but from different starting materials one can obtain
compound X as a mixture of several stereoisomers (Scheme 2 depicts only one of them)
Experiment 2
wwwShimiPediair
7
O
ClO
Cl NH
A
NH
OH
Cl
OP
OP
O
OO
OO
C19H22N2O
TBPP =Bn
Bn
Bn
Bn
Scheme 1
MeOHMeO-
F
H2 Pd-C
1 n-BuLi2TBPP3 H2 Pd-C
[ B ]C
DE
LiAlH4
CH3
O
CH3
N O
O
NH
OCH
3Chiral
Scheme 2
342
1
X
Problems
a Draw structural formulas for compounds A-F and 4-hydroxygramine
b Draw structural formulas of the starting materials for the synthesis of compound X
c Compound X could be obtained from the same starting materials but in a stereochemically pure
form (as depicted in Scheme 2) Choose which conditions should be applied to achieve this
I the reaction temperature should be increased
II the reagents should be mixed in a certain order
III the reaction should be carried out in the presence of L-proline
IV the reaction mixture should be stirred always in the same direction
d Determine the absolute configuration of all stereogenic centers in compound X
e Explain the reactivity of the C-3 position of indole by means of appropriate mesomeric structures
TASK 5
The Diels-Alder Reaction
Cycloaddition reactions including Diels-Alder reaction in most cases are stereospecific what
means that geometry of substrates (eg dienophile or diene) determine the structure of the formed
product
X
Y Y
X+
General Scheme of the Diels-Alder Reaction
I The carboxylic acid with molecular formula C4H4O4 as well as its dimethyl esters (A B and C)
exist as two geometric isomers and one regioisomer and all of them undergo the Diels-Alder
reaction The reaction of geometric isomers A and B mixture with diene D composed entirely of
carbon and hydrogen atoms provides diastereomeric products E1 E2 and F Ozonolysis of E and F
mixture with zinc dust form product G (diastereomeric mixture) and subsequent reduction with
LiAlH4 excess leads to the mixture of diastereomers of polihydroxylic alcohol H
wwwShimiPediair
8
OHOH
OHOHLiAlH
4A i B
D E1 + E2 i F(diastereomers)
1) O3
2) ZnG
Hmixture of three diastereomers
1) O3
2) Zn
ID
J
mixture of three diastereomers
Compounds A and B after ozonolysis with zinc dust form the same product I which also undergo
cycloaddition reaction with diene D The reaction leads to the cyclic ether J The molecular mass of
J is equal to the sum of mass of reactants D and I
II Stereochemistry in Diels-Alder reaction Isomer A in the reaction with diene D forms the
mixture of two achiral diastereomers E1 and E2 The reduction of diastereomeric mixture of E1 and
E2 with hydrogen on palladium catalyst leads to only one product K
The Diels-Alder reaction of isomer B with diene D provides exclusively racemic mixture of
cycloadduct F The product F after ozonolysis with zinc dust and subsequent reduction with LiAlH4
excess leads to one diastereomer of polihydroxylic alcohol H (as a racemic mixture)
Problems
I
a Draw the structure of geometric isomer A or B and structure of regioisomer C
b Draw the structure of diene D
c Determine general structure of Diels-Alder product EF (no stereochemistry required)
d Draw the general structure of ozonolysis product G (no stereochemistry required)
e Draw the structure of compound I and general structure of J (no stereochemistry required)
II
a Determine unambiguously geometry of isomers A and B
b Draw stereochemical structures of compounds E (E1 and E2) and their reduction product K
c Draw the stereochemical structure of compound F and stereochemical structure of
corresponding diastereomer of polihydroxylic alcohol H
d Draw all possible stereoisomers of the product J Determine enantiomers and diastereomers
SOLUTIONS
SOLUTION OF TASK 1
a The total concentration of cyclam cL = [L] + [HL+] + [H2L
2+] + [H3L
3+] + [H4L
4+] Using acidic
dissociation constants after rearrangement one can write
cL = [L]1 + [H+]Ka4 + [H
+]2(Ka4 Ka3) + [H
+]3(Ka4 Ka3 Ka2) + [H
+]
4(Ka4 Ka3 Ka2 Ka1)
Taking into account the values of dissociation constants it can be assumed that at pH = 7 the
form H2L2+
predominates in the solution ie the above equation can be simplified to the form
cL = [L][H+]
2(Ka3 Ka4) thus [L] = cL Ka3 Ka4 [H
+]
2 After introducing numerical results for
cL = 001 M and [H+] = 10
7 M the obtained [L] = 9
10
11 M
wwwShimiPediair
9
OP
O
PO
P
O O
O
P
b Because stability constant = [ML]([M][L]) the ratio rsquo = cL[L] Basing
on previous data cL[L] = 001 910
11 = 11
10
8
c Because [ML][M] = 999 01 rsquo = [ML]([M]cL) = 999 (01 001) = 1
10
5
Thus = rsquo11
10
8 = 11
10
13
d = [ML]([M][L]) = [ML][H+]2([M]cL Ka3 Ka4) Assuming that a deposit of
hydroxide does not precipitate Ks0 = [M][OH]
2 the above equation can be written in the form
= [ML][H+]
2[OH
]
2(cL Ks0 Ka3 Ka4) = KW
2 ([ML]cL) ( Ks0 Ka3 Ka4)
e For [ML] = cL = 001 M using appropriate constants the minimal values of fulfilling the
conditions from (d) are 410
12 for the Cu
2+ complex and 6
10
8 for the Ni
2+ complex Comparison
of these results with experimental data shows that Cu2+
and Ni2+
complexes obey the conditions
from (c) and (d)
Because for Ni2+
- cyclam complex rsquo = [ML]([M]cL) for [ML] = cL [M] = 1rsquo is obtained
For the Ni2+
complex rsquo = 210
22 11
10
8 = 2
10
14 Thus [M] = [Ni
2+] = 1 (2
10
14) ie
[Ni2+
] = 5
10
15 M
2-fold dilution results in the same value of the [ML]cL ratio therefore Ni2+
ions concentration does
not change as well
SOLUTION OF TASK 2
a White phosphorus is built of P4 molecules Depending on the reaction conditions it may form
P4On (n = 6 7 8 9 or 10) oxides in the reaction with oxygen whose structures stem from P4
molecules structure Freezing point depression of benzene allows one to calculate the molality of
oxide A solution (cmA) and consequently its molar mass
0840125
430A
t
t
m
E
Tc molkg czyli 220
08400100
1850A
mAb
A
cm
mM gmol
This value corresponds to P4O6 molar mass (21988 gmol) Thus compound A is
phosphorus(III) oxide P4O6 Phosphorus reaction with oxygen proceeds according to the
following equation
P4 + 3O2 P4O6
b The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule There is
one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs
This structure agrees with 31
P NMR spectrum indicating all phosphorus nuclei are chemically
equivalent
c P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation
P4O6 + 6H2O 4H3PO3
d Phosphoric(III) acid contains PndashH bond and is a diprotic acid of an average strength It forms
therefore two series of salts and sparsely soluble barium salt containing HPO32minus
anion will be
formed with the excess of barium hydroxide
H3PO3 + Ba(OH)2 3 + 2H2O
HPO32minus
anion exhibits tetrahedral structure (sp3 hybridisation of phosphorus) There are three
oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron wwwShimiPediair
10
Nickel coordination sphere
O
H
O
PO
O
PO
H
O
or
2 2
e Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base
in chemical reactions Therefore Lewis adduct is formed in the reaction with nickel carbonyl in
which phosphorus forms a chemical bond with nickel replacing one of the CO ligands Assuming
that the only gaseous product of the reaction is CO the molar ratio of the evolved CO to the used
P4O6 equals
0642 g 126 g
002292 000573 4 12801 gmol 21988 gmol
which indicates that four nickel atoms
were bound by a P4O6 molecule The reaction proceeds according to the following equation
4Ni(CO)4 + P4O6 P4O6middot4Ni(CO)3 + 4CO
The formula of compound C is P4O6middot4Ni(CO)3 (or P4O6[Ni(CO)3]4) The molar mass of
compound C equals 79076 gmol Such a reaction course is confirmed by the approximate molar
mass of compound C calculated from the reaction product mass
4 6
C
P O
45785 gmol
000573
CmM
n
f It follows from the fact that all of phosphorus nuclei are chemically
equivalent in the molecule of compound C that every phosphorus atom is
bound to one nickel atom and three oxygen atoms The coordination
centres in the form of nickel atoms satisfy the 18 electron rule (10 valence
electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons
coming from phosphorus) Therefore the structure of nickel coordination
sphere is tetrahedral and ligands bound to nickel atom are localised in
vertices of a slightly deformed tetrahedron
g Similarly as in the reaction with nickel carbonyl P4O6 phosphorus oxide serves as a Lewis base
in the reaction with B2H6 forming an adduct with acid (BH3) As a result of the reaction with
water the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and
H3BO3 boric acid (acid E) The liberated gas is molecular hydrogen and its quantity depends on
the amount of reacted diborane
2
2 6
H 3
B H 3
800595 10
6 2241 10 6m
Vn
V
mole which correspond to B2H6 mass equal to
2 6 2 6 2 6
3
B H B H B H 0595 10 2767 00165m n M g The molar ratio of P4O6 to B2H6 in the Lewis
adduct equals 3(0147 00165) g 0595 10 0594 0595 11
21988 gmol
Only two phosphorus atoms are involved in bond formation with BH3 and the formula of
compound D is P4O6 middot 2BH3 (or P4O6 middot
B2H6) The hydrolysis reaction proceeds according to the
following equation
P4O6 middot 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2
P Ni
C
C
CO
OO
O
O
O
wwwShimiPediair
11
h The reaction between P4O6 and B2H6 leading to the formation of compound D
(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition
Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid
The reaction product compound D is a Lewis adduct The yield of compound D formation
reaction equals
4 6 4 6 4 6
D D
P O P O P O
267 24755100 100 100 770
308 21988
Dn m Mw
n m M
i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron
B P
O
O
OH
HH
or
B P
O
O
OH
HH
SOLUTION OF TASK 3
a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows
clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means
that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the
range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex
and in the range c[E]0 ltlt 05 towards the E2M complex
b The complex formation reactions occurring at the equilibrium are as follows
(k1) E + M EM and (k2) 2E + M E2M
The complex formation equilibrium constants k1 and k2 also referred to as the complex
stability constants can be written as
)()2]E([]T[]E[
]ET[β
k21kk21k0
k11
ccccc
c
(1)
)()2E]([]T[]E[
]TE[β
k2k1
2
k21k0
k2
2
22
ccccc
c
(2)
c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be
determined by introducing the simplifications suggested in Note 1 In equation (1) we assume
that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2
in the first term of the denominator and get
cc
c
)]E([β
k10
1k1
(3)
In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0
and in the second term of the denominator we can neglect subtraction of ck1 and we get
)(]E[
βk2
2
0
k22
cc
c
(4)
Then from equations (3) and (4) we determine
1
101k
β1
β]E[
c
cc (5)
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
wwwShimiPediair
14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
wwwShimiPediair
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
wwwShimiPediair
6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
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10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
4
Table 1 Results of experiment 1 [E]0 = 0219 moldm3
[NaSCN][12C4] δobs ppm
0
0045
0096
0150
0240
0300
2761
2846
2937
3025
3190
3282
Experiment 2 In all measurements the initial concentration of the crown ether was kept constant at
[E]0 = 0050 moldm3 while the thiocyanate concentration varied but was much higher than that of
the emerging complex The sodium ion concentrations c and the corresponding resultant chemical
shifts δobs are given in Table 2
Table 2 Results of experiment 2 [E]0 = 0050 moldm3
Total Na+ ion concentration
c (moldm3)
Observed chemical shift
δobs ppm
0204
0271
0359
0495
0646
0768
4437
4558
4660
4757
4821
4856
Experiment 3 The spectrum of the pure 12C4 ether was recorded again at temperature ndash50degC then
a small amount of NaSCN was added and the spectrum was recorded again The spectrum showed
distinctly resolved ether and complex peaks (the fast exchange reaction does not occur here) It
turned out that the chemical shift of the pure ether virtually did not change with decreasing
temperature One can therefore assume that the chemical shift of the k2 complex that is stable in the
presence of a large excess of ether is the same at both 23degC and at ndash50degC and is δk2 = 4575 ppm
Using the results of the three experiments described above determine the stoichiometry of complexes
that are formed in the system calculate their stability constants and characteristic chemical shift of the
k1 complex by following the directions given below
In the solution one should assume the following symbols for the major quantities occurring in the problem
[E]0 ndash total concentration of the crown ether constant in each experiment
[Na+] = c ndash total concentration of sodium ions varying during titration
δE ndash chemical shift of the pure 12C4 crown ether
δk1 δk2 ndash chemical shifts of two complexes
ck1 ck2 ndash concentrations of EnM complexes respectively in equilibrium
To simplify the notation of equations the following symbols can be also introduced
Eobsobs Ek1k1 and Ek2k2
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5
Problems
a Based on the plot in Fig 2 determine the stoichiometric coefficients of the resulting complexes Justify
the answer with a short comment
b Give chemical equations for reactions taking place in the system and the formulae for stability
constants of the complexes by writing them down in a form containing the initial ether concentration
[E]0 and the concentration of sodium ions c
c Using appropriate equations determine complex concentrations ck1 and ck2 by introducing the
simplifications proposed in Note 1
d Give an equation describing the dependence of the averaged chemical shift δobs on corresponding
chemical shifts of the ether δE (data) and the two complexes δk1 (this should be treated as an unknown)
and δk2 (data) Present the equation as a dependence of corresponding differences of chemical shifts that
is obs k1 and k2 Write the two equations in two forms corresponding to the limiting cases by
introducing corresponding complex concentrations determined in Direction c
e Calculate β2 from the simplified equation from Direction d for the limiting case [E]0 gtgt c by
determining the slope of the straight line (δobsndashδE) = f(c) from two selected points from Table 1 and using
known value of δk2 (see Note 2)
f Solve the equation from Direction d in the high Na+ concentration limit (c) by converting the equation
so as to get a linear dependence of 1(δobs ndash δE) on 1c Calculate β1 and δk1 constants (see Note 3)
Notes
1 The equations for equilibrium constants can be simplified if the experiment conditions can
be considered as the limiting ones
In experiment 1 [E]0 gtgt c so the equilibrium is strongly shifted towards the k2 complex and
we can assume that [E]0 gtgt ck2 gtgt ck1 asymp 0
In experiment 2 [E]0 ltlt c so the equilibrium is strongly shifted towards the k1 complex and
we assume that c gtgt ck1 gtgt ck2 asymp 0
Corresponding approximations are best introduced by determining ck1 and ck2 as functions of
complex stability constants β1 and β2
2 Fig 3 shows linear dependence of (δobs ndash δE) on c in the limit [E]0 gtgt c The plot should facilitate the
selection of points of which the coordinates allow to calculate the slope of the straight line
Fig 3
Experiment 1
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6
3 To solve the problem in the high Na+ concentration limit ([E]0 ltlt c) it is necessary to
ldquolineariserdquo the dependence of the corresponding difference of chemical shifts (δobsndashδE) on
varying [Na+] = c concentration The dependence of 1(δobsndashδE) on 1c should be linear
(1(δobsndashδE) = a(1c) + b) in the limit of large values of the ratio c[E]0 ie in the conditions of
experiment 2 Fig 4 shows the relationship plotted using the data from Table 2
Fig 4
TASK 4
Biologically active indole derivatives
Tryptamines are derivatives of indole (a heterocyclic compound consisting of two condensed rings
a benzene ring and a pyrrole ring) abundantly present in living organisms To tryptamines belong
inter alia neurotransmiters as important as melatonin or serotonin Due to that also other
tryptamine derivatives are not neutral to human organism and may have medicinal use or exert
hallucinogenic effects
The phosphorylated tryptamine derivative F initially isolated from plant sources has been obtained
through the reaction sequence depicted in Scheme 1 The starting material for the synthesis was 4-
hydroxyindole It is known that the most reactive position of indole is the C-3 within the pyrrole
ring which easily undergoes aromatic electrophilic substitution but may also be involved in
reactions typical for enamines
First the phenol group of 4-hydroxyindole was protected by reacting it with benzyl chloride in the
presence of sodium methanolate This led to compound A which was then subjected to Friedel-
Crafts acylation with oxalyl chloride leading to compound B Subsequently B without isolation
was reacted with dimethylamine which resulted in compound C In the next step C was reduced to
D (molecular formula C19H22N2O) by means of LiAlH4 and then catalytically hydrogenated to yield
compound E and toluene as a by-product In the last step E was converted into F by applying
following reagents 1 n-buthyllithium 2 tetrabenzyl pyrophosphate (TBPP) 3 catalytic
hydrogenation (PdC H2)
A homologue of compound E known as 4-hydroxygramine and having molecular formula of
C11H14N2O may be obtained in a multi-component reaction (Mannich reaction) by reacting
compound A with formaldehyde and dimethylamine followed by a catalytic hydrogenation
Exploiting a similar multi-component reaction but from different starting materials one can obtain
compound X as a mixture of several stereoisomers (Scheme 2 depicts only one of them)
Experiment 2
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7
O
ClO
Cl NH
A
NH
OH
Cl
OP
OP
O
OO
OO
C19H22N2O
TBPP =Bn
Bn
Bn
Bn
Scheme 1
MeOHMeO-
F
H2 Pd-C
1 n-BuLi2TBPP3 H2 Pd-C
[ B ]C
DE
LiAlH4
CH3
O
CH3
N O
O
NH
OCH
3Chiral
Scheme 2
342
1
X
Problems
a Draw structural formulas for compounds A-F and 4-hydroxygramine
b Draw structural formulas of the starting materials for the synthesis of compound X
c Compound X could be obtained from the same starting materials but in a stereochemically pure
form (as depicted in Scheme 2) Choose which conditions should be applied to achieve this
I the reaction temperature should be increased
II the reagents should be mixed in a certain order
III the reaction should be carried out in the presence of L-proline
IV the reaction mixture should be stirred always in the same direction
d Determine the absolute configuration of all stereogenic centers in compound X
e Explain the reactivity of the C-3 position of indole by means of appropriate mesomeric structures
TASK 5
The Diels-Alder Reaction
Cycloaddition reactions including Diels-Alder reaction in most cases are stereospecific what
means that geometry of substrates (eg dienophile or diene) determine the structure of the formed
product
X
Y Y
X+
General Scheme of the Diels-Alder Reaction
I The carboxylic acid with molecular formula C4H4O4 as well as its dimethyl esters (A B and C)
exist as two geometric isomers and one regioisomer and all of them undergo the Diels-Alder
reaction The reaction of geometric isomers A and B mixture with diene D composed entirely of
carbon and hydrogen atoms provides diastereomeric products E1 E2 and F Ozonolysis of E and F
mixture with zinc dust form product G (diastereomeric mixture) and subsequent reduction with
LiAlH4 excess leads to the mixture of diastereomers of polihydroxylic alcohol H
wwwShimiPediair
8
OHOH
OHOHLiAlH
4A i B
D E1 + E2 i F(diastereomers)
1) O3
2) ZnG
Hmixture of three diastereomers
1) O3
2) Zn
ID
J
mixture of three diastereomers
Compounds A and B after ozonolysis with zinc dust form the same product I which also undergo
cycloaddition reaction with diene D The reaction leads to the cyclic ether J The molecular mass of
J is equal to the sum of mass of reactants D and I
II Stereochemistry in Diels-Alder reaction Isomer A in the reaction with diene D forms the
mixture of two achiral diastereomers E1 and E2 The reduction of diastereomeric mixture of E1 and
E2 with hydrogen on palladium catalyst leads to only one product K
The Diels-Alder reaction of isomer B with diene D provides exclusively racemic mixture of
cycloadduct F The product F after ozonolysis with zinc dust and subsequent reduction with LiAlH4
excess leads to one diastereomer of polihydroxylic alcohol H (as a racemic mixture)
Problems
I
a Draw the structure of geometric isomer A or B and structure of regioisomer C
b Draw the structure of diene D
c Determine general structure of Diels-Alder product EF (no stereochemistry required)
d Draw the general structure of ozonolysis product G (no stereochemistry required)
e Draw the structure of compound I and general structure of J (no stereochemistry required)
II
a Determine unambiguously geometry of isomers A and B
b Draw stereochemical structures of compounds E (E1 and E2) and their reduction product K
c Draw the stereochemical structure of compound F and stereochemical structure of
corresponding diastereomer of polihydroxylic alcohol H
d Draw all possible stereoisomers of the product J Determine enantiomers and diastereomers
SOLUTIONS
SOLUTION OF TASK 1
a The total concentration of cyclam cL = [L] + [HL+] + [H2L
2+] + [H3L
3+] + [H4L
4+] Using acidic
dissociation constants after rearrangement one can write
cL = [L]1 + [H+]Ka4 + [H
+]2(Ka4 Ka3) + [H
+]3(Ka4 Ka3 Ka2) + [H
+]
4(Ka4 Ka3 Ka2 Ka1)
Taking into account the values of dissociation constants it can be assumed that at pH = 7 the
form H2L2+
predominates in the solution ie the above equation can be simplified to the form
cL = [L][H+]
2(Ka3 Ka4) thus [L] = cL Ka3 Ka4 [H
+]
2 After introducing numerical results for
cL = 001 M and [H+] = 10
7 M the obtained [L] = 9
10
11 M
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9
OP
O
PO
P
O O
O
P
b Because stability constant = [ML]([M][L]) the ratio rsquo = cL[L] Basing
on previous data cL[L] = 001 910
11 = 11
10
8
c Because [ML][M] = 999 01 rsquo = [ML]([M]cL) = 999 (01 001) = 1
10
5
Thus = rsquo11
10
8 = 11
10
13
d = [ML]([M][L]) = [ML][H+]2([M]cL Ka3 Ka4) Assuming that a deposit of
hydroxide does not precipitate Ks0 = [M][OH]
2 the above equation can be written in the form
= [ML][H+]
2[OH
]
2(cL Ks0 Ka3 Ka4) = KW
2 ([ML]cL) ( Ks0 Ka3 Ka4)
e For [ML] = cL = 001 M using appropriate constants the minimal values of fulfilling the
conditions from (d) are 410
12 for the Cu
2+ complex and 6
10
8 for the Ni
2+ complex Comparison
of these results with experimental data shows that Cu2+
and Ni2+
complexes obey the conditions
from (c) and (d)
Because for Ni2+
- cyclam complex rsquo = [ML]([M]cL) for [ML] = cL [M] = 1rsquo is obtained
For the Ni2+
complex rsquo = 210
22 11
10
8 = 2
10
14 Thus [M] = [Ni
2+] = 1 (2
10
14) ie
[Ni2+
] = 5
10
15 M
2-fold dilution results in the same value of the [ML]cL ratio therefore Ni2+
ions concentration does
not change as well
SOLUTION OF TASK 2
a White phosphorus is built of P4 molecules Depending on the reaction conditions it may form
P4On (n = 6 7 8 9 or 10) oxides in the reaction with oxygen whose structures stem from P4
molecules structure Freezing point depression of benzene allows one to calculate the molality of
oxide A solution (cmA) and consequently its molar mass
0840125
430A
t
t
m
E
Tc molkg czyli 220
08400100
1850A
mAb
A
cm
mM gmol
This value corresponds to P4O6 molar mass (21988 gmol) Thus compound A is
phosphorus(III) oxide P4O6 Phosphorus reaction with oxygen proceeds according to the
following equation
P4 + 3O2 P4O6
b The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule There is
one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs
This structure agrees with 31
P NMR spectrum indicating all phosphorus nuclei are chemically
equivalent
c P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation
P4O6 + 6H2O 4H3PO3
d Phosphoric(III) acid contains PndashH bond and is a diprotic acid of an average strength It forms
therefore two series of salts and sparsely soluble barium salt containing HPO32minus
anion will be
formed with the excess of barium hydroxide
H3PO3 + Ba(OH)2 3 + 2H2O
HPO32minus
anion exhibits tetrahedral structure (sp3 hybridisation of phosphorus) There are three
oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron wwwShimiPediair
10
Nickel coordination sphere
O
H
O
PO
O
PO
H
O
or
2 2
e Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base
in chemical reactions Therefore Lewis adduct is formed in the reaction with nickel carbonyl in
which phosphorus forms a chemical bond with nickel replacing one of the CO ligands Assuming
that the only gaseous product of the reaction is CO the molar ratio of the evolved CO to the used
P4O6 equals
0642 g 126 g
002292 000573 4 12801 gmol 21988 gmol
which indicates that four nickel atoms
were bound by a P4O6 molecule The reaction proceeds according to the following equation
4Ni(CO)4 + P4O6 P4O6middot4Ni(CO)3 + 4CO
The formula of compound C is P4O6middot4Ni(CO)3 (or P4O6[Ni(CO)3]4) The molar mass of
compound C equals 79076 gmol Such a reaction course is confirmed by the approximate molar
mass of compound C calculated from the reaction product mass
4 6
C
P O
45785 gmol
000573
CmM
n
f It follows from the fact that all of phosphorus nuclei are chemically
equivalent in the molecule of compound C that every phosphorus atom is
bound to one nickel atom and three oxygen atoms The coordination
centres in the form of nickel atoms satisfy the 18 electron rule (10 valence
electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons
coming from phosphorus) Therefore the structure of nickel coordination
sphere is tetrahedral and ligands bound to nickel atom are localised in
vertices of a slightly deformed tetrahedron
g Similarly as in the reaction with nickel carbonyl P4O6 phosphorus oxide serves as a Lewis base
in the reaction with B2H6 forming an adduct with acid (BH3) As a result of the reaction with
water the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and
H3BO3 boric acid (acid E) The liberated gas is molecular hydrogen and its quantity depends on
the amount of reacted diborane
2
2 6
H 3
B H 3
800595 10
6 2241 10 6m
Vn
V
mole which correspond to B2H6 mass equal to
2 6 2 6 2 6
3
B H B H B H 0595 10 2767 00165m n M g The molar ratio of P4O6 to B2H6 in the Lewis
adduct equals 3(0147 00165) g 0595 10 0594 0595 11
21988 gmol
Only two phosphorus atoms are involved in bond formation with BH3 and the formula of
compound D is P4O6 middot 2BH3 (or P4O6 middot
B2H6) The hydrolysis reaction proceeds according to the
following equation
P4O6 middot 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2
P Ni
C
C
CO
OO
O
O
O
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11
h The reaction between P4O6 and B2H6 leading to the formation of compound D
(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition
Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid
The reaction product compound D is a Lewis adduct The yield of compound D formation
reaction equals
4 6 4 6 4 6
D D
P O P O P O
267 24755100 100 100 770
308 21988
Dn m Mw
n m M
i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron
B P
O
O
OH
HH
or
B P
O
O
OH
HH
SOLUTION OF TASK 3
a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows
clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means
that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the
range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex
and in the range c[E]0 ltlt 05 towards the E2M complex
b The complex formation reactions occurring at the equilibrium are as follows
(k1) E + M EM and (k2) 2E + M E2M
The complex formation equilibrium constants k1 and k2 also referred to as the complex
stability constants can be written as
)()2]E([]T[]E[
]ET[β
k21kk21k0
k11
ccccc
c
(1)
)()2E]([]T[]E[
]TE[β
k2k1
2
k21k0
k2
2
22
ccccc
c
(2)
c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be
determined by introducing the simplifications suggested in Note 1 In equation (1) we assume
that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2
in the first term of the denominator and get
cc
c
)]E([β
k10
1k1
(3)
In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0
and in the second term of the denominator we can neglect subtraction of ck1 and we get
)(]E[
βk2
2
0
k22
cc
c
(4)
Then from equations (3) and (4) we determine
1
101k
β1
β]E[
c
cc (5)
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
wwwShimiPediair
14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
wwwShimiPediair
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
wwwShimiPediair
6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
5
Problems
a Based on the plot in Fig 2 determine the stoichiometric coefficients of the resulting complexes Justify
the answer with a short comment
b Give chemical equations for reactions taking place in the system and the formulae for stability
constants of the complexes by writing them down in a form containing the initial ether concentration
[E]0 and the concentration of sodium ions c
c Using appropriate equations determine complex concentrations ck1 and ck2 by introducing the
simplifications proposed in Note 1
d Give an equation describing the dependence of the averaged chemical shift δobs on corresponding
chemical shifts of the ether δE (data) and the two complexes δk1 (this should be treated as an unknown)
and δk2 (data) Present the equation as a dependence of corresponding differences of chemical shifts that
is obs k1 and k2 Write the two equations in two forms corresponding to the limiting cases by
introducing corresponding complex concentrations determined in Direction c
e Calculate β2 from the simplified equation from Direction d for the limiting case [E]0 gtgt c by
determining the slope of the straight line (δobsndashδE) = f(c) from two selected points from Table 1 and using
known value of δk2 (see Note 2)
f Solve the equation from Direction d in the high Na+ concentration limit (c) by converting the equation
so as to get a linear dependence of 1(δobs ndash δE) on 1c Calculate β1 and δk1 constants (see Note 3)
Notes
1 The equations for equilibrium constants can be simplified if the experiment conditions can
be considered as the limiting ones
In experiment 1 [E]0 gtgt c so the equilibrium is strongly shifted towards the k2 complex and
we can assume that [E]0 gtgt ck2 gtgt ck1 asymp 0
In experiment 2 [E]0 ltlt c so the equilibrium is strongly shifted towards the k1 complex and
we assume that c gtgt ck1 gtgt ck2 asymp 0
Corresponding approximations are best introduced by determining ck1 and ck2 as functions of
complex stability constants β1 and β2
2 Fig 3 shows linear dependence of (δobs ndash δE) on c in the limit [E]0 gtgt c The plot should facilitate the
selection of points of which the coordinates allow to calculate the slope of the straight line
Fig 3
Experiment 1
wwwShimiPediair
6
3 To solve the problem in the high Na+ concentration limit ([E]0 ltlt c) it is necessary to
ldquolineariserdquo the dependence of the corresponding difference of chemical shifts (δobsndashδE) on
varying [Na+] = c concentration The dependence of 1(δobsndashδE) on 1c should be linear
(1(δobsndashδE) = a(1c) + b) in the limit of large values of the ratio c[E]0 ie in the conditions of
experiment 2 Fig 4 shows the relationship plotted using the data from Table 2
Fig 4
TASK 4
Biologically active indole derivatives
Tryptamines are derivatives of indole (a heterocyclic compound consisting of two condensed rings
a benzene ring and a pyrrole ring) abundantly present in living organisms To tryptamines belong
inter alia neurotransmiters as important as melatonin or serotonin Due to that also other
tryptamine derivatives are not neutral to human organism and may have medicinal use or exert
hallucinogenic effects
The phosphorylated tryptamine derivative F initially isolated from plant sources has been obtained
through the reaction sequence depicted in Scheme 1 The starting material for the synthesis was 4-
hydroxyindole It is known that the most reactive position of indole is the C-3 within the pyrrole
ring which easily undergoes aromatic electrophilic substitution but may also be involved in
reactions typical for enamines
First the phenol group of 4-hydroxyindole was protected by reacting it with benzyl chloride in the
presence of sodium methanolate This led to compound A which was then subjected to Friedel-
Crafts acylation with oxalyl chloride leading to compound B Subsequently B without isolation
was reacted with dimethylamine which resulted in compound C In the next step C was reduced to
D (molecular formula C19H22N2O) by means of LiAlH4 and then catalytically hydrogenated to yield
compound E and toluene as a by-product In the last step E was converted into F by applying
following reagents 1 n-buthyllithium 2 tetrabenzyl pyrophosphate (TBPP) 3 catalytic
hydrogenation (PdC H2)
A homologue of compound E known as 4-hydroxygramine and having molecular formula of
C11H14N2O may be obtained in a multi-component reaction (Mannich reaction) by reacting
compound A with formaldehyde and dimethylamine followed by a catalytic hydrogenation
Exploiting a similar multi-component reaction but from different starting materials one can obtain
compound X as a mixture of several stereoisomers (Scheme 2 depicts only one of them)
Experiment 2
wwwShimiPediair
7
O
ClO
Cl NH
A
NH
OH
Cl
OP
OP
O
OO
OO
C19H22N2O
TBPP =Bn
Bn
Bn
Bn
Scheme 1
MeOHMeO-
F
H2 Pd-C
1 n-BuLi2TBPP3 H2 Pd-C
[ B ]C
DE
LiAlH4
CH3
O
CH3
N O
O
NH
OCH
3Chiral
Scheme 2
342
1
X
Problems
a Draw structural formulas for compounds A-F and 4-hydroxygramine
b Draw structural formulas of the starting materials for the synthesis of compound X
c Compound X could be obtained from the same starting materials but in a stereochemically pure
form (as depicted in Scheme 2) Choose which conditions should be applied to achieve this
I the reaction temperature should be increased
II the reagents should be mixed in a certain order
III the reaction should be carried out in the presence of L-proline
IV the reaction mixture should be stirred always in the same direction
d Determine the absolute configuration of all stereogenic centers in compound X
e Explain the reactivity of the C-3 position of indole by means of appropriate mesomeric structures
TASK 5
The Diels-Alder Reaction
Cycloaddition reactions including Diels-Alder reaction in most cases are stereospecific what
means that geometry of substrates (eg dienophile or diene) determine the structure of the formed
product
X
Y Y
X+
General Scheme of the Diels-Alder Reaction
I The carboxylic acid with molecular formula C4H4O4 as well as its dimethyl esters (A B and C)
exist as two geometric isomers and one regioisomer and all of them undergo the Diels-Alder
reaction The reaction of geometric isomers A and B mixture with diene D composed entirely of
carbon and hydrogen atoms provides diastereomeric products E1 E2 and F Ozonolysis of E and F
mixture with zinc dust form product G (diastereomeric mixture) and subsequent reduction with
LiAlH4 excess leads to the mixture of diastereomers of polihydroxylic alcohol H
wwwShimiPediair
8
OHOH
OHOHLiAlH
4A i B
D E1 + E2 i F(diastereomers)
1) O3
2) ZnG
Hmixture of three diastereomers
1) O3
2) Zn
ID
J
mixture of three diastereomers
Compounds A and B after ozonolysis with zinc dust form the same product I which also undergo
cycloaddition reaction with diene D The reaction leads to the cyclic ether J The molecular mass of
J is equal to the sum of mass of reactants D and I
II Stereochemistry in Diels-Alder reaction Isomer A in the reaction with diene D forms the
mixture of two achiral diastereomers E1 and E2 The reduction of diastereomeric mixture of E1 and
E2 with hydrogen on palladium catalyst leads to only one product K
The Diels-Alder reaction of isomer B with diene D provides exclusively racemic mixture of
cycloadduct F The product F after ozonolysis with zinc dust and subsequent reduction with LiAlH4
excess leads to one diastereomer of polihydroxylic alcohol H (as a racemic mixture)
Problems
I
a Draw the structure of geometric isomer A or B and structure of regioisomer C
b Draw the structure of diene D
c Determine general structure of Diels-Alder product EF (no stereochemistry required)
d Draw the general structure of ozonolysis product G (no stereochemistry required)
e Draw the structure of compound I and general structure of J (no stereochemistry required)
II
a Determine unambiguously geometry of isomers A and B
b Draw stereochemical structures of compounds E (E1 and E2) and their reduction product K
c Draw the stereochemical structure of compound F and stereochemical structure of
corresponding diastereomer of polihydroxylic alcohol H
d Draw all possible stereoisomers of the product J Determine enantiomers and diastereomers
SOLUTIONS
SOLUTION OF TASK 1
a The total concentration of cyclam cL = [L] + [HL+] + [H2L
2+] + [H3L
3+] + [H4L
4+] Using acidic
dissociation constants after rearrangement one can write
cL = [L]1 + [H+]Ka4 + [H
+]2(Ka4 Ka3) + [H
+]3(Ka4 Ka3 Ka2) + [H
+]
4(Ka4 Ka3 Ka2 Ka1)
Taking into account the values of dissociation constants it can be assumed that at pH = 7 the
form H2L2+
predominates in the solution ie the above equation can be simplified to the form
cL = [L][H+]
2(Ka3 Ka4) thus [L] = cL Ka3 Ka4 [H
+]
2 After introducing numerical results for
cL = 001 M and [H+] = 10
7 M the obtained [L] = 9
10
11 M
wwwShimiPediair
9
OP
O
PO
P
O O
O
P
b Because stability constant = [ML]([M][L]) the ratio rsquo = cL[L] Basing
on previous data cL[L] = 001 910
11 = 11
10
8
c Because [ML][M] = 999 01 rsquo = [ML]([M]cL) = 999 (01 001) = 1
10
5
Thus = rsquo11
10
8 = 11
10
13
d = [ML]([M][L]) = [ML][H+]2([M]cL Ka3 Ka4) Assuming that a deposit of
hydroxide does not precipitate Ks0 = [M][OH]
2 the above equation can be written in the form
= [ML][H+]
2[OH
]
2(cL Ks0 Ka3 Ka4) = KW
2 ([ML]cL) ( Ks0 Ka3 Ka4)
e For [ML] = cL = 001 M using appropriate constants the minimal values of fulfilling the
conditions from (d) are 410
12 for the Cu
2+ complex and 6
10
8 for the Ni
2+ complex Comparison
of these results with experimental data shows that Cu2+
and Ni2+
complexes obey the conditions
from (c) and (d)
Because for Ni2+
- cyclam complex rsquo = [ML]([M]cL) for [ML] = cL [M] = 1rsquo is obtained
For the Ni2+
complex rsquo = 210
22 11
10
8 = 2
10
14 Thus [M] = [Ni
2+] = 1 (2
10
14) ie
[Ni2+
] = 5
10
15 M
2-fold dilution results in the same value of the [ML]cL ratio therefore Ni2+
ions concentration does
not change as well
SOLUTION OF TASK 2
a White phosphorus is built of P4 molecules Depending on the reaction conditions it may form
P4On (n = 6 7 8 9 or 10) oxides in the reaction with oxygen whose structures stem from P4
molecules structure Freezing point depression of benzene allows one to calculate the molality of
oxide A solution (cmA) and consequently its molar mass
0840125
430A
t
t
m
E
Tc molkg czyli 220
08400100
1850A
mAb
A
cm
mM gmol
This value corresponds to P4O6 molar mass (21988 gmol) Thus compound A is
phosphorus(III) oxide P4O6 Phosphorus reaction with oxygen proceeds according to the
following equation
P4 + 3O2 P4O6
b The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule There is
one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs
This structure agrees with 31
P NMR spectrum indicating all phosphorus nuclei are chemically
equivalent
c P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation
P4O6 + 6H2O 4H3PO3
d Phosphoric(III) acid contains PndashH bond and is a diprotic acid of an average strength It forms
therefore two series of salts and sparsely soluble barium salt containing HPO32minus
anion will be
formed with the excess of barium hydroxide
H3PO3 + Ba(OH)2 3 + 2H2O
HPO32minus
anion exhibits tetrahedral structure (sp3 hybridisation of phosphorus) There are three
oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron wwwShimiPediair
10
Nickel coordination sphere
O
H
O
PO
O
PO
H
O
or
2 2
e Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base
in chemical reactions Therefore Lewis adduct is formed in the reaction with nickel carbonyl in
which phosphorus forms a chemical bond with nickel replacing one of the CO ligands Assuming
that the only gaseous product of the reaction is CO the molar ratio of the evolved CO to the used
P4O6 equals
0642 g 126 g
002292 000573 4 12801 gmol 21988 gmol
which indicates that four nickel atoms
were bound by a P4O6 molecule The reaction proceeds according to the following equation
4Ni(CO)4 + P4O6 P4O6middot4Ni(CO)3 + 4CO
The formula of compound C is P4O6middot4Ni(CO)3 (or P4O6[Ni(CO)3]4) The molar mass of
compound C equals 79076 gmol Such a reaction course is confirmed by the approximate molar
mass of compound C calculated from the reaction product mass
4 6
C
P O
45785 gmol
000573
CmM
n
f It follows from the fact that all of phosphorus nuclei are chemically
equivalent in the molecule of compound C that every phosphorus atom is
bound to one nickel atom and three oxygen atoms The coordination
centres in the form of nickel atoms satisfy the 18 electron rule (10 valence
electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons
coming from phosphorus) Therefore the structure of nickel coordination
sphere is tetrahedral and ligands bound to nickel atom are localised in
vertices of a slightly deformed tetrahedron
g Similarly as in the reaction with nickel carbonyl P4O6 phosphorus oxide serves as a Lewis base
in the reaction with B2H6 forming an adduct with acid (BH3) As a result of the reaction with
water the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and
H3BO3 boric acid (acid E) The liberated gas is molecular hydrogen and its quantity depends on
the amount of reacted diborane
2
2 6
H 3
B H 3
800595 10
6 2241 10 6m
Vn
V
mole which correspond to B2H6 mass equal to
2 6 2 6 2 6
3
B H B H B H 0595 10 2767 00165m n M g The molar ratio of P4O6 to B2H6 in the Lewis
adduct equals 3(0147 00165) g 0595 10 0594 0595 11
21988 gmol
Only two phosphorus atoms are involved in bond formation with BH3 and the formula of
compound D is P4O6 middot 2BH3 (or P4O6 middot
B2H6) The hydrolysis reaction proceeds according to the
following equation
P4O6 middot 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2
P Ni
C
C
CO
OO
O
O
O
wwwShimiPediair
11
h The reaction between P4O6 and B2H6 leading to the formation of compound D
(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition
Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid
The reaction product compound D is a Lewis adduct The yield of compound D formation
reaction equals
4 6 4 6 4 6
D D
P O P O P O
267 24755100 100 100 770
308 21988
Dn m Mw
n m M
i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron
B P
O
O
OH
HH
or
B P
O
O
OH
HH
SOLUTION OF TASK 3
a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows
clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means
that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the
range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex
and in the range c[E]0 ltlt 05 towards the E2M complex
b The complex formation reactions occurring at the equilibrium are as follows
(k1) E + M EM and (k2) 2E + M E2M
The complex formation equilibrium constants k1 and k2 also referred to as the complex
stability constants can be written as
)()2]E([]T[]E[
]ET[β
k21kk21k0
k11
ccccc
c
(1)
)()2E]([]T[]E[
]TE[β
k2k1
2
k21k0
k2
2
22
ccccc
c
(2)
c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be
determined by introducing the simplifications suggested in Note 1 In equation (1) we assume
that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2
in the first term of the denominator and get
cc
c
)]E([β
k10
1k1
(3)
In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0
and in the second term of the denominator we can neglect subtraction of ck1 and we get
)(]E[
βk2
2
0
k22
cc
c
(4)
Then from equations (3) and (4) we determine
1
101k
β1
β]E[
c
cc (5)
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
wwwShimiPediair
14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
wwwShimiPediair
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
wwwShimiPediair
6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
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9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
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10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
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6
3 To solve the problem in the high Na+ concentration limit ([E]0 ltlt c) it is necessary to
ldquolineariserdquo the dependence of the corresponding difference of chemical shifts (δobsndashδE) on
varying [Na+] = c concentration The dependence of 1(δobsndashδE) on 1c should be linear
(1(δobsndashδE) = a(1c) + b) in the limit of large values of the ratio c[E]0 ie in the conditions of
experiment 2 Fig 4 shows the relationship plotted using the data from Table 2
Fig 4
TASK 4
Biologically active indole derivatives
Tryptamines are derivatives of indole (a heterocyclic compound consisting of two condensed rings
a benzene ring and a pyrrole ring) abundantly present in living organisms To tryptamines belong
inter alia neurotransmiters as important as melatonin or serotonin Due to that also other
tryptamine derivatives are not neutral to human organism and may have medicinal use or exert
hallucinogenic effects
The phosphorylated tryptamine derivative F initially isolated from plant sources has been obtained
through the reaction sequence depicted in Scheme 1 The starting material for the synthesis was 4-
hydroxyindole It is known that the most reactive position of indole is the C-3 within the pyrrole
ring which easily undergoes aromatic electrophilic substitution but may also be involved in
reactions typical for enamines
First the phenol group of 4-hydroxyindole was protected by reacting it with benzyl chloride in the
presence of sodium methanolate This led to compound A which was then subjected to Friedel-
Crafts acylation with oxalyl chloride leading to compound B Subsequently B without isolation
was reacted with dimethylamine which resulted in compound C In the next step C was reduced to
D (molecular formula C19H22N2O) by means of LiAlH4 and then catalytically hydrogenated to yield
compound E and toluene as a by-product In the last step E was converted into F by applying
following reagents 1 n-buthyllithium 2 tetrabenzyl pyrophosphate (TBPP) 3 catalytic
hydrogenation (PdC H2)
A homologue of compound E known as 4-hydroxygramine and having molecular formula of
C11H14N2O may be obtained in a multi-component reaction (Mannich reaction) by reacting
compound A with formaldehyde and dimethylamine followed by a catalytic hydrogenation
Exploiting a similar multi-component reaction but from different starting materials one can obtain
compound X as a mixture of several stereoisomers (Scheme 2 depicts only one of them)
Experiment 2
wwwShimiPediair
7
O
ClO
Cl NH
A
NH
OH
Cl
OP
OP
O
OO
OO
C19H22N2O
TBPP =Bn
Bn
Bn
Bn
Scheme 1
MeOHMeO-
F
H2 Pd-C
1 n-BuLi2TBPP3 H2 Pd-C
[ B ]C
DE
LiAlH4
CH3
O
CH3
N O
O
NH
OCH
3Chiral
Scheme 2
342
1
X
Problems
a Draw structural formulas for compounds A-F and 4-hydroxygramine
b Draw structural formulas of the starting materials for the synthesis of compound X
c Compound X could be obtained from the same starting materials but in a stereochemically pure
form (as depicted in Scheme 2) Choose which conditions should be applied to achieve this
I the reaction temperature should be increased
II the reagents should be mixed in a certain order
III the reaction should be carried out in the presence of L-proline
IV the reaction mixture should be stirred always in the same direction
d Determine the absolute configuration of all stereogenic centers in compound X
e Explain the reactivity of the C-3 position of indole by means of appropriate mesomeric structures
TASK 5
The Diels-Alder Reaction
Cycloaddition reactions including Diels-Alder reaction in most cases are stereospecific what
means that geometry of substrates (eg dienophile or diene) determine the structure of the formed
product
X
Y Y
X+
General Scheme of the Diels-Alder Reaction
I The carboxylic acid with molecular formula C4H4O4 as well as its dimethyl esters (A B and C)
exist as two geometric isomers and one regioisomer and all of them undergo the Diels-Alder
reaction The reaction of geometric isomers A and B mixture with diene D composed entirely of
carbon and hydrogen atoms provides diastereomeric products E1 E2 and F Ozonolysis of E and F
mixture with zinc dust form product G (diastereomeric mixture) and subsequent reduction with
LiAlH4 excess leads to the mixture of diastereomers of polihydroxylic alcohol H
wwwShimiPediair
8
OHOH
OHOHLiAlH
4A i B
D E1 + E2 i F(diastereomers)
1) O3
2) ZnG
Hmixture of three diastereomers
1) O3
2) Zn
ID
J
mixture of three diastereomers
Compounds A and B after ozonolysis with zinc dust form the same product I which also undergo
cycloaddition reaction with diene D The reaction leads to the cyclic ether J The molecular mass of
J is equal to the sum of mass of reactants D and I
II Stereochemistry in Diels-Alder reaction Isomer A in the reaction with diene D forms the
mixture of two achiral diastereomers E1 and E2 The reduction of diastereomeric mixture of E1 and
E2 with hydrogen on palladium catalyst leads to only one product K
The Diels-Alder reaction of isomer B with diene D provides exclusively racemic mixture of
cycloadduct F The product F after ozonolysis with zinc dust and subsequent reduction with LiAlH4
excess leads to one diastereomer of polihydroxylic alcohol H (as a racemic mixture)
Problems
I
a Draw the structure of geometric isomer A or B and structure of regioisomer C
b Draw the structure of diene D
c Determine general structure of Diels-Alder product EF (no stereochemistry required)
d Draw the general structure of ozonolysis product G (no stereochemistry required)
e Draw the structure of compound I and general structure of J (no stereochemistry required)
II
a Determine unambiguously geometry of isomers A and B
b Draw stereochemical structures of compounds E (E1 and E2) and their reduction product K
c Draw the stereochemical structure of compound F and stereochemical structure of
corresponding diastereomer of polihydroxylic alcohol H
d Draw all possible stereoisomers of the product J Determine enantiomers and diastereomers
SOLUTIONS
SOLUTION OF TASK 1
a The total concentration of cyclam cL = [L] + [HL+] + [H2L
2+] + [H3L
3+] + [H4L
4+] Using acidic
dissociation constants after rearrangement one can write
cL = [L]1 + [H+]Ka4 + [H
+]2(Ka4 Ka3) + [H
+]3(Ka4 Ka3 Ka2) + [H
+]
4(Ka4 Ka3 Ka2 Ka1)
Taking into account the values of dissociation constants it can be assumed that at pH = 7 the
form H2L2+
predominates in the solution ie the above equation can be simplified to the form
cL = [L][H+]
2(Ka3 Ka4) thus [L] = cL Ka3 Ka4 [H
+]
2 After introducing numerical results for
cL = 001 M and [H+] = 10
7 M the obtained [L] = 9
10
11 M
wwwShimiPediair
9
OP
O
PO
P
O O
O
P
b Because stability constant = [ML]([M][L]) the ratio rsquo = cL[L] Basing
on previous data cL[L] = 001 910
11 = 11
10
8
c Because [ML][M] = 999 01 rsquo = [ML]([M]cL) = 999 (01 001) = 1
10
5
Thus = rsquo11
10
8 = 11
10
13
d = [ML]([M][L]) = [ML][H+]2([M]cL Ka3 Ka4) Assuming that a deposit of
hydroxide does not precipitate Ks0 = [M][OH]
2 the above equation can be written in the form
= [ML][H+]
2[OH
]
2(cL Ks0 Ka3 Ka4) = KW
2 ([ML]cL) ( Ks0 Ka3 Ka4)
e For [ML] = cL = 001 M using appropriate constants the minimal values of fulfilling the
conditions from (d) are 410
12 for the Cu
2+ complex and 6
10
8 for the Ni
2+ complex Comparison
of these results with experimental data shows that Cu2+
and Ni2+
complexes obey the conditions
from (c) and (d)
Because for Ni2+
- cyclam complex rsquo = [ML]([M]cL) for [ML] = cL [M] = 1rsquo is obtained
For the Ni2+
complex rsquo = 210
22 11
10
8 = 2
10
14 Thus [M] = [Ni
2+] = 1 (2
10
14) ie
[Ni2+
] = 5
10
15 M
2-fold dilution results in the same value of the [ML]cL ratio therefore Ni2+
ions concentration does
not change as well
SOLUTION OF TASK 2
a White phosphorus is built of P4 molecules Depending on the reaction conditions it may form
P4On (n = 6 7 8 9 or 10) oxides in the reaction with oxygen whose structures stem from P4
molecules structure Freezing point depression of benzene allows one to calculate the molality of
oxide A solution (cmA) and consequently its molar mass
0840125
430A
t
t
m
E
Tc molkg czyli 220
08400100
1850A
mAb
A
cm
mM gmol
This value corresponds to P4O6 molar mass (21988 gmol) Thus compound A is
phosphorus(III) oxide P4O6 Phosphorus reaction with oxygen proceeds according to the
following equation
P4 + 3O2 P4O6
b The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule There is
one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs
This structure agrees with 31
P NMR spectrum indicating all phosphorus nuclei are chemically
equivalent
c P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation
P4O6 + 6H2O 4H3PO3
d Phosphoric(III) acid contains PndashH bond and is a diprotic acid of an average strength It forms
therefore two series of salts and sparsely soluble barium salt containing HPO32minus
anion will be
formed with the excess of barium hydroxide
H3PO3 + Ba(OH)2 3 + 2H2O
HPO32minus
anion exhibits tetrahedral structure (sp3 hybridisation of phosphorus) There are three
oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron wwwShimiPediair
10
Nickel coordination sphere
O
H
O
PO
O
PO
H
O
or
2 2
e Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base
in chemical reactions Therefore Lewis adduct is formed in the reaction with nickel carbonyl in
which phosphorus forms a chemical bond with nickel replacing one of the CO ligands Assuming
that the only gaseous product of the reaction is CO the molar ratio of the evolved CO to the used
P4O6 equals
0642 g 126 g
002292 000573 4 12801 gmol 21988 gmol
which indicates that four nickel atoms
were bound by a P4O6 molecule The reaction proceeds according to the following equation
4Ni(CO)4 + P4O6 P4O6middot4Ni(CO)3 + 4CO
The formula of compound C is P4O6middot4Ni(CO)3 (or P4O6[Ni(CO)3]4) The molar mass of
compound C equals 79076 gmol Such a reaction course is confirmed by the approximate molar
mass of compound C calculated from the reaction product mass
4 6
C
P O
45785 gmol
000573
CmM
n
f It follows from the fact that all of phosphorus nuclei are chemically
equivalent in the molecule of compound C that every phosphorus atom is
bound to one nickel atom and three oxygen atoms The coordination
centres in the form of nickel atoms satisfy the 18 electron rule (10 valence
electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons
coming from phosphorus) Therefore the structure of nickel coordination
sphere is tetrahedral and ligands bound to nickel atom are localised in
vertices of a slightly deformed tetrahedron
g Similarly as in the reaction with nickel carbonyl P4O6 phosphorus oxide serves as a Lewis base
in the reaction with B2H6 forming an adduct with acid (BH3) As a result of the reaction with
water the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and
H3BO3 boric acid (acid E) The liberated gas is molecular hydrogen and its quantity depends on
the amount of reacted diborane
2
2 6
H 3
B H 3
800595 10
6 2241 10 6m
Vn
V
mole which correspond to B2H6 mass equal to
2 6 2 6 2 6
3
B H B H B H 0595 10 2767 00165m n M g The molar ratio of P4O6 to B2H6 in the Lewis
adduct equals 3(0147 00165) g 0595 10 0594 0595 11
21988 gmol
Only two phosphorus atoms are involved in bond formation with BH3 and the formula of
compound D is P4O6 middot 2BH3 (or P4O6 middot
B2H6) The hydrolysis reaction proceeds according to the
following equation
P4O6 middot 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2
P Ni
C
C
CO
OO
O
O
O
wwwShimiPediair
11
h The reaction between P4O6 and B2H6 leading to the formation of compound D
(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition
Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid
The reaction product compound D is a Lewis adduct The yield of compound D formation
reaction equals
4 6 4 6 4 6
D D
P O P O P O
267 24755100 100 100 770
308 21988
Dn m Mw
n m M
i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron
B P
O
O
OH
HH
or
B P
O
O
OH
HH
SOLUTION OF TASK 3
a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows
clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means
that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the
range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex
and in the range c[E]0 ltlt 05 towards the E2M complex
b The complex formation reactions occurring at the equilibrium are as follows
(k1) E + M EM and (k2) 2E + M E2M
The complex formation equilibrium constants k1 and k2 also referred to as the complex
stability constants can be written as
)()2]E([]T[]E[
]ET[β
k21kk21k0
k11
ccccc
c
(1)
)()2E]([]T[]E[
]TE[β
k2k1
2
k21k0
k2
2
22
ccccc
c
(2)
c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be
determined by introducing the simplifications suggested in Note 1 In equation (1) we assume
that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2
in the first term of the denominator and get
cc
c
)]E([β
k10
1k1
(3)
In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0
and in the second term of the denominator we can neglect subtraction of ck1 and we get
)(]E[
βk2
2
0
k22
cc
c
(4)
Then from equations (3) and (4) we determine
1
101k
β1
β]E[
c
cc (5)
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
wwwShimiPediair
14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
wwwShimiPediair
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
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6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
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8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
7
O
ClO
Cl NH
A
NH
OH
Cl
OP
OP
O
OO
OO
C19H22N2O
TBPP =Bn
Bn
Bn
Bn
Scheme 1
MeOHMeO-
F
H2 Pd-C
1 n-BuLi2TBPP3 H2 Pd-C
[ B ]C
DE
LiAlH4
CH3
O
CH3
N O
O
NH
OCH
3Chiral
Scheme 2
342
1
X
Problems
a Draw structural formulas for compounds A-F and 4-hydroxygramine
b Draw structural formulas of the starting materials for the synthesis of compound X
c Compound X could be obtained from the same starting materials but in a stereochemically pure
form (as depicted in Scheme 2) Choose which conditions should be applied to achieve this
I the reaction temperature should be increased
II the reagents should be mixed in a certain order
III the reaction should be carried out in the presence of L-proline
IV the reaction mixture should be stirred always in the same direction
d Determine the absolute configuration of all stereogenic centers in compound X
e Explain the reactivity of the C-3 position of indole by means of appropriate mesomeric structures
TASK 5
The Diels-Alder Reaction
Cycloaddition reactions including Diels-Alder reaction in most cases are stereospecific what
means that geometry of substrates (eg dienophile or diene) determine the structure of the formed
product
X
Y Y
X+
General Scheme of the Diels-Alder Reaction
I The carboxylic acid with molecular formula C4H4O4 as well as its dimethyl esters (A B and C)
exist as two geometric isomers and one regioisomer and all of them undergo the Diels-Alder
reaction The reaction of geometric isomers A and B mixture with diene D composed entirely of
carbon and hydrogen atoms provides diastereomeric products E1 E2 and F Ozonolysis of E and F
mixture with zinc dust form product G (diastereomeric mixture) and subsequent reduction with
LiAlH4 excess leads to the mixture of diastereomers of polihydroxylic alcohol H
wwwShimiPediair
8
OHOH
OHOHLiAlH
4A i B
D E1 + E2 i F(diastereomers)
1) O3
2) ZnG
Hmixture of three diastereomers
1) O3
2) Zn
ID
J
mixture of three diastereomers
Compounds A and B after ozonolysis with zinc dust form the same product I which also undergo
cycloaddition reaction with diene D The reaction leads to the cyclic ether J The molecular mass of
J is equal to the sum of mass of reactants D and I
II Stereochemistry in Diels-Alder reaction Isomer A in the reaction with diene D forms the
mixture of two achiral diastereomers E1 and E2 The reduction of diastereomeric mixture of E1 and
E2 with hydrogen on palladium catalyst leads to only one product K
The Diels-Alder reaction of isomer B with diene D provides exclusively racemic mixture of
cycloadduct F The product F after ozonolysis with zinc dust and subsequent reduction with LiAlH4
excess leads to one diastereomer of polihydroxylic alcohol H (as a racemic mixture)
Problems
I
a Draw the structure of geometric isomer A or B and structure of regioisomer C
b Draw the structure of diene D
c Determine general structure of Diels-Alder product EF (no stereochemistry required)
d Draw the general structure of ozonolysis product G (no stereochemistry required)
e Draw the structure of compound I and general structure of J (no stereochemistry required)
II
a Determine unambiguously geometry of isomers A and B
b Draw stereochemical structures of compounds E (E1 and E2) and their reduction product K
c Draw the stereochemical structure of compound F and stereochemical structure of
corresponding diastereomer of polihydroxylic alcohol H
d Draw all possible stereoisomers of the product J Determine enantiomers and diastereomers
SOLUTIONS
SOLUTION OF TASK 1
a The total concentration of cyclam cL = [L] + [HL+] + [H2L
2+] + [H3L
3+] + [H4L
4+] Using acidic
dissociation constants after rearrangement one can write
cL = [L]1 + [H+]Ka4 + [H
+]2(Ka4 Ka3) + [H
+]3(Ka4 Ka3 Ka2) + [H
+]
4(Ka4 Ka3 Ka2 Ka1)
Taking into account the values of dissociation constants it can be assumed that at pH = 7 the
form H2L2+
predominates in the solution ie the above equation can be simplified to the form
cL = [L][H+]
2(Ka3 Ka4) thus [L] = cL Ka3 Ka4 [H
+]
2 After introducing numerical results for
cL = 001 M and [H+] = 10
7 M the obtained [L] = 9
10
11 M
wwwShimiPediair
9
OP
O
PO
P
O O
O
P
b Because stability constant = [ML]([M][L]) the ratio rsquo = cL[L] Basing
on previous data cL[L] = 001 910
11 = 11
10
8
c Because [ML][M] = 999 01 rsquo = [ML]([M]cL) = 999 (01 001) = 1
10
5
Thus = rsquo11
10
8 = 11
10
13
d = [ML]([M][L]) = [ML][H+]2([M]cL Ka3 Ka4) Assuming that a deposit of
hydroxide does not precipitate Ks0 = [M][OH]
2 the above equation can be written in the form
= [ML][H+]
2[OH
]
2(cL Ks0 Ka3 Ka4) = KW
2 ([ML]cL) ( Ks0 Ka3 Ka4)
e For [ML] = cL = 001 M using appropriate constants the minimal values of fulfilling the
conditions from (d) are 410
12 for the Cu
2+ complex and 6
10
8 for the Ni
2+ complex Comparison
of these results with experimental data shows that Cu2+
and Ni2+
complexes obey the conditions
from (c) and (d)
Because for Ni2+
- cyclam complex rsquo = [ML]([M]cL) for [ML] = cL [M] = 1rsquo is obtained
For the Ni2+
complex rsquo = 210
22 11
10
8 = 2
10
14 Thus [M] = [Ni
2+] = 1 (2
10
14) ie
[Ni2+
] = 5
10
15 M
2-fold dilution results in the same value of the [ML]cL ratio therefore Ni2+
ions concentration does
not change as well
SOLUTION OF TASK 2
a White phosphorus is built of P4 molecules Depending on the reaction conditions it may form
P4On (n = 6 7 8 9 or 10) oxides in the reaction with oxygen whose structures stem from P4
molecules structure Freezing point depression of benzene allows one to calculate the molality of
oxide A solution (cmA) and consequently its molar mass
0840125
430A
t
t
m
E
Tc molkg czyli 220
08400100
1850A
mAb
A
cm
mM gmol
This value corresponds to P4O6 molar mass (21988 gmol) Thus compound A is
phosphorus(III) oxide P4O6 Phosphorus reaction with oxygen proceeds according to the
following equation
P4 + 3O2 P4O6
b The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule There is
one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs
This structure agrees with 31
P NMR spectrum indicating all phosphorus nuclei are chemically
equivalent
c P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation
P4O6 + 6H2O 4H3PO3
d Phosphoric(III) acid contains PndashH bond and is a diprotic acid of an average strength It forms
therefore two series of salts and sparsely soluble barium salt containing HPO32minus
anion will be
formed with the excess of barium hydroxide
H3PO3 + Ba(OH)2 3 + 2H2O
HPO32minus
anion exhibits tetrahedral structure (sp3 hybridisation of phosphorus) There are three
oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron wwwShimiPediair
10
Nickel coordination sphere
O
H
O
PO
O
PO
H
O
or
2 2
e Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base
in chemical reactions Therefore Lewis adduct is formed in the reaction with nickel carbonyl in
which phosphorus forms a chemical bond with nickel replacing one of the CO ligands Assuming
that the only gaseous product of the reaction is CO the molar ratio of the evolved CO to the used
P4O6 equals
0642 g 126 g
002292 000573 4 12801 gmol 21988 gmol
which indicates that four nickel atoms
were bound by a P4O6 molecule The reaction proceeds according to the following equation
4Ni(CO)4 + P4O6 P4O6middot4Ni(CO)3 + 4CO
The formula of compound C is P4O6middot4Ni(CO)3 (or P4O6[Ni(CO)3]4) The molar mass of
compound C equals 79076 gmol Such a reaction course is confirmed by the approximate molar
mass of compound C calculated from the reaction product mass
4 6
C
P O
45785 gmol
000573
CmM
n
f It follows from the fact that all of phosphorus nuclei are chemically
equivalent in the molecule of compound C that every phosphorus atom is
bound to one nickel atom and three oxygen atoms The coordination
centres in the form of nickel atoms satisfy the 18 electron rule (10 valence
electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons
coming from phosphorus) Therefore the structure of nickel coordination
sphere is tetrahedral and ligands bound to nickel atom are localised in
vertices of a slightly deformed tetrahedron
g Similarly as in the reaction with nickel carbonyl P4O6 phosphorus oxide serves as a Lewis base
in the reaction with B2H6 forming an adduct with acid (BH3) As a result of the reaction with
water the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and
H3BO3 boric acid (acid E) The liberated gas is molecular hydrogen and its quantity depends on
the amount of reacted diborane
2
2 6
H 3
B H 3
800595 10
6 2241 10 6m
Vn
V
mole which correspond to B2H6 mass equal to
2 6 2 6 2 6
3
B H B H B H 0595 10 2767 00165m n M g The molar ratio of P4O6 to B2H6 in the Lewis
adduct equals 3(0147 00165) g 0595 10 0594 0595 11
21988 gmol
Only two phosphorus atoms are involved in bond formation with BH3 and the formula of
compound D is P4O6 middot 2BH3 (or P4O6 middot
B2H6) The hydrolysis reaction proceeds according to the
following equation
P4O6 middot 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2
P Ni
C
C
CO
OO
O
O
O
wwwShimiPediair
11
h The reaction between P4O6 and B2H6 leading to the formation of compound D
(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition
Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid
The reaction product compound D is a Lewis adduct The yield of compound D formation
reaction equals
4 6 4 6 4 6
D D
P O P O P O
267 24755100 100 100 770
308 21988
Dn m Mw
n m M
i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron
B P
O
O
OH
HH
or
B P
O
O
OH
HH
SOLUTION OF TASK 3
a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows
clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means
that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the
range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex
and in the range c[E]0 ltlt 05 towards the E2M complex
b The complex formation reactions occurring at the equilibrium are as follows
(k1) E + M EM and (k2) 2E + M E2M
The complex formation equilibrium constants k1 and k2 also referred to as the complex
stability constants can be written as
)()2]E([]T[]E[
]ET[β
k21kk21k0
k11
ccccc
c
(1)
)()2E]([]T[]E[
]TE[β
k2k1
2
k21k0
k2
2
22
ccccc
c
(2)
c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be
determined by introducing the simplifications suggested in Note 1 In equation (1) we assume
that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2
in the first term of the denominator and get
cc
c
)]E([β
k10
1k1
(3)
In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0
and in the second term of the denominator we can neglect subtraction of ck1 and we get
)(]E[
βk2
2
0
k22
cc
c
(4)
Then from equations (3) and (4) we determine
1
101k
β1
β]E[
c
cc (5)
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
wwwShimiPediair
14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
wwwShimiPediair
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
wwwShimiPediair
6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
8
OHOH
OHOHLiAlH
4A i B
D E1 + E2 i F(diastereomers)
1) O3
2) ZnG
Hmixture of three diastereomers
1) O3
2) Zn
ID
J
mixture of three diastereomers
Compounds A and B after ozonolysis with zinc dust form the same product I which also undergo
cycloaddition reaction with diene D The reaction leads to the cyclic ether J The molecular mass of
J is equal to the sum of mass of reactants D and I
II Stereochemistry in Diels-Alder reaction Isomer A in the reaction with diene D forms the
mixture of two achiral diastereomers E1 and E2 The reduction of diastereomeric mixture of E1 and
E2 with hydrogen on palladium catalyst leads to only one product K
The Diels-Alder reaction of isomer B with diene D provides exclusively racemic mixture of
cycloadduct F The product F after ozonolysis with zinc dust and subsequent reduction with LiAlH4
excess leads to one diastereomer of polihydroxylic alcohol H (as a racemic mixture)
Problems
I
a Draw the structure of geometric isomer A or B and structure of regioisomer C
b Draw the structure of diene D
c Determine general structure of Diels-Alder product EF (no stereochemistry required)
d Draw the general structure of ozonolysis product G (no stereochemistry required)
e Draw the structure of compound I and general structure of J (no stereochemistry required)
II
a Determine unambiguously geometry of isomers A and B
b Draw stereochemical structures of compounds E (E1 and E2) and their reduction product K
c Draw the stereochemical structure of compound F and stereochemical structure of
corresponding diastereomer of polihydroxylic alcohol H
d Draw all possible stereoisomers of the product J Determine enantiomers and diastereomers
SOLUTIONS
SOLUTION OF TASK 1
a The total concentration of cyclam cL = [L] + [HL+] + [H2L
2+] + [H3L
3+] + [H4L
4+] Using acidic
dissociation constants after rearrangement one can write
cL = [L]1 + [H+]Ka4 + [H
+]2(Ka4 Ka3) + [H
+]3(Ka4 Ka3 Ka2) + [H
+]
4(Ka4 Ka3 Ka2 Ka1)
Taking into account the values of dissociation constants it can be assumed that at pH = 7 the
form H2L2+
predominates in the solution ie the above equation can be simplified to the form
cL = [L][H+]
2(Ka3 Ka4) thus [L] = cL Ka3 Ka4 [H
+]
2 After introducing numerical results for
cL = 001 M and [H+] = 10
7 M the obtained [L] = 9
10
11 M
wwwShimiPediair
9
OP
O
PO
P
O O
O
P
b Because stability constant = [ML]([M][L]) the ratio rsquo = cL[L] Basing
on previous data cL[L] = 001 910
11 = 11
10
8
c Because [ML][M] = 999 01 rsquo = [ML]([M]cL) = 999 (01 001) = 1
10
5
Thus = rsquo11
10
8 = 11
10
13
d = [ML]([M][L]) = [ML][H+]2([M]cL Ka3 Ka4) Assuming that a deposit of
hydroxide does not precipitate Ks0 = [M][OH]
2 the above equation can be written in the form
= [ML][H+]
2[OH
]
2(cL Ks0 Ka3 Ka4) = KW
2 ([ML]cL) ( Ks0 Ka3 Ka4)
e For [ML] = cL = 001 M using appropriate constants the minimal values of fulfilling the
conditions from (d) are 410
12 for the Cu
2+ complex and 6
10
8 for the Ni
2+ complex Comparison
of these results with experimental data shows that Cu2+
and Ni2+
complexes obey the conditions
from (c) and (d)
Because for Ni2+
- cyclam complex rsquo = [ML]([M]cL) for [ML] = cL [M] = 1rsquo is obtained
For the Ni2+
complex rsquo = 210
22 11
10
8 = 2
10
14 Thus [M] = [Ni
2+] = 1 (2
10
14) ie
[Ni2+
] = 5
10
15 M
2-fold dilution results in the same value of the [ML]cL ratio therefore Ni2+
ions concentration does
not change as well
SOLUTION OF TASK 2
a White phosphorus is built of P4 molecules Depending on the reaction conditions it may form
P4On (n = 6 7 8 9 or 10) oxides in the reaction with oxygen whose structures stem from P4
molecules structure Freezing point depression of benzene allows one to calculate the molality of
oxide A solution (cmA) and consequently its molar mass
0840125
430A
t
t
m
E
Tc molkg czyli 220
08400100
1850A
mAb
A
cm
mM gmol
This value corresponds to P4O6 molar mass (21988 gmol) Thus compound A is
phosphorus(III) oxide P4O6 Phosphorus reaction with oxygen proceeds according to the
following equation
P4 + 3O2 P4O6
b The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule There is
one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs
This structure agrees with 31
P NMR spectrum indicating all phosphorus nuclei are chemically
equivalent
c P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation
P4O6 + 6H2O 4H3PO3
d Phosphoric(III) acid contains PndashH bond and is a diprotic acid of an average strength It forms
therefore two series of salts and sparsely soluble barium salt containing HPO32minus
anion will be
formed with the excess of barium hydroxide
H3PO3 + Ba(OH)2 3 + 2H2O
HPO32minus
anion exhibits tetrahedral structure (sp3 hybridisation of phosphorus) There are three
oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron wwwShimiPediair
10
Nickel coordination sphere
O
H
O
PO
O
PO
H
O
or
2 2
e Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base
in chemical reactions Therefore Lewis adduct is formed in the reaction with nickel carbonyl in
which phosphorus forms a chemical bond with nickel replacing one of the CO ligands Assuming
that the only gaseous product of the reaction is CO the molar ratio of the evolved CO to the used
P4O6 equals
0642 g 126 g
002292 000573 4 12801 gmol 21988 gmol
which indicates that four nickel atoms
were bound by a P4O6 molecule The reaction proceeds according to the following equation
4Ni(CO)4 + P4O6 P4O6middot4Ni(CO)3 + 4CO
The formula of compound C is P4O6middot4Ni(CO)3 (or P4O6[Ni(CO)3]4) The molar mass of
compound C equals 79076 gmol Such a reaction course is confirmed by the approximate molar
mass of compound C calculated from the reaction product mass
4 6
C
P O
45785 gmol
000573
CmM
n
f It follows from the fact that all of phosphorus nuclei are chemically
equivalent in the molecule of compound C that every phosphorus atom is
bound to one nickel atom and three oxygen atoms The coordination
centres in the form of nickel atoms satisfy the 18 electron rule (10 valence
electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons
coming from phosphorus) Therefore the structure of nickel coordination
sphere is tetrahedral and ligands bound to nickel atom are localised in
vertices of a slightly deformed tetrahedron
g Similarly as in the reaction with nickel carbonyl P4O6 phosphorus oxide serves as a Lewis base
in the reaction with B2H6 forming an adduct with acid (BH3) As a result of the reaction with
water the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and
H3BO3 boric acid (acid E) The liberated gas is molecular hydrogen and its quantity depends on
the amount of reacted diborane
2
2 6
H 3
B H 3
800595 10
6 2241 10 6m
Vn
V
mole which correspond to B2H6 mass equal to
2 6 2 6 2 6
3
B H B H B H 0595 10 2767 00165m n M g The molar ratio of P4O6 to B2H6 in the Lewis
adduct equals 3(0147 00165) g 0595 10 0594 0595 11
21988 gmol
Only two phosphorus atoms are involved in bond formation with BH3 and the formula of
compound D is P4O6 middot 2BH3 (or P4O6 middot
B2H6) The hydrolysis reaction proceeds according to the
following equation
P4O6 middot 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2
P Ni
C
C
CO
OO
O
O
O
wwwShimiPediair
11
h The reaction between P4O6 and B2H6 leading to the formation of compound D
(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition
Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid
The reaction product compound D is a Lewis adduct The yield of compound D formation
reaction equals
4 6 4 6 4 6
D D
P O P O P O
267 24755100 100 100 770
308 21988
Dn m Mw
n m M
i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron
B P
O
O
OH
HH
or
B P
O
O
OH
HH
SOLUTION OF TASK 3
a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows
clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means
that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the
range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex
and in the range c[E]0 ltlt 05 towards the E2M complex
b The complex formation reactions occurring at the equilibrium are as follows
(k1) E + M EM and (k2) 2E + M E2M
The complex formation equilibrium constants k1 and k2 also referred to as the complex
stability constants can be written as
)()2]E([]T[]E[
]ET[β
k21kk21k0
k11
ccccc
c
(1)
)()2E]([]T[]E[
]TE[β
k2k1
2
k21k0
k2
2
22
ccccc
c
(2)
c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be
determined by introducing the simplifications suggested in Note 1 In equation (1) we assume
that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2
in the first term of the denominator and get
cc
c
)]E([β
k10
1k1
(3)
In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0
and in the second term of the denominator we can neglect subtraction of ck1 and we get
)(]E[
βk2
2
0
k22
cc
c
(4)
Then from equations (3) and (4) we determine
1
101k
β1
β]E[
c
cc (5)
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
wwwShimiPediair
14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
wwwShimiPediair
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
wwwShimiPediair
6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
9
OP
O
PO
P
O O
O
P
b Because stability constant = [ML]([M][L]) the ratio rsquo = cL[L] Basing
on previous data cL[L] = 001 910
11 = 11
10
8
c Because [ML][M] = 999 01 rsquo = [ML]([M]cL) = 999 (01 001) = 1
10
5
Thus = rsquo11
10
8 = 11
10
13
d = [ML]([M][L]) = [ML][H+]2([M]cL Ka3 Ka4) Assuming that a deposit of
hydroxide does not precipitate Ks0 = [M][OH]
2 the above equation can be written in the form
= [ML][H+]
2[OH
]
2(cL Ks0 Ka3 Ka4) = KW
2 ([ML]cL) ( Ks0 Ka3 Ka4)
e For [ML] = cL = 001 M using appropriate constants the minimal values of fulfilling the
conditions from (d) are 410
12 for the Cu
2+ complex and 6
10
8 for the Ni
2+ complex Comparison
of these results with experimental data shows that Cu2+
and Ni2+
complexes obey the conditions
from (c) and (d)
Because for Ni2+
- cyclam complex rsquo = [ML]([M]cL) for [ML] = cL [M] = 1rsquo is obtained
For the Ni2+
complex rsquo = 210
22 11
10
8 = 2
10
14 Thus [M] = [Ni
2+] = 1 (2
10
14) ie
[Ni2+
] = 5
10
15 M
2-fold dilution results in the same value of the [ML]cL ratio therefore Ni2+
ions concentration does
not change as well
SOLUTION OF TASK 2
a White phosphorus is built of P4 molecules Depending on the reaction conditions it may form
P4On (n = 6 7 8 9 or 10) oxides in the reaction with oxygen whose structures stem from P4
molecules structure Freezing point depression of benzene allows one to calculate the molality of
oxide A solution (cmA) and consequently its molar mass
0840125
430A
t
t
m
E
Tc molkg czyli 220
08400100
1850A
mAb
A
cm
mM gmol
This value corresponds to P4O6 molar mass (21988 gmol) Thus compound A is
phosphorus(III) oxide P4O6 Phosphorus reaction with oxygen proceeds according to the
following equation
P4 + 3O2 P4O6
b The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule There is
one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs
This structure agrees with 31
P NMR spectrum indicating all phosphorus nuclei are chemically
equivalent
c P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation
P4O6 + 6H2O 4H3PO3
d Phosphoric(III) acid contains PndashH bond and is a diprotic acid of an average strength It forms
therefore two series of salts and sparsely soluble barium salt containing HPO32minus
anion will be
formed with the excess of barium hydroxide
H3PO3 + Ba(OH)2 3 + 2H2O
HPO32minus
anion exhibits tetrahedral structure (sp3 hybridisation of phosphorus) There are three
oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron wwwShimiPediair
10
Nickel coordination sphere
O
H
O
PO
O
PO
H
O
or
2 2
e Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base
in chemical reactions Therefore Lewis adduct is formed in the reaction with nickel carbonyl in
which phosphorus forms a chemical bond with nickel replacing one of the CO ligands Assuming
that the only gaseous product of the reaction is CO the molar ratio of the evolved CO to the used
P4O6 equals
0642 g 126 g
002292 000573 4 12801 gmol 21988 gmol
which indicates that four nickel atoms
were bound by a P4O6 molecule The reaction proceeds according to the following equation
4Ni(CO)4 + P4O6 P4O6middot4Ni(CO)3 + 4CO
The formula of compound C is P4O6middot4Ni(CO)3 (or P4O6[Ni(CO)3]4) The molar mass of
compound C equals 79076 gmol Such a reaction course is confirmed by the approximate molar
mass of compound C calculated from the reaction product mass
4 6
C
P O
45785 gmol
000573
CmM
n
f It follows from the fact that all of phosphorus nuclei are chemically
equivalent in the molecule of compound C that every phosphorus atom is
bound to one nickel atom and three oxygen atoms The coordination
centres in the form of nickel atoms satisfy the 18 electron rule (10 valence
electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons
coming from phosphorus) Therefore the structure of nickel coordination
sphere is tetrahedral and ligands bound to nickel atom are localised in
vertices of a slightly deformed tetrahedron
g Similarly as in the reaction with nickel carbonyl P4O6 phosphorus oxide serves as a Lewis base
in the reaction with B2H6 forming an adduct with acid (BH3) As a result of the reaction with
water the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and
H3BO3 boric acid (acid E) The liberated gas is molecular hydrogen and its quantity depends on
the amount of reacted diborane
2
2 6
H 3
B H 3
800595 10
6 2241 10 6m
Vn
V
mole which correspond to B2H6 mass equal to
2 6 2 6 2 6
3
B H B H B H 0595 10 2767 00165m n M g The molar ratio of P4O6 to B2H6 in the Lewis
adduct equals 3(0147 00165) g 0595 10 0594 0595 11
21988 gmol
Only two phosphorus atoms are involved in bond formation with BH3 and the formula of
compound D is P4O6 middot 2BH3 (or P4O6 middot
B2H6) The hydrolysis reaction proceeds according to the
following equation
P4O6 middot 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2
P Ni
C
C
CO
OO
O
O
O
wwwShimiPediair
11
h The reaction between P4O6 and B2H6 leading to the formation of compound D
(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition
Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid
The reaction product compound D is a Lewis adduct The yield of compound D formation
reaction equals
4 6 4 6 4 6
D D
P O P O P O
267 24755100 100 100 770
308 21988
Dn m Mw
n m M
i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron
B P
O
O
OH
HH
or
B P
O
O
OH
HH
SOLUTION OF TASK 3
a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows
clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means
that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the
range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex
and in the range c[E]0 ltlt 05 towards the E2M complex
b The complex formation reactions occurring at the equilibrium are as follows
(k1) E + M EM and (k2) 2E + M E2M
The complex formation equilibrium constants k1 and k2 also referred to as the complex
stability constants can be written as
)()2]E([]T[]E[
]ET[β
k21kk21k0
k11
ccccc
c
(1)
)()2E]([]T[]E[
]TE[β
k2k1
2
k21k0
k2
2
22
ccccc
c
(2)
c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be
determined by introducing the simplifications suggested in Note 1 In equation (1) we assume
that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2
in the first term of the denominator and get
cc
c
)]E([β
k10
1k1
(3)
In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0
and in the second term of the denominator we can neglect subtraction of ck1 and we get
)(]E[
βk2
2
0
k22
cc
c
(4)
Then from equations (3) and (4) we determine
1
101k
β1
β]E[
c
cc (5)
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
wwwShimiPediair
14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
wwwShimiPediair
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
wwwShimiPediair
6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
10
Nickel coordination sphere
O
H
O
PO
O
PO
H
O
or
2 2
e Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base
in chemical reactions Therefore Lewis adduct is formed in the reaction with nickel carbonyl in
which phosphorus forms a chemical bond with nickel replacing one of the CO ligands Assuming
that the only gaseous product of the reaction is CO the molar ratio of the evolved CO to the used
P4O6 equals
0642 g 126 g
002292 000573 4 12801 gmol 21988 gmol
which indicates that four nickel atoms
were bound by a P4O6 molecule The reaction proceeds according to the following equation
4Ni(CO)4 + P4O6 P4O6middot4Ni(CO)3 + 4CO
The formula of compound C is P4O6middot4Ni(CO)3 (or P4O6[Ni(CO)3]4) The molar mass of
compound C equals 79076 gmol Such a reaction course is confirmed by the approximate molar
mass of compound C calculated from the reaction product mass
4 6
C
P O
45785 gmol
000573
CmM
n
f It follows from the fact that all of phosphorus nuclei are chemically
equivalent in the molecule of compound C that every phosphorus atom is
bound to one nickel atom and three oxygen atoms The coordination
centres in the form of nickel atoms satisfy the 18 electron rule (10 valence
electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons
coming from phosphorus) Therefore the structure of nickel coordination
sphere is tetrahedral and ligands bound to nickel atom are localised in
vertices of a slightly deformed tetrahedron
g Similarly as in the reaction with nickel carbonyl P4O6 phosphorus oxide serves as a Lewis base
in the reaction with B2H6 forming an adduct with acid (BH3) As a result of the reaction with
water the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and
H3BO3 boric acid (acid E) The liberated gas is molecular hydrogen and its quantity depends on
the amount of reacted diborane
2
2 6
H 3
B H 3
800595 10
6 2241 10 6m
Vn
V
mole which correspond to B2H6 mass equal to
2 6 2 6 2 6
3
B H B H B H 0595 10 2767 00165m n M g The molar ratio of P4O6 to B2H6 in the Lewis
adduct equals 3(0147 00165) g 0595 10 0594 0595 11
21988 gmol
Only two phosphorus atoms are involved in bond formation with BH3 and the formula of
compound D is P4O6 middot 2BH3 (or P4O6 middot
B2H6) The hydrolysis reaction proceeds according to the
following equation
P4O6 middot 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2
P Ni
C
C
CO
OO
O
O
O
wwwShimiPediair
11
h The reaction between P4O6 and B2H6 leading to the formation of compound D
(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition
Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid
The reaction product compound D is a Lewis adduct The yield of compound D formation
reaction equals
4 6 4 6 4 6
D D
P O P O P O
267 24755100 100 100 770
308 21988
Dn m Mw
n m M
i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron
B P
O
O
OH
HH
or
B P
O
O
OH
HH
SOLUTION OF TASK 3
a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows
clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means
that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the
range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex
and in the range c[E]0 ltlt 05 towards the E2M complex
b The complex formation reactions occurring at the equilibrium are as follows
(k1) E + M EM and (k2) 2E + M E2M
The complex formation equilibrium constants k1 and k2 also referred to as the complex
stability constants can be written as
)()2]E([]T[]E[
]ET[β
k21kk21k0
k11
ccccc
c
(1)
)()2E]([]T[]E[
]TE[β
k2k1
2
k21k0
k2
2
22
ccccc
c
(2)
c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be
determined by introducing the simplifications suggested in Note 1 In equation (1) we assume
that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2
in the first term of the denominator and get
cc
c
)]E([β
k10
1k1
(3)
In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0
and in the second term of the denominator we can neglect subtraction of ck1 and we get
)(]E[
βk2
2
0
k22
cc
c
(4)
Then from equations (3) and (4) we determine
1
101k
β1
β]E[
c
cc (5)
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
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14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
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5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
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3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
wwwShimiPediair
6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
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10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
11
h The reaction between P4O6 and B2H6 leading to the formation of compound D
(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition
Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid
The reaction product compound D is a Lewis adduct The yield of compound D formation
reaction equals
4 6 4 6 4 6
D D
P O P O P O
267 24755100 100 100 770
308 21988
Dn m Mw
n m M
i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron
B P
O
O
OH
HH
or
B P
O
O
OH
HH
SOLUTION OF TASK 3
a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows
clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means
that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the
range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex
and in the range c[E]0 ltlt 05 towards the E2M complex
b The complex formation reactions occurring at the equilibrium are as follows
(k1) E + M EM and (k2) 2E + M E2M
The complex formation equilibrium constants k1 and k2 also referred to as the complex
stability constants can be written as
)()2]E([]T[]E[
]ET[β
k21kk21k0
k11
ccccc
c
(1)
)()2E]([]T[]E[
]TE[β
k2k1
2
k21k0
k2
2
22
ccccc
c
(2)
c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be
determined by introducing the simplifications suggested in Note 1 In equation (1) we assume
that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2
in the first term of the denominator and get
cc
c
)]E([β
k10
1k1
(3)
In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0
and in the second term of the denominator we can neglect subtraction of ck1 and we get
)(]E[
βk2
2
0
k22
cc
c
(4)
Then from equations (3) and (4) we determine
1
101k
β1
β]E[
c
cc (5)
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
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14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
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5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
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3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
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6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
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7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
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8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
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9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
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10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
12
and 2
2
0
2
2
0k2
β]E[1
β]E[
cc (6)
d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the
complexes δk1 and δk2 can be written as follows
k2
0
k1k1
0
k1
0
k2k10
[E]2
[E][E]
2[E]δ
cδ
cδ
ccδ Eobs
(7)
remembering that [E]0 = [E] + ck1 + 2 ck2
By performing corresponding multiplications reducing terms in equation (7) and introducing
substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2
we obtain
k2
0
k2k1
0
k1obs
]E[2
]E[
cc (8)
Equation (8) should be written as two equations for limiting conditions that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain
k1
1
11obs_k
β1
β
c
c (9)
and k2
2
2
0
20obs_k2
β]E[1
β]E[2
c (10)
e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10) written as a function of c
c
2
2
0
k220obs_k2
β]E[1
β]E[2
(11)
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor
m can be determined from the point that is farthest away from the 00 point but still
belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates
of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3
and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine
β2 and having substituted the data we get β2 = 201 (moldm3)-2
f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals as suggested by Note 4 (by
rising both sides of the equation to the power minus1)
Ek11Ek11
1
Eobs_k1
11
β
11
β
β11
cc
c (12)
And convert to the form
Ek1Ek11
1
Eobs_k1
11
β
11
c (13)
The plot in Fig 4 shows ideal linearity of data Eobs_k1
1
= a(1c)+b so comparing
corresponding parameters of the straight line with equation (13) we get the following
relationships
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13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
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14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
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5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
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3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
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4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
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6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
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8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
13
)(β
1
Ek11 a and
Ek1
1
b
For two extreme points from Table 2 we calculate corresponding reciprocals as given in the
Table below
c (moldm3) (1c) (moldm
3)-1
δobs ppm (1(δobs
ndash δE)
ppm
-1
0204 4902 4437 0597
0768 1302 4856 0477
Then we calculate parameters of the straight line a = 00332 ppm-1(moldm
3) and b = 0434 ppm
-1
and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and
β1= 131 (moldm3)-1
SOLUTION OF TASK 4
a A
B
C
D
E
F
4-hydroksygramina
b Reactants for the synthesis of X
c Conditions for
stereoselective
reaction III
d Absolute
configuration of X
3S 4S
e Reactivity of position C-3 of indole comes from high electron density in this position In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring which would be energetically unfavorable
struktura I struktura II struktura III
wwwShimiPediair
14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
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5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
wwwShimiPediair
6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
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7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
14
SOLUTION OF TASK 5
Part I
I a
CO2Me
MeO2C
CO2Me
CO2Me
or
(A or B)
I a CO
2MeMeO
2C
C I b D
I c
CO2Me
CO2Me
CO2Me
CO2Me
or
E or F
I d
CO2Me
CO2Me
O
OG
I e
OO
OMeI
I e
O
CO2Me
O
CO2Me
or
J
Part II
II a CO
2Me
MeO2C
CO2Me
CO2Me
A B
(Z)-isomer (E)-isomer
II b
CO2Me
CO2Me
CO2Me
CO2Me
H
CO2Me
CO2Me
H
H
H
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
+
+
+
cycloadducts cis E (E1 i E2)
two diastereomers (achiral)
(from ester A)
or
or
II c
OHOH
OHOH
OHOH
OHOHor
diastereomer of compound H
obtained from cykloadduct F
II b
CO2Me
CO2Me
CO2Me
CO2Me
K IIc
CO2Me
CO2Me
H
CO2Me
H
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
MeO2C
MeO2C
H
CO2Me
H
MeO2C
enantiomers
cycloadduct trans F
(obtained from ester B)
or
or
wwwShimiPediair
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
wwwShimiPediair
6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its
hydroxyl groups are substituted with chloride ions and as a result its composition is variable
A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved
in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA
solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was
adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and
boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling
the solution was transferred to flask A and water was added to the graduation mark
A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was
dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and
water was added to the graduation mark
Glassware and reagents at your disposal
burette two Erlenmeyer flasks with ST
beaker 25 mL volumetric pipette
graduated cylinder 10 mL volumetric pipette
small funnel wash bottle with distilled water
ca 0050 molL KSCN solution
ca 0025 molL MgCl2 solution
Reagents at disposal of all participants
EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10
ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution
eriochrome black T mixed with NaCl and a spatula chloroform with a pipette
Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet
Additional information The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2minus
complex but the aluminium complex is inert The eriochrome black T
magnesium complex is much less stable than the MgY2minus
complex The AgNO3 solution is acidified
with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso wwwShimiPediair
2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
wwwShimiPediair
3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
wwwShimiPediair
4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
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5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
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6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
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7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
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8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
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10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
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2
Problems
a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem
b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem the procedures given below and the available reagents
c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem
d Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a and b
e Give the determined concentrations of solutions form task d
f Determine the percentage of aluminium in PAC
g Determine the percentage of chloride ions in PAC
h Determine the stoichiometric formula of PAC Al(OH)xCly
i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)
j Why is chloroform introduced into the flask during chloride ions determination Is it necessary
to do it when determining bromide ions Justify your answer
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to
the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes
correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer
solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue
Repeat the titration
Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrantrsquos volume has to be read The titration must not be continued even though the
analyte solution may recover the blue colour
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the
expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the
range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the
flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down)
Repeat the titration
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3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
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4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
wwwShimiPediair
5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
wwwShimiPediair
6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
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7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
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8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
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9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
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10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
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3
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions surface active
agents (surfactants) are used diversely in analytics The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (eg complexes with metal titration indicators) they may be
decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate
between cationic anionic and non-ionic surfactants
The solutions of substances given in the table below are placed in test tubes labelled A-J The
solutions concentrations are also given
Tes
t tu
bes
A-J
Substance Concentration
Iron(III) chloride 210-5
molL
Mercury(II) nitrate 210-5
molL
Dithizone HDz 210-4
molL
Eriochrome cyanine R ECR 210-4
molL
Safranin T SFT 210-4
molL
Rose Bengal RB 210-4
molL
Potassium palmitate PK 110-2
molL
Sodium dodecyl sulphate SDS 110-2
molL
Triton X-100 TX 110-2
molL
Cetyltrimethylammonium chloride CTA 110-2
molL
Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-
spectrophotometric determination of mercury or silver (it forms orange and yellow chelates
respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic
surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100
poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant
Glassware and reagents at your disposal
8 empty test tubes
6 polyethylene Pasteur pipettes
wash bottle with distilled water
sulphuric acid 1molL
indicator paper
You may use the solutions for problem 1
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4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
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5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
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6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
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7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
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8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
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9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
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10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
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4
Problems
a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo
colour and their pH and carrying out simple tests for the presence of surfactants
b Derive a plan that will allow you to identify substances in the solutions
c Identify the substances present in the solutions in test tubes A-J using the available reagents and
the given procedure
d Give justification for your identification confirming it with two observations
e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator
if the analysed solution contains cationic surfactants Justify your answer with appropriate
observations
Investigating the influence of surfactants on coloured systems
Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is
taking place in the solution Carry out a blind test for comparison
USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS
Eriochrome cyanine R ECR Rose Bengal RB
Safranin T SFT Dithizone H2Dz
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5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
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6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
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7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
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8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
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9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
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10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
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5
SOLUTION OF PROBLEM 1
a Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions
reacted with some of the EDTA forming AlY The remaining EDTA has
to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the
sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and
the relative volumes of the flask and pipette one may write
22 14504 MgClEDTAMgClEDTAAl cVcnnn
b Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic In order to determine the amount of chloride
ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution Taking into account the relative volumes of the flask and pipette one gets the
following relationship
)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33
c Equations of the reactions taking place during the chemical analysis
PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2
3
yx
Aluminium ions reaction with EDTA H2AlYYHAl 2
2
3 OHOHH 2
Titration of EDTA excess with magnesium chloride solution
2HMgYMgYH 222
2 OHNHOHNHH 244 HMgInHInMg2
Reactions taking place during chloride ions determination
AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN
m1 = 02718 g
m2 = 05895 g S = 0800
Titration Titrant volumes mL
Conc of MgCl2
Determination of Al
Conc of AgNO3 and
KSCN
Determination of Cl
V0 2000 1990 average 1995
V1 1000 990 average 995
V3 1870 1880 average 1875
V2 1605 1615 average 1610
EDTA concentration
004990 molL
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6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
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7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
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8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
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9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
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10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
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6
d Derivation of the formulae for solutions concentrations
Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of
MgCl2 solution 0
EDTA
0
MgCl
V
c01
V
n2
2
MgClc
Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)
solution using V3 mL of KSCN solution
KSCNClAgNO nnn3
KSCN3MgClKSCNMgClAgNO cVc102nn2c25223
From the information in the problem KSCNAgNO cS
25c25
3
S
KSCN
AgNO
cc
3
KSCN3MgCl
KSCN cVc102S
c25
2
3
MgCl
3
MgCl
KSCNVS-25
cS102
V-S
25
c102c 22
e Concentrations of MgCl2 KSCN and AgNO3 solutions
2MgClc 002501 molL cKSCN = 004002 molL
3AgNOc 005003 molL
f Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 02718 g
mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489
g Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 05895 g
mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918
h Stoichiometric formula of PAC
PAC formula ndash Al(OH)xCly x + y = 3
molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232
y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149
i Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed acidifying the solution 32
2 AlYH 2HAlY
H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is
shifted towards AlY complex formation
j Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is
more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN
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7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
7
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions
Substance Substance
A Potassium palmitate PK F Iron(III) chloride
B Sodium dodecyl sulphate SDS G Mercury(II) nitrate
C Triton X-100 TX H Eriochrome cyanine R ECR
D Cetyltrimethylammonium chloride CTA I Safranin T SFT
E Dithizone HDz J Rose Bengal RB
a Probable arrangement of substance in test tubes A-J
Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions
They are in test tubes A B C D F and G
Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube
A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I
and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking
which is visible for test tubes B C and D
b Identification plan
Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in
water ndash the only one from surfactants) with coloured solution -
dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its
colour to light yellow
Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash
formation of the red-orange precipitate soluble in any surfactant
The other colourless not frothing solution containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet
solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant
Remaining TX does not form precipitates with any of the dyes
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
8
c and d identification of substances in test tubes A-J and justification
Identification Justification
A PK
Opalescent and slightly alkaline solution
Froths when shook with distilled water
+ K rarr precipitate the only one of surfactants
+ MgCl2 or running water rarr white precipitate
B SDS
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution
RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes
C TX
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Dissolves dithizone mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX rarr no changes
D CTA
Colourless and neutral solution
Upon shaking with distilled or running water froths abundantly
+ K rarr no changes
Fe-ECR (violet) + CTA rarr blue the only one of surfactants
RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution
E HDz
Orange and slightly alkaline solution
+ K rarr brown precipitate soluble in TX (SDS CTA)
Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)
F FeCl3
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr orange solution
+ ECR (tt H) rarr violet solution + CTA rarr blue solution
+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)
G Hg(NO3)2
Colourless and slightly acidic solution does not froth
+ KSCN (problem 1) rarr no changes
+ ECR (tt H) rarr no changes
+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)
wwwShimiPediair
9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
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9
Identification Justification
H ECR
Orange and neutral solution
+ K rarr no changes
+ Fe(III) rarr vilet solution + CTA rarr blue solution
+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution
I SFT
Red and neutral solution
+ K rarr no changes
+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution
+ CTA rarr no changes
+ TX rarr no changes
J RB
Red and neutral solution
+ K rarr turns colourless
+ SDS rarr no changes
+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution
+ TX rarr no changes
e Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue just as at the end of magnesium titration with EDTA solution
Used abbreviations + K ndash addition of sulphuric acid
+ CTA ndash addition of cetyltrimethyl ammonium solution
+ SDS ndash addition of sodium dodecyl sulphate solution
+ Fe(III) ndash addition of iron(III) chloride solution
+ Hg(II) ndash addition of mercury(II) nitrate solution
+ TX ndash addition of Triton X-100 solution
tt ndash test tube
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10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
wwwShimiPediair
10
Comments to the solution of task 2
b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of
the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the
orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes
do not change their colour upon acidification One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions
as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming
precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash
Triton X-100
c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash
formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate
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