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PHYSICAL CHEMISTRY LAB MANUAL
EXPERIMENT: 1
CONDUCTOMETRIC TITRATION OF A STRONG ACID WITH A STONG BASE
AIM:
To determine the concentration of a strong acid by a
conductometric titration using a strong base
INTRODUCTION:
The conductivity of an electrolytic solution depends on 1. ionic
concentration and 2. Ionic mobility. The changes in the number and
nature of the ions during an acid base titration, which affect the
total conductance of the solution, can be utilized for the
determination of end point. Now let us consider an acid base
titration in which 50.0 ml of 0.1 N hydrochloric acid is titrated with
1.0N sodium hydroxide solution. In aqueous solution a strong acid
can be considered to be completely in ionized form.
Although both chloride and hydrogen ions contribute to the total
conductance of the solution, hydrogen ion has a major share due to
its high mobility (349.8 mhos.cm2.mol-1) when compared to chloride
ion (76.4 mhos.cm2.mol-1). When a small portion of a strong alkali
such as sodium hydroxide is added to the solution, the OH - ion of
the alkali reacts with the hydrogen ion forming water. Further, a
sodium ion is introduced in its place. It means addition of sodium
hydroxide results in a replacement of high mobile proton with a low
mobile sodium ion (50.1 mhos.cm2.mol-1). This results in a steep
decrease of conductance until the equivalence point is reached.
1
PHYSICAL CHEMISTRY LAB MANUAL
After the equivalence point the excess of sodium hydroxide causes
the conductance to rise a little bit slowly when compared to the
decrease before the equivalence point. Both sodium ions and
hydroxyl ions contribute to the conductance after the equivalence
point. The major contribution, however, is due to OH -, which has an
ionic mobility of 198.6. mhos.cm2.mol-1. Therefore the titration
curve contains two lines intersecting at the end point. As
conductance is sensitive to dilution, a volume correction needs to be
applied to the meter readings. The volume of the analyte solution
goes on increasing with the addition of sodium hydroxide solution
during the titration.. If ‘C’ is the measured conductance, ‘V’ is the
initial volume of the analyte solution and ‘v’ is the volume of sodium
hydroxide added, then the corrected conductance is given by,
The corrected conductance in milli ohms should be plotted against
volume of the sodium hydroxide for detecting the end point of the
titration.
SOLUTIONS:
1. Sodium hydroxide solution : Prepare 250 ml of
approximately 1.0 N Sodium hydroxide by dissolving around
10 gms of analytical grade sodium hydroxide (gm.equivalent
weight=40.0) in carbon dioxide free distilled/ deionized water
2
PHYSICAL CHEMISTRY LAB MANUAL
or conductivity water. Use a rough balance for weighing
sodium hydroxide.
2. Oxalic Acid (gm.equivalent weight = 63.03) solution:
Prepare 100 ml of 1.0 N oxalic acid solution by dissolving
exactly 6.303 gms of analytical grade oxalic acid in distilled
water and make it up to the mark. Calculate the strength of
the oxalic acid solution using the actual weight of the
substance transferred in to the flask if it is not exactly 6.303
gms.
3. Hydrochloric acid solution: Prepare 100.0 ml of
approximately 1.0 N hydrochloric acid stock solution by
diluting concentrated hydrochloric acid (11.4N).
PROCEDURE:
Standardize sodium hydroxide solution using standard oxalic
acid solution with phenolphthalein as indicator.
Also standardize the hydrochloric acid using the standard
sodium hydroxide solution with phenolphthalein as indicator.
Prepare 100 ml of 0.1N hydrochloric acid solution by the exact
dilution of the stock solution.
Switch on the conductivity meter for at least one hour before
taking any measurements.
Standardize the conductivity meter (using internal standard)
following the instructions given in the Instruction Manual
supplied by the manufacturer along with the Instrument.
Clean the conductivity cell thoroughly with distilled water and
then with conductivity water.
Prepare the experimental solution by taking 25.0 ml each of
0.1N hydrochloric acid and conductivity water with a burette
in to a 100 ml beaker. Mix the contents of the beakers
thoroughly with a glass rod.
3
PHYSICAL CHEMISTRY LAB MANUAL
Insert the conductivity cell in to the experimental solution and
note down the meter reading after selecting an appropriate
range using the range switch of the meter.
Fill the burette with the standardized solution of sodium
hydroxide and titrate the experimental solution by adding 0.2-
0.5 ml portions.
Note down the meter readings each time after thoroughly
stirring the contents of the beaker.
Tabulate the results in the Table format shown below
Concentration of the titrant (Sodium hydroxide)= ________ N
Total initial volume of the Analyte solution, V= ________ ml
S.N
o.
Volume of Sodium
hydroxide added,
v ml
Observed
Conductance
in milli mhos,
Corrected
conductance,
C’= C X (V+v)/V
milli mhos
1 0.0
2 1.0
3 2.0
4 3.0
etc.
Draw a graph of corrected conductance versus volume of
sodium hydroxide added.
Join the points linearly before and after the equivalence point
and extend them to get the intersection point.
Note down the volume of the sodium hydroxide ‘x’
corresponding to the intersection point (end point).
Calculate the concentration of hydrochloric acid using the
formula V1.N1=V2.N2, where V1 is the volume of hydrochloric
acid taken = 25.0 ml, N1 is the concentration of H Cl to be
4
PHYSICAL CHEMISTRY LAB MANUAL
determined, V2 is the volume of sodium hydroxide at the end
point = x ml and N2 is the strength of the standardized sodium
hydroxide
EXPERIMENT: 2
CONDUCTOMETRIC TITRATION OF A MIXTURE OF STRONG
ACID AND WEAK ACID WITH A STRONG BASE
AIM: To determine the concentrations of strong acid and weak acid
in a mixture by a conductometric titration using a strong base.
INTRODUCTION:
When a mixture of strong acid and weak acid is taken, the
strong acid exists almost completely in the ionic form as shown
below.
Whereas as the weak acid such as acetic acid mostly exists in the
un-dissociated molecular form. In the presence of the strong acid,
due to common ion effect, the dissociation is further suppressed and
the molecule almost exists in the un dissociated molecular form.
Therefore, the initial conductivity of a mixture of strong acid and
weak acid is only due to the ions of the strong acid. The initial
conductivity of the solution is therefore, high, due to the high
mobility and abundance of the hydrogen ions in solution. When
sodium hydroxide is added during the titration, these hydrogen ions
are replaced with less mobile sodium ions resulting in a rapid
decrease of conductance. This happens until all the strong acid is
neutralized. Further addition of sodium hydroxide results in the
formation of
5
PHYSICAL CHEMISTRY LAB MANUAL
sodium acetate due to the neutralization of acetic acid. As sodium
acetate exists in ionic form, its formatioAn raises the conductance of
the solution slightly due to the low ionic mobilities of both Na+ (50.1
mhos.cm2.mol-1) and CH3COO- (40.9 mhos.cm2.mol-1) ions. This
continues until all the acetic acid is neutralized. The excess alkali
added over and above this again raises the conductance but now
with different and high rate due to the relatively high mobility of
hydroxyl ions. The titration curve therefore contains three linear
portions with two intersection points. The first intersection point ‘x’
corresponds to the neutralization of the strong acid and the second
at ‘y’ corresponds to the total of strong and weak acids. From the
values of ‘x’ and ‘y’ the concentrations of both strong and weak
acids in a mixture can be determined.
SOLUTIONS:
1. Sodium hydroxide solution: Prepare 250 ml of approximately
1.0 N Sodium hydroxide by dissolving around 10 gms of
analytical grade sodium hydroxide (gm.equivalent
weight=40.0) in carbon dioxide free distilled/ deionized water
or conductivity water. Use a rough balance for weighing
sodium hydroxide.
2. Oxalic Acid (gm.equivalent weight = 63.03) solution: Prepare
100 ml of 1.0 N oxalic acid solution by dissolving exactly
6.303 gms of analytical grade oxalic acid in distilled water and
make it up to the mark. Calculate the strength of the oxalic
6
PHYSICAL CHEMISTRY LAB MANUAL
acid solution using the actual weight of the substance
transferred in to the flask if it is not exactly 6.303 gms.
3. Hydrochloric acid solution: Prepare 100.0 ml of approximately
1.0 N hydrochloric acid stock solution by diluting concentrated
hydrochloric acid (11.4N).
4. Acetic acid solution: Prepare 100 ml of stock solution of
approximately 1.0 N acetic acid using conductivity water. This
can be done by the dilution of approximately 6.0 ml (use
measuring jar) of glacial acetic acid (17.4N) to 100 ml using
conductivity water.
PROCEDURE:
1. Standardize sodium hydroxide solution using standard oxalic
acid solution with phenolphthalein as indicator.
2. Standardize the stock solutions of hydrochloric and acetic acid
using the standardized sodium hydroxide solution.
3. Prepare 0.1 N solutions of both hydrochloric and acetic acid by
exact dilution of the stock solutions.
4. Switch on the conductivity meter for at least one hour before
taking any measurements.
5. Standardize the conductivity meter (using internal standard)
following the instructions given in the Instruction Manual
supplied by the manufacturer along with the Instrument.
6. Clean the conductivity cell thoroughly with distilled water and
then with conductivity water.
7. Prepare the experimental solution by taking 25.0 ml each of
0.1N hydrochloric acid and 0.1N acetic acid with a burette in
7
PHYSICAL CHEMISTRY LAB MANUAL
to a 100 ml beaker. Mix the contents of the beakers
thoroughly with a glass rod.
8. Insert the conductivity cell in to the experimental solution and
note down the meter reading after selecting an appropriate
range using the range switch of the meter.
9. Fill the burette with the standardized solution of sodium
hydroxide and titrate the experimental solution by adding 0.2-
0.5 ml portions. Note down the meter readings each time
after thoroughly stirring the contents of the beaker.
10. Tabulate the results in the Table format shown below
Concentration of the titrant (Sodium hydroxide) = ____________
N
Total initial volume of the Analyte solution, V= ________
S.N
o.
Volume of Sodium
hydroxide added,
v ml
Observed
Conductance
in milli mhos,
Corrected
conductance,
C’= C X (V+v)/V
milli mhos
1 0.0
2 1.0
3 2.0
4 3.0
etc.
Draw a graph of corrected conductance versus volume of
sodium hydroxide added.
Join the points in the three portions of the curve linearly and
extend them to get the intersection points.
Note down the volumes of the sodium hydroxide ‘x’ and ‘y’
corresponding to the intersection points.
8
PHYSICAL CHEMISTRY LAB MANUAL
Calculate the strengths of the strong acid and weak acid in the
mixture as follows
EXPERIMENT : 3
THE POTENTIOMETRIC TITRATION OF FERROUS AMMONIUM
SULPHATE WITH POTASSIUM DICHROMATE
AIM:
To determine the concentration of a ferrous ammonium sulphate
solution, will be determined quantitatively by a potentiometric
titration against a standard solution of potassium dichromate.
THEORY:
Potassium dichromate is a strong oxidizing agent. Its principle
advantage is its availability as a primary standard. The long-term
stability of its solutions is another advantage for its use in red-ox
titrations. It is not however, as strong an oxidizing agent as MnO4- or
Ce4+. Its reduction half reaction is,
On the other hand, the solutions of ferrous ammonium sulphate
[Fe(NH4)2(SO4)2.6H2O] are normally very susceptible to aerial
oxidation, but when prepared in 0.5 to 1.0 M sulphuric acid may
remain stable for as long as a month. Its reduction half reaction is,
9
PHYSICAL CHEMISTRY LAB MANUAL
The potential of the cell, given by,
PtFe3+, Fe2+Cr2O72-,Cr3+,H+Pt
Ecell = Ecathode – Eanode = 1.33-1.771 =0.559 V
The spontaneity of a chemical reaction requires G= -nFE, to be
negative. If the emf of the cell is positive, G, is negative and the
reaction is spontaneous. This indicates that the reaction between Cr
(VI) and Fe (II) in which Fe (II) is oxidized to Fe (III) and Cr (VI) is
reduced to Cr (III) is possible. Potentiometric methods of analysis
can be employed to detect the end point of the titration. The
electrochemical cell before the equivalence point is
Hg(liq) Hg2Cl2(solid), KCl (aq, satd.) Fe(II), Fe(III) Pt
Note that the Pt cathode is an inert indicator electrode that carries
electrons to the reduction half-cell. The electrode itself does not
undergo either oxidation or reduction. The reference electrode is a
saturated calomel electrode. The cell potential before the
equivalence point depends on the relative concentrations of the
oxidized and reduced forms of the iron.
Where, is the standard reduction potential (=0.771V) of
Fe(III)/ Fe(II) system
After the equivalence point the indicator electrode assumes the
potential of chromium system. The potential of the cell after the
equivalence point can therefore be represented by the equation,
Where, is the standard reduction potential (= 1.33V) of the
chromium system. The end point of the titration is indicated by a
large change (jump) in the cell potential due to the difference in the
standard reduction potentials of iron and chromium systems.
10
PHYSICAL CHEMISTRY LAB MANUAL
SOLUTIONS:
1. Ferrous ammonium sulphate solution: Prepare 100.0 ml
of 0.1N ferrous ammonium sulphate solution in 1.0 N sulphuric
acid as follows. Weigh accurately by the method of difference
about 3.9 gms of ferrous ammonium sulphate in to a 100 ml
volumetric flask. Add 5 ml of 1:1 sulphuric acid to it and make
it up to the mark with distilled water after a proper dissolution
of the substance. The concentration of the solution can be
calculated using the equation, 0.1 X W /3.9214, where ‘W’ is
the weight of the substance.
2. Potassium Dichromate solution: Prepare 250 ml of 0.1 N
potassium dichromate solution as follows. Weigh accurately
about 1.2 gms of the substance in to a 250 ml volumetric flask
and make it up to the mark with distilled water after proper
dissolution of the sample. The concentration of the solution
can be calculated using the equation, 0.1 X W/ 1.226, where
‘W’ is the weight of the substance.
3. 1:1 Sulphuric acid solution: Prepare 50 ml of 1:1 sulphuric
acid solution as follows: Mix approximately 25.0 ml of
concentrated sulphuric acid and 25.0 ml of water taking the
following precautions. Always add acid to water (not the
reverse) slowly in small portions with constant stirring and
cooling
PROCEDURE:
Prepare the experimental solution as follows: Take 10.00 ml of
ferrous ammonium sulphate solution using a burette in to a
100 ml beaker and add 5 ml of 1:1 sulphuric acid and 35 ml of
water with measuring jar. Mix the solution well using a glass
rod or magnetic stirrer
11
PHYSICAL CHEMISTRY LAB MANUAL
Wash the external surface of the saturated calomel electrode
(reference electrode) using distilled water
Now introduce both the electrodes in to the experimental
solution and connect them to the potentiometer, platinum
electrode to the red colored terminal and reference electrode
to the black terminal.
Fill the Burette with 0.1N potassium dichromate solution
Now perform a Pilot or Rough titration to locate the
equivalence point, by adding 1.0 ml portions of dichromate to
the analyte solution and noting down the potentiometer
readings in milli volts after thorough stirring of the solution. A
large change in the cell potential indicates the position of the
equivalence point. Note down the corresponding burette
readings.
Now an accurate titration can be performed by repeating all
the steps above but adding only 0.1 ml portions of dichromate
in the volume range where jump in cell potential has occurred
in the previous pilot titration.
Tabulate the results of the titration in the format shown in the
observations
Draw a sigmoid graph of observed cell potential versus
volume of the dichromate added. Note down the position (x)
of the end point. Also draw a first derivative graph i.e. E/V
versus Vaverage and note down the end point (x), which is the
volume of the dichromate corresponding to the maximum of
the curve.
The concentration of the ferrous ammonium sulphate solution
can now be determined using the equation, V1.N1=V2.N2 i.e.
12
PHYSICAL CHEMISTRY LAB MANUAL
[Fe(II)]= x X 0.1/10.0 , where ‘x’ is the volume of the
dichromate at the end point.
OBSERVATIONS:
For both Pilot and Accurate titrations use the following Table format
for recording your observations.
Volume of Ferrous ammonium sulphate: ________ml
VOLUME OF K2Cr2O7
ADDED, ml (V)
POTENTIAL OF THE
ELECTROCHEMICAL CELL
AGAINST SCE, milli volts, (E)
E/V Vaverage
0.0
1.0
2.0
etc.
SAMPLE DATA AND RESULTS:
Volume of Ferrous ammonium sulphate: 10.0 ml
VOLUME OF K2Cr2O7 ( 0.1
N) ADDED, ml (V)
POTENTIAL OF THE
ELECTROCHEMICAL CELL
AGAINST SCE, milli volts,
(E)
E/V Vaverage
0.00 370
1.00 391 21 0.50
2.00 408 17 1.50
3.00 425 17 2.50
4.00 437 12 3.50
5.00 449 12 4.50
6.00 462 13 5.50
7.00 476 14 6.50
8.00 503 27 7.50
8.10 507 40 8.05
8.20 512 50 8.15
8.30 518 60 8.25
8.40 526 80 8.35
13
PHYSICAL CHEMISTRY LAB MANUAL
8.50 536 100 8.45
8.60 565 290 8.55
8.70 691 1260 8.65
8.80 717 260 8.75
8.90 728 110 8.85
9.00 737 90 8.95
10.00 766 29 9.50
SIGMOID PLOT
14
PHYSICAL CHEMISTRY LAB MANUAL
FIRST DERIVATIVE PLOT
Concentration of ferrous ammonium sulphate = 0.1 X 8.65/10.0 =
0.08650 N
EXPERIMENT: 4
KINETICS OF ESTER HYDROLYSIS
AIM: Determination of relative strengths of acids by Ester hydrolysis
INTRODUCTION: The hydrolysis of an ester, for example ethyl
acetate can be given by, is catalyzed by acids.
The rate of hydrolysis is proportional to the concentration of the
acid. Thus a measure of the rate (rate constant) in the presence of
different concentrations of acids can be used to study the relative
15
PHYSICAL CHEMISTRY LAB MANUAL
strengths of acids. The acid catalyzed hydrolysis of ester follows
pseudo first order and thus the rate constant is given by,
where ‘a’ is the initial concentration of the ester and ‘(a-x)’ is the
concentration of un-hydrolyzed ester at time ‘t’. The reaction is
followed by titrating fixed volume of the reaction mixture at fixed
intervals of time with sodium hydroxide. Titre value is a measure of
acetic acid produced with time. Let ‘T0’ is the titre value at the start
of the reaction,’Tt ‘ is the titre value at different time intervals of the
reaction is started and ‘T’ is the titre value when the reaction is
assumed to be completed.. Then ‘T-T0’ corresponds to ‘a’ and ‘T-
Tt’ corresponds to ‘a-x’. Hence the rate constant ‘k’ is
The hydrolysis is conducted at two different acid concentrations.
The corresponding rate constants k1 and k2 are obtained. The ration
‘k1/k2 ‘gives relative strengths of acids.
SOLUTIONS/ CHEMICALS:
1. Sodium hydroxide solution: Prepare 500 ml of 1N sodium
hydroxide solution by dissolving 20 gms of sodium hydroxide
(use rough balance) in carbon dioxide free distilled water.
2. Hydrochloric acid solution: Prepare 250 ml each of 1N and 2N
hydrochloric acid solutions by exact dilution of a pre-
standardized (~4N ) hydrochloric acid solution.
3. Ethyl acetate
4. Ice cold water
5. Phenolphthalein Indicator
PROCEDURE:
16
PHYSICAL CHEMISTRY LAB MANUAL
Measurement of T0 : Take 10.0 ml of 1.0N HCl in to a 250 ml
conical flask. Add 30 ml distilled water and a drop of
phenolphthalein indicator. Titrate with 1.0N sodium hydroxide
until a permanent pink color is obtained. Note down the
volume of sodium hydroxide consumed as the value of the T0.
This is to be repeated for 2.0 N HCl also.
Study of the kinetics: Take 100.0 ml of 1.0N HCl in a reaction
bottle. Keep the bottle in a constant temperature water bath
to attain thermal equilibrium. Add 5.0 ml of ethyl acetate to
the contents of the reaction bottle simultaneously starting a
stop-watch when the pipette is half empty. Swirl the contents
thoroughly to obtain a uniform solution. For every ten minutes
withdraw 10.0 ml of reaction mixture in to a 250 ml conical
flask containing 40 ml of ice-cold water and one drop of
phenolphthalein indicator and titrate with standard sodium
hydroxide. This is to be continued for 70 mts. The titre values
for different time intervals represent ‘Tt ‘ values. Preserve the
remaining reaction mixture for 24 hrs and then titrate again
10.0 ml of reaction mixture with sodium hydroxide. Note down
the volume of the alkali consumed as T value.
Using T0, Tt and T values thus obtained calculate he rate
constant. Repeat the same procedure for 2N HCl .
Tabulate the observations in the format shown below
Concentration of acid = ______________
Titre value at zero time (T0) = __________________
Titre value at infinite time (T) = _______________
TIME
(Minutes)
Tt (T- Tt) log(T- Tt)
17
PHYSICAL CHEMISTRY LAB MANUAL
Average, k = _____________ min -
1
Graphical method: Make a plot of log(T- Tt) versus Time ‘t’as
shown. Calculate the slope of the straight line obtained. Slope
multiplied with 2,303 gives the rate constant.
If ‘k1’ and ‘k2’ are the rate constants obtained for 1N and 2N
hydrochloric acid solutions respectively, the relative strengths
of acids is given by k1/ k2.
EXPERIMENT: 5
DETERMINATION OF EQUILIBRIUM CONSTANT OF
POTASSIUM TRIODIDE EQUILIBRIUM
AIM:
Determination of the equilibrium constant of the equilibrium,
.
THEORY: Potassium iodide combines with iodine to form potassium
tri iodide, the equilibrium between KI, I2 and KI3 is represented by,
18
PHYSICAL CHEMISTRY LAB MANUAL
and the corresponding equilibrium constant Ke is given by,
Where the subscript ‘e’ stands for the equilibrium molar
concentrations. The concentrations at equilibrium can be calculated
using the distribution law (or partition law). As an example take
iodine as a solute and the two immiscible liquids as water and
chloroform. Iodine distributes between these two and one can
observe that iodine distributes between water and chloroform is a
definite ratio. This constant ratio is described as partition coefficient
( ) of Iodine between chloroform and water and expressed as
Partition coefficient =
i.e Partition coefficient =
This ratio (Which is found to be 125) is independent of the
total amount of dissolved iodine. If I2 and also KI are added to the
binary liquid system chloroform and water, value as expressed
above can be assumed to remain the same in spite of KI3 formation
according to the reaction in aqueous phase. This facilitates in
obtaining the
equilibrium concentration of I2, KI and KI3 and thus equilibrium
constant of the above equilibrium can be obtained.
Solutions/Chemicals
1. 250ml of 0.1M KI : It is prepared by weighing of KI and
dissolving in water in a 250ml volumetric flask and making it
up to the mark with distilled water.
2. 250ml of 0.1 N sodium thiosulfate : (Hypo) : Dissolve of
sodium thiosulfate using distilled water in a 250ml volumetric
flask and make it up to the mark. It is standardized against
standard dichromate solution.
19
PHYSICAL CHEMISTRY LAB MANUAL
3. 1%. Iodine solution (100ml) : Take 1g. of iodine in a beaker
and add 15 – 20ml of chloroform and stir it with a glass rod.
Transfer the I2 solution formed in to a 100ml volumetric flask.
Then add another 15-20ml of chloroform to the solid iodine
left in the beaker and transfer the I2 solution formed leaving
the un dissolved solid I2 in the beaker. Repeat the procedure
until all the 100ml chloroform is added.
4. 250ml of 0.1N K 2 Cr2 O7 : Dissolve of K2 Cr2 O7 using distilled
water in a 250ml volumetric flask and the solution is made up
to the mark.
5. 100ml of 0.5M oxalic acid : It is prepared by weighing and
dissolving of oxalic acid in distilled water in a 100ml
volumetric flask.
6. 100ml of 2N Hydrochloric acid : About 20ml of concentrated
Hydrochloric acid is diluted to 100ml using distilled water to
give approximately 2N Hydrochloric acid.
7. 1% starch solution : 100ml of distilled water is taken in a
beaker and set to boil. When it is about to boil 1g.of starch is
made in to a paste and put in to the beaker.
8. 5% potassium iodide : 250ml of 5% KI is prepared by
dissolving 12.5g of KI in 250ml distilled water.
Procedure :
Take 50ml of 0.1N KI in a 200ml bottle and add 15ml of 1%
Iodine (prepared in chloroform) and 10ml of chloroform. Close the
bottle with a cap and seal it, with liquid wax. Shake the bottle for 45
minutes, remove the wax seal and take the contents in to a
separating funnel. The organic layer and the aqueous layer are
separated and they are separately titrated with 0.01N Hypo solution
to find out iodine present in them following the procedure described
below.
20
PHYSICAL CHEMISTRY LAB MANUAL
Titration of organic layer : 5ml of organic layer is taken in to a
250ml conical flask, add 10ml of 5% potassium iodine and add 10ml
distilled water. The contents are titrated against 0.01N hypo
solution from a burette. When the contents in the conical flask
appear light yellow add 0.5ml of freshly prepared starch solution.
Then it turns to blue in color and continue the titration to see it is
colorless. The color change from blue to colorless is the end point.
Volume of hypo consumed is noted. Repeat the titration until
concurrent values are obtained.
Titration aqueous layer : Take 10ml of aqueous layer in to a
250ml conical flask and add 20ml distilled water. Titrate the
contents against 0.01N hypo solution from a burette. When the
contents appear light yellow in color add 0.5ml of freshly prepared
starch solution and continue the titration and till it turns from blue
to colorless indicating end point. Volume of hypo consumed is
noted.
The entire experiment described above is repeated by taking
different amounts of iodine and chloroforms ;
Calculation :
Concentration of iodine in organic layer =
Where V1 is volume of hypo consumed for 5ml organic layer
Concentration of iodine in aqueouslayer =
V2 is volume of aqueous layer taken
C2 = [I2]e + [KI3]e …………….(3)
[I2]e = (Where 125 is the partition coefficient of iodine
between Chloroform and water)
From equation (3) [KI3]e = C2 – [I2]e = C2 -
Concentration of KI consumed in KI3 formation is equal to [KI3]e
21
PHYSICAL CHEMISTRY LAB MANUAL
[KI]e = [KI]initial – [KI3]e
[KI]initial = 0.05M
[KI]e = 0.05 -
Equilibrium constant for
is
Ke =
Substituting [I2]e, [KI3]e and [KI]e from equations (4) , (5) and (6)
respectively
Equation (7) becomes
C1 and C2, calculated using (1) and (2) are substituted and thus K
the equilibrium constant can be calculated.
The above experiment can be used to find out the unknown
concentrations of KI after knowing Ke value from equation (8). The
above described procedure is used, C1 and C2 are obtained. 0.05 in
equation (8) is replaced with x (to be determined). Now with Ke
known and C1 and C2 obtained from the experiment x can be
calculated.
EXPERIMENT: 6
DETERMINATION OF CRITICAL SOLUTION TEMPERATURE
OF PHENOL-WATER SYSTEM
AIM:
To determine the critical solution temperature of phenol-water
system.
22
PHYSICAL CHEMISTRY LAB MANUAL
INTRODUCTION:
Some liquids are completely miscible and some are completely
immiscible. In between these extremes, there is an important type
of system consisting of two liquids which are partially miscible. An
example of this is phenol and water. If a little phenol is added to
water at room temperature, it dissolves completely, but if the
addition is continued a point is reached when no further dissolution
possible and two liquid layers form. One of these layers is a
saturated solution of phenol in water and the other one is water in
phenol. The two layers in equilibrium are called conjugate solutions.
Rise in temperature can cause mutual solubilities of phenol and
water and there exists a temperature at which phenol and water are
miscible in all proportions. This is known as critical solution
temperature of phenol-water system. At this temperature and
beyond this temperature phenol and water are miscible in all
proportions.
Determination of critical solution Temperature :
CHEMICALS:
80% phenol : 80 gms of phenol is dissolved in distilled water and
100gms of a solution of phenol in water is obtained.
APPARATUS:
1. A boiling tube made of corning glass. It is of about 3cm in
diameter and of about 30ml capacity.
2. 600 ml beaker of corning glass
3. Two glass stirrers
4. A sensitive thermometer
5. A spirit lamp
PROCEDURE:
Take the 600ml beaker with sufficient volume of water and place it
over a wire gauge kept over a tripod stand. A glass stirrer is kept in
to the beaker and the boiling tube is suspended in to this beaker of
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PHYSICAL CHEMISTRY LAB MANUAL
water. A small glass stirrer is kept in the boiling tube. The sensitive
thermometer is suspended in to the boiling tube. Phenol-water
mixture is taken in the boiling tube and the miscibility temperature
is found by heating the water bath with a spirit lamp. The detailed
procedure consists of taking 10ml of 80% phenol and 2ml of water
in the boiling tube. At room temperature this composition does not
give a clear solution and cloudiness is observed. Heat the water
bath with a spirit lamp. Cloudiness disappears. Note down the
clearing temperature. Put off the spirit lamp and allow it cool down.
Cloudiness reappears. Note down the clearing temperature. Average
of clearing and clouding temperatures is taken as mean miscibility
temperature. Now the composition is varied by adding another
1.0ml of water in to the same boiling tube which already contains
10ml phenol and 2ml water. Heat the water bath and find the
clearing and clouding temperatures and the mean miscibility
temperature. It is continued up to 10ml addition of water. Remove
the apparatus and wash the boiling tube, the stirrers and the
thermometer. Repeat the above procedure, now fixing volume of
water as 10ml and varying phenol each time by 1.0ml. Mean
miscibility temperature for each composition is recorded. Now the
density of phenol is determined using specific gravity bottle and the
percentage of phenol (wt/wt) in all compositions is calculated using
the formula
% phenol (wt/wt) =
V1 and V2 are volumes of phenol and water, d1 and d2 are respective
densities.
Now a graph is drawn using mean miscibility temperature versus %
phenol (wt/wt) as shown here.
A curve is obtained with a maximum. The maximum in the curve
represents the critical solution temperature. At this temperature and
beyond this temperature one can see phenol and water are miscible
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PHYSICAL CHEMISTRY LAB MANUAL
in all proportions. The critical solution temperature (CST) and %
phenol corresponding to CST are noted.
EXPERIMENT: 7
EFFECT OF AN ELECTROLYTE (NaCl ) ON THE MISCIBILITY
TEMPERATURE OF PHENOL – WATER SYSTEM
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PHYSICAL CHEMISTRY LAB MANUAL
Addition of a foreign substance to the phenol – water system has a
definite effect over the miscibility temperature of phenol and water.
This property can be used to determine unknown concentrations of
sodium chloride.
PROCEDURE:
About 250ml of N/10 sodium chloride is made and diluted to give
N/20, N/30, N/40, N/50 and N/60 solutions of sodium chloride. 5ml of
each of these is mixed with 5ml of 80%. Phenol in the boiling tube
and the miscibility temperature are found in each case. A graph is
made between miscibility temperature and concentration of NaCl. A
straight line with a positive slope is obtained as shown here.
The above study of effect of NaCl can be used to estimate the
concentration of NaCl in a given solution. The procedure consists of
taking 5ml of the sample and add 5ml of 80% phenol and find the
miscibility temperature. The above graph is used as a calibration
graph to read the unknown concentrations.
Caution : Phenol is highly corrosive and so take care not to come in
direct contact with even a phenol – water mixture. If any thing falls
on the skin wash with plenty of water and put some sodium
bicarbonate over that part.
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