PRECIPITATOMETRICPRECIPITATOMETRIC(PRECIPITATION)(PRECIPITATION)

TITRATIONTITRATIONAn application method of

Inorganic Pharmaceutical Analysis

Lecturer : Dr. Tutus GusdinarPharmacochemistry Research Group

School of Pharmacy INSTITUT TEKNOLOGI BANDUNG

Precipitatometric TitrationPrecipitatometric Titration• Compared to acid-base or reduction-oxidation

titrations the precipitatometric titration has no muchtitrations the precipitatometric titration has no much more methods

• Difficult to select a suitable indicatorDifficult to select a suitable indicator• Difficult to obtain an accurate precipitate composition,

the coprecipitation effect is oftenly occured.p p ySolubility = a saturated concentration of analyte (crysstaline or solid form) dissolved in solvent at a(crysstaline or solid form) dissolved in solvent at a defined temperature.

BaSO4( ) Ba2+ + SO42-BaSO4(p) Ba + SO4

Solubility Product Constant : Ksp = [Ba2+] [SO42-]

(at equilibrium state)(at equilibrium state)Solubility product

Saturated solution is occured when a continuous addition of a substance is performed into its solvent until dissolution precess is terminated or it can beprecess is terminated, or it can be achieved by increasing ionic concentration ntil the precipitate formationuntil the precipitate formation.

Solubility is influenced by temperature, solvent properties and other ionssolvent properties, and other ions existed in a solution.

Solubilitys

y

Saturated

X ( t ti f i )X (concentration of ion)

Factors influencing the solubility

1. Temperature2. Solvent properties2. Solvent properties3. Common ions4 Ionic activity4. Ionic activity5. pH6. Hydrolysis7. Metal hydroxydeeta yd o yde8. Complex compound formation

1. Temperature effect

Most of inorganic salts increase its solubility by increasing temperature Itssolubility by increasing temperature. Its better to use hot/warm solution for filtering

d hi i it t E tiand washing precipitate. Exception : precipitates which are slighty soluble in hot/warm solution (e.g. Hg2Cl2, MgNH4PO4) could be filtered after g 4 4)previously stored in refrigerator.

2. Solvent effect

Most of inorganic salts dissolve in water but not in organic solvent. Water molecule has a higher dipole moment and could be attracted by cations or anions to form hydrate ions Like hydrogen ion forming a hydrated ion H O+ freeions. Like hydrogen ion forming a hydrated ion,H3O+, free energy released by ion–solvent interaction could increase attractive ionic force to precipitate more solid lattice.attractive ionic force to precipitate more solid lattice.Crysstaline ions have no attractive force into organic solvents, its solubility is smaller than those in water. In ychemical analysis, solubility difference could be used as basic of separation of many compounds. Example : mixed f C (NO ) d S (NO ) b t d i l tof Ca(NO3)2 and Sr(NO3)2 can be separated in solvent

mixture containing alcohol and eter, which yields a soluble Ca(NO3)2 and an insoluble Sr(NO3)2Ca(NO3)2 and an insoluble Sr(NO3)2.

3 Common ion effect3. Common ion effectPrecipitate dissolves more easily in water than in solution

t i i i F l i A Cl l bilitcontaining common ions. For example, in AgCl, solubility product of [Ag+][Cl-] < its constant of solubility product (Ksp AgCl = 1x10-10) in pure water where [Ag+]=[Cl-] = 1x10-5 M;AgCl = 1x10 ) in pure water, where [Ag ]=[Cl ] = 1x10 M;when AgNO3 is added upon [Ag+] = 1x10-4 M, the [Cl-] decreases into 1x10-6 M and reaction shifts to the right sidedecreases into 1x10 M, and reaction shifts to the right side as : Ag+ + Cl- AgClThere is salt addition to the precipitate while concentration p pof Cl- decreases.This technique of common ion addition is oftenly used for :

1) completion of precipitation process2) precipitate washing with a solution containing

common ion effect

If there is exessed common ions in a solution, the precipitate solubility will be greater than estimated Ksp, then common ion addition must be limited up to 10%.Example : Calculate molar concentration of CaF2 dissolved

in a) water; b) CaCl2 0,01 M; c) NaF 0,01 M. Ksp CaF2 = 4 10 11 H d l i i l t d4x10-11. Hydrolysis is neglected.

CaF2(p) Ca2+ + 2 F-

) S C 2a) Solubility s = [Ca2+], then [F-] = 2s[Ca2+][F-] = Ksps . (2s)2 = 4x10-11, then s = 2,1 x 10-4 M

b) [Ca2+] = (0,01+s) ; [F-] = 2s; then s = 3,2 x 10-5 Mc) [Ca2+] = s ; [F-] = (0,01+s); then s = 4 x 10-7 MCommon ions decrease precipitate solubility, [F-] effect is greater than [Ca2+] effect.

Solubility of Ag-halide in Na-halide at 18 oCSolubility of AgX, M

1

10-2

1

10-4

AgBr

AgCl

8

10-6

AgI

g

10-10

10-8

10

10-8 10-6 10-4 10-2 1

Concentration of NaX, M

4 C i ff t4. Common ion effect

Most of precipitated compounds increase its solubility in a solution containing substance not reacted to precipitatesolution containing substance not reacted to precipitate ions. This phenomena is called as ionic activity effect or diverse ion effect or neutral salt effect For example solubilitydiverse ion effect or neutral salt effect. For example, solubility of AgCl and BaSO4 in KNO3 solution.

5 5[KNO3] (M) [AgCl]x10-5 M [BaSO4]x10-5 M

0,000 (water) 1,00 1,00

0 001 1 04 1 210,001 1,04 1,21

0,005 1,08 1,48

0 010 1 12 1 70

Δ=12% Δ=70%

0,010 1,12 1,70

Molarity is an ionic activity occured in a high diluted solution, more concentrated solution decreases faster the activitycoefficient (f), caused by greater different charge ofattractive ionic force. The ionic effectivity (in equlibrium state) decreases as well, and addition of precipitate is required to recover ionic activity.

aAg+ . aCl- = Kosp (Ksp at a defined state of ionic activity)

fAg+[Ag+] . fCl- [Cl-] = Kosp[Ag+][Cl-] = Kosp / fAg+ . fCl- = Ksp

The lower activity coefficient of both ion yields the greatermolar concentration of product. Increased BaSO4 solubility isgreater than solubility of AgCl, or ionic activity coefficient of divalent ion is smaller than those of univalent ion.In highly diluted solution the f = 1 and the Ksp = Kosp.

Relative augmentation of the solubilityRelative augmentation of the solubilityof AgCl and BaSO4 in KNO3 solution

s/soo

1 6

1,7

1,5

1,6BaSO4

1 3

1,4

1,2

1,3

AgCl

1 0

1,1

KNO3, M1,0

0,001 0,005 0,01

Example : Calculate molar solubility of BaSO4 in KNO3

0,01 M solution using activity coefficient calculated by Debye-Huckel equation. Solution of KNO3 (1:1) has ionic strength equal to its molarity of 0,01M. Read from the table : fBa

2+ = 0,667 ; fSO4

2- = 0,659.

Ksp = 1,00 x 10-10 / 0,667 x 0,659 = 2,27 x 10-10 = s2

then s = 1,51 x 10-5 M.

For it’s performed in neglected very low ionic solubility, the activity effect is not a serious problem in the chemical analysis. Precipitation process in high ionic concentration is quiet rare. Hence, ionic activity influence gives no significant error.

5. pH effectSolubility of weak acid salt depends on pH of the solution.For example : oxalic, sulfide, hydroxyde, carbonate,phosphate.Proton reacts with the anion to form weak acid, and increases salt solubility.a) Monovalent salt : MA(p) M+ + A-

HA + H2O H3O+ + A-2 3

Analytical concentration Ca = [A-] + [HA] = [A-]{[H3O+]+Ka}/KaFraction of A- : [A-]/Ca = Ka / {[H3O+]+Ka = α1[ ] {[ 3 ] 1

[A-] = α1 .CaSubstituted to the Ksp = [M+][A-] = [M+]. α1.Cap [ ][ ] [ ] 1

Ksp/α1 = Keff = [M+].CaKeff = Effective equilibrium constant, varied on the pHKeff Effective equilibrium constant, varied on the pH

for pH depends to α1.

b) Divalent salt :MA2 M2+ + 2 A2-

H2A + 2 H2O 2 H3O+ + A2-

Kef = Ksp/α12 = [M2+] Ca2

[A2-] = α2 . Ca

α2 = Ka1.Ka2 / { [H3O+]+[H3O+]Ka1+Ka1Ka2 }

Kef = Ksp/α2 = [M2+] . Ca

Molar concentration of iron species in a ferric hydroxyde solution as function of pH in room temperaturesolution as function of pH in room temperature

log Clog C-1 Fe3+

FeOH2+

[Fe3+][OH-]3 = Ksp = 2 x 10-39

[FeOH2+][H+]/[Fe3+] = 9 x 10-4

[Fe(OH)2+][H+]2/[Fe3+] = 5 x 10-7

-3

-2[Fe(OH)2 ][H ] /[Fe ] = 5 x 10[Fe2(OH)2

4+][H+]2/[Fe3+]2 = 1,1 x 10-3

-4Fe(OH)2

+

-6

-5

-7Fe2(OH)2

4+

pH0 1 2 3 4 5

Solubility of HgS at 20 oC as function of pH in asolution containing total sulfide H S + HS-solution containing total sulfide H2S + HS

log [Hg]total, log [Hg(HS)2], dst log [H2S],log [HS-]

-6-2

H2S HS-

-3Hg total-7

4

Hg total

-8

-4

9

-5Hg.HS2- HgS2

2- Hg(HS)2

pH

-94 5 6 7 8 9 10

6. Hydrolysis effect6 yd o ys s e ect

A weak acid salt dissolved in water changes pH of the solution.MA M+ + A-

A- + H2O HA + OH-2

A very weak acid HA has lower Ka and an insoluble MA has ylower Ksp. At a lower [A-] hydrolysis reaction is completed.Depend to the Ksp it should show two extreeme conditions :p pa) A very low solubility of precipitate where pH is not changed

by hydrolysis reaction.b) A high solubility of precipitate where OH- ion produced from

water molecule is neglected.

7. Metal hydroxyde effect

As occured by hydrolysis effect, when a metal hydroxyde dissolved in water the pH will be not changeddissolved in water, the pH will be not changed.

M(OH)2 M2+ + 2 OH-

OH + H O H O+ + OHOH- + H2O H3O+ + OH-

[M2+][OH-]2 = Ksp[H O+][OH ] K[H3O+][OH-] = KwCharge balance : 2 [M2+] + [H3O+] = [OH-]

M l l bilit b l l td f th 3 tiMolar solubility can be calculatd from these 3 equations.When M(OH)2 dissolved then [OH-] increases, this anion will hif di i i i h l f (H O i f d)shift water dissociation reaction to the left (H2O is formed) :

M(OH)2 (p) M2+ + 2 OH-

2H2O H3O+ + OH-

Depend to solubility of OH- it should show two extreemeDepend to solubility of OH it should show two extreeme conditions :a) A very low solubility of precipitate where pH is nota) A very low solubility of precipitate where pH is not

changed by the reaction.[H3O+] = [OH-] = 1 0 x 10-7[H3O ] = [OH ] = 1,0 x 10

Ksp = [M2+][OH-]2

s = Ksp / (1 0 x 10-7)2s = Ksp / (1,0 x 10 )b) A high solubility of precipitate increases [OH-], but

[H3O+] is very low (neglected).[H3O ] is very low (neglected).Charge balance of these equation is either 2[M2+] = [OH-] or [OH-] = 2s[ ] [ ]

Ksp = [M2+][OH-]2 = s (2s)2

33

s = Ksp/4

8. Complex compound formation effectSlightly soluble salt solubility is influenced by a compound forming complex to the metal cation Complexing ion could beforming complex to the metal cation. Complexing ion could be an anion or a neutral molecule which is common or diverse to the precipitate; e g hydrolysis effect of complexing ion of OH-the precipitate; e.g. hydrolysis effect of complexing ion of OH .Example : NH3 is used for separing Ag from Hg.Ag+ + NH Ag(NH )+ K = 2 3 x 103Ag + NH3 Ag(NH3) K1 = 2,3 x 103

Ag(NH3)+ + NH3 Ag(NH3) 2+ K2 = 6,0 x 103

Non-complexed silver fraction (β2) can be calculated as follow:β = 1 / { 1 + K [NH ] + K K [NH ]2 } = [Ag+] / Cβ2 = 1 / { 1 + K1[NH3] + K1K2[NH3]2 } = [Ag+] / CAg

Ksp = [Ag+][Cl-] = β2 CAg [Cl-]Ksp/β K C [Cl ]Ksp/β2 = Kef = CAg [Cl-]

Example : Calculate molar solubility of AgCl in the solution of NH 0 01 M (as a final concentration of free ammoniaof NH3 0,01 M (as a final concentration of free ammonia solution). Ksp AgCl = 1,0 x 10-10. Stability constant K1 = 2,3 x 103 and K2 = 6,0 x 103.2

β2 = 1 / {1 + 2,3 + 103(10-2) + 1,4 x 107 (10-2)2 = 7,1 x 10-4β2 { , ( ) , ( ) ,Keff = 1,0 x 10-10 / 7,1 x 10-4 = 1,4 x 10-7

s = CAg = [Cl-] Ag [ ]s2 = 1,4 x 10-7 , and s = 3,4 x 10-4 M

In existed precipitating ions, most of precipitate could form soluble complex compound In the first step the solubilitysoluble complex compound. In the first step, the solubility decreases into the minimum caused by common ion effect, but then it increases after formating complex compound inbut then it increases after formating complex compound inenough quantity.

AgCl forms soluble complex with Ag+ and Cl- :AgCl + Cl- AgCl -AgCl + Cl- AgCl2-

AgCl2- + Cl- AgCl32-

A Cl + A + A Cl+AgCl + Ag+ Ag2Cl+

C r e of AgCl sol bilit in sol tion of NaCl and AgNOCurve of AgCl solubility in solution of NaCl and AgNO3

(AgCl is more soluble in AgNO3 0,1 M and NaCl 1 M than in water)

-1 -2 -3 -4 -4 -3 -2 -1

log[Cl-] log[Ag+]-3

-4

-55

-6

log[Cl-]log[Ag+] -7

Titration CurveIn a titration of 50 ml of NaCl 0.10 M solution with a solution of AgNO3 0 10 M calculate chloride ion concentrationsof AgNO3 0.10 M, calculate chloride ion concentrations during the titration and plot a titration curve of pCl vs ml of AgNO3. Ksp AgCl = 10 x 10-10.3

Condition before titration : [Cl-] = 0.10 M or pCl = 1.00[ ] p

After addition of 10 ml AgNO3 :g 3

Ag+ + Cl- AgCl (p)initial 1.00 mmol 5.00 mmolchange -1.0 mmol -1.0 mmolequilibrium - 4.0 mmolequilibrium 4.0 mmol

[Cl-] = 4.00 mmol / 60.0 ml = 0.067 M or pCl = 1.17

After addition of 49.9 ml AgNO3 :Ag+ + Cl- AgCl (p)Ag+ + Cl- AgCl (p)

initial 4.99 mmol 5.00 mmolh 4 99 l 4 99 lchange - 4.99 mmol - 4.99 mmol

equilibrium - 0.01 mmol[Cl ] 0 01 l / 99 9 l 1 0 10 4 M Cl 4 00[Cl-] = 0.01 mmol / 99.9 ml = 1.0 x 10-4 M or pCl = 4.00

A h E i l P iAt the Equivalent Point :Ag+ + Cl- AgCl (p)

initial 5.00 mmol 5.00 mmolchange - 5.00 mmol - 5.00 mmolequilibrium - -

[Ag+] = [Cl-] then [Ag+][Cl-] = Ksp = 1.0 x 10-10

[Cl-] = 1.0 x 10-5 or pCl = 5.00

After addition of 60 0 ml AgNO :After addition of 60.0 ml AgNO3 :Ag+ + Cl- AgCl (p)

initial 6 00 mmol 5 00 mmolinitial 6.00 mmol 5.00 mmolchange - 5.00 mmol - 5.00 mmolequilibrium 1 00 mmolequilibrium 1.00 mmol -

[Ag+] = 1.00 mmol / 110 ml = 9.1 x 10-3 MpAg = 2 04 or pCl = 10 00 2 04 = 7 96pAg = 2.04 or pCl = 10.00 – 2.04 = 7.96

Generally for halide salts :A + + X A X ( )Ag+ + X- AgX (p)

Equilibrium constant : K = 1 / [Ag+][X-] = 1 / KspS ll th K hi h th K f tit ti tiSmaller the Ksp, higher the K of a titration reaction.

ARGENTOMETRIC

AgI

ARGENTOMETRICTITRATION CURVES

AgBr13

AgClX

11

10Ksp AgCl = 1 x 10-10

KspAgBr = 2 x 10-12pX 10

8

KspAgBr = 2 x 10 12

KspAgI = 1 x 10-16

6

4

10 20 30 40 50 60 70 80

2

ml of AgNO3

10 20 30 40 50 60 70 80

Titration FeasibilityAs in an acid-base titration, the feasible K value of precipitation reaction could be calculated from its ionic concentrations. Example :

A solution of 50 ml of NaX 0.10 M is titrated with 50 ml of A NO 0 10 M C l l h K d K f A X hAgNO3 0.10 M. Calculate the K and Ksp of AgX when addition of 49.95 ml of the titrant could achieve a completed stoichiometric ionic reaction where the pX changed in 2 00stoichiometric ionic reaction, where the pX changed in 2.00 units at the addition of 2 drops (0,10 ml) of titrant solution.

Consider NaX as a complete dissolved salt the reaction is :Consider NaX as a complete dissolved salt, the reaction is :Ag+ + X- AgX (p)

K = 1/Ksp An addition of one drop of titrant performed beforeK 1/Ksp. An addition of one drop of titrant performed beforethe equivalent point consumes 4.995 mmol of Ag+, and dissolved Ag+ needed to reach the equivalent point is :dissolved Ag needed to reach the equivalent point is :(50 x 0.10) – 4.995 mmol = 0.005 mmol.

[X-] = 0.005 mmol / 99.95 ml = 5 x 10-5 M or pX = 4,30a) If ΔpX = 2 00 then pX = 6 30 or [X ] = 5 x 10 7 Ma) If ΔpX = 2,00 then pX = 6.30 or [X-] = 5 x 10-7 MThe titration consumed 50.00 ml of titrant means one drop

f il iexcess of silver ion :[Ag+] = 0.05 x 0.10 mmol / 100.05 ml = 5 x 10-5 M.K 1 / {(5 10 5)(5 10 7)} 4 1010K = 1 / {(5 x 10-5)(5 x 10-7)} = 4 x 1010

Ksp = 1 / (4 x 1010) = 2,5 x 10-11 M

b) If we consider ΔpX = 1.00 the K = 4 x 109 will be obtained.

In the titration of X- with Ag+ ions, the ΔpX value at the equivalent point depends on the analyte and titrant concentrationspoint depends on the analyte and titrant concentrations.

These phenomena are equal to those in acid-base titration.Before equivalent point : smaller the [X-], higher the pX.

At the equivelent point : smaller the [X-], smaller the ΔpX.

At a smaller titrant concentation, the lower ,curve branchement after the equivelent point as well as the lower of ΔpX at thepoint, as well as the lower of ΔpX at the equivalent point.

To obtain a good feasible titration of Cl-To obtain a good feasible titration of Cl(analyte) with Ag+ (titrant) solutions, it isrequired both concentrations of analyterequired both concentrations of analyte and titrant smaller than 0.10 M.

Methods inMethods in precipitatometric titrationp p

Argentometric methodArgentometric methodMercurimetric methodKolthoff titration

Argentometric titration

Argentometry is the most usefull method ofArgentometry is the most usefull method of precipitatometric titration, as it caused of very low solubility product of halide (or pseudohalide) salts i esolubility product of halide (or pseudohalide) salts,i.e.

Ksp AgCl = 1,82 . 10-10 Ksp AgCN = 2,2 . 10-16

Ksp AgCNS = 1,1 . 10-12 Ksp AgI = 8,3 . 10-17

Ksp AgBr = 5,0 . 10-13

Th 3 h i f d i d i iThere are 3 techniques of end point determination• method of Mohr (indicator : chromic potassium)• method of Volhard (indicator : ferric salt)• method of Fajans (indicator : fluorosceince)method of Fajans (indicator : fluorosceince)

ARGENTOMETRY – MOHRMohr titration is used for determination of halide or pseudohalide in a solution Chromate ion (CrO 2-) ispseudohalide in a solution. Chromate ion (CrO4 ) is added to serve as indicator. At the end point the chromate ion is combined with silver ion to form thechromate ion is combined with silver ion to form the sparingly soluble, red, silver chromate, Ag2CrO4.

Ksp Ag CrO = 1 2 10-12 mol3 L-3Ksp Ag2CrO4 = 1,2 . 10-12 mol3.L-3

Ksp AgCl = 1,82 . 10-10 mol2.L-2

[ Please consider the stoichiometric unit of these ionic reactions ]

Although the solubility product constant (Ksp) of AgCrO4 is close to the Ksp of silver (pseudo)halida, g 4 ( )these silver salts have different solubility.

Mohr titration has to be performed at a neutral or weak basic solution of pH 7-9 (or pH 6-10).

In a lower pH (acid) solution the chromate-In a lower pH (acid) solution the chromate-dichromate equilibrium decreases the sensitivity of [CrO 2-] then inhibits formation of Ag CrO[CrO4 ], then inhibits formation of Ag2CrO4precipitate.2 CrO 2- + 2 H+ Cr O 2- + H O2 CrO4

2- + 2 H+ Cr2O72- + H2O

Chromate Dichromate

In a higher pH (basic) solution, Ag2O will be formed as a black precipitate.

Ag+ + Cl- AgCl (p)Ag + Cl AgCl (p)Ag+ + CrO4

2- Ag2CrO4 (p) red

Solubility of Ag2CrO4 > Solubility of AgCl(8 4 x 10-5 M) (1 35 x 10-5 M)(8.4 x 10 M) (1.35 x 10 M)

If Ag+ solution is added to a Cl- solution containing of small quantity of CrO4

2- then AgClcontaining of small quantity of CrO4 , then AgCl will firstly precipitated, while Ag2CrO4 has not yet, and [Ag+] increases progressively untilyet, and [Ag ] increases progressively until solubility product of the ions reach the value of Ksp Ag2CrO4 (2,0 x 10-12) to form red precipitate.Ksp Ag2CrO4 (2,0 x 10 ) to form red precipitate.

At th i l t i t A Cl 5 00At the equivalent point : pAg = pCl = 5.00

[A +][C O 2 ] 2 00 10 12[Ag+][CrO42-] = 2.00 x 10-12

Then [CrO42-] = 2.00x10-12 / (1.0x10-5)2 = 0.02 M

This chromate concentration is too high as ll l f C O 2 di t b i lyellow colour of CrO4

2- disturb visual observation of Ag2CrO4 (red). In laboratory work

ll t 0 005 /d 0 01 M fwe use usually at 0.005 s/d 0.01 M for minimizing titration error. And it could be

t d b i di t bl k tit ti bcorrected by an indicator blank titration, or by AgNO3 solution standized to a pure chloride salt ( l diti t th l tit ti )(equal condition to the sample titration) .

Mohr titration is limited at pH 6-10 (or pH 7-9). I i ti i b i l ti ill bIonic reaction in basic solution will be :Ag+ + OH- 2AgOH Ag2O + H2OIn acid sol tion [CrO 2 ] ill decrease ntil onl aIn acid solution, [CrO4

2-] will decrease until only a few of ionized HCrO4

-, as the reaction will be followed with the dichromate formation :followed with the dichromate formation :2H+ + CrO4

2- 2HCrO4- Cr2O7

2- + H2O(chromate) (dichromate)

If the [CrO42-] is too low ( < 0,005 M) then the

reaction needs excessive addition of [Ag+] toreaction needs excessive addition of [Ag+] to precipitate Ag2CrO4; this might be a titration errorerror.Dichromate ion, Cr2O7

2-, can not be used as indicator as Ag Cr O precipitate is easilyindicator as Ag2Cr2O7 precipitate is easily dissolved in this ionic solution.

Mohr method can be used for determining Br-gand CN- in a weak basic solution, but it is not feasible for I- and CNS- as of the salt precipitate adsorption.

Ag+ can not be titrated with Cl- solution using indicator of CrO4

2-, because Ag2CrO4 will be 4 , g2 4early obtained and slowly dissolves near the equivalent point. This shlould be overcome by q p yperforming a back titration technique : to the solution of Ag+ we add standardized excessive gCl- solution, then the excess of Cl- is titrated with standardized Ag+ using indicator of CrO4

2-g g 4.

ARGENTOMETRY - VOLHARDVolhard titration is an indirect (back titration) technique which is used if reaction is too slow or iftechnique which is used if reaction is too slow or if there is no appropriate indicator selected for determining the equivalent point.determining the equivalent point.Titration principle :Silver solution is added excessively to a (pseudo)halideSilver solution is added excessively to a (pseudo)halideBr- + Ag+ AgBr (precipitate)

excess

After reaction has completed, the precipitate is filtered, then the filtrate is titrated with a standardized thiocyanate solution.Ag+ + SCN- AgSCN (solution)

Fe(III) indicator reacts with thiocyanate ion to form a red colour solution :red colour solution :

Fe3+ + SCN- [Fe(SCN)]2+

Th ti i id diti i b iThe reaction requires acid condition, as in basic solution the ferric ions form Fe(OH)3 precipitate.

39 3 3Ksp Fe(OH)3 = 2.10-39 mol3L-3

( [Fe3+] = 10-2 M is usually used)( [ ] y )

Example : A solution of KBr is titrated with Volhard procedure requires p p qaddition of 100 ml of excessed AgNO3 0,095 M, then titrated with 18,3 ml of KSCN 0,100 M using a Fe3+ indicator. Calculate Br- concentration in the initial solutionthe initial solution.

Volhard method is very common for ion determination based on Ag+ and Cl- reactions, as of very low precipitate salt solubility Acid condition isvery low precipitate salt solubility. Acid condition is required to avoid hydrolysis of Fe3+ indicator. In a

l l i f l i (H 2neutral solution, some of colour cations (Hg2+, Co2+, Ni2+, Cu2+) will precipitate and disturb the stoichiometric reaction.Volhard method can be used as a direct titration ofVolhard method can be used as a direct titration ofAg+ with CNS- as well as back titration of Cl-, Br- and I d t i ti Th B d I t b di t bI- determination. These Br- and I- can not be disturb by the CNS- as solubility of AgBr is equal to those of AgCNS, while solubility of AgI < solbility of AgCNS.

Titration error of Cl- determination would happen ifAgCl reacts with CNS- :AgCl reacts with CNS :

AgCl(s) + CNS- AgCNS + Cl-

Solubility of AgCNS < solubility of AgCl, then the reaction in above will shift to the left side of the reaction, and the result of Cl- analysis will be lower But this could be prevented by filtering AgCllower. But this could be prevented by filtering AgCl precipitate or nitrobenzen (poison !) addition before titration with CNS-. Nitrobenzen becomes a lipid layer between precipitate and CNS- solution.

ARGENTOMETRY – FAJANSFajans titration use adsorption indicators, i.e. organiccompounds which is adsorbed into col loidalprecipitate surface during the titration processes.Example : Fluoresence in form of its fluorescenate (yellowish

green) anion react with Ag+ to form an intensive red precipitate which is adsorbed to AgCl precipitate surface caused by ionic pair interaction.Fi t t f tit ti tit i Fi l t f tit ti A +I dFirst step of titration titrasi Final step of titration : Ag+Ind-

Cl- Ag+ Ind-

Cl-

Cl

ClCl-

Cl-

Cl-

Cl-

Ag+Ag+

Ag+

Ag+

Ag+

Ind-

Ind-AgCl

Cl-

Cl-Cl- AgCl Ag+Ag+

Ag+

Ind-

Ind-

Adsorption of colour organic compound on a precipitate surface could induce intramolecular electronic shift then change solution colourelectronic shift, then change solution colour. This phenomenon is usually used for end titration d i f il l i i idetection of silver salt precipitation.Before Eq.Pt. :

(AgCl).Cl- M+

Primary layer Secondary layer Excess of Cl-

After Eq.Pt. :q(AgCl).Ag+ X-

Primary layer Secondary layer Excess of Ag+y y y y g

A precipitate tends to adsorp easily ions formatingA precipitate tends to adsorp easily ions formatingslighty soluble salts with ions of the precipitate lattice.Then, Ag+ or Cl- could be more easily adsorb by AgCl precipitate than by Na+ or NO3

-. These anions in thep p y 3

solution will be attracted by the precipitate to form anionic secondary layerionic secondary layer.

Fl i i k i id f fl i t iFluorescein is a weak organic acid, forms a fluoresceinate ion which is not adsorbed by colloidal AgCl precipitate during

f Cl B t i th l ti ith f A + thexcess of Cl-. But in the solution with more excess of Ag+ the fluoresceinate anion will be adsorbed to form a Ag+ shielding l f ll d b h th l ti l t i klayer, followed by change the solution colour to pink.

The factors that must be consider in choosing an adsorption indicator :1) At the equivalent point don’t let the AgCl1) At the equivalent point, don t let the AgCl

precipitate grow too fast to form a big coagulant,for this could sharply decrease adsorptivity offor this could sharply decrease adsorptivity of the precipitate surface to indicator molecules. But if it was happened we can overcome byBut if it was happened, we can overcome by addition of dextrin molecules (as a protective colloid) into the solution so that increasecolloid) into the solution, so that increase particles dispersion, the colour change will be reversible and after equivalent point we canreversible and after equivalent point we can perform back titration using a standardized chloride (Cl-) solutionchloride (Cl ) solution.

2) Adsorption of indicator should be happen near and more fastly at the equivalent point. A bad indicator performance will result a too strong adsorption process resulting which can substitute ion adsorbed before the equivalent point.

3) The pH must be weel controlled for maintainingindication ionic concentration asam, in basic or acidic solution. For example, fluorescein (Ka = 10-7) in a solution having pH > 7 will release small quantity of fluoreseinate ion, hence we can not observe colour change of the indicator.Fluorescein is used feasibly in a solution of pH 7-10, difluorescein (Ka=10-4) is at pH 4-10.

4) It is recommended to choose ionic form of4) It is recommended to choose ionic form of indicator with ionic charge in opposite of titrant ion Adsorption of indicator can not happenion. Adsorption of indicator can not happen before an excess of titrant ion.I tit ti f A + l ti ith Cl th l i l tIn a titration of Ag+ solution with Cl- methyl violet (as chloride salt of an organic base) could be

d i di t f h iused as an indicator of choice. Before an excess of Cl- in the solution that resulting a negative charge on colloid layer, cation should not be adsorbed. For this condition, we can use dichlorofluorescein, but it must be added just close to the equivalent point.

Adsorption Indicators

INDICATOR ANALYTE TITRANT REACTION CONDITION

Diklorofluorescein Cl- Ag+ pH = 4Fluorescein Cl- Ag+ pH = 7 – 8 Eosin Br-, I-, SCN- Ag+ pH = 2Thorin SO4

2- Ba2+ pH = 1,5 – 3,5Bromcresol green SCN- Ag+ pH = 4 – 5Bromcresol green SCN Ag pH 4 5Methyl violet Ag+ Cl- acid solutionRhodamin 6G Ag+ Br- HNO3 upto 0,3 MO th h T Pb2+ C O 2 t l 0 02 M lOrthochrome T Pb2+ CrO4

2- neutral 0,02 M solnBromphenol blue Hg2

2+ Cl- solution of 0,1 M

List of Precipitation Titrations

ANALYTE TITRANT INDICATOR METHOD

Cl-, Br- AgNO3 K2CrO4 Mohr

Cl- Br- I- SCN- AgNO3 Adsorption FajansCl ,Br ,I ,SCN AgNO3 Adsorption Fajans

Br-,I-,SCN-,AsO43- AgNO3 + KSCN Fe(III) Volhard

(not filtered)Cl-,CN-,CO3

2-,S2- AgNO3 + KSCN Fe(III) VolhardC2O4

2-,CrO42- (filtered)

F- Th(IV) Alizarin FajansSO4

2- BaCl2 Tetrahydroxyquinoline FajansPO4

3- PbAc2 Dibromofluorescein Fajans C O 2 PbA Fl i F jCrO4

2- PbAc2 Fluorescein FajansAg+ KSCN Fe(III) VolhardZn2+ K4Fe(CN)6 Diphenylamine FajansHg 2+ NaCl Bromphenol blue FajansHg2

2+ NaCl Bromphenol blue Fajans

MERCURIMETRIC TITRATIONHg2+ + 2 Cl- HgCl2 (as for other halides)

When halide ions is titrated with mercuric nitrate solution,[Hg2+] is not found at the equivalent point caused ofprecipitation of HgCl2 during the titration process.After equivalent point, [Hg2+] increases, react with indicator to form a Hg-indicator complex, e.g. Nitropruside form white precipitate, acid solution of diphenylcarbazide ordiphenylcarbazon in forms intensive violet colour solution.

Mercurimetric titration requires a blanc titration:0 17 ml of Hg(NO ) 0 1 N for 50 ml of HgCl 0 05 N0,17 ml of Hg(NO3)2 0,1 N for 50 ml of HgCl2 0,05 N.

This blanc titration volume varies as [HgCl2]EquivalentPointFor the excess of [Hg2+] reacts with HgCl2 :For the excess of [Hg ] reacts with HgCl2 :

HgCl2 + Hg2+ 2 HgCl+

KOLTHOFF TITRATIONDetermination of Zn2+ (as titrant) which is precipitated with standard solution of K-Ferrocyanide

2 K4Fe(CN)6 + 3 Zn2+ K2Zn3[Fe(CN)6]2 + 6 K+4 ( )6 2 3[ ( )6]2

potassium ferro (II) cyanide potassium zink ferro (II) cyanide

Titration end point is detected by using external indicator such as uranylnitrateTitration end point is detected by using external indicator such as uranylnitrate, ammonium molybdate, FeCl3, etc, which needs a special technical skill. So it isbetter to use internal indicators such as diphenylamine, diphenylbenzidine,diphenylamine sulfonate etcdiphenylamine sulfonate, etc.A redox reaction of Fe2+ Fe3+ has rduction potential (at 30 oC) as follow :

E = Eo + 0,060 log [Fe(CN)63-] / [Fe(CN)6

4-]

Acidic solution of ferro-ferric cyanide has much lower reduction potentialcompared to those required to oxydize the indicator, forms intensive coloured of oxidized form. When Zn2+ is added to this solution then a Zn-ferrocyanide will be formed, followed with increasing reduction potential for Fe(CN)6

4- removed from the solution. After Fe(CN)6

4- completely reacted, a sharp increase of reduction ( )6 p y , ppotential which is followed by appearing blue colour of oxidized form of indicator, caused by excess of Zn2+.in the solution.

The ENDThe END