2. PRECIPITIMETRIC TITRATIONS1
PRECIPITIMETRIC TITRATIONS
Ashraf M. Mahmoud, Ph.D.

3. Theory: based on the determination of some compounds or ions through the formation of insoluble salt (precipitate) by using precipitating agents. .
Requirement for application: the reaction should be rapid and the precipitate must be practically insoluble (KSP = or lower).
Term
Parts of solvents required per one part of solute
Very soluble
Freely soluble
Soluble
Sparingly soluble
Slightly soluble
Very slightly soluble
Practically insoluble
Less than one part
1-10 parts
10-30 parts
parts
100-1,000 parts
1,000-10,000 parts
>10,000 parts

4. Solubility product constant (KSP)
AB AB A- + B+
AB A- + B+
Solid Sol. molecules Ions
Forward reaction (Vf depends only on the temperature):
Vf = Kf (at constant temperature)
Backward reaction (Vb depends on the conc. of both ions) at constant temp. :
Vb = Kb [A-] [B+]
At equilibrium: Vf = Vb
Vf = Kb [A-][B+]
Ksp = Kf / Kb = Vf / Kb = [A- ] [ B+ ]
Generally for AnBm salt:
KSPAnBm = [Am-]n [Bn+]m
e.g For AgCl, Ksp = [Ag+] [Cl-],
For Ag2CrO4, Ksp = [Ag+]2[CrO42-],
For Pb3(PO4)2 , Ksp = [Pb2+]3[PO43-]2

5. Factors affecting solubility
1. Common ion: depress the solubility ( enhances precipitation .
2. Diverse ion: increase solubility; due to interionic attraction which stabilize
the ions of the precipitate in their ionic (i.e. not molecular) form.
3. Temperature: usually slightly increases the solubility (except PbCl2).
4. Complex formation: usually increases the solubility.
AgCl + 2 Cl [AgCl3]2-
AgCl + 2 CN [Ag(CN)2]- + Cl-
AgCl + 2 NH [Ag(NH3)2]+ + Cl-
5. Solvent: inorganic salts more soluble in water than organic solvents.
Ca(NO3)2 dissolves in alcohol-ether mixture, however Sr(NO3)2 does not.
6. pH: salts of weak acids dissolve at low pH (high H+ conc.)

6. Fractional Precipitation
Precipitation of Cl- and I- (initial conc. 0.1 M) by Ag+
(KSP is and for AgCl and Agl respectively).
Precipitation of Agl begins when: [Ag+] = KSPAgI / [I-] = / 10-1 = M.
Precipitation of AgCl starts when: [Ag+] = KSPAgCl / [Cl-] = / 10-1 = 10-9 M.
All Agl is precipitated before AgCl starts to precipitate.
Actually, both ions will be precipitated simultaneously when:
[Ag+] = KSPAgI/[I-] = KSPAgCl/[Cl-], i.e. when
[Cl-]/[I-] = KSPAgCl/ KSPAgI = /10-16 = 106
i.e. when the ratio of [I-]: [Cl-] = 1 : 106

7. Titration Curves in Precipitimetric Titrations
Volume of AgNO3
Cl-
Br- I-
It is a plot of the precipitated ion conc. against volume of
titrant added.
Titration of 100 ml 0.1 N Cl- with 0.1 N AgNO3:
1. The initial pCl- = -log [Cl-] = -log 0.1 = 1.0
2. Upon adding 50 ml of 0.1 N AgNO3,
[Cl-] = (500.1) / 150 = 3.310-2 M, & pCl- = 1.48
3. Upon adding 90 ml of 0.1 N AgNO3,
[Cl-] = (100.1) / 190 = 5.310-3 M, & pCl- = 2.28
4. At the equivalence point; upon adding 100 ml of 0.1 N AgNO3,
[Cl-] = [Ag+] = (KSPAgCl)1/2 = (1.210-10 ) 1/2 = 1.1 & pCl- = pAg+ = 4.96
5. After the equivalence point, upon adding ml of 0.1 N AgNO3,
[Ag+] = (0.10.1) / = 5 & pAg+ = 4.3
But, pKSPAgCl = pAg+ + pCl- = 24.96 = So, pCl- = 9.92 – 4.3 = 5.62
The sharpness of the end point and the magnitude of the inflection:
Directly related to the initial conc. & inversely related to Ksp

8. Application of precipitimetry4
1-Determination of chloride by Mohr’s method (formation of a secondary colored ppt.)
This method is used for the determination of halides in neutral or slightly alkaline medium (pH = 6.5-9) using potassium chromate as indicator.
NaCl + AgNO3 = AgCl + NaNO3
White ppt.
2AgNO3 + K2CrO4 = Ag2CrO KNO3
Brick red ppt.

9. 5
Why…?
1] Conc. of pot. Chromate is critical ?
To prevent precipitation of silver chromate until complete precipitation of AgCl as Ksp of silver chromate is less than Ksp of AgCl.
2] not acidic medium?
To prevent Formation of Ag2Cr2O7 (Soluble).
3] not strong alkaline medium?
To prevent Formation of AgOH (silver hydroxide) then Ag2O (silver oxide, black ppt) which may be precipitated before Ag2CrO4 and interfere with E.p. detection
4] If ammonium salts present, pH sholdn’t exceed 8?
Free NH3 will be produced and dissolve AgCl forming Ag[NH3]2Cl complex.
5] This method can’t be used for detn. of I- and SCN-?
a- AgI and AgSCN adsorb chromate ions so strongly False E.p.
b- AgI being yellow in color, interfere with E.p. detection.

10. 2-Determination of Bromide by Volhard’s method
1] Silver can be determined in acidic med.[nitric acid] with thiocyanate as a titrante and ferric alum as indicator. [ Volhard‛s method].
2] In this method, can applied halides & cyanide in acidic medium.
3] For bromide, treating it with excess AgNO3
and back titrated AgNO3 with thiocyanate.

11. AgNO3 + NH4SCN = AgSCN + NH4NO3
Principle
KBr + AgNO3 = AgBr KNO3
known excess pale yellow ppt.
AgNO3 + NH4SCN = AgSCN + NH4NO3
white ppt.
Fe3+ + SCN- = [Fe (SCN)]++
blood red coloured soln.

12. NOTES Why…? 1] Acidic med.? ♠ favorable to formation of the
blood red complex.
♠ to dissolve the silver salts of
carbonate and phosphate.
2] Not HCl ? Due to formation of AgCl ppt.
or H2SO4 ? Form. of slightly dissociate Ag2SO4
3] Not basic med. ? Form. of AgOH then Ag2O
[ black ppt.]

13. 4] If applied to chloride , AgCl should be removed to prevent reaction of thiocyanate with it [ Ksp of AgSCN < Ksp of AgCl].
Removed by one of these three methods:
1- Filtered off.
2- Boiling to co-agulate
3- Addition of nitrobenzene, act as a coating sub.
arround AgCl ppt.

14. Determination of a mixture of chloride & iodide ions by Fajan’s method
Principle:-
NO NO3-
NO3-
Na Na+
Na+
Ag Ag+
Ag+
Cl Cl-
Cl-
Primary adsorbed layer
AgCl
AgCl
Secondary adsorbed layer
1- During titration
2- At the end point

15.  Fl- Fl- Ag+ Ag+ AgCl Ag+ Fl- Pink complex
3- Competition bet. Fluorescein and nitrate ions at the end point
a- Choride and iodide ions differ in the ease with which they can be adsorbed on the corresponding silver halide.
b- So iodide can be determined by titration with standard silver nitrate in the presence of eosin as indicator.

16. c- Total can be determined by a similar titration in the presence of fluorescein as indicator.
d- Chloride is obtained by difference.
Notes
1] Anionic part of the indicator will compete with nitrate ion in forming the secondary layer. Why?
As its silver salt is less soluble in water than silver nitrate and so it is preferentially adsorbed and forms a colored complex with silver ion on the surface of the precipitate.
2] Eosin cannot be used for detn. of chloride:
Because eosinate anion is strongly adsorbed by AgCl ppt. Before the equivalence point.

17. Good Luck
Ashraf M. Mahmoud