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PRECIPITIMETRIC TITRATIONS

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2 PRECIPITIMETRIC TITRATIONS
1 PRECIPITIMETRIC TITRATIONS Ashraf M. Mahmoud, Professor

3 Theory of precipitimetric titrations
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). Solubility product constant (KSP) AB (s) AB (aq) A- + B+ AB A- + B+ Solid Sol. molecules Ions 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

4 Factors affecting solubility 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.)

5 Fractional Precipitation Fractional Precipitation
When a mixture of two ions are titrated, the less soluble precipitate will be precipitated first When the equivalence point of the less soluble precipitate is obtained, the concentration of titrant suddenly increases and then the more soluble salt begins to precipitate Therefore, we should observe 2 breaks in the titration curve, one corresponding to each equivalence point

6 Fractional Precipitation
Titration curve for precipitation of Cl- and I- (initial conc. 0.1 M) by AgNO3 soln. (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. Because we add AgNO3 gradually from the burret, the conc. of Ag+ starts low and increase gradually, therefore, all Agl is precipitated firstly while Cl- ions remains in solution. Then, as [Ag+] increases, 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
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 Volume of AgNO3 Cl- Br- I-

8 [X] = [initial conc. of X ]* [dilution]
Examples M * V / n = M * V / n [X] = [initial conc. of X ]* [dilution] Example [1]: in a titration of 25.0 mL of 0.1 M [I- ] with 0.05 M Ag+ and monitor Ag+: Ag I- = AgI (ppt) The expected equivalence volume of titrant Veq. point : M I- * V I- / n = M Ag+ * V Ag+ / n V eq. point = 25 x 0.1 / = 50 mL M * V / n = M * V / n

9 Examples Step 1 25.0 mL of 0.04132 M Hg2(NO3)2 ≠ 0.05789 M KIO3 ,
Ksp for Hg2(NO3)2 is 1.3x10-18 Calculate the concentration of Hg22+ in solution (a) after addition of mL KIO3, (b) after addition of mL KIO3, and (c) at the equivalence point Step 1 M * V / n = M * V / n * 25 / 1 = * V / 2

10 Examples Calculate the concentration of Hg22+ in solution
(a) after addition of mL KIO3, * V /1 = 34 * /2 V of consumed Hg2(NO3)2 = [Hg2(NO3)2] = ( )* / ( ) = M (b) after addition of mL KIO3 [after end point] M * V / n = M * V / n the precipitation is complete and we have gone past the equivalence point

11 Examples Ksp Hg2(IO3)2 = [IO3-]2 * [Hg22+]
1.3 x = [2.9 x 10-4 ]2 * [Hg22+] (c) at the equivalence point

12 Application of precipitimetry
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.

13 Application of precipitimetry
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.

14 Application of precipitimetry
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.

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

16 Application of precipitimetry
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.]

17 Application of precipitimetry
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.

18 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

19  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.

20 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.

21 Good Luck Ashraf M. Mahmoud


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