1. 1 PRECIPITATION REACTIONS Solubility of Salts Section 18.4

2. 2 Metal Chloride Salts These products are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions.

3. 3 Analysis of Silver Group The products are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) = Ag + (aq) + Cl - (aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED.

4. 4 Analysis of Silver Group AgCl(s) = Ag + (aq) + Cl - (aq) When solution is SATURATED, expt. shows that [Ag + ] = 1.67 x 10 -5 M. This is equivalent to the SOLUBILITY of AgCl. What is [Cl - ]? [Cl - ] is equivalent to the AgCl solubility.

5. 5 AgCl(s) = Ag + (aq) + Cl - (aq) Saturated solution has [Ag + ] = [Cl - ] = 1.67 x 10 -5 M Use this to calculate K c K c = [Ag + ] [Cl - ] = (1.67 x 10 -5 )(1.67 x 10 -5 ) = 2.79 x 10 -10

6. 6 K c = [Ag + ] [Cl - ] = 2.79 x 10 -10 Because this is the product of “solubilities”, we call it K sp = solubility product constant See Table 18.2 and Appendix J Solubility Product Constant

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8. 8 Lead(II) Chloride PbCl 2 (s) = Pb 2+ (aq) + 2 Cl - (aq) K sp = 1.9 x 10 -5

9. 9 Solution Solubility = [Pb 2+ ] = 1.30 x 10 -3 M [I - ] = ? [I - ] = 2 x [Pb 2+ ] = 2.60 x 10 -3 M Solubility of Lead(II) Iodide Consider PbI 2 dissolving in water PbI 2 (s) = Pb 2+ (aq) + 2 I - (aq) Calculate K sp if solubility = 0.00130 M

10. 10 Calculate K sp K sp = [Pb 2+ ] [I - ] 2 = X{2 X} 2 K sp = 4 X 3 = 4[Pb 2+ ] 3 Solubility of Lead(II) Iodide = 4 (solubility) 3 K sp = 4 (1.30 x 10 -3 ) 3 = 8.8 x 10 -9

11. 11 Precipitating an Insoluble Salt Hg 2 Cl 2 (s) = Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 If [Hg 2 2+ ] = 0.010 M, what [Cl - ] is req’d to just begin the precipitation of Hg 2 Cl 2 ? That is, what is the maximum [Cl - ] that can be in solution with 0.010 M Hg 2 2+ without forming Hg 2 Cl 2 ?

12. 12 Precipitating an Insoluble Salt Hg 2 Cl 2 (s) = Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 Recognize that. K sp = product of maximum ion concs. Precip. begins when product of ion concs. EXCEEDS the K sp.

13. 13 K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 Solution [Cl - ] that can exist when [Hg 2 2+ ] = 0.010 M, If this conc. of Cl - is just exceeded, Hg 2 Cl 2 begins to precipitate.

14. 14 Hg 2 Cl 2 (s) = Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 Now use [Cl - ] = 1.0 M. What is the value of [Hg 2 2+ ] at this point? Solution [Hg 2 2+ ] = K sp / [Cl - ] 2 = K sp / (1.0) 2 = 1.1 x 10 -18 M The concentration of Hg 2 2+ has been reduced by 10 16 !

15. 15 The Common Ion Effect

16. 16 The Common Ion Effect Adding an ion “common” to an equilibrium causes the equilibrium to shift back to reactant.

17. 17 Common Ion Effect PbCl 2 (s) = Pb 2+ (aq) + 2 Cl - (aq) K sp = 1.9 x 10 -5

18. 18 Calculate the solubility of BaSO 4 BaSO 4 (s) = Ba 2+ ( aq ) + SO 4 2- ( aq ) (a)In pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 The Common Ion Effect

19. 19 K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) = Ba 2+ (aq) + SO 4 2- (aq) Solution Solubility = [Ba 2+ ] = [SO 4 2- ] = x K sp = [Ba 2+ ] [SO 4 2- ] = x 2 x = (K sp ) 1/2 = 1.1 x 10 -5 M Solubility in pure water = 1.1 x 10 -5 M BaSO 4 in pure water

20. 20 Solution Solubility in pure water = 1.1 x 10 -5 mol/L. Now starting with 0.010 M Ba 2+. Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSO 4 be less than or greater than in pure water?___ BaSO 4 in in 0.010 M Ba(NO 3 ) 2. BaSO 4 (s) = Ba 2+ (aq) + SO 4 2- (aq)

21. 21 Solution [Ba 2+ ][SO 4 2- ] initial change equilib. The Common Ion Effect + y 0.010 0 + y 0.010 + yy

22. 22 K sp = [Ba 2+ ] [SO 4 2- ] = (0.010 + y) (y) Because y < 1.1 x 10 -5 M (pure), 0.010 + y is about equal to 0.010. Therefore, K sp = 1.1 x 10 -10 = (0.010)(y) y = 1.1 x 10 -8 M = solubility in presence of added Ba 2+ ion. The Common Ion Effect Solution

23. 23 SUMMARY Solubility in pure water = x = 1.1 x 10 -5 M Solubility in presence of added Ba 2+ = 1.1 x 10 -8 M Le Chatelier’s Principle is followed! Add to the right: equilibrium goes to the left The Common Ion Effect BaSO 4 (s) = Ba 2+ (aq) + SO 4 2- (aq)

24. 24 Separating Metal Ions Cu 2+, Ag +, Pb 2+ K sp Values AgCl1.8 x 10 -10 PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14 K sp Values AgCl1.8 x 10 -10 PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14

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26. 26 Separating Salts by Differences in K sp A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. Which precipitates first? K sp for Ag 2 CrO 4 = 9.0 x 10 -12 K sp for PbCrO 4 = 1.8 x 10 -14 Solution The substance whose K sp is first exceeded precipitates first.. The ion requiring the lesser amount of CrO 4 2- ppts. first.

27. 27 Separating Salts by Differences in K sp [CrO 4 2- ] to ppt. PbCrO 4 = K sp / [Pb 2+ ] = 1.8 x 10 -14 / 0.020 = 9.0 x 10 -13 M Solution Calculate [CrO 4 2- ] required by each ion. [CrO 4 2- ] to ppt. Ag 2 CrO 4 = K sp / [Ag + ] 2 = 9.0 x 10 -12 / (0.020) 2 = 2.3 x 10 -8 M PbCrO 4 precipitates first

28. 28 A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. PbCrO 4 ppts. first. K sp (Ag 2 CrO 4 )= 9.0 x 10 -12 K sp (PbCrO 4 ) = 1.8 x 10 -14 How much Pb 2+ remains in solution when Ag + begins to precipitate? Separating Salts by Differences in K sp

29. 29 Separating Salts by Differences in K sp We know that [CrO 4 2- ] = 2.3 x 10 -8 M to begin to ppt. Ag 2 CrO 4. What is the Pb 2+ conc. at this point? [Pb 2+ ] = K sp / [CrO 4 2- ] = 1.8 x 10 -14 / 2.3 x 10 -8 M = 7.8 x 10 -7 M Lead ion has dropped from 0.020 M to < 10 -6 M

30. 30 Separating Salts by Differences in K sp Add CrO 4 2- to solid PbCl 2. The less soluble salt, PbCrO 4, precipitates PbCl 2 (s) + CrO 4 2- = PbCrO 4 + 2 Cl - SaltK sp PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14

31. 31 Separating by K sp PbCl 2 (s) + CrO 4 2- = PbCrO 4 + 2 Cl - SaltK sp PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14 PbCl 2 (s) = Pb 2+ + 2 Cl - K 1 = K sp Pb 2+ + CrO 4 2- = PbCrO 4 K 2 = 1/K sp K net = K 1 K 2 = 9.4 x 10 8 Net reaction is product-favored

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