1. PRECIPITATION REACTIONS Chapter 19 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

2. 2 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Analysis of Silver Group All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions.

3. 3 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Analysis of Silver Group Although all salts formed in this experiment 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 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 - ]? This is also equivalent to the AgCl solubility.

5. 5 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Analysis of Silver Group 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 ) = (1.67 x 10 -5 )(1.67 x 10 -5 ) = 2.79 x 10 -10 = 2.79 x 10 -10

6. 6 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Analysis of Silver Group AgCl(s) Ag + (aq) + Cl - (aq) 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 19.2 and Appendix J

7. 7 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Lead(II) Chloride PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) K sp = 1.9 x 10 -5

8. 8 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Consider PbI 2 dissolving in water PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) Calculate K sp if solubility = 0.00130 M Solution Solubility = [Pb 2+ ] = 1.30 x 10 -3 M [I - ] = _______ ? Solubility of Lead(II) Iodide

9. 9 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Consider PbI 2 dissolving in water PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) Calculate K sp if solubility = 0.00130 M Solution 1.Solubility = [Pb 2+ ] = 1.30 x 10 -3 M [I - ] = 2 x [Pb 2+ ] = 2.60 x 10 -3 M [I - ] = 2 x [Pb 2+ ] = 2.60 x 10 -3 M Solubility of Lead(II) Iodide

10. 10 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Consider PbI 2 dissolving in water PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) Calculate K sp if solubility = 0.00130 M Solution 1.Solubility = [Pb 2+ ] = 1.30 x 10 -3 M [I - ] = 2 x [Pb 2+ ] = 2.60 x 10 -3 M [I - ] = 2 x [Pb 2+ ] = 2.60 x 10 -3 M 2.K sp = [Pb 2+ ] [I - ] 2 Solubility of Lead(II) Iodide

11. 11 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Consider PbI 2 dissolving in water PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) Calculate K sp if solubility = 0.00130 M Solution 1.Solubility = [Pb 2+ ] = 1.30 x 10 -3 M [I - ] = 2 x [Pb 2+ ] = 2.60 x 10 -3 M [I - ] = 2 x [Pb 2+ ] = 2.60 x 10 -3 M 2.K sp = [Pb 2+ ] [I - ] 2 = [Pb 2+ ] {2 [Pb 2+ ]} 2 = [Pb 2+ ] {2 [Pb 2+ ]} 2 = 4 [Pb 2+ ] 3 = 4 [Pb 2+ ] 3 Solubility of Lead(II) Iodide

12. 12 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Consider PbI 2 dissolving in water PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) Calculate K sp if solubility = 0.00130 M Solution 2.K sp = 4 [Pb 2+ ] 3 = 4 (solubility) 3 K sp = 4 (1.30 x 10 -3 ) 3 = 8.8 x 10 -9 K sp = 4 (1.30 x 10 -3 ) 3 = 8.8 x 10 -9 Solubility of Lead(II) Iodide

13. 13 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 ?

14. 14 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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.

15. 15 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 Solution [Cl - ] that can exist when [Hg 2 2+ ] = 0.010 M,

16. 16 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 Solution [Cl - ] that can exist when [Hg 2 2+ ] = 0.010 M,

17. 17 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 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.

18. 18 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 Now raise [Cl - ] to 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 = K sp / (1.0) 2 = 1.1 x 10 -18 M The concentration of Hg 2 2+ has been reduced by 10 16 !

19. 19 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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

20. 20 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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.

21. 21 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 Calculate [CrO 4 2- ] required by each ion.

22. 22 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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 = 1.8 x 10 -14 / 0.020 = 9.0 x 10 -13 M [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 = 9.0 x 10 -12 / (0.020) 2 = 2.3 x 10 -8 M PbCrO 4 precipitates first. PbCrO 4 precipitates first. 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

23. 23 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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

24. 24 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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? Solution 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? Separating Salts by Differences in K sp

25. 25 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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. 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? Solution [Pb 2+ ] = K sp / [CrO 4 2- ] = 1.8 x 10 -14 / 2.3 x 10 -8 M = 7.8 x 10 -7 M = 7.8 x 10 -7 M Separating Salts by Differences in K sp

26. 26 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 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. 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? Solution [Pb 2+ ] = K sp / [CrO 4 2- ] = 1.8 x 10 -14 / 2.3 x 10 -8 M = 7.8 x 10 -7 M = 7.8 x 10 -7 M Lead ion has dropped from 0.020 M to < 10 -6 M Separating Salts by Differences in K sp

27. 27 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Common Ion Effect Adding an Ion “Common” to an Equilibrium

28. 28 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution Solubility in pure water = [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 The Common Ion Effect

29. 29 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution Solubility in pure water = 1.1 x 10 -5 mol/L Now dissolve BaSO 4 in water already containing 0.010 M Ba 2+. The Common Ion Effect

30. 30 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution Solubility in pure water = 1.1 x 10 -5 mol/L. Now dissolve BaSO 4 in water already containing 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?___ The Common Ion Effect

31. 31 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution [Ba 2+ ][SO 4 2- ] [Ba 2+ ][SO 4 2- ]initialchangeequilib. The Common Ion Effect

32. 32 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution [Ba 2+ ][SO 4 2- ] [Ba 2+ ][SO 4 2- ] initial0.010 0 change+ y + y equilib.0.010 + y y The Common Ion Effect

33. 33 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution K sp = [Ba 2+ ] [SO 4 2- ] = (0.010 + y) (y) Because y < 1.1 x 10 -5 M (= x, the solubility in pure water), this means 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

34. 34 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution 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! The Common Ion Effect