1 PRECIPITATION REACTIONS Solubility of Salts Section 18.4

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

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

6 K c = [Ag + ] [Cl - ] = 2.79 x 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 Lead(II) Chloride PbCl 2 (s) = Pb 2+ (aq) + 2 Cl - (aq) K sp = 1.9 x 10 -5

9 Solution Solubility = [Pb 2+ ] = 1.30 x M [I - ] = ? [I - ] = 2 x [Pb 2+ ] = 2.60 x 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 = M

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 ) 3 = 8.8 x 10 -9

11 Precipitating an Insoluble Salt Hg 2 Cl 2 (s) = Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x = [Hg 2 2+ ] [Cl - ] 2 If [Hg 2 2+ ] = 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 M Hg 2 2+ without forming Hg 2 Cl 2 ?

12 Precipitating an Insoluble Salt Hg 2 Cl 2 (s) = Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x = [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 K sp = 1.1 x = [Hg 2 2+ ] [Cl - ] 2 Solution [Cl - ] that can exist when [Hg 2 2+ ] = M, If this conc. of Cl - is just exceeded, Hg 2 Cl 2 begins to precipitate.

14 Hg 2 Cl 2 (s) = Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 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 M The concentration of Hg 2 2+ has been reduced by !

15 The Common Ion Effect

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

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

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

19 K sp for BaSO 4 = 1.1 x 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 M Solubility in pure water = 1.1 x M BaSO 4 in pure water

20 Solution Solubility in pure water = 1.1 x mol/L. Now starting with 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 M Ba(NO 3 ) 2. BaSO 4 (s) = Ba 2+ (aq) + SO 4 2- (aq)

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

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

23 SUMMARY Solubility in pure water = x = 1.1 x M Solubility in presence of added Ba 2+ = 1.1 x 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 Separating Metal Ions Cu 2+, Ag +, Pb 2+ K sp Values AgCl1.8 x PbCl x PbCrO x K sp Values AgCl1.8 x PbCl x PbCrO x

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26 Separating Salts by Differences in K sp A solution contains 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 K sp for PbCrO 4 = 1.8 x Solution The substance whose K sp is first exceeded precipitates first.. The ion requiring the lesser amount of CrO 4 2- ppts. first.

27 Separating Salts by Differences in K sp [CrO 4 2- ] to ppt. PbCrO 4 = K sp / [Pb 2+ ] = 1.8 x / = 9.0 x 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 / (0.020) 2 = 2.3 x M PbCrO 4 precipitates first

28 A solution contains 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 K sp (PbCrO 4 ) = 1.8 x How much Pb 2+ remains in solution when Ag + begins to precipitate? Separating Salts by Differences in K sp

29 Separating Salts by Differences in K sp We know that [CrO 4 2- ] = 2.3 x 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 / 2.3 x M = 7.8 x M Lead ion has dropped from M to < M

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 Cl - SaltK sp PbCl x PbCrO x

31 Separating by K sp PbCl 2 (s) + CrO 4 2- = PbCrO Cl - SaltK sp PbCl x PbCrO x PbCl 2 (s) = Pb 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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