1. 1 lSome combinations of ions in solution form insoluble salts. Recall the solubility rules from Chapter 4. lEven insoluble salts dissolve to a small extent. lSubstances that are considered “insoluble” are more appropriately termed “slightly” or “sparingly” soluble. lThese substances establish an equilibrium system between the solid and its ions in solution. lCa(OH) 2 equilibrium 15.6 Solubility Equilibria

2. 2 lSolubility of insoluble salts is described by the solubility product constant, K sp. AgCl(s) ⇌ Ag + (aq) + Cl - (aq) AgCl(s) ⇌ Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] = 1.70 x 10 - 10 Ag 2 S(s) ⇌ 2Ag + (aq) + S 2 - (aq) Ag 2 S(s) ⇌ 2Ag + (aq) + S 2 - (aq) K sp = [Ag + ] 2 [S 2 - ] = 1.0 x 10 - 51 lUse to determine conditions for precipitation

3. 3 Solubility lSolubility (s) = molar concentration of dissolved salt; ion concentrations are related to this by their coefficients. lWe can solve solubility problems just like ANY other equilibrium problems. lUse an I.C.E. Equilibrium Analysis!!!

4. 4 Solubility lAfter setting up the balanced equation, can solve for S using an ice chart. lHint: The only difference that you will see in solubility problems is that the unknown concentration is designated with an ‘s’ instead of an ‘x’. lPractice: #75

5. 5 Determining K sp from Solubility l1.What is the K sp value of Calcium fluoride, given that the molar solubility of Calcium fluoride is 2.1 x 10 -4 M?

6. 6 Determining K sp from Solubility Can solve by setting up an ICE table: CaF 2 (s) ⇌ Ca 2+ (aq) + 2 F - (aq) Can solve by setting up an ICE table: CaF 2 (s) ⇌ Ca 2+ (aq) + 2 F - (aq) Initial Conc.0.00 M0.00 M Change + s+ 2s Equil. Conc. s2s K sp = [Ca 2+ ][F - ] 2 = (s)(2s) 2 K sp = [Ca 2+ ][F - ] 2 = (s)(2s) 2 s = 2.1 x 10 -4 M so Ksp = 4s 3 = 3.7 x 10 -11 s = 2.1 x 10 -4 M so Ksp = 4s 3 = 3.7 x 10 -11 lPractice #77

7. 7 Calculating Solubility l1. Determine the molar solubility of iron(III) hydroxide. (K sp = 1.1 x 10 -36 )

8. 8 Calculating Solubility Can solve by setting up an ICE table: Fe(OH) 3 (s) ⇌ Fe 3+ (aq) + 3OH - (aq) Can solve by setting up an ICE table: Fe(OH) 3 (s) ⇌ Fe 3+ (aq) + 3OH - (aq) Initial Conc.0.00 M0.00 M Change + s+ 3s Equil. Conc. s3s K sp = [Fe 3+ ][OH - ] 3 = (s)(3s) 3 = 1.1 x 10 - 36 K sp = [Fe 3+ ][OH - ] 3 = (s)(3s) 3 = 1.1 x 10 - 36 27s 4 = 1.1 x 10 - 36 s 4 = 4.07 x 10 - 38 s = 4.49 x 10 - 10 M

9. 9 Calculating Solubility lIf solubility, S, is known, can reverse this procedure and calculate K sp : S = 4.49 x 10 - 10 M [Fe 3 + ] = 4.49 x 10 - 10 M [OH - ] = 3 x 4.49 x 10 - 10 = 1.35 x 10 - 9 K sp = [Fe 3 + ][OH - ] 3 = (4.49 x 10 - 10 )(1.35 x 10 - 9 ) 3 so, K sp = 1.1 x 10 - 36

10. 10 Calculating Solubility lPractice: #81 lQuestions about problems in Notes??

11. 11 Comparing Solubility using K sp lWe can directly compare only those salts that have the same exponents in the solubility product expression. lIf that condition is met, the lower the value of K sp, the less soluble the salt. lIf K sp expressions are different, K sp can be used to calculate the solubility, which can be compared for any salts.

12. 12 15.6 Factors that Affect Solubility lCan modify solubility to dissolve minerals and ores, to precipitate ions from solution, to separate and purify ions. lExamples: lremove hardness from water by adding Na 2 CO 3 remove Ag + from water by adding Cl - to recover Ag remove Ag + from water by adding Cl - to recover Ag ldissolve Cu(OH) 2. CuCO 3 to mine Cu lseparate the rare earth ions separate U 4 + from Th 4 + separate U 4 + from Th 4 + lseparate the halide ions

13. 13 Common Ion Effect AgCl(s) ⇌ Ag + (aq) + Cl - (aq) AgCl(s) ⇌ Ag + (aq) + Cl - (aq) lCommon ion effect: Adding a common ion to a solution of a slightly soluble salt will cause the solubility of the salt to decrease. add more Cl - to precipitate more Ag + from solution. add more Cl - to precipitate more Ag + from solution.

14. 14 Common Ion Effect lSaturated solution of AgCl: K sp = 1.70 x 10 - 10 = [Ag + ][Cl - ] = S 2 S = [Ag + ] = [Cl - ] = 1.30 x 10 - 5 M Effect of adding 0.100 M Cl - Effect of adding 0.100 M Cl - (>> 1.30 x 10 - 5, so [Cl - ]= 0.100 M) 1.70 x 10 - 10 = [Ag + ](0.100) [Ag + ] = 1.70 x 10 - 9 M

15. 15 pH & Insoluble Basic Salts lMetal hydroxide solubility depends on the pH. Mg(OH) 2 (s) ⇌ Mg 2 + (aq) + 2OH - (aq) Mg(OH) 2 (s) ⇌ Mg 2 + (aq) + 2OH - (aq) lAdjust the pH to adjust the solubility. AgCN(s) ⇌ Ag + (aq) + CN - (aq) AgCN(s) ⇌ Ag + (aq) + CN - (aq) K sp = 1.6 x 10 - 14 lAdd HNO 3 (why not HCl?) to dissolve: CN - (aq) + H 3 O + (aq)  HCN(aq) + H 2 O(l) CN - (aq) + H 3 O + (aq)  HCN(aq) + H 2 O(l) K = 2.5 x 10 9

16. 16 lAdd the two equations (multiply the Ks): AgCN(s) + H 3 O + (aq) ⇌ Ag + (aq) + HCN(aq) + H 2 O(l) K = 1.6 x 10 - 14 x 2.5 x 10 9 = 4.0 x 10 - 5 = [Ag + ][HCN]/[H 3 O + ] = [Ag + ][HCN]/[H 3 O + ] lAdding acid shifts the equilibrium towards products. pH & Insoluble Basic Salts

17. 17 General Effects of pH on Solubility lIf the anion acts like a base, salt will be more soluble in acidic solutions. lIf the cation acts like an acid, salt will be more soluble in basic solutions. lIf neither ion acts like an acid / base, the solubility will be unaffected by pH.

18. 18 15.7 Precipitation and Separation of Ions lTo precipitate an ion, we must add sufficient common ion for Q sp to exceed K sp. lTo determine if a salt will precipitate, compare the Ksp value to the Qsp value (ratio of products to reactants when not necessarily at equilibrium). lQ < K ; no precipitate, more will dissolve lQ = K ; solution is saturated; no precipitate, no more will dissolve lQ > K ; precipitate will form

19. 19 Precipitation of Ions lExample: How much Cl - must be in solution to begin precipitation of Ag + from a 0.100 M Ag + solution. K sp = 1.70 x 10 - 10 K sp = 1.70 x 10 - 10 = (0.100)[Cl - ] K sp = 1.70 x 10 - 10 = (0.100)[Cl - ] [Cl - ] = 1.70 x 10 - 10 /0.100 = 1.70 x 10 - 9 M To dissolve AgNO 3 in water, with no precipitation of AgCl, we must have very pure water (no Cl - in the water).

20. 20 Precipitation of Ions lExample: Will a solution containing 1.0 x 10 - 6 M Cl - and 1.5 x 10 - 4 M Ag + form a precipitate of AgCl? K sp = 1.70 x 10 - 10 lAnswer: Q sp = (1.0 x 10 - 6 )(1.5 x 10 - 4 ) = 1.5 x 10 - 10 Qsp < K sp, so AgCl will not precipitate.

21. 21 Precipitation of Ions lGroup Work: Want to separate 0.100 M Pb 2 + from 0.100 M Ag +, using Cl -. How much Ag + is left when the Pb 2 + begins to precipitate? K sp of PbCl 2 = 1.6 x 10 - 5 K sp of AgCl = 1.70 x 10 - 10 19m09vd1

22. 22 Precipitation of Ions For PbCl 2, K sp = 1.6 x 10 - 5 = (0.100)[Cl - ] 2 For PbCl 2, K sp = 1.6 x 10 - 5 = (0.100)[Cl - ] 2 [Cl - ] = 0.0126 M to begin precipitating PbCl 2. ForAgCl, K sp = 1.70 x 10 - 10 = [Ag + ](0.0126) ForAgCl, K sp = 1.70 x 10 - 10 = [Ag + ](0.0126) [Ag + ] = 1.35 x 10 - 8 M left when PbCl 2 begins to precipitate. % Ag + left = (1.35 x 10 - 8 M) / 0.100 M x 100% % Ag + left = (1.35 x 10 - 8 M) / 0.100 M x 100% = 1.35 x 10 - 5 % = 1.35 x 10 - 5 % l Questions about problems in notes??