2. 1 Solubility Equilibria Solubility Product Constant K sp for saturated solutions at equilibrium

3. 2 Comparing values.

4. 3 Solubility Product (K sp ) = [products] x /[reactants] y but..... reactants are in solid form, so K sp =[products] x i.e. A 2 B 3 (s)  2A 3+ + 3B 2– K sp =[A 3+ ] 2 [B 2– ] 3 Given: AgBr(s)  Ag + + Br – In a saturated solution @25 o C, the [Ag + ] = [Br – ]= 5.7 x 10 –7 M. Determine the K sp value.

5. 4 Problem: A saturated solution of silver chromate was to found contain 0.022 g/L of Ag 2 CrO 4. Find K sp Eq. Expression: Ag 2 CrO 4 (s)  2Ag + + CrO 4 2– K sp = [Ag + ] 2 [CrO 4 2– ] So we must find the concentrations of each ion and then solve for K sp.

6. 5 Problem: A saturated solution of silver chromate was to contain 0.022 g/L of Ag 2 CrO 4. Find K sp Eq. Expression: Ag 2 CrO 4 (s)  2Ag + + CrO 4 2– K sp = [Ag + ] 2 [CrO 4 2– ] Ag + : 0.022 g Ag 2 CrO 4 L 1.33 x 10 –4 CrO 4 –2 : 0.022g Ag 2 CrO 4 L 6.63 x 10 –5

7. 6 Problems working from K sp values. Given: K sp for MgF 2 is 6.4 x 10 –9 @ 25 o C Find: solubility in mol/L and in g/L MgF 2 (s)  Mg 2+ + 2F – K sp = [Mg 2+ ][F – ] 2 I. C. E. N/A 0 0 N/A +x +2x K sp = [x][2x] 2 = 4x 3 6.4 x 10 –9 = 4x 3 now for g/L: 7.3 x 10 –2

8. 7 The common ion effect “Le Chatelier” overhead fig 17.16 What is the effect of adding NaF? CaF 2 (s)  Ca 2+ + 2F -

9. 8 Solubility and pH CaF 2 (s)  Ca 2+ + 2F – Add H + (i.e. HCl) 2F – + H +  HF

10. 9 Solubility and pH Mg(OH) 2 (s)  Mg 2+ + 2OH – Adding NaOH? Adding HCl?

11. 10 The common ion effect “Le Chatelier” Why is AgCl less soluble in sea water than in fresh water? AgCl(s)  Ag + + Cl – Seawater contains NaCl

12. 11 Problem: The solubility of AgCl in pure water is 1.3 x 10 – 5 M. What is its solubility in seawater where the [Cl – ] = 0.55 M? (K sp of AgCl = 1.8 x 10 –10 ) AgCl(s)  Ag + + Cl – I. C. E. N/A 0 0.55 N/A +x +x N/A +x 0.55 + x K sp = [Ag + ][Cl – ] K sp = [x][0.55 + x] try dropping this x K sp = 0.55x 1.8 x 10 –10 = 0.55x x = 3.3 x 10 –10 = [Ag + ]=[AgCl] “AgCl is much less soluble in seawater”

13. 12 more Common ion effect: a. What is the solubility of CaF 2 in 0.010 M Ca(NO 3 ) 2 ? K sp (CaF 2 ) = 3.9 x 10 –11 CaF 2 (s)  Ca 2+ + 2F – [Ca 2+ ] [F – ] I. C. E. 0.010 0 +x +2x 0.010 + x 2x K sp = [0.010 + x][2x] 2  [0.010][2x] 2 = 0.010(4x 2 ) 3.9 x 10 –11 = 0.010(4x 2 ) x = 3.1 x 10 –5 M Ca 2+ from CaF 2 so = M of CaF 2 Now YOU determine the solubility of CaF 2 in 0.010 M NaF. K sp =[Ca 2+ ][F - ] 2

14. 13 Answer:3.9 x 10 –7 M Ca 2+ CaF 2 (s)  Ca 2+ + 2F – 0 0.010 +x 2x x 0.010 + 2x K sp = [x][0.010 + 2x] 2 3.9 x 10 -11 =x(0.010) 2  x(0.010) 2 x = 3.9 x 10 -7 What does x tell us

15. 14 Reaction Quotient (Q): will a ppt. occur? Use Q (also called ion product) and compare to K sp Q < K sp reaction goes Q = K sp Equilibrium Q > K sp reaction goes

16. 15 Problem: A solution is 1.5 x 10 –6 M in Ni 2+. Na 2 CO 3 is added to make the solution 6.0 x 10 –4 M in CO 3 2–. K sp (NiCO 3 ) = 6.6 x 10 –9. Will NiCO 3 ppt? NiCO 3  Ni 2+ + CO 3 2– K sp = [Ni 2+ ][CO 3 2– ] Q = [Ni 2+ ][CO 3 2– ] Q = [1.5 x 10 –6 ][6.0 x 10 –4 ] = 9.0 x 10 –10 Q < K sp no ppt. We must compare Q to K sp.

17. 16 Problem: 0.50 L of 1.0 x 10 –5 M Pb(OAc) 2 is combined with 0.50 L of 1.0 x 10 –3 M K 2 CrO 4. a. Will a ppt occur? K sp (PbCrO 4 ) = 1.8 x 10 –14 Pb(OAc) 2 (aq) + K 2 CrO 4 (aq)  PbCrO 4 (s) + 2KOAc(aq) then: PbCrO 4 (s)  Pb 2+ + CrO 4 2– K sp = [Pb 2+ ][CrO 4 2– ] [Pb 2+ ]: 0.50 L 1 L 5.0 x 10 –6 [CrO 4 2- ]: 0.50 L 5.0 x 10 -4 Q = [5.0 x 10 -6 ][5.0 x 10 -4 ] = 2.5 x 10 -9 compare to K sp : Q > K sp so a ppt. will occur 1 L

18. 17 b. find the Eq. conc. of Pb 2+ remaining in solution after the PbCrO 4 precipitates. Since [Pb 2+ ] = 5.0 x 10 -6 and [CrO 4 2- ] = 5.0 x 10 -4 and there is a 1:1 stoichiometry, Pb 2+ is the limiting reactant. PbCrO 4 (s)  Pb 2+ + CrO 4 2– I. (after ppt.) C. E. 0 5.0 x 10 -4 - 5.0 x 10 -6 = 5.0 x 10 -4 +x +x x 5.0 x 10 –4 + x K sp = [x][5.0 x 10 –4 + x] Try dropping the “+ x” term. K sp (PbCrO 4 ) = 1.8 x 10 –14 1.8 x 10 –14 = x(5.0 x 10 -4 ) x = [Pb 2+ ] = 3.6 x 10 –11 This is the concentration of Pb 2+ remaining in solution.

19. 18 Complex ion formation: Ag + (aq) + NH 3 (aq)  Ag(NH 3 ) + (aq) AgCl(s)  Ag + + Cl – Ag(NH 3 ) + (aq) + NH 3 (aq)  Ag(NH 3 ) 2 + (aq) Ag + (aq) + 2NH 3 (aq)  {Ag(NH 3 ) 2 } + (aq) formation or stability constant: K sp = 1.8 x 10 –10 For Cu 2+ : Cu 2+ + 4NH 3  [Cu(NH 3 ) 4 ] 2+ (aq) K 1 x K 2 x K 3 x K 4 = K f = 6.8 x 10 12

20. 19 Solubility and complex ions: Problem: How many moles of NH 3 must be added to dissolve 0.050 mol of AgCl in 1.0 L of H 2 O? (K sp AgCl = 1.8 x 10 –10 ; K f [Ag(NH 3 ) 2 ] + = 1.6 x 10 7 ) AgCl(s)  Ag + + Cl – Ag + (aq) + 2NH 3 (aq)  Ag(NH 3 ) 2 + (aq) sum of RXNS: AgCl(s) + 2NH 3  Ag(NH 3 ) 2 + (aq) + Cl – = 2.9 x 10 –3 Now use K overall to solve the problem: K overall = 2.9 x 10 –3 = [NH 3 ] eq = 0.93 but..... How much NH 3 must we add? [NH 3 ] total = 0.93 + (2 x 0.050) = 1.03 M 2 ammonia’s for each Ag +

21. 20 Fractional Precipitation: “ppting” one ion at a time. Compounds must have different K sp values (i.e. different solubilities) Example: K sp CdS = 3.9 x 10 –29 and K sp NiS = 3.0 x 10 –21 ? Which will ppt. first? least soluble Problem: A solution is 0.020 M in both Cd 2+ and Ni 2+. Just before NiS begins to ppt., what conc. of Cd 2+ will be left in solution? Approach: Find conc. of S 2– ion when Ni 2+ just begins to ppt. since Cd 2+ will already be ppting. Then use this S 2– conc. to find Cd 2+. NiS(s)  Ni 2+ + S 2– K sp = 3.0 x 10 –21 = [0.020][S 2– ] [S 2– ] = 1.5 x 10 –19 M when Ni 2+ just begins to ppt. So: CdS(s)  Cd 2+ + S 2– K sp = 3.9 x 10 –29 = [Cd 2+ ][1.5 x 10 –19 ] [Cd 2+ ] = 2.6 x 10 –10 M when NiS starts to ppt.