1. CHE1102, Chapter 17 Learn, 1 Chapter 17 Solubility and Simultaneous Equilibria

2. CHE1102, Chapter 17 Learn, 2 Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s)Ba 2+ (aq) + SO 4 2  (aq) Saturated solution – no more solute will dissolve

3. CHE1102, Chapter 17 Learn, 3 Solubility Product Constant The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2  ] – equilibrium constant for ionic compounds that are only slightly soluble

4. CHE1102, Chapter 17 Learn, 4 K sp is not the same as solubility. Solubility generally expressed as (g/L), (g/mL), or in mol/L (M). –K sp - Only one value for given solid at a given temperature Temperature dependence Solubility and thus K sp changes with T Solubility Products

5. CHE1102, Chapter 17 Learn, 5 Solubility Equilibria When ionic salt dissolves in water It dissociates into separate hydrated ions Initially, no ions in solution CaF 2 (s)  Ca 2+ (aq) + 2F – (aq) As dissolution occurs, ions build up and collide Ca 2+ (aq) + 2F – (aq)  CaF 2 (s) At equilibrium CaF 2 (s) Ca 2+ (aq) + 2F – (aq) We now have a saturated solution K sp = [Ca + ][F – ] 2

6. CHE1102, Chapter 17 Learn, 6 Writing K sp Equilibrium Laws AgCl(s) Ag + (aq) + Cl – (aq) K sp = [Ag + ][Cl – ] PbI 2 (s) Pb 2+ (aq) + 2I – (aq) K sp = [Pb 2+ ][I – ] 2 Ag 2 CrO 4 (s) 2Ag + (aq) + CrO 4 2– (aq) K sp = [Ag + ] 2 [CrO 4 2– ] AuCl 3 (s) Au 3+ (aq) + 3Cl – (aq) K sp = [Au 3+ ][Cl – ] 3

7. CHE1102, Chapter 17 Learn, 7 – # moles of ionic solid that dissolves per 1 L DI water to form a saturated solution at 25  C 6.7 x 10 - 5 moles of CaCO 3 (s) will dissolve per 1 L DI water producing 6.7 x 10 - 5 M Ca 2+ (aq) and 6.7 x 10 - 5 M CO 3 2 - (aq) Determine the final equilibrium concentration of Ca 2+ (aq) and CO 3 2 - (aq) in a saturated solution Molar Solubility

8. CHE1102, Chapter 17 Learn, 8 Given Solubilities, Calculate K sp At 25 °C, the solubility of AgCl is 1.34 × 10 –5 M. Calculate the solubility product for AgCl. AgCl(s) Ag + (aq) + Cl – (aq) K sp = [Ag + ][Cl – ] K sp = (1.34 × 10 –5 )(1.34 × 10 –5 ) K sp = 1.80 × 10 –10 AgCl (s) Ag + (aq) +Cl – (aq) I0.00 C E 1.34 × 10 –5 M

9. CHE1102, Chapter 17 Learn, 9 Learning Check The solubility of calcium fluoride, CaF 2, in pure water is 2.15 × 10 ‑ 4 M. What is the value of K sp ? A. 1.99 × 10 –11 B. 3.98 × 10 –11 C. 9.94 × 10 –12 D. 1.85 × 10 -7 CaF 2 Ca 2+ + 2F – [Ca 2+ ] = (2.15 × 10 –4 ) [F - ] = 2(2.15 × 10 –4 ) K sp = [Ca][F] 2 = (2.15 × 10 –4 ) (4.3 x 10 -4 ) 2 K sp = 3.98 × 10 –11

10. CHE1102, Chapter 17 Learn, 10 Given K sp, Calculate Solubility What is the molar solubility of CuI in water? Determine the equilibrium concentrations of Cu + and I – CuI(s) Cu + (aq) + I – (aq) K sp = [Cu + ][I – ] K sp = 1.3  10 –12

11. CHE1102, Chapter 17 Learn, 11 Molar Solubilities from K sp Solve K sp expression K sp = 1.3 × 10 –12 = (x)(x) x 2 = 1.3 × 10 –12 x = 1.1 × 10 –6 M = calculated molar solubility of CuI = [Cu + ] = [I – ] Conc (M)CuI(s)Cu + (aq) +I – (aq) Initial Conc. 0.00 Change Equilibrium Conc. +x+x+x+x xx No entries No entries No entries

12. CHE1102, Chapter 17 Learn, 12 Given K sp, Calculate Solubilities Calculate the solubility of CaF 2 in water at 25 °C, if K sp = 3.4 × 10 –11. CaF 2 (s) Ca 2+ (aq) + 2F – (aq) 1. Write equilibrium law K sp = [Ca 2+ ][F – ] 2 2. Construct concentration table Conc (M) CaF 2 (s) Ca 2+ (aq) 2F – (aq) Initial Conc.(No entries0.00 Changein this Equilibrium Conc.column) +2x +x+x 2x2xx

13. CHE1102, Chapter 17 Learn, 13 Molar Solubilities from K sp 3. Solve the K sp expression K sp = [Ca 2+ ][F – ] 2 = (x) (2x) 2 3.4 × 10 –11 = 4x 3 x = 2.0 × 10 –4 M = molar solubility of CaF 2 [Ca 2+ ] = x = 2.0 × 10 –4 M [F – ] = 2x = 2(2.0 × 10 –4 M) = 4.0 × 10 –4 M

14. CHE1102, Chapter 17 Learn, 14 Factors Affecting Solubility The Common-Ion Effect –If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease: BaSO 4 (s)Ba 2+ (aq) + SO 4 2  (aq)

15. CHE1102, Chapter 17 Learn, 15 pH –If a substance has a basic anion, it will be more soluble in an acidic solution. –Substances with acidic cations are more soluble in basic solutions. Factors Affecting Solubility

16. CHE1102, Chapter 17 Learn, 16 Common Ion Effect What happens if another salt, containing one of the ions in our insoluble salt, is added to a solution? Consider PbI 2 (s) Pb 2+ (aq) + 2I – (aq) –Saturated solution of PbI 2 in water Add KI PbI 2 (yellow solid) precipitates out –Why? Le Chatelier’s Principle Add product I – Equilibrium moves to left and solid PbI 2 forms

17. CHE1102, Chapter 17 Learn, 17 Learning Check What effect would adding copper(II) nitrate have on the solubility of CuS? A. The solubility would increase B. The solubility would decrease C. The solubility would not change

18. CHE1102, Chapter 17 Learn, 18 Common Ion Effect A.What is the molar solubility of Ag 2 CrO 4 in pure water? B.What is the molar solubility of Ag 2 CrO 4 in 0.10 M AgNO 3 ? C.What is the molar solubility of Ag 2 CrO 4 in 0.10 M Na 2 CrO 4 ? Ag 2 CrO 4 (s) 2Ag + (aq) + CrO 4 2– (aq) K sp = [Ag + ] 2 [CrO 4 2– ] = 1.1 × 10 –12 Consider three cases

19. CHE1102, Chapter 17 Learn, 19 Common Ion Effect A. What is the solubility of Ag 2 CrO 4 in pure water? Ag 2 CrO 4 (s)2Ag + (aq) +CrO 4 2– (aq) I(No entries0.00 M Cin this Ecolumn) +x+x+2x x2x2x K sp = [Ag + ] 2 [CrO 4 2– ] = (2x) 2 (x) = 1.1  10 –12 = 4x 3 x = Solubility of Ag 2 CrO 4 = 6.5 x 10 -5 M [CrO 4 2– ] = x = 6.5 x 10 -5 M [Ag + ] = 2x = 1.3 × 10 –4 M

20. CHE1102, Chapter 17 Learn, 20 Common Ion Effect B. What is the molar solubility of Ag 2 CrO 4 in 0.10 M AgNO 3 solution? K sp = 1.1 × 10 –12 K sp = 1.1 × 10 –12 = (0.10 M) 2 [x] x = Solubility of Ag 2 CrO 4 = 1.1 × 10 –10 M [Ag + ] = 0.10 M [CrO 4 2– ] = 1.1 × 10 –10 M Ag 2 CrO 4 (s) 2Ag + (aq) +CrO 4 2– (aq) I(No entries0.10 M0.00 Cin this Ecolumn) +x+x+2x x ≈0.10

21. CHE1102, Chapter 17 Learn, 21 Common Ion Effect C. What is the solubility of Ag 2 CrO 4 in 0.100 M Na 2 CrO 4 ? Ag 2 CrO 4 (s) 2Ag + (aq) +CrO 4 2– (aq) I(No entries0.00 M0.10 M Cin this Ecolumn) +x+x+2x 2x2x≈0.10 K sp = (2x) 2 (0.10) = 1.1  10 –12 = 4x 2 (0.10) x = Solubility of Ag 2 CrO 4 = 1.66 × 10 –6 M [CrO 4 2– ] = 0.10 M [Ag + ] = 2x = 3.3 × 10 –6 M

22. CHE1102, Chapter 17 Learn, 22 Common Ion Effect What have we learned about the solubility of silver chromate? A.Dissolving it in pure water the solubility was 3.0 × 10 –4 M B.Dissolving it in AgNO 3 solution solubility was 1.1 × 10 –10 M C.Dissolving it in Na 2 CrO 4 solution solubility was 3.2 × 10 –5 M Common ion appearing the most in the formula of the precipitate decreases the solubility the most

23. CHE1102, Chapter 17 Learn, 23 Will a Precipitate Form? In a solution, –If Q = K sp, the system is at equilibrium and the solution is saturated. –If Q < K sp, more solid can dissolve until Q = K sp. –If Q > K sp, the salt will precipitate until Q = K sp.

24. CHE1102, Chapter 17 Learn, 24 Predicting Precipitation Will a precipitate of PbI 2 form if 100.0 mL of 0.0500 M Pb(NO 3 ) 2 are mixed with 200.0 mL of 0.100 M NaI? PbI 2 (s) Pb 2+ (aq) + 2I – (aq) K sp = [Pb 2+ ][I – ] 2 = 9.8 × 10 –9 Strategy for solving 1.Calculate concentrations in mixture prepared 2.Calculate Q sp = [Pb 2+ ][I – ] 2 3.Compare Q sp to K sp

25. CHE1102, Chapter 17 Learn, 25 Predicting Precipitation Step 1. Calculate concentrations –V total = 100.0 mL + 200.0 mL = 300.0 mL –[Pb 2+ ] = 1.67 × 10 –2 M –[I – ] = 6.67 × 10 –2 M

26. CHE1102, Chapter 17 Learn, 26 Predicting Precipitation Step 2. Calculate Q sp –Q sp = [Pb 2+ ][I – ] 2 = (1.67 × 10 –2 M)(6.67 × 10 –2 M) 2 –Q sp =7.43 × 10 –5 Step 3. Compare Q sp and K sp –Q sp = 7.43 × 10 –5 –K sp = 9.8 × 10  9 –Q sp > K sp so precipitation will occur

27. CHE1102, Chapter 17 Learn, 27 pH and Solubility Mg(OH) 2 (s) Mg 2+ (aq) + 2OH – (aq) –Increase OH – shift equilibrium to left –Add H + shift equilibrium to right –Le Châtelier’s Principle Ag 3 PO 4 (s) 3Ag + (aq) + PO 4 3– (aq) –Add H + increase solubility –H + (aq) + PO 4 3– (aq)  HPO 4 2– (aq) AgCl(s) Ag + (aq) + Cl – (aq) –Adding H + has no effect on solubility Why? –Cl – is very, very weak base, so neutral anion –So adding H + doesn’t effect Cl – concentration