1. Chapter 17 Solubility and Simultaneous Equilibria

2. Solubility Products
Consider the equilibrium that exists in a saturated solution of BaSO4 in water:
BaSO4(s)
Ba2+(aq) + SO42(aq)
A saturated solution is one in which no more solute can dissolve.
Saturated solution – no more solute will dissolve

3. Solubility Product Constant
– equilibrium constant for ionic compounds that are only slightly soluble
The equilibrium constant expression for this equilibrium is
Ksp = [Ba2+] [SO42]
Ksp is the equilibrium constant for ionic compounds that are only slightly soluble.

4. Solubility Products Ksp is not the same as solubility.
Solubility generally expressed as (g/L), (g/mL), or in mol/L (M).
Ksp - Only one value for a given solid at a given temperature
Temperature dependence
Solubility and thus Ksp changes with T
The solubility of barium sulfate is how much of this material will dissolve in a given amount of solution. The solubility product, in the case of barium sulfate will be the square of that [Ba][SO4].

5. Solubility Equilibria
When ionic salt dissolves in water It dissociates into separate hydrated ions Initially, no ions in solution CaF2(s)  Ca2+(aq) + 2F–(aq) As dissolution occurs, ions build up and collide Ca2+(aq) + 2F–(aq)  CaF2(s) At equilibrium CaF2(s) Ca2+(aq) + 2F–(aq) We now have a saturated solution Ksp = [Ca+][F–]2

6. Writing Ksp Equilibrium Laws
AgCl(s) Ag+(aq) + Cl–(aq) Ksp = [Ag+][Cl–]
PbI2(s) Pb2+(aq) + 2I–(aq) Ksp = [Pb2+][I–]2
Ag2CrO4(s) Ag+(aq) + CrO42–(aq) Ksp = [Ag+]2[CrO42–]
AuCl3(s) Au3+(aq) + 3Cl–(aq) Ksp = [Au3+][Cl–]3
Note that the solid does not appear in the Ksp expression because it is has a constant concentration.

7. Molar Solubility
– # moles of ionic solid that dissolves per 1 L DI water to form a saturated solution at 25 C
Determine the final equilibrium concentration of Ca2+ (aq) and CO32- (aq) in a saturated solution
Molar solubility always equals “X”
6.7 x 10-5 moles of CaCO3 (s) will dissolve per 1 L DI water producing 6.7 x M Ca2+ (aq) and 6.7 x M CO32- (aq)

8. Given Solubilities, Calculate Ksp
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) Ksp = [Ag+][Cl–]
AgCl(s)
Ag+(aq) +
Cl–(aq) I
0.00
0. 00 C E
1.34 × 10–5 M
1.34 × 10–5 M
1.34 × 10–5 M
1.34 × 10–5 M
Ksp = (1.34 × 10–5)(1.34 × 10–5)
Ksp = 1.80 × 10–10

9. Learning Check
The solubility of calcium fluoride, CaF2, in pure water is 2.15 × 10‑4 M. What is the value of Ksp? A × 10–11 B × 10–11 C × 10–12 D × 10-7 CaF2 Ca2+ + 2F– [Ca2+] = (2.15 × 10–4) [F-] = 2(2.15 × 10–4) Ksp = [Ca][F]2 = (2.15 × 10–4) (4.3 x 10-4)2 Ksp = 3.98 × 10–11

10. Given Ksp, 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)
Ksp = [Cu+][I–]
Ksp = 1.3  10–12

11. Molar Solubilities from Ksp
Solve Ksp expression
Ksp = 1.3 × 10–12 = (x)(x)
x2 = 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.
No entries No entries
No entries +x +x x x

12. Given Ksp, Calculate Solubilities
Calculate the solubility of CaF2 in water at 25 °C, if Ksp = 3.4 × 10–11.
CaF2(s) Ca2+(aq) + 2F– (aq)
1. Write equilibrium law
Ksp = [Ca2+][F–]2
2. Construct concentration table
Conc (M)
CaF2(s)
Ca2+(aq)
2F–(aq)
Initial Conc.
(No entries
0.00
Change
in this
Equilibrium Conc.
column) +x
+2x x 2x

13. Molar Solubilities from Ksp
3. Solve the Ksp expression
Ksp = [Ca2+][F–]2 = (x) (2x)2
3.4 × 10–11 = 4x3
x = 2.0 × 10–4 M = molar solubility of CaF2
[Ca2+] = x = 2.0 × 10–4 M
[F–] = 2x = 2(2.0 × 10–4 M) = 4.0 × 10–4 M

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:
BaSO4(s)
Ba2+(aq) + SO42(aq)

15. Factors Affecting SolubilitypH
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.

16. Common Ion Effect
What happens if another salt, containing one of the ions in our insoluble salt, is added to a solution?
Consider PbI2(s) Pb2+(aq) + 2I–(aq)
Saturated solution of PbI2 in water
Add KI
PbI2 (yellow solid) precipitates out
Why?
Le Chatelier’s Principle
Add product I–
Equilibrium moves to left and solid
PbI2 forms

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. Common Ion Effect Consider three cases
What is the molar solubility of Ag2CrO4 in pure water?
What is the molar solubility of Ag2CrO4 in 0.10 M AgNO3?
What is the molar solubility of Ag2CrO4 in 0.10 M Na2CrO4?
Ag2CrO4(s) Ag+(aq) + CrO42–(aq)
Ksp = [Ag+]2[CrO42–] = 1.1 × 10–12
x = 9.0 10–10 M

19. Common Ion Effect A. What is the solubility of Ag2CrO4 in pure water?
Ag2CrO4(s)
2Ag+(aq) +
CrO42–(aq) I
(No entries
0.00 M C
in this E
column)
+2x +x 2x x
Ksp = [Ag+]2[CrO42–] = (2x)2(x) = 1.1  10–12 = 4x3
x = 9.0 10–10 M
x = Solubility of Ag2CrO4 = 6.5 x 10-5 M
[CrO42–] = x = 6.5 x 10-5 M
[Ag+] = 2x = 1.3 × 10–4 M

20. Common Ion Effect
B. What is the molar solubility of Ag2CrO4 in 0.10 M AgNO3 solution? Ksp = 1.1 × 10–12
Ksp = 1.1 × 10–12 = (0.10 M)2[x]
x = Solubility of Ag2CrO4 = 1.1 × 10–10 M
[Ag+] = 0.10 M
[CrO42–] = 1.1 × 10–10 M
Ag2CrO4(s)
2Ag+(aq) +
CrO42–(aq) I
(No entries
0.10 M
0.00 C
in this E
column)
+2x +x
≈0.10 x

21. Common Ion Effect Ag2CrO4(s) 2Ag+(aq) + CrO42–(aq) I (No entries
C. What is the solubility of Ag2CrO4 in M Na2CrO4?
Ag2CrO4(s)
2Ag+(aq) +
CrO42–(aq) I
(No entries
0.00 M
0.10 M C
in this E
column)
+2x +x 2x
≈0.10
Ksp = (2x)2(0.10) = 1.1  10–12 = 4x2(0.10)
x = Solubility of Ag2CrO4 = 1.66 × 10–6 M
[CrO42–] = 0.10 M
[Ag+] = 2x = 3.3 × 10–6 M

22. Common Ion Effect
What have we learned about the solubility of silver chromate?
Dissolving it in pure water the solubility was 3.0 × 10–4 M
Dissolving it in AgNO3 solution solubility was 1.1 × 10–10 M
Dissolving it in Na2CrO4 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. Will a Precipitate Form?
In a solution,
If Q = Ksp, the system is at equilibrium and the solution is saturated.
If Q < Ksp, more solid can dissolve until Q = Ksp.
If Q > Ksp, the salt will precipitate until Q = Ksp.

24. Predicting Precipitation
Will a precipitate of PbI2 form if mL of M Pb(NO3)2 are mixed with mL of M NaI?
PbI2(s) Pb2+(aq) + 2I–(aq)
Ksp = [Pb2+][I–]2 = 9.8 × 10–9
Strategy for solving
Calculate concentrations in mixture prepared
Calculate Qsp = [Pb2+][I–]2
Compare Qsp to Ksp

25. Predicting Precipitation
Step 1. Calculate concentrations
Vtotal = mL mL = mL
[Pb2+] = 1.67 × 10–2 M
[I–] = 6.67 × 10–2 M

26. Predicting Precipitation
Step 2. Calculate Qsp
Qsp = [Pb2+][I–]2 = (1.67 × 10–2 M)(6.67 × 10–2 M)2
Qsp =7.43 × 10–5
Step 3. Compare Qsp and Ksp
Qsp = 7.43 × 10–5
Ksp = 9.8 × 109
Qsp > Ksp so precipitation will occur

27. pH and Solubility Mg(OH)2(s) Mg2+(aq) + 2OH–(aq)
Increase OH– shift equilibrium to left
Add H+ shift equilibrium to right
Le Châtelier’s Principle
Ag3PO4(s) Ag+(aq) + PO43–(aq)
Add H+ increase solubility
H+(aq) + PO43–(aq)  HPO42–(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