1. Solubility Equilibria Review Section 17.4

2. The Solubility Product Constant, Ksp
Ksp - the equilibrium constant for a solid dissolving in water.
S - the solubility (maximum that will dissolve) in g/L or mol/L
BaSO4(s)
Ba2+(aq) + SO42−(aq) 
Ksp = [Ba2+] [SO42−]
* Remember, solids are not in equilibrium expressions!

3. Converting Between Ksp and Solubility
Ag2CrO4 (s) is dissolved in water. At equilibrium, the molar solubility* of Ag2CrO4 (s) is 6.5 x 10-5 M. Calculate Ksp.
Ag2CrO4(s)  2 Ag+ + CrO42−
Ksp = [Ag+] 2 [CrO42−]
Ksp = [1.3 x 10-4 ] 2 [6.5 x 10-5 ] = 1.1 x 10-12
* Molar solubility: Maximum number of moles that dissolve per liter of solution.

4. Converting Between Ksp and Solubility
The Ksp for CaF2 is 3.9 x at 25°C. Calculate the molar solubility of CaF2.
CaF2 (s)  Ca F− I -- C +x
+2x E x 2x
Ksp = [Ca 2+] [F−]2 = (x)(2x)2 = 4x3 = 3.9 x 10-11
* Molar solubility: Maximum number of moles that dissolve per liter of solution.

5. Solubility – Common Ions
Adding a common ion DECREASES solubility.
Calculate the molar solubility of CaF2 in a solution containing M Ca(NO3)2. Ksp = 3.9 x at 25°C
CaF2 (s)  Ca F− I --
0.020 M C +x
+2x E
~ M 2x
Ksp = [Ca 2+] [F−]2 = (0.020)(2x)2 = 3.9 x 10-11
* Molar solubility: Maximum number of moles that dissolve per liter of solution.

6. Factors Affecting Solubility - pH
Adding a strong acid will INCREASE solubility if the anion is associated with a weak acid.
© 2009, Prentice-Hall, Inc.

7. Factors Affecting Solubility - pH
STRONG ACIDS
HI → H+ + I–
HCl → H+ + Cl–
HBr → H+ + B–
HNO3 → H+ + NO3–
HClO3 → H+ + ClO3–
HClO4 → H+ + ClO4–
H2SO4 → 2 H+ + SO42-
If the anion in the salt is NOT one of the anions associated with a strong acid, then the addition of acid (H+) will INCREASE the solubility of the salt.
© 2009, Prentice-Hall, Inc.

8. Factors Affecting Solubility - pH
Assume strong acid completely dissociates into H+ and Cl–
Cl– is a spectator ion
Added H+ reacts with anion of salt (F–) to reform weak acid
Dissolving reaction shifts RIGHT (increase solubility)
CaF2(s) ⇄ Ca F –
2F – + 2H+ ⇄ 2HF (aq)
CaF2(s) + 2H+ ⇄ Ca2+ +2HF (aq)
© 2009, Prentice-Hall, Inc.

9. Factors Affecting Solubility - pH
Assume strong acid completely dissociates into H+ and anion
Strong acid anion is a spectator ion
Added H+ reacts with OH– to reform weak acid
Dissolving reaction shifts RIGHT (increase solubility)
Mg(OH)2(s) ⇄ Mg OH –
H2O is a weak acid: 2OH – + 2H+ ⇄ 2H2O (aq)
Mg(OH)2(s) + 2H+ ⇄
Mg2+ + 2H2O(aq)

10. “Will a precipitate form” = Q problem
Fe(OH)3 (s)  Fe3+ (aq) + 3 OH- (aq) Ksp = 4.87 x 10-17
If 500 mL M Fe(NO3)3 is mixed with 500. mL M NaOH, will a precipitate form?
Na+ and NO3- are always soluble so they are spectator ions. The possible insoluble salt would be Fe(OH)3. New volume is 500 mL mL = 1000 mL.
M1V1 = M2V2 M2= M1V1 / V2
[Fe3+] = (0.0004M)x(500mL/1000mL) = M
[OH-] = (0.0002M)x(500mL/1000mL) = M
Q = [Fe3+][OH-]3 = (0.0002)(0.0001)3 = 2 x 10-16
Q > Ksp so the rxn will shift left
YES! A precipitate (Fe(OH)3) will form.