1. Aim # 14: How can we determine the solubility of slightly soluble salts?
H.W. # 14
Study pp. 759 – (up to the common ion effect) Ans. ques. p. 781 # 19,21,22,27
Do Now: Zumdahl (8th ed.)
p. 741 # 90,87

2. I For the system PbCl2(s) Pb2+ (aq) + 2Cl-(aq) Ksp = [Pb2+][Cl-]2
The equilibrium constant for the equilibrium expression between a solid salt and its ions in solution is called the solubility product constant (or solubility constant).
It is represented by the symbol Ksp.
Problem: Write the expression for the solubility product constant for a) calcium chloride b) barium carbonate

3. Ans: a) Ca3(PO4)2 (s) 3Ca2+(aq) + 2PO43-(aq)
at 250C, Ksp = [Ca2+]3[PO43-]2 = 1.3 x 10-32
b) BaCO3(s) = Ba2+(aq) + CO32-(aq)
at 250C, Ksp = [Ba2+][CO32-] = 1.6 x 10-9

4. Selected Solubility Product Constants, Ksp, at Room Temperature
Compound Formula Ksp
aluminum hydroxide Al(OH) x 10-33
barium carbonate BaCO x 10-9
barium chromate BaCrO x 10-11
barium sulfate BaSO x 10-9
cadmium sulfide CdS x 10-28
calcium carbonate CaCO x 10-9
calcium sulfate CaSO x 10-5
copper(II) iodide CuI x 10-12
iron(II) sulfide FeS x 10-19
lead(II) choride PbCl x 10-5
lead(II) chromate PbCrO x 10-16
lead(II) sulfate PbSO x 10-8
lead(II) sulfide PbS x 10-29
magnesium hydroxide Mg(OH) x 10-12
silver bromide AgBr x 10-13
silver chloride AgCl x 10-10
silver chromate Ag2CrO x 10-12
silver iodide AgI x 10-17

5. The values of Ksp change with temperature
The values of Ksp change with temperature. The values given in reference tables are usually for 250C
The solubility of a salt (the concentrations of its ions) changes with changes in concentrations of other solutes- salts with common ions, as well as with pH.
However, for a given temperature,
Ksp DOESN’T CHANGE.
II Solubility and Ksp
Problem: Find the solubility product constant, Ksp, for lead (II) bromide at 250C. Its solubility is 1.0 x 10-2 M.

6. Ans: PbBr2(s) Pb2+(aq) + 2Br-(aq) Ksp = [Pb2+][Br-]2
[Br-] = 2[Pb2+] = 2(1.0 x 10-2)
Ksp = (1.0 x 10-2)(2.0 x 10-2)2
Ksp = 4.0 x 10-6
Note: Units are omitted for Ksp values
Problem: Calculate the Ksp value for bismuth sulfide, Bi2S3, which has a solubility of 1.0 x M at 250C.

7. Ans: Bi2S3(s) 2Bi3+(aq) + 3s2-(aq)
Ksp = [Bi3+][S2-]
Ksp = [2(1.0 x 10-15)]2[3(1.0 x 10-15)]3
Ksp = 1.1 x 10-73
B. Calculating solubility from Ksp
Problem: The Ksp value for copper(II) iodate, Cu(IO3)2, is 1.4 x 10-7 at 250c. Calculate its solubility at 250C.
Ans: Cu(IO3)2(s) Cu2+(aq) + 2IO3-(aq)
Ksp = [Cu2+][IO3-]2 = 1.4 x 10-7
Let [Cu2+] = x
then [IO3-] = 2x

8. 1.4 x 10-7 = (x)(2x)2 = 4x3
x = 3√ 3.5 x = 3.3 x 10-3 mol/L
Problem: The value of Ksp for cerium hydroxide, Ce(OH)3,
is 1.5 x What is the molar solubility of Ce(OH)3 in a solution that contains 0.10 M NaOH?
Ans: Ce(OH)3(s) Ce3+(aq) + 3OH-(aq)
Ksp = [Ce3+][OH-]3 = 1.5 x 10-20
NaOH is a strong base, and dissociates completely.
from NaoH: [OH-] = 0.10 M from Ce(OH)3: [Ce3+] = x
[OH-] = 3x

9. 1.5 x = x( x)3
since x is small, [OH-] ≈ x = x(.10)3 = .001x x = 1.5 x = [Ce3+] = [Ce(OH)3]
What would the solubility be if Ce(OH)3 was the only
species in solution?
Ans: x = (x)(3x)3 = 27x x4 = 5.6 x 10-22
x = [Ce3+] = [Ce(OH)3] = 4.9 x 10-6

10. Practice Problems Zumdahl (8th ed.) p. 766 # 24,26,28 p. 769 # 89
Note: Since Ksp is a constant, adding OH- ion to the solution (from NaOH) decreases the amount of Ce ion which can remain in solution (and the solubility of Ce(OH)3).
Practice Problems
Zumdahl (8th ed.) p. 766 # 24,26, p. 769 # 89