CHM 2046C Module 12 Chapter 18 Name: __ Answers__
I. Estimating Salt Solubility from Ksp 10 points
Given the following insoluble salts, estimate the salt’s molar solubility from the Ksp :
1. Cu(OH)2 (s)
+ HOH (l) ß à Cu2+ (aq) + 2 OH1- (aq)
Ksp = 2.2 x 10 -20
Ksp = [Cu2+][OH1-]2
If x = amount (moles) of Copper II hydroxide that dissolves
in a liter of water, then
2.2 x 10 -20
= [x][2x]2 = 4x3
x3 = 5.5 x 10 -21
Therefore the molar solubility from Ksp is
x = 1.7 x 10 -7 mol/L
2. MgCO3 (s) +
HOH (l) ß à Mg2+ (aq) +
CO32- (aq)
Ksp = [Mg2+][ CO3 2-] Ksp = 6.8 x 10 -6
If x = amount (moles) of Magnesium carbonate that dissolves
in a liter of water, then
6.8 x 10 -6= [x][x] =
x2
x2 = 6.8 x 10 -6
Therefore the molar solubility from Ksp is
x = 2.6 x 10 -3 mol/L
3. Al2(SO4)3 (s)
+ HOH (l) ß à 2 Al3+ (aq) + 3 SO42- (aq)
Ksp = 9.8 x 10 -21
Ksp = [Al3+]2[ SO42-]3
If x = amount (moles) of Aluminum sulfate that dissolves in
a liter of water, then
9.8 x 10 -21= [2x]2[3x]3 =
108x5
x5 = 9.1 x 10 -23 = 910 x 10 -25
Therefore the molar solubility from Ksp is
x = 3.9 x 10 -5 mol/L
4. Cu2S (s) + HOH (l) ß à 2 Cu1+ (aq) + S2- (aq)
Ksp = 2.2 x 10 -48
Ksp = [Cu1+][S2-] Ksp = 2.2 x 10 -48
If x = amount (moles) of Copper I sulfide that dissolves in
a liter of water, then
2.2 x 10 -48
= [2x]2 [x] = 4x3
x3 = 550 x 10 -51
Therefore the molar solubility from Ksp is
x = 8.2 x 10 -17 mol/L
McMurry Reference:
Read Section 15.11 pages
619-622. Study Worked Example 15.10 page 521. Do Problem 15.23 on page 521.
Try Problems 15.96 and
15.97 page 642 at the end of the chapter
Kotz Reference:
See Section 18.4 Relating Solubility and Ksp pages 875-878. Review worked examples 18.9 and 18.10 pages
876-877. Work Exercise 18.13 page 878. At the end of the chapter there are
eight additional exercises: #45-#52
pages 896-7.
Additional Worked Example:
