CH 117 Spring 2015
Worksheet 16
1. Calculate the molar solubility and solubility in g/L of silver phosphate (Ksp = 8.9 x 10-17).
Use ICE chart to calculate molar solubility (x corresponds to the molar solubility of
the solid).
Ag3PO4 (s)
3 Ag+
PO43
Initial
n/a
0
0
Change
n/a
+3x
+x
Equilibrium
n/a
3x
x
Ksp = 8.9 x 10-17 = (3x)3 * x = 27x4  x = 4.26 x 10-5 M
In grams/L  4.26 x 10-5 mol/L * molar mass of Ag3PO4 (418.6 g/mol) = .018 g/L
2. Describe how increasing or decreasing the pH of a solution will affect the solubility of a
salt.
Only conjugates of weak acids and bases will affect solubility – conjugates of strong
acids or bases have NO effect on the solubility of a salt.
If you have a basic salt, lowering the pH will increase the solubility of the salt.
Increasing the pH will decrease the solubility.
If you have an acidic salt, lowering the pH will decrease the solubility of the salt.
Increasing the pH will increase the solubility of the salt.
3. Calculate the molar solubility of barium sulfate in a 0.02 M sodium sulfate solution. The
Ksp of barium sulfate is 1.1 x 10-10.
This is a common ion effect problem – when a salt is dissolved in a solution that
contains either a cation or anion already present in the salt. Common ions act to
decrease solubility generally.
In this case the common ion is sulfate – present both in the salt and the solution in
which it is dissolved.
Set these calculations up the same way as before, the only difference is that you will
start out with some of one of the ions instead of 0.
Initial
Change
Equilibrium
BaSO4 (s)
n/a
n/a
n/a

Ba2+
0
+x
x
SO420.02 M
+x
.02+x
Ksp = 1.1 x 10-10 = x(.02+x)  use the 5% rule to ignore the x in parentheses, since
.02 is so much larger than K.
1.1 x 10-10 = .02x  x = 5.5 x 10-9 M