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Solving Problems: A Chemistry Handbook
Chemistry: Matter and Change
177
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER 17
Cu(OH) 2 (s) 3 Cu 2ϩ (aq) ϩ 2OH Ϫ (aq)
The coefficient of Cu 2ϩ is 1, and the coefficient of OH Ϫ is 2, so the
following is the solubility product constant expression.
K sp ϭ [Cu 2ϩ ][OH Ϫ ] 2
Tabulated K sp values may be used to calculate the molar solubility of
a sparingly soluble ionic compound and also to calculate ion concentrations in a saturated solution. The following example problem
illustrates these calculations.
Example Problem 17-5
Calculating Molar Solubility and Ion Concentration from K sp
The K sp for lead(II) fluoride (PbF 2 ) is 3.3 ϫ 10 Ϫ8 at 25°C. Use this
K sp value to calculate the following.
a. The solubility in mol/L of PbF 2
b. The fluoride ion concentration in a saturated solution of PbF 2
a. Write the balanced equation for the solubility equilibrium, and
write the K sp expression.
PbF 2 (s) 3 Pb 2ϩ (aq) ϩ 2F Ϫ (aq)
K sp ϭ [Pb 2ϩ ][F Ϫ ] 2 ϭ 3.3 ϫ 10 Ϫ8
The moles of Pb 2ϩ ions in solution equal the moles of PbF 2 that
dissolved. Therefore, let [Pb 2ϩ ] equal s, where s represents the
solubility of PbF 2 . Because there are two F Ϫ ions for every
Pb 2ϩ ion, [F Ϫ ] ϭ 2s. Substitute these terms into the K sp expression and solve for s.
(s)(2s) 2 ϭ 3.3 ϫ 10 Ϫ8
(s)(4s 2 ) ϭ 3.3 ϫ 10 Ϫ8
4s 3 ϭ 3.3 ϫ 10 Ϫ8
s 3 ϭ
ϭ8.25 ϫ 10 Ϫ9
Here three digits are retained for accuracy, but the final answer
will be rounded to two digits.
s ϭ [Pb 2ϩ ] ϭ ͙
3 8.25 ϫ
ෆ 10 Ϫ9
ෆ ϭ 2.0 ϫ 10 Ϫ3 mol/L
3.3 ϫ 10 Ϫ8
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