Question

If the solubility product of \(\mathrm{AB_2}\) is \(3.20\times10^{-11}\ \mathrm{M^3}\), then the solubility of \(\mathrm{AB_2}\) in pure water is ______ \(\times10^{-4}\ \mathrm{mol\ L^{-1}}\).
[Assuming that neither kind of ion reacts with water]

Hint:

For \(\mathrm{AB_2}\) with solubility s, \(K_{sp} = s\,(2s)^2 = 4s^3\).

Answer 1

Correct Answer: 2

\(\mathrm{AB_2 \rightleftharpoons A^{2+} + 2B^-}\). If the solubility is s, then \([\mathrm{A^{2+}}] = s\) and \([\mathrm{B^-}] = 2s\).
\(K_{sp} = s\,(2s)^2 = 4s^3 = 3.20\times10^{-11}\)
\(s^3 = 8\times10^{-12}\), so \(s = 2\times10^{-4}\ \mathrm{mol\ L^{-1}}\).
The blank is therefore 2.
Answer: 2