Question:
If the solubility product of $\mathrm{AB}_{2}$ is $3.20 \times 10^{-11} \mathrm{M}^{3}$, then the solubility of $\mathrm{AB}_{2}$ in pure water is________$---^{\times} 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$. [Assuming that neither kind of ion reacts with water]
Solution:
$\mathrm{AB}_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{A}_{\text {(aq.) }}^{+2}+2 \mathrm{~B}_{\text {(aq.) }}^{-}: \mathrm{K}_{\text {sp }}$
$\mathrm{K}_{\mathrm{SP}}=\mathrm{S}^{1} \times(2 \mathrm{~s})^{2}=4 \mathrm{~s}^{3}$
$3.2 \times 10^{-11}=4 \times \mathrm{S}^{3}$
$\mathrm{S}=2 \times 10^{-4} \mathrm{M} / \mathrm{L}$