Question

pH of a saturated solution of $Ca(OH)_2$ is $9$ . The solubility product $(K_{sp})$ of $Ca(OH)_2$ is:

• A. $0.125\times 10^{-15}$
• B. $0.5\times10^{-10}$
• C. $0.5\times10^{-15}$
• D. $0.25\times10^{-10}$

Answer 1

Answer: (C)

$CO(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^{-}$
$pH = 9$ Hence $pOH = 14 - 9 = 5$
Hence $[Ca^{2+} ] = \frac{10^{-5}}{2}$
Thus $K_{sp} = [Ca^{2+} ][OH^{-}]^2$
$ = \left( \frac{10^{-5}}{2}\right) (10^{-5})^2$
$ = 0.5 \times 10^{-15}$