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. $0.5\times 10^{-10}$
• C. $0.5\times 10^{-15}$
• D. $0.25\times 10^{-10}$

Answer 1

Answer: (C)

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