Question

pH of saturated solution of $Ca(OH{)}_2$ is $9$. the solubility product $\left(K_{sp}\right)$ of $Ca(OH{)}_2$ is:

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

Answer 1

Answer: (A)

$0.5\times {10}^{-15}$

$Ca(OH{)}_2\rightleftharpoons C{a}^{2+}+2O{H}^{-}$

$pH=9$

Hence, $pOH=14-9=5$

$\left[O{H}^{-}\right]={10}^{-5}M$

Hence $\left[C{a}^{2+}\right]=\frac{{10}^{-5}}{2}$

Thus $K_{sp}=\left[C{a}^{2+}\right][O{H}^{-}{]}^2$

$=\left(\frac{{10}^{-5}}{2}\right)({10}^{-5}{)}^2$

$=0.5\times {10}^{-15}$