Question

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

$Ca{\left(OH\right)}_2\rightleftharpoons \underset{\left(S\right)}{C{a}^{+2}}+\underset{\left(2S\right)}{2O{H}^{-}}$
$pH$ of saturated solution $=9$
$pH=9,\left[H^{+}\right]=10^{-9}$
Hence $\left[O{H}^{-}\right]=\frac{10^{-14}}{10^{-9}}=10^{-5}$
$2s=10^{-5},s=\frac{10^{-5}}{2}$
$k_{sp}$ for $Ca{\left(OH\right)}_2\Rightarrow 4s^3$
$\Rightarrow 4\times {\left(\frac{{\left(10\right)}^{-5}}{2}\right)}^3$
$=\frac{10^{-15}}{2}=0.5\times 10^{-15}$