Question

pH of a saturated solution of $Ca\left(O H\right)_{2}$ is $9$ . The solubility product $\left(K_{s p}\right)$ of $Ca\left(O H\right)_{2}$ 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)

$Ca\left(O H\right)_{2}\rightleftharpoons\underset{\left(S\right)}{C a^{+ 2}}+\underset{\left(2 S\right)}{2 O H^{-}}$
$pH$ of saturated solution $=9$
$pH=9, \, \left[H^{+}\right]=10^{- 9}$
Hence $\left[O H^{-}\right]=\frac{10^{- 14}}{10^{- 9}}=10^{- 5}$
$2s=10^{- 5}, \, s=\frac{10^{- 5}}{2}$
$k_{s p}$ for $Ca\left(O H\right)_{2}\Rightarrow 4s^{3}$
$\Rightarrow 4\times \left(\frac{\left(10\right)^{-5}}{2}\right)^{3}$
$=\frac{10^{-1 5}}{2}=0.5\times 10^{- 15}$