Question

The solubility product of a sparingly soluble metal hydroxide $M ( OH )_{2}$ at $298 \,K$ is $5 \times 10^{-16} mol ^{3} d m^{-9}$. The $pH$ value of its aqueous and saturated solution is

• A. 5
• B. 9
• C. 11.5
• D. 8.5

Answer 1

Answer: (B)

$M(OH)_2 \leftrightharpoons \underset{s}{M^+} + \underset{2s}{2OH^-}$
$K_{s p}=(s)(2 s)^{2}=4 s^{3}=5 \times 10^{-16}$
$\therefore s=\sqrt[3]{\frac{5 \times 10^{-6}}{4}}=5 \times 10^{-6}$
Conc. of $O H^{-}=2 \times 5 \times 10^{-6}$
$=10^{-5} mol \,dm ^{-3}$
$p O H=-\log \left[O H^{-}\right]$
$=-\log 10^{-5}=5$
$p H=14-p O H$
$=14-5=9 .$