Question

The solubility product of a sparingly soluble metal hydroxide $M(OH{)}_2$ at $298K$ is $5\times {10}^{-16}mo{l}^{3}d{m}^{-9}$. The $pH$ value of its aqueous and saturated solution is:

• A. $5$
• B. $9$
• C. $11.5$
• D. $2.5$

Answer 1

The correct option is B $9$

$M(OH{)}_2\rightleftharpoons M^{2+}+2O{H}^{-}$

$K_{sp}=\left(s\right)(2s{)}^2=4s^3=5\times {10}^{-16}$

$\therefore s=\sqrt[3]{3\frac{5\times {10}^{-16}}{4}}=\sqrt[3]{3\frac{500\times {10}^{-18}}{4}}=5\times {10}^{-6}$

Conc. of $O{H}^{-}=2s=2\times 5\times {10}^{-6}$$={10}^{-5}mold{m}^{-3}$

And, $pOH=-\log \left[O{H}^{-}\right]$

$=-\log {10}^{-5}=5$

$\therefore pH=14-pOH=14-5=9$.

Hence, the correct answer is option $\text{B}$.