Question

If the solubility product of $MOH$ is $1\times {10}^{-10}mo{l}^{2}d{m}^{-2}.$ Then the $pH$ of its aqueous solution will be:

• A. $12$
• B. $9$
• C. $6$
• D. $3$

Answer 1

Answer: (B)

$9$

Solubility product of $MOH$ is ${10}^{-10}$ ${mol}^2$ ${dm}^{-2}$.

$K_{sp}={10}^{-10}{mol}^2{dm}^{-2}$

$MOH\rightleftharpoons M^{+}+{OH}^{-}$

$K_{sp}\left(M{\left(OH\right)}_2\right)={10}^{-10}=S^2$

$S={10}^{-5}$

$\Rightarrow \left[{OH}^{-}\right]={10}^{-5}$

$pOH=5(\because pOH=-log\left[{OH}^{-}\right])$

$pH=9(\because pH=14-pOH)$

$\therefore$ $pH$ of aqeous solution is $9$.