Question

The $pH$ at which $M(OH{)}_2$ will begin to precipitate from a solution containing $0.10M{M}^{2+}$ ions $[K_{sp}$ of $M(OH{)}_2=1\times {10}^{-9}{M}^3]$ is:

• A. $7$
• B. $12$
• C. $9$
• D. $10$

Answer 1

The correct option is D $10$

$Mg(OH{)}_2\leftrightharpoons M{g}^{+2}+2O{H}^{-}$

At equilibrium $c_1$ $c_2$

Given $c_1=0.1M$

Also, $k_{sp}=c_1\times c_2^2$

${10}^{-9}=0.1c_2^2$

$c_2^2={10}^{-8}$

$c_2^2={10}^{-4}$

$\left[O{H}^{-}\right]={10}^{-4}$

$\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}$

$\left[H^{+}\right]\times {10}^{-4}={10}^{-14}$

$\left[H^{+}\right]={10}^{-10}$

$pH=-\log \left[H^{+}\right]$

$pH=-\log \left[{10}^{-10}\right]$

$pH=10$