Question

At ${25}^0$C, the solubility product of $Mg(OH{)}_2$ is 1.0 x ${10}^{-11}$. At which pH, will $M{g}^{2+}$ ions start precipiating in the form of $Mg(OH{)}_2$ from a solution of 0.001 M $M{g}^{2+}$ ions?

• A. 8
• B. 9
• C. 10
• D. 11

Answer 1

The correct option is C 10

$K_{sp}$ of ${Mg\left(OH\right)}_2=\left[{Mg}^{2+}\right]{\left[{OH}^{-}\right]}^2$

${Mg\left(OH\right)}_2\rightleftharpoons {Mg}^{2+}+2{OH}^{-}$

$\left[{OH}^{-}\right]=\sqrt{\frac{K_{sp}}{\left[{Mg}^{2+}\right]}}$ given, $K_{sp}=1.0\times {10}^{-4}$

$=\sqrt{\frac{{10}^{-11}}{{10}^{-3}}}$ $\left[{Mg}^{2+}\right]={10}^{-3}$

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

$pOH=4$

$pH=14-4=10$

at $pH=10$, ${Mg}^{2+}$ ions starts precipitating.