Question

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

• A. 9
• B. 10
• C. 11
• D. 8

Answer 1

The correct option is B 10

Given,
$K_{sp}=1.0\times {10}^{-11},[M{g}^{2+}=0.001M$

$M{g}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)\rightleftharpoons Mg(OH{)}_2$

For this, we know that
$K_{sp}=\left[M{g}^{2+}\right][O{H}^{-}{]}^2\Rightarrow [O{H}^{-}]=\sqrt{\frac{K_{sp}}{\left[M{g}^{2+}\right]}}=\sqrt{\frac{1.0\times {10}^{-11}}{0.001}}$

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


Also, $pH+pOH=14$
$pH+4=14$
$\Rightarrow pH=10$
So, $M{g}^{2+}$ ions start precipitating in the form of $Mg(OH{)}_2$ at $pH=10$.