Question

The solubility of $Mg(OH{)}_2$ in a particular buffer is found to be $0.95g{L}^{-1}.$ What is the pH of the buffer solution? $\left[K_{sp}\right(Mg\left(OH{)}_2\right)=1.8\times {10}^{-11}]$

• A. $6.52$
• B. $4.48$
• C. $9.52$
• D. $8.52$

Answer 1

The correct option is C $9.52$

Solubility of $Mg(OH{)}_2=0.95g{L}^{-1}=\frac{0.95}{58}mol{L}^{-1}$
$\therefore \left[M{g}^{2+}\right]=0.0164M$

$Mg\left(O{H}_2\right)\left(s\right)\rightleftharpoons M{g}^{2+}\left(aq\right)+2\left[O{H}^{-}\right]\left(aq\right)$
$K_{sp}=\left[M{g}^{2+}\right][O{H}^{-}{]}^2$
$\Rightarrow 1.8\times {10}^{-11}=\left[0.0164\right][O{H}^{-}{]}^2$
$\Rightarrow \left[O{H}^{-}\right]=3.31\times {10}^{-5}M$

$\Rightarrow pOH=-log\left[O{H}^{-}\right]=4.48\Rightarrow \therefore pH=14-4.48=9.52$