Question

$K_{sp}Mg(OH{)}_2$ is $4.0\times {10}^{-12}$. The number of moles of $M{g}^{2+}$ ions in one litre of its saturated solution in 0.1 M NaOH is :

• A. $4.0\times {10}^{-10}$
• B. $1.0\times {10}^{-4}$
• C. $2.0\times {10}^{-6}$
• D. $8.0\times {10}^{-6}$

Answer 1

The correct option is A

$4.0\times {10}^{-10}$

$\left[O{H}^{-}\right]=[O{H}^{-}{]}_{NaOH}+[O{H}^{-}{]}_{Mg(OH{)}_2}=0.1+2s\approx 0.1M(2X<<0.1)K_{sp}\left[Mg\right(OH{)}_2]=[M{g}^{2+}\left]\right[O{H}^{-}{]}^2=\left[M{g}^{2+}\right]\times {0.1}^2=4.0\times {10}^{12}\Rightarrow \left[M{g}^{2+}\right]=4.0\times {10}^{-12}M$