Question

Find out the solubility of $Mg(OH{)}_2$ in $0.1MNaOH$ solution.
[Given that the solubility product of $Mg(OH{)}_2$ is $1.8\times {10}^{-11}$]

• A. $2\times {10}^{-8}M$
• B. $1\times {10}^{-13}M$
• C. $1.8\times {10}^{-9}M$
• D. $2\times {10}^{-13}M$

Answer 1

The correct option is C $1.8\times {10}^{-9}M$

• $NaOH\to \underset{0.1M}{N{a}^{+}}+\underset{0.1M}{O{H}^{-}}$
• $Mg(OH{)}_2\to {\underset{s}{Mg}}^{2+}+\underset{2s+0.1}{2O{H}^{-}}$

$K_{sp}=\left[M{g}^{2+}\right][O{H}^{–}{]}^2$
$\Rightarrow 1.8\times {10}^{–11}=s(2s+0.1{)}^2$
$1.8\times {10}^{–11}=s\times (0.1{)}^2$
$\therefore s=1.8\times {10}^{–9}M$