Question

Find out the solubility of $Mg(OH{)}_2$ in $0.1\text{M}NaOH$. Given that solubility product of $Mg(OH{)}_2$ is $1.8\times {10}^{–11}$.

• A. $3.24\times {10}^{-10}\text{M}$
• B. $1.8\times {10}^{-9}\text{M}$
• C. $3.24\times {10}^{-9}\text{M}$
• D. $1.8\times {10}^{-10}\text{M}$

Answer 1

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

Let the solubility of $Mg(OH{)}_2$ be ${\text{s mol L}}^{–1}$
$Mg(OH{)}_2\rightleftharpoons M{g}^{2+}+2O{H}^{-}$
$s2s+0.1$
[Since $s<<0.1$ therefore $2s+0.1\approx 0.1$]
$K_{\text{sp}}=\left[M{g}^{2+}\right][O{H}^{-}{]}^2$
$1.8\times {10}^{-11}=\left(s\right)(0.1{)}^2$
$s=1.8\times {10}^{-9}\text{M}$