Question

A solution has $0.05M$ ${Mg}^{2+}$ and $0.05M$ ${NH}_3$. Calculate the concentration of ${NH}_{4}Cl$ required to prevent the formation of $Mg{\left(OH\right)}_2$ in this solution. $K_{sp}$ of $Mg{\left(OH\right)}_2=9.0\times {10}^{-12}$ and ionisation constant of ${NH}_3=1.8\times {10}^{-5}$

Answer 1

The maximum concentration of $\left[{OH}^{-}\right]$ ions that will precipitate $Mg{\left(OH\right)}_2$ is calculated by applying the equation
$K_{sp}=\left[{Mg}^{2+}\right]{\left[{OH}^{-}\right]}^2$
${\left[{OH}^{-}\right]}^2=\frac{K_{sp}}{\left[{Mg}^{2+}\right]}=\frac{9.0\times {10}^{-12}}{0.05}=1.8\times {10}^{-10}$
or $\left[{OH}^{-}\right]=1.34\times {10}^{-5}M$
${NH}_3$ is present in solution in the form of ${NH}_{4}OH$
${NH}_3+H_{2}O\rightleftharpoons {NH}_{4}OH\rightleftharpoons {NH}_4^{+}+{OH}^{-}$
$0.05$ $0.05$
The ionisation of ${NH}_{4}OH$ is suppressed by the addition of ${NH}_{4}Cl$ (strong electrolyte)
$K_{{NH}_3}=K_{{NH}_{4}OH}=\frac{\left[{NH}_4^{+}\right]\left[{OH}^{-}\right]}{\left[{NH}_{4}OH\right]}$
Whole of the concentration of ${NH}_4^{+}$ ions is provided by ${NH}_{4}Cl$.
$\left[{NH}_4^{+}\right]=\frac{K_{{NH}_{4}OH}\times \left[{NH}_{4}OH\right]}{\left[{OH}^{-}\right]}=\frac{1.8\times {10}^{-5}\times 0.05}{1.34\times {10}^{-5}}=0.067M$
i.e., $\left[{NH}_{4}Cl\right]=0.067M$