Question

A solution has $0.05\text{M}M{g}^{2+}$ and $0.05\text{M}N{H}_3$. Calculate the concentration of $N{H}_{4}Cl$ required to prevent the formation of $Mg(OH{)}_2$ in this solution. $K_{sp}\text{of}Mg(OH{)}_2=9.0\times {10}^{-12}$ and ionization constant of $N{H}_3=1.8\times {10}^{-5}$
(Given, $\sqrt{1.8}=1.34$)

• A. $0.067\text{M}$
• B. $0.67\text{M}$
• C. $0.8\text{M}$
• D. $0.08\text{M}$

Answer 1

The correct option is A $0.067\text{M}$

The maximum concentration of $\left[O{H}^{-}\right]$ ions that precipitate $Mg(OH{)}_2$ is calculated by equation
$K_{sp}=\left[M{g}^{2+}\right][O{H}^{-}{]}^2$
$[O{H}^{-}{]}^2=\frac{K_{sp}}{\left[M{g}^{2+}\right]}=\frac{9.0\times {10}^{-12}}{0.05}=1.8\times {10}^{-10}$
$\left[O{H}^{-}\right]=1.34\times {10}^{-5}\text{M}$
$N{H}_3$ is present in the solution in the form of $N{H}_{4}OH$
$\underset{0.05}{N{H}_3}+H_{2}O\rightleftharpoons \underset{0.05}{N{H}_{4}OH}\rightleftharpoons N{H}_4^{+}+O{H}^{-}$
The ionization of $N{H}_{4}OH$ is suppressed by addition of $N{H}_{4}Cl\left(\text{strong electrolyte}\right)$
$K_{N{H}_3}=K_{N{H}_{4}OH}=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}$
Whole of the concentration of $N{H}_4^{+}$ ions is provided by $N{H}_{4}Cl$
$\left[N{H}_4^{+}\right]=\frac{K_{N{H}_{4}OH}\times \left[N{H}_{4}OH\right]}{\left[O{H}^{-}\right]}$
$=\frac{1.8\times {10}^{-5}\times 0.05}{1.34\times {10}^{-5}}=0.067\text{M}$
$\Rightarrow \left[N{H}_{4}Cl\right]=0.067\text{M}$