Question

What is the $K_c$ for the reaction:

${Mg{\left(OH\right)}_2}_{\left(s\right)}+2{{NH}_4^{+}}_{\left(aq\right)}\rightleftharpoons 2{{NH}_3}_{\left(g\right)}+2{H_{2}O}_{\left(aq\right)}+{Mg}_{\left(aq\right)}^{2+}$

(Given that the solubility product for $Mg(OH{)}_2,K_{sp}$ is $6\times {10}^{-12}mo{l}^3{L}^{-3}$ and

the base dissociation constant for $N{H}_{4}OH,K_b$ is $1.8\times {10}^{-5}mol{L}^{-1}$)

• A. $2.05\times {10}^{-2}$
• B. $1.65\times {10}^{-2}$
• C. $1.85\times {10}^{-2}$
• D. None of these

Answer 1

Answer: (C)

$1.85\times {10}^{-2}$

${Mg{\left(OH\right)}_2}_{\left(s\right)}+2{{NH}_4^{+}}_{\left(aq\right)}\rightleftharpoons 2{{NH}_3}_{\left(g\right)}+2{H_{2}O}_{\left(aq\right)}+{Mg}_{\left(aq\right)}^{2+}$

For the given reaction-

$K_c=\frac{{\left[{NH}_3\right]}^2{\left[H_{2}O\right]}^2\left[{Mg}^{2+}\right]}{{\left[{NH}_4^{+}\right]}^2}=\frac{{\left[{NH}_{4}OH\right]}^2\left[{Mg}^{2+}\right]}{{\left[{NH}_4^{+}\right]}^2}.........\left(i\right)$

For ${NH}_{4}OH$, a weak base

$K_b=\frac{\left[{NH}_4^{+}\right]\left[{OH}^{-}\right]}{\left[{NH}_{4}OH\right]}........\left(ii\right)$

By Eqs. $\left(i\right)$ and $\left(ii\right)$

$K_c\times {\left(K_b\right)}^2=\left[{Mg}^{2+}\right]{\left[{OH}^{-}\right]}^2=K_{{sp}_{{Mg\left(OH\right)}_2}}$

$\therefore K_c=\frac{K_{sp}}{{\left(K_b\right)}^2}=\frac{6\times {10}^{-12}}{{(1.8\times {10}^{-5})}^2}=1.85\times {10}^{-2}$

Hence, the correct option is C.