Question

Freshly precipitated $Al$ and $Mg$ hydroxides are stirred vigorously in a buffer solution containing $0.25M$ of ${NH}_{4}Cl$ and $0.05M$ of ${NH}_{4}OH$. $\left[{Al}^{3+}\right]$ and $\left[{Mg}^{2+}\right]$ in solution are:
Given:

$K_b$ for ${NH}_{4}OH=1.8\times {10}^{-5}$

$K_{{sp}_{Al{\left(OH\right)}_3}}=6\times {10}^{-32}$

$K_{{sp}_{Mg{\left(OH\right)}_2}}=8.9\times {10}^{-12}$

• A. $0.656M$
• B. $0.686M$
• C. $0.698M$
• D. none of these

Answer 1

The correct option is B $0.686M$

For buffer solution having ${NH}_{4}Cl=0.25M$ and ${NH}_{4}OH=0.05M$

$pOH=-\log K_b+\log \frac{\left[ConjugateAcid\right]}{\left[Base\right]}$

$pOH=-\log [1.8\times {10}^{-5}]+\log \frac{0.25}{0.25}$

$\therefore$ $\left[{OH}^{-}\right]=3.6\times {10}^{-6}M$

Freshly precipitated $Al{\left(OH\right)}_3$ $Mg{\left(OH\right)}_3$ are stirred vigorously in it

$\therefore$ $\left[{Al}^{3+}\right]{\left[{OH}^{-}\right]}^3=K_{{sp}_{Al{\left(OH\right)}_3}}$

$\left[{Al}^{3+}\right]{[3.6\times {10}^{-6}]}^3=6\times {10}^{-32}$

$\left[{Al}^{3+}\right]=1.28\times {10}^{-15}M$

Similarly for $Mg{\left(OH\right)}_2$:

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

$\left[{Mg}^{2+}\right]{[3.6\times {10}^{-6}]}^2=8.9\times {10}^{-12}$

$\left[{Mg}^{2+}\right]=0.686M$