Question

Freshly precipitated aluminium and magnesium hydroxides are stirred vigorously in a buffer solution containing $0.05mol{L}^{-1}$ of ${NH}_{4}OH$ and $0.25mol{L}^{-1}$ of ${NH}_{4}Cl$. Calculate the concentration of aluminium and magnesium ions in solution.
$K_{sp}Al{\left(OH\right)}_3=6\times {10}^{-32};K_b{NH}_{4}OH=1.8\times {10}^{-5}$
$K_{sp}Mg{\left(OH\right)}_2=6\times {10}^{-10}$

Answer 1

Given:$Acid=0.05M,\left[N{H}_{4}Cl\right]=0.25M$

$K_b\left(N{H}_{4}OH\right)=1.8\times {10}^{-5}$

From Henderson's equation, we calculate the pH of solution mixture of acid and conjugate bases.

$pOH=\log \frac{\left[Salt\right]}{\left[Acid\right]}-\log K_b$

$-\log \left[{OH}^{-}\right]=\log \frac{0.25}{0.15}-\log \left(1.8\times {10}^{-5}\right)$
$\left[{OH}^{-}\right]=0.36\times {10}^{-5}M$

Equilibrium reaction for Aluminium precipitate:

$A{l}^{3+}+3O{H}^{-}\rightleftharpoons Al{\left(OH\right)}_3$

$K_{sp}=\frac{\left[A{l}^{3+}\right]{\left[O{H}^{-}\right]}^3}{Al{\left(OH\right)}_3}$
$\left[{Al}^{3+}\right]=\frac{K_{sp}Al{\left(OH\right)}_3}{{\left[{OH}^{-}\right]}^3}=1.29\times {10}^{-10}M$

Equilibrium precipitate for magnesium:

$M{g}^{2+}+2O{H}^{-}\rightleftharpoons Mg{\left(OH\right)}_2{K}_{sp}=\frac{\left[M{g}^{2+}\right]{\left[O{H}^{-}\right]}^2}{Mg{\left(OH\right)}_2}$

$\left[{Mg}^{2+}\right]=\frac{K_{sp}Mg{\left(OH\right)}_2}{{\left[{OH}^{-}\right]}^2}=46.3M$