Question

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

Answer 1

$pOH={\text{log}}_{10}\frac{\left[\text{Salt}\right]}{\left[\text{Base}\right]}-\text{log}{K}_b$
$-{\text{log}}_{10}\left[O{H}^{-}\right]={\text{log}}_{10}0.25-{\text{log}}_{10}\frac{1.8\times {10}^{-5}}{0.05}$
${\text{log}}_{10}\left[O{H}^{-}\right]={\text{log}}_{10}\frac{1.8\times {10}^{-5}}{5}$
$\left[O{H}^{-}\right]=0.36\times {10}^{-5}{\text{mol L}}^{-1}$
$\left[A{l}^{3+}\right]=\frac{K_{sp}\left(Al\right(OH{)}_3)}{[O{H}^{-}{]}^3}$
$=\frac{6\times {10}^{-32}}{(0.36\times {10}^{-5}{)}^3}$
$=1.29\times {10}^{-10}{\text{mol L}}^{-1}$
$\left[M{g}^{2+}\right]=\frac{K_{sp}\left(Mg\right(OH{)}_2)}{[O{H}^{-}{]}^2}$
$=\frac{6\times {10}^{-10}}{(0.36\times {10}^{-5}{)}^2}=46.3{\text{mol L}}^{-1}$