Chemical and Ionic Equilibrium 2 Question 71
70. Freshly precipitated aluminium and magnesium hydroxides are stirred vigorously in a buffer solution containing $0.25 \mathrm{~mol} / \mathrm{L}^{-}$of $\mathrm{NH}_{4} \mathrm{Cl}$ and $0.05 \mathrm{M}$ of ammonium hydroxide. Calculate the concentration of aluminium and magnesium ions in solution.
$$ \begin{aligned} K_{b}\left[\mathrm{NH}{4} \mathrm{OH}\right] & =1.8 \times 10^{-5} \ K{\text {sp }}\left[\mathrm{Mg}(\mathrm{OH}){2}\right] & =8.9 \times 10^{-12} \ K{\text {sp }}\left[\mathrm{Al}(\mathrm{OH})_{3}\right] & =6 \times 10^{-32} \end{aligned} $$
(1989, 3M)
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Solution:
- $\mathrm{pOH}$ of buffer solution $=\mathrm{p} K_{b}+\log \frac{\left[\mathrm{NH}{4}^{+}\right]}{\left[\mathrm{NH}{4} \mathrm{OH}\right]}$
$$ \begin{aligned} & =-\log \left(1.8 \times 10^{-5}\right)+\log \frac{0.25}{0.05}=5.44 \ {\left[\mathrm{OH}^{-}\right] } & =3.6 \times 10^{-6} \mathrm{M} \ {\left[\mathrm{Al}^{3+}\right] } & =\frac{K_{\text {sp }}}{\left[\mathrm{OH}^{-}\right]^{3}}=\frac{6 \times 10^{-32}}{\left(3.6 \times 10^{-6}\right)^{3}}=1.28 \times 10^{-15} \mathrm{M} \ {\left[\mathrm{Mg}^{2+}\right] } & =\frac{K_{\text {sp }}}{\left[\mathrm{OH}^{-}\right]^{2}}=\frac{8.9 \times 10^{-12}}{\left(3.6 \times 10^{-6}\right)^{2}}=0.68 \mathrm{M} \end{aligned} $$
