Chemical and Ionic Equilibrium - Result Question 56
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 M of ammonium hydroxide. Calculate the concentration of aluminium and magnesium ions in solution.
$\begin{aligned} K_b\left[\mathrm{NH}4 \mathrm{OH}\right] & =1.8 \times 10^{-5} \\ K{\text {sp }}\left[\mathrm{Mg}(\mathrm{OH})2\right] & =8.9 \times 10^{-12} \\ K{\text {sp }}\left[\mathrm{Al}(\mathrm{OH})_3\right] & =6 \times 10^{-32}\end{aligned}$
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Solution:
$\begin{aligned} & \text { pOH of buffer solution }=\mathrm{p} K_b+\log \frac{\left[\mathrm{NH}4^{+}\right]}{\left[\mathrm{NH}4 \mathrm{OH}\right]} \\ &=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{0.25}{0.05}=5.44 \\ & {\left[\mathrm{OH}^{-}\right] }=3.6 \times 10^{-6} \mathrm{M} \\ & {\left[\mathrm{Al}^{3+}\right] }=\frac{K{\text {sp }}}{\left[\mathrm{OH}^{-}\right]^3}=\frac{6 \times 10^{-32}}{\left(3.6 \times 10^{-6}\right)^3}=1.28 \times 10^{-15} \mathrm{M} \\ & {\left[\mathrm{Mg}^{2+}\right] }=\frac{K{\text {sp }}}{\left[\mathrm{OH}^{-}\right]^2}=\frac{8.9 \times 10^{-12}}{\left(3.6 \times 10^{-6}\right)^2}=0.68 \mathrm{M} \end{aligned}$
BETA
