Chemical and Ionic Equilibrium - Result Question 56
73. The solubility of $\mathrm{Mg}(\mathrm{OH})_2$ in pure water is $9.57 \times 10^{-3} \mathrm{~g} / \mathrm{L}$. Calculate its solubility (in $\mathrm{g} / \mathrm{L}$ ) in $0.02$ $ \mathrm{M} $ $\mathrm{Mg}\left(\mathrm{NO}_3\right)_2$ solution.
(1986, 5M)
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Solution:
In pure water, solubility $=\frac{9.57}{58} \times 10^{-3} \mathrm{M} $
$=1.65 \times 10^{-4} \mathrm{M} $
$ K_{\text {sp }}=4 S^3=4\left(1.65 \times 10^{-4}\right)^3=1.8 \times 10^{-11} $
$\text { In } 0.02 \hspace{1mm} \mathrm{M} \hspace{1mm}\mathrm{Mg}\left(\mathrm{NO}_3\right)_2 \text {; } $
solubility of $\mathrm{Mg}(\mathrm{OH})2 =\sqrt{\frac{K{\mathrm{sp}}}{\left[\mathrm{Mg}^{2+}\right]}} \times \frac{1}{2} $
$=1.5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} $
$=1.5 \times 10^{-5} \times 58 \mathrm{~g} \mathrm{~L}^{-1} $
$=8.7 \times 10^{-4} \mathrm{~g} \mathrm{~L}^{-1}$
BETA
