Chemical and Ionic Equilibrium 2 Question 75
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.
$(1986,5 \mathrm{M})$
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Solution:
- In pure water, solubility $=\frac{9.57}{58} \times 10^{-3} \mathrm{M}$
$$ =1.65 \times 10^{-4} \mathrm{M} $$
$K_{\text {sp }}=4 S^{3}=4\left(1.65 \times 10^{-4}\right)^{3}=1.8 \times 10^{-11}$
In $0.02 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}{3}\right){2}$; solubility of $\mathrm{Mg}(\mathrm{OH}){2}=\sqrt{\frac{K{\text {sp }}}{\left[\mathrm{Mg}^{2+}\right]}} \times \frac{1}{2}$
$$ \begin{aligned} & =1.5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \ & =1.5 \times 10^{-5} \times 58 \mathrm{~g} \mathrm{~L}^{-1} \ & =8.7 \times 10^{-4} \mathrm{~g} \mathrm{~L}^{-1} \end{aligned} $$
