Chemical and Ionic Equilibrium - Result Question 131
####73. The solubility of $Mg(OH) _2$ in pure water is $9.57 \times 10^{-3} g / L$. Calculate its solubility (in $g / L$ ) in $0.02 M Mg\left(NO _3\right) _2$ solution.
$(1986,5 M)$
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Solution:
- In pure water, solubility $=\frac{9.57}{58} \times 10^{-3} M$
$$ =1.65 \times 10^{-4} M $$
$K _{\text {sp }}=4 S^{3}=4\left(1.65 \times 10^{-4}\right)^{3}=1.8 \times 10^{-11}$
In $0.02 M Mg\left(NO _3\right) _2$; solubility of $Mg(OH) _2=\sqrt{\frac{K _{\text {sp }}}{\left[Mg^{2+}\right]}} \times \frac{1}{2}$
$$ \begin{aligned} & =1.5 \times 10^{-5} mol L^{-1} \ & =1.5 \times 10^{-5} \times 58 g L^{-1} \ & =8.7 \times 10^{-4} g L^{-1} \end{aligned} $$
