Question

The solubility of $Mg{\left(OH\right)}_2$ in pure water is $9.57\times {10}^{-3}g{L}^{-1}$. Calculate its solubility in $g{L}^{-1}$ in $0.02M$ $Mg{\left({NO}_3\right)}_2$ solution.

Answer 1

Solubility of $Mg{\left(OH\right)}_2$ in pure water
$=9.57\times {10}^{-3}g{L}^{-1}$
$=\frac{9.57\times {10}^{-3}}{Mol.mass}mol{L}^{-1}=\frac{9.57\times {10}^{-3}}{58}=1.65\times {10}^{-4}mol{L}^{-1}$
Further, $Mg{\left(OH\right)}_2\rightleftharpoons {Mg}^{2+}+2{OH}^{-}$
$K_s=\left[{Mg}^{2+}\right]{\left[{OH}^{-}\right]}^2=S\times {\left(2S\right)}^2=4\times {(1.65\times {10}^{-4})}^3=17.9685\times {10}^{-12}$
Let $S^{\prime }$ be the solubility of $Mg{\left(OH\right)}_2$ in presence of $Mg{\left({NO}_3\right)}_2$
$\left[{Mg}^{2+}\right]=(S^{\prime }+c)=(S^{\prime }+0.02)$
$\left[{OH}^{{}^{\prime }}\right]=2S^{\prime }$
So, $K_s=(S^{\prime }+0.02){\left(2S^{\prime }\right)}^2$
$17.9685\times {10}^{-12}=4{\left(S^{\prime }\right)}^2(S^{\prime }+0.02)$
$\frac{17.9685\times {10}^{-12}}{4}={\left(S^{\prime }\right)}^3+0.02{\left(S^{\prime }\right)}^2$ [neglecting ${\left(S^{\prime }\right)}^3$
$S^{\prime }=14.9868\times {10}^{-6}mol{L}^{-1}$
solubility of $Mg{\left(OH\right)}_2$ in $g{litre}^{-1}=S^{\prime }\times M=8.69\times {10}^{-4}g{L}^{-1}$