Question

The solubility of $Mg(OH{)}_2$, in pure water is $9.57\times {10}^{-3}g{L}^{-1}$. Calculate its solubility in $g{L}^{-1}$ in $0.02\text{M}Mg(N{O}_3{)}_2.$

• A. $8.69\times {10}^{-6}$
• B. $8.69\times {10}^{-4}$
• C. $14.98\times {10}^{-6}$
• D. $14.98\times {10}^{-4}$

Answer 1

The correct option is B $8.69\times {10}^{-4}$

$\text{Solubility of Mg}$ ${\text{OH}}_2$ in pure water
$=9.57\times {10}^{-3}g{\text{L}}^{-1}$
$=\frac{9.57\times {10}^{-3}}{\text{Mol. mass}}\text{mol}{\text{L}}^{-1}$
$=\frac{9.57\times {10}^{-3}}{58}=1.65\times {10}^{-4}\text{mol}{\text{L}}^{-1}$

$Mg(OH{)}_2$ $\rightleftharpoons {\text{Mg}}^{2+}+{\text{OH}}^{-}$
$\text{S}\text{2S}$

${\text{K}}_{sp}=\left[{\text{Mg}}^{2+}\right]{\left[{\text{OH}}^{-}\right]}^2$
$=\text{S}\times {\left(2\text{S}\right)}^2=4{\text{S}}^3$
$=4\times {(1.65\times {10}^{-4})}^3$
$=17.9685\times {10}^{-12}$

$\text{Let S' be the solubility of}\text{Mg}{\left(\text{OH}\right)}_2\text{in presence of Mg}{\left({\text{NO}}_3\right)}_2$

$\text{As}\left[{\text{Mg}}^{2+}\right]=\left(\text{S'+c}\right)=\left(\text{S'+0.02}\right)$

$\left[\text{OH'}\right]=2\text{S'}$

$\text{So}{\text{K}}_s=\left(\text{S'+0.002}\right){\left(\text{2S'}\right)}^2$

$17.9685\times {10}^{-12}=4{\left(\text{S'}\right)}^2\left(\text{S'+0.02}\right)$

$\frac{17.9685\times {10}^{-12}}{4}={\left(\text{S'}\right)}^3+0.02{\left(\text{S'}\right)}^2$
$\text{On neglecting}{\left(S^{\prime }\right)}^3\text{we get},$
$4.4921\times {10}^{-12}=0.02{\left(\text{S'}\right)}^2$
${\left(\text{S'}\right)}^2=\frac{4.4921}{0.02}\times {10}^{-12}$ $\text{S'}=14.9868\times {10}^{-6}\text{mol}{\text{L}}^{-1}$
$\text{Solubility of Mg}{\left(\text{OH}\right)}_2\text{in g}{\text{L}}^{-1}=\text{S'}\times \text{M}$
$=14.9868\times {10}^{-6}\times 58$
$=8.69\times {10}^{-4}\text{g}{\text{L}}^{-1}$