Question

The solubility of $Mg(OH{)}_2$ in pure water is $9.57\times {10}^{-3}g/L$. Its solubility in $g/L$ in 0.02M $Mg(N{O}_3{)}_2$ solution is:

• A. $1.5\times {10}^{-4}$
• B. $8.63\times {10}^{-4}$
• C. $0.5\times {10}^{-3}$
• D. $0.5\times {10}^{-5}$

Answer 1

The correct option is B $8.63\times {10}^{-4}$

The solubility of $Mg(OH{)}_2$ in pure water is $9.57\times {10}^{-3}g/L$

or $\frac{9.57\times {10}^{-3}}{58.3}mol/L=1.64\times {10}^{-4}mol/L$.

The expression for the solubility product is as given below.

$K_{sp}=s(4s{)}^2=4s^3=4(1.64\times {10}^{-4}{)}^3=1.76\times {10}^{-11}$

In the presence of 0.02M $Mg(N{O}_3{)}_2$ solution, the expression for the solubility product becomes,

$K_{sp}=(0.02+s)(2s{)}^2=4s^3+0.08s^2$.

Hence, $1.76\times {10}^{-11}=4s^3+0.08s^2$ or $s=1.48\times {10}^{-5}moles/L$.

The solubility in $g/L$ will be $1.48\times {10}^{-5}moles/L\times 58.3=8.63\times {10}^{-4}$.