Question

The solubility product of $Mg(OH{)}_2$ is $1.2\times {10}^{-11}$ . The solubility of this compound is gram per $100c{m}^3$ of solution is :

• A. $1.4\times {10}^{-4}$
• B. $8.36\times {10}^{-4}$
• C. $0.816$
• D. $1.4$

Answer 1

Answer: (B)

$8.36\times {10}^{-4}$

The solubility product of $Mg(OH{)}_2$ is $1.2\times {10}^{-11}$.

The expression for the solubility product is as given below.

$K_{sp}=\left[M{g}^{2+}\right][O{H}^{-}{]}^2=s(2s{)}^2=4s^3=1.2\times {10}^{-11}$.

Hence, $s=(\frac{1.2\times {10}^{-11}}{4}{)}^{\left(\frac{1}{3}\right)}=1.44\times {10}^{-4}$ moles/litre

The solubility in g/L will be $1.44\times {10}^{-4}moles/L\times 58=8.16\times {10}^{-3}$ grams/litre.

The solubility in $grams/100c{m}^3$ will be $8.36\times {10}^{-3}\times 0.1=8.36\times {10}^{-4}grams/100c{m}^3$.