Question

The solubility of $Mg(OH{)}_2$ in water at ${25}^{\circ }\text{C}$ is $0.00912\text{g}$ per litre. Find $K_{sp}$ for $Mg(OH{)}_2$ assuming complete ionization.

• A. $1.52\times {10}^{-9}$
• B. $1.52\times {10}^{-13}$
• C. $1.52\times {10}^{-11}$
• D. $3.04\times {10}^{-11}$

Answer 1

The correct option is A $1.52\times {10}^{-9}$

$\begin{array}{ccccccc}Mg(OH{)}_2& \rightleftharpoons & M{g}^{2+}& +& 2O{H}^{-}& & \\ \text{1 mole}& & \text{1 mole}& & \text{2 mole}& & \end{array}$
$K_{sp}=\left[M{g}^{2+}\right][O{H}^{-}{]}^2\dots \dots (i)$
Concentration of $Mg(OH{)}_2=\frac{0.00912\text{g/L}}{58.3\text{g/mol}}$
$=1.56\times {10}^{-4}\text{mol/L}$
Since one mole of $Mg(OH{)}_2$ yields one mole of $M{g}^{2+}$ ion and two moles of $O{H}^{-}$ ion, then
$\left[M{g}^{2+}\right]=1.56\times {10}^{-4}\text{mol/L}$
$\left[O{H}^{-}\right]=2\times 1.56\times {10}^{-4}=3.12\times {10}^{-4}\text{mol/L}$
On putting the values in equation (i)
$K_{sp}=(1.56\times {10}^{-4})\times (3.12\times {10}^{-4}{)}^2$
$=1.52\times {10}^{-11}$ for $Mg(OH{)}_2$ at ${25}^{\circ }\text{C}$