Question

Solubility of $M{X}_2$-type electrolytes is $0.5\times {10}^{-4}$ mol/L. Then find out $K_{sp}$ of electrolyte :

• A. $5\times {10}^{-12}$
• B. $25\times {10}^{-10}$
• C. $1\times {10}^{-13}$
• D. $5\times {10}^{-13}$

Answer 1

The correct option is D $5\times {10}^{-13}$

$\underset{0.5\times {10}^{-4}M}{\underset{Solubility}{M{X}_2}}\to \underset{0.5\times {10}^{-4}M}{M^{2+}}+\underset{2\times 0.5\times {10}^{-4}M}{2X^{-}}$

(on 100 % ionisation)
$\therefore K_{sp}$ of $M{X}_2=\left[M^{2+}\right][X^{-}{]}^2$

$=(0.5\times {10}^{-4})(1.0\times {10}^{-4}{)}^2$

$=5\times {10}^{-13}$

Hence, option D is correct.