Question

Solubility product constant $\left(K_{sp}\right)$ of salts of types $MX,M{X}_2$ and $M{X}_3$ at temperature T are $4.0\times {10}^{-8},3.2\times {10}^{-14}$ and $2.7\times {10}^{-15}$ respectively. Solubilities $\left(mold{m}^{-3}\right)$ of the salts at temperature 'T' are in the order :

• A. $MX>M{X}_2>M_{3}X$
• B. $M_{3}X>M{X}_2>MX$
• C. $M{X}_2>M_{3}X>MX$
• D. $MX>M_{3}X>M{X}_2$

Answer 1

The correct option is D $MX>M_{3}X>M{X}_2$

Let 's' be the solubility$MX\left(s\right)\rightleftharpoons M^{+}\left(aq\right)+X^{-}\left(aq\right)$
Then solubility product $K_{sp}=\left[M^{+}\right]\left[X^{-}\right]-ss$

$K_{sp}=s^2$
or $s=\sqrt{K_{sp}}$
$s=\sqrt{4\times {10}^{-8}}=2\times {10}^{-4}$

Similarly, for $M{X}_2$
$M{X}_2\left(s\right)\rightleftharpoons M^{2+}\left(aq\right)+2X^{-}\left(aq\right)-s2s$

$K_{sp}=\left[M^{2+}\right][X^{-}{]}^2$
$K_{sp}=s\times (2s{)}^2=4s^3$
or $s={\left[\frac{K_{sp}}{4}\right]}^{\frac{1}{3}}$
$s={\left[\frac{3.2\times {10}^{-14}}{4}\right]}^{\frac{1}{3}}={[\frac{32}{4}\times {10}^{-15}]}^{\frac{1}{3}}$
$s={[8\times {10}^{-15}]}^{\frac{1}{3}}$
$s=2\times {10}^{-5}$

and for $M{X}_3$
$M{X}_3\left(s\right)\rightleftharpoons M^{3+}\left(aq\right)+3X^{-}\left(aq\right)-s3s$

$K_{sp}=\left[M^{3+}\right][X^{-}{]}^3$
$K_{sp}=s\times 3s^3=27s^4$
or $s={\left[\frac{K_{sp}}{27}\right]}^{\frac{1}{4}}$
$s={\left[\frac{2.7\times {10}^{-15}}{27}\right]}^{\frac{1}{4}}$
$s={\left[{10}^{-16}\right]}^{\frac{1}{4}}$
$s={10}^{-4}$
Thus the solubilities are in the order $MX>M{X}_3>M{X}_2$
i.e the correct answer is (d).