Question

Solubility product constants $\left(K_{sp}\right)$ of salts of types $MX,M{X}_2,M_{3}X$ at temperature $T$ are $4.0\times {10}^{-8},3.2\times {10}^{-4}$ and $2.7\times {10}^{-13}$ respectively. Solubility ($mol$ ${dm}^{-3}$) 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 $MX>M_{3}X>M{X}_2$

$MX\rightleftharpoons M^{+}+X^{-}$

$S$ $S$

Solubility $\left(S_1\right)$ of $MX=\sqrt{K_{sp}}=\sqrt{4\times {10}^{-8}}=2\times {10}^{-4}M$

$M{X}_2\rightleftharpoons M^{+}+2X^{-}$

$S$ $2S$

$K_{sp}=\left(S{)}^1\right(2S{)}^2$

$=4S^3$

$S=3\sqrt{\frac{k_{sp}}{4}}$

$=\sqrt{\frac{3.2\times {10}^{-4}}{4}}$

$K_{sp}=2\times {10}^{-5}$

$K_{sp}\rightleftharpoons {\underset{3S}{3M}}^{+}+{\underset{S}{X}}^{3-}$

$K_{sp}=(3s{)}^3{s}^1={275}^4$

$S={\left(\frac{2.7\times {10}^{-15}}{27}\right)}^{1/4}$

$=({10}^{-16}{)}^{1/4}$

$S={10}^{-4}M$

$=1\times {10}^{-4}M$