Question

Three sparingly soluble salts $M_{2}X,MX$ and $M{X}_3$ have same value of solubility product. Their solubilities follow the order

• A. $M{X}_3>MX>M_{2}X$
• B. $MX>M{X}_3>M_{2}X$
• C. $MX>M_{2}X>M{X}_3$
• D. $M{X}_3>M_{2}X>MX$

Answer 1

Answer: (D)

Let solubility of $M_{2}X$ is $S_1,MX$ is $S_2$ and $M{X}_3$ is $S_3$
$S_1={\left(K_{sp}/4\right)}^{-\frac{1}{3}};S_2={\left(K_{sp}\right)}^{1/2}$
$S_3={\left(K_{sp}/27\right)}^{1/4}$
As $K_{sp}$ Let solubility of $10^{-12}$ (say) In that case
$S_1={\frac{1}{3}}_{\sqrt{4}},S_2=10^6$ and $S_3=\frac{1}{3_{\sqrt{3}}}10^{-3}$
Clearly $S_3>S_1>S_2$
$\therefore$ The correct order of solubilities is
$M{X}_3>M_{2}X>M_{1}X$