Question

Solubility products of the salts of types $MX$, $M{X}_2$$M_{3}X$ at temperature $T$ are $4\times {10}^{-8}$, $3.2\times {10}^{-14}$ and $27\times {10}^{-16}$ respectively. Solubility $\left(mold{m}^{-3}\right)$ of the salts at $T$ 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 A $MX>M{X}_2>M_{3}X$

Let $S_1,S_2,S_3$ denote solubility of $MX,M{X}_2$ and $M{X}_3$ respectively.

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

$S_1$ $S_1$ $\Rightarrow K_{sp}=S_1^2=4\times {10}^{-8}$

$\therefore S_1=2\times {10}^{--4}$

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

$S_2$ $2S_2$

$K_{sp}=4S_2^3=32\times {10}^{-15}$

$\Rightarrow S_2=2\times {10}^{-5}=0.2\times {10}^{-4}$

$M_{3}X\to 3M^{+}+X^{-}$

$3S_3$ $S_3$

$K_{sp}=27S_3^4=27\times {10}^{-16}$

$\therefore S_3={10}^{-4}$

$S_1>S_3>S_2$

Order of solubility is $MX>M_{3}X>M{X}_2$