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# Three sparingly solube salts $M_2V, MX and MX_3$ have same value of solubility product. Their solubilities follow the order

$\begin{array}{1 1} (a) MX_3 > MX > M_2X \\ (b) MX > MX_3 > M_2X \\ (c) MX > M_2X > MX_3 \\(d) MX_3 > M_2X > MX \end{array}$

Answer: $MX_3 > M_2X > MX$

Given, Solubility of $M_2X$ is $S_1$. $MX$ is $S_2$ and $MX_3$ is $S_3$

$S_1 = (\frac{K_{sp}}{4})^{-\frac{1}{3}}$
$S_2 = (K_{sp})^\frac{1}{2}$
$S_3 = (\frac{K_{sp}}{27})^{\frac{1}{4}}$

As $K_{sp}$ is generally of the order of $10^{-12}$ (say)
In that case

$S_1 = \sqrt[\frac{1}{3}]{4}$
$S_2 = 10^{-6}$
$S_3 = \sqrt[\frac{1}{3}]{3}$

Clearly, $S_3 > S_2 > S_1$
$\therefore$ The correct order of solubilities is

$MX_3 > M_2X > MX$