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Q.
Three sparingly soluble salts $M_2X, MX$ and $MX_3$ have same value of solubility product. Their solubilities follow the order
Equilibrium
Solution:
Let solubility of $M_{2}X$ is $S_{1}, MX$ is $S_{2}$ and $MX_{3}$ 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
$MX_{3}>\,M_{2}X>\,M_{1}X$