Question

Three sparingly soluble salts $M_2X, MX$ and $MX_3$ have same value of solubility product. Their solubilities follow the order

• A. $MX_3 > MX > M_2X$
• B. $MX > MX_3 > M_2X$
• C. $MX > M_2X > MX_3$
• D. $MX_3 > M_2X > MX$

Answer 1

Answer: (D)

$MX_3 > M_2X > MX$

Answer 2

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 $MX_{3}>\,M_{2}X>\,M_{1}X$