Question

Solubility products of the salts of types $M{X}_1,M{X}_2,M_{3}X$ at temperature are $4\times {10}^{-14}$, $3.2\times {10}^{-14}$ 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

Answer: (B)

$M_{3}X>M{X}_2>MX$

For the compound $M{X}_1$, let the solubility be $S_1$, then

$\left[S_1\right]\left[S_1\right]=4\times {10}^{-14}$

=> $S_1=2\times {10}^{-7}$

For the compound $M{X}_2$, let the solubility be $S_2$, then

$\left[S_2\right][2S_1{]}^2=3.2\times {10}^{-14}$

=> $4{S_2}^3=3.2\times {10}^{-14}$

=> ${S_2}^3=0.8\times {10}^{-14}$

=> ${S_2}^3=8\times {10}^{-15}$

=> $S_2=2\times {10}^{-5}$

For the compound $M_{3}X$, let the solubility be $S_3$, then

$\left[3S_3{]}^3\right[S_3]=27\times {10}^{-16}$

=> $27[S_3{]}^4=27\times {10}^{-16}$

=>$S_3={10}^{-4}$

Therefore, $S_3>S_2>S_1$