QJEEAdv08Ch97a6: JEE Advanced 2008 Chemistry - Class 11 JEEAdv08Ch97a6 Solution: Solubility product constants (K ) of salts of types MX, MX and…

MX > M $_3$ X > MX $_2$

$\Rightarrow~~\mathrm{MX} \to$ $\mathrm{\mathop {{M^ + }}\limits_S + \mathop {{M^ - }}\limits_S}$

Solubiliity, (S $_1$ ) of

$MX = \sqrt {{K_{sp}}} = \sqrt {4 \times {{10}^{ - 8}}}$

$= 2 \times {10^{ - 4}}\,M$

$\Rightarrow M{X_2} \to \mathop {{M^ + }}\limits_S + \mathop {2{X^ - }}\limits_S$

${K_{sp}} = {(s)^1}{(2s)^2} = 4{s^3}$

$S = {\left( {{{3.2 \times {{10}^{ - 14}}} \over 4}} \right)^{{1 \over 3}}}$

${K_{sp}} = 2 \times {10^{ - 5}}\,M$

$\Rightarrow {M_3}X \to \mathop {3{M^ + }}\limits_{3S} + \mathop {{X^{3 - }}}\limits_S$

${K_{sp}} = {(3s)^3}{s^1} = 27{s^4}$

$S = {\left( {{{2.7 \times {{10}^{ - 15}}} \over {27}}} \right)^{{1 \over 9}}}$

$= {({10^{ - 16}})^{{1 \over 2}}}$

$S = {10^{ - 4}}\,M = 1 \times {10^{ - 14}}\,M$

Hence, option (D) is correct.

MX > M $_3$ X > MX $_2$