Q. The molar solubility (in mol $L^{- 1}$) of a sparingly soluble salt $MX_{4}$ is ‘$s$’. The corresponding solubility product is $K_{sp}.s$ is given in terms of $K_{sp}$ by the relation
A) $s = (\frac{K_{sp}1}{28)}^{\frac{1}{4}}$
B) $s = \left( 128\ K_{sp} \right)^{\frac{1}{4}}$
C) $s = \left( 256\ K_{sp} \right)^{\frac{1}{5}}$
D) $s = \left( \frac{K_{sp}}{256} \right)^{\frac{1}{5}}$
Explanation
(d)
For the solute $A_{x}B_{y} \rightleftharpoons x\ A + y\ B$
$K_{sp} = x^{x}y^{y}(s)^{x + y}$
$MX_{4} \rightleftharpoons M^{4 +} + 4X^{-}$
$x = 1,y = 4$
$\therefore$ $K_{sp} = (4)^{4}(1)^{1}(s)^{5} = 256s^{5}$
$\therefore$ $s = \left( \frac{K_{sp}}{256} \right)^{\frac{1}{5}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.