JEE Main 2026 — Apr 5, Shift 1 — Chemistry solution

JEE Main2026Apr 5, Shift 1Chemistry

$M_{3} A_{2}$ is a sparingly soluble salt of molar mass $y$ g mol $^{- 1}$ and solubility $x$ g L $^{- 1}$ . The ratio of the molar concentration of the anion ( $A^{3 -}$ ) to the solubility product of the salt is

A
$\frac{1}{54} \cdot \frac{y ^{4}}{x ^{4}}$
B
$\frac{y ^{5}}{108 x ^{4}}$
C
$108 \cdot \frac{x ^{5}}{y ^{5}}$
D
$\frac{1}{108} \cdot \frac{y ^{4}}{x ^{4}}$

Solution

Molar solubility $s = x / y$ mol/L. Dissociation: $M_{3} A_{2} \to 3 M^{2 +} + 2 A^{3 -}$ gives $[M^{2 +}] = 3 s$ and $[A^{3 -}] = 2 s$ .

$K_{s p} = (3 s)^{3} (2 s)^{2} = 108 s^{5}$ .

Required ratio: $\frac{[ A ^{3 -} ]}{K _{s p}} = \frac{2 s}{108 s ^{5}} = \frac{1}{54 s ^{4}} = \frac{1}{54} \cdot \frac{y ^{4}}{x ^{4}}$ .

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