$\mathrm{M}_{3} \mathrm{~A}_{2}$ is a sparingly soluble salt of molar mass y $\mathrm{g} \mathrm{~mol}^{-1}$ and solubility $\times \mathrm{g} \mathrm{L}^{-1}$. The ratio of the molar concentration of the anion ( $\mathrm{A}^{3-}$ ) 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) $\frac{1}{108} \frac{y^{4}}{x^{4}}$

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Equilibrium

M_{3} A_{2}

is a sparingly soluble salt of molar mass y

g mol^{- 1}

and solubility

\times g L^{- 1}

. The ratio of the molar concentration of the anion (

A^{3 -}

) to the solubility product of the salt is

Answer: (A)

Let the molar solubility be

s

. Since the molar mass is

y g mo l^{- 1}

and the solubility is

x g L^{- 1}

s = \frac{x}{y} mol L^{- 1}

. For

M_{3} A_{2} (s) ⇌ 3 M^{2 +} + 2 A^{3 -}

the equilibrium concentrations are

[M^{2 +}] = 3 s

and

[A^{3 -}] = 2 s

, so

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

. The required ratio is

\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}}

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option