Chemistry Q55 – JEE-MAIN 2026

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

Clue 1 of 3: Dissociation Equilibrium

Write the balanced dissociation equation for the sparingly soluble salt $M_{3} A_{2}$ and define molar solubility $S$ .

🧠 Full Solution Path

Step 1: Define Molar Solubility and

K_{s p}

expression✦ Active

The dissociation of the salt $M_{3} A_{2}$ can be represented as:

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

Let $S$ be the molar solubility in mol L $^{- 1}$ . Then, the equilibrium concentrations are $[M^{2 +}] = 3 S$ and $[A^{3 -}] = 2 S$ . The solubility product $K_{s p}$ is given by:

K_{s p} = [M^{2 +}]^{3} [A^{3 -}]^{2} = (3 S)^{3} (2 S)^{2} = (27 S^{3}) (4 S^{2}) = 108 S^{5}

Step 2: Convert given solubility to molar solubility○ Expand

The given solubility is $x$ g L $^{- 1}$ and the molar mass is $y$ g mol $^{- 1}$ . Molar solubility $S$ is calculated as:

S = \frac{{solubility (g L}^{- 1})}{{molar mass (g mol}^{- 1})} = \frac{x}{y} {mol L}^{- 1}

Step 3: Calculate the ratio of anion concentration to

K_{s p}

○ Expand

The molar concentration of the anion $A^{3 -}$ is $[A^{3 -}] = 2 S = 2 (\frac{x}{y})$ . Substituting $S$ into the $K_{s p}$ expression:

K_{s p} = 108 S^{5} = 108 {(\frac{x}{y})}^{5}

Now, calculate the required ratio:

\frac{[A^{3 -}]}{K_{s p}} = \frac{2 (x / y)}{108 (x / y)^{5}} = \frac{2}{108} \frac{1}{(x / y)^{4}} = \frac{1}{54} \frac{y^{4}}{x^{4}}

This matches option 1.

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Chemistry Question 55 – JEE-MAIN 2026

Clue 1 of 3: Dissociation Equilibrium

🧠 Full Solution Path