JEE Main 2026 — Chemistry Physical Chemistry

2026mcqhard

$M_3 A_2$ 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

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

The molar solubility $S$ of the salt is given by $S = \dfrac{x}{y}$ mol L $^{-1}$ .

The dissociation of the sparingly soluble salt $M_3A_2$ is:

$M_3A_2 \rightleftharpoons 3M^{2+} + 2A^{3-}$

The molar concentration of the anion $A^{3-}$ is:

$[A^{3-}] = 2S$

The solubility product $K_{sp}$ of the salt is:

$K_{sp} = [M^{2+}]^3 [A^{3-}]^2 = (3S)^3 (2S)^2 = 27S^3 \times 4S^2 = 108S^5$

The ratio of the molar concentration of the anion to the solubility product is:

$\dfrac{[A^{3-}]}{K_{sp}} = \dfrac{2S}{108S^5} = \dfrac{1}{54S^4}$

Substituting $S = \dfrac{x}{y}$ into the expression, we get:

$\text{Ratio} = \dfrac{1}{54 \left(\dfrac{x}{y}\right)^4} = \dfrac{1}{54} \cdot \dfrac{y^4}{x^4}$

Answer: $\dfrac{1}{54} \cdot \dfrac{y^4}{x^4}$

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2026 — Chemistry Physical Chemistry

2026mcqhard

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

The molar solubility $S$ of the salt is given by $S = \dfrac{x}{y}$ mol L $^{-1}$ .

The dissociation of the sparingly soluble salt $M_3A_2$ is:

$M_3A_2 \rightleftharpoons 3M^{2+} + 2A^{3-}$

The molar concentration of the anion $A^{3-}$ is:

$[A^{3-}] = 2S$

The solubility product $K_{sp}$ of the salt is:

$K_{sp} = [M^{2+}]^3 [A^{3-}]^2 = (3S)^3 (2S)^2 = 27S^3 \times 4S^2 = 108S^5$

The ratio of the molar concentration of the anion to the solubility product is:

$\dfrac{[A^{3-}]}{K_{sp}} = \dfrac{2S}{108S^5} = \dfrac{1}{54S^4}$

Substituting $S = \dfrac{x}{y}$ into the expression, we get:

$\text{Ratio} = \dfrac{1}{54 \left(\dfrac{x}{y}\right)^4} = \dfrac{1}{54} \cdot \dfrac{y^4}{x^4}$

Answer: $\dfrac{1}{54} \cdot \dfrac{y^4}{x^4}$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

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Keep practicing Physical Chemistry

JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry