The solubility product constants of Ag2CrO4 and AgBr are 32x and 4y respectively at 298 K. The value of (molarity of Ag2CrO4 / molarity of AgBr) can be expressed as :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q56 – JEE-MAIN 2026

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Home Chemistry EquilibriumQuestion #417

The solubility product constants of

A g_{2} C r O_{4}

and

A g B r

are

32 x

and

4 y

respectively at 298 K. The value of

(\frac{molarity of A g_{2} C r O_{4}}{molarity of A g B r})

can be expressed as :

Clue 1 of 3: Molar Solubility

The molarity of a sparingly soluble salt in its saturated solution is defined as its molar solubility, denoted by 's'.

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Ninja Strategy•Stoichiometry-Solubility Relationship

Recognize that for a salt $A_{m} B_{n}$ , solubility $s \propto (K_{s p})^{1 / (m + n)}$ . For Ag₂CrO₄, $s \propto (K_{s p})^{1 / 3}$ , and for AgBr, $s \propto (K_{s p})^{1 / 2}$ . This immediately determines the powers of x and y in the ratio.

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Step 1: Determine the Molar Solubility of Ag₂CrO₄✦ Active

For silver chromate, $A g_{2} C r O_{4}$ , the dissociation equilibrium is $A g_{2} C r O_{4} (s) ⇌ 2 A g^{+} (a q) + C r O_{4}^{2 -} (a q)$ . If its molar solubility is $s_{1}$ , the ion concentrations are $[A g^{+}] = 2 s_{1}$ and $[C r O_{4}^{2 -}] = s_{1}$ .

K_{s p} (A g_{2} C r O_{4}) = [A g^{+}]^{2} [C r O_{4}^{2 -}] = (2 s_{1})^{2} (s_{1}) = 4 s_{1}^{3}

Given that $K_{s p} = 32 x$ , we can solve for $s_{1}$ :

4 s_{1}^{3} = 32 x ⟹ s_{1}^{3} = 8 x ⟹ s_{1} = \sqrt[3]{8 x} = 2 \sqrt[3]{x}

Step 2: Determine the Molar Solubility of AgBr○ Expand

For silver bromide, $A g B r$ , the dissociation equilibrium is $A g B r (s) ⇌ A g^{+} (a q) + B r^{-} (a q)$ . If its molar solubility is $s_{2}$ , the ion concentrations are $[A g^{+}] = s_{2}$ and $[B r^{-}] = s_{2}$ .

K_{s p} (A g B r) = [A g^{+}] [B r^{-}] = (s_{2}) (s_{2}) = s_{2}^{2}

Given that $K_{s p} = 4 y$ , we can solve for $s_{2}$ :

s_{2}^{2} = 4 y ⟹ s_{2} = \sqrt{4 y} = 2 \sqrt{y}

Step 3: Calculate the Ratio of Molarities○ Expand

The required value is the ratio of the molar solubility of $A g_{2} C r O_{4}$ to that of $A g B r$ .

\frac{molarity of A g_{2} C r O_{4}}{molarity of A g B r} = \frac{s_{1}}{s_{2}} = \frac{2 \sqrt[3]{x}}{2 \sqrt{y}} = \frac{\sqrt[3]{x}}{\sqrt{y}}

This corresponds to option 4.

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Chemistry Question 56 – JEE-MAIN 2026

Clue 1 of 3: Molar Solubility

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