The solubility product constants of Ag_2CrO_4 and AgBr are 32x and 4y respectively at 298 K. The value of ( molarity of Ag_2CrO_4 molarity of AgBr ) can be expressed as :

Correct answer: D. [3] x y Let the solubility of Ag_2CrO_4 be S_1 . For Ag_2CrO_4 leftharpoons 2Ag^+ + CrO_4^ 2- K_ sp = (2S_1)^2(S_1) = 4S_1^3 4S_1^3 = 32x S_1^3 = 8x S_1 = 2 [3] x Let the solubility of AgBr be S_2 . For AgBr leftharpoons Ag^+ + Br^- K_ sp = S_2^2 S_2^2 = 4y S_2 = 2 y The ratio of their molarities is: (S_1)/(S_2) = 2 [3] x 2 y = [3] x y Answer: [3] x y

JEE Main 20262 AprilMorning Shift

Exam: JEE Main
Year: 2026
Date: 2 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Ionic Equilibrium

JEE Main 2026 · 2 April · Morning Shift · Chemistry · Ionic Equilibrium

The solubility product constants of Ag_2CrO_4 and AgBr are 32x and...

The solubility product constants of

Ag_2CrO_4 and AgBr

are

32x and 4y

respectively at

298

K. The value of \left(\dfrac{\text{molarity of } Ag_2CrO_4}{\text{molarity of } AgBr}\right)$ can be expressed as :

Options

A $\dfrac{2\sqrt[3]{x}}{y}$
B $2\sqrt{\dfrac{x}{y}}$
C $\sqrt{\dfrac{x}{y}}$
D $\dfrac{\sqrt[3]{x}}{\sqrt{y}}$

Correct answer

D. $\dfrac{\sqrt[3]{x}}{\sqrt{y}}$

Step-by-step solution

Let the solubility of

Ag_2CrO_4

S_1

.

For

Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}

K_{sp} = (2S_1)^2(S_1) = 4S_1^3

4S_1^3 = 32x \Rightarrow S_1^3 = 8x \Rightarrow S_1 = 2\sqrt[3]{x}

Let the solubility of

AgBr

S_2

.

For

AgBr \rightleftharpoons Ag^+ + Br^-

K_{sp} = S_2^2

S_2^2 = 4y \Rightarrow S_2 = 2\sqrt{y}

The ratio of their molarities is:

\dfrac{S_1}{S_2} = \dfrac{2\sqrt[3]{x}}{2\sqrt{y}} = \dfrac{\sqrt[3]{x}}{\sqrt{y}}

Answer:

\dfrac{\sqrt[3]{x}}{\sqrt{y}}

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