Given is a concentrated solution of a weak electrolyte A x B y of concentration ' c ' and dissociation constant ' K '. the degree of dissociation is given by:

(frac K c x+y-1 x x y y ) 1/x+y

(frac c x+y-1 x x y y K ) 1/x+y

Chemistry · Ionic Equilibrium

JEE Main 2026 — 6 April, Evening Shift — Question 50

Given is a concentrated solution of a weak electrolyte $\mathrm{A}_{x} \mathrm{~B}_{y}$ of concentration ' c ' and dissociation constant ' K '. the degree of dissociation is given by:

Option A:
$\left[K \times c^{x+y-1} x^{x} y^{y}\right]^{x+y}$
Option B:
$\left(\frac{K}{c^{x+y-1} x^{x} y^{y}}\right)^{\frac{1}{x+y}}$
Correct
Option C:
$\left(\frac{c^{x+y-1} x^{x} y^{y}}{K}\right)^{x+y}$
Option D:
$\left(\frac{c^{x+y-1} x^{x} y^{y}}{K}\right)^{\frac{1}{x+y}}$

Answer: B

Step-by-step solution

$\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{y}} \rightleftharpoons \mathrm{xA}^{\mathrm{y}+}+\mathrm{yB}^{\mathrm{x}-}$ $\mathrm{C}(1-\alpha) \quad \mathrm{xC} \alpha \quad \mathrm{yC} \alpha$ $\mathrm{K}=\frac{(\mathrm{xC} \alpha)^{\mathrm{x}}(\mathrm{xC} \alpha)^{\mathrm{y}}}{\mathrm{C}(1-\alpha)}$ As $\alpha \ll 1 \Rightarrow(1-\alpha) \simeq 1$ $K=x^{x} y^{y} C^{x+y-1} \alpha^{x+y}$ $\alpha=\left(\frac{K}{C^{x+y-1} x^{x} y^{y}}\right)^{\frac{1}{x+y}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Fundamental Definitions