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

(B) $\left(\frac{K}{c^{x+y-1} x^{x} y^{y}}\right)^{\frac{1}{x+y}}$

(A) $\left[K \times c^{x+y-1} x^{x} y^{y}\right]^{x+y}$

(C) $\left(\frac{c^{x+y-1} x^{x} y^{y}}{K}\right)^{x+y}$

(D) $\left(\frac{c^{x+y-1} x^{x} y^{y}}{K}\right)^{\frac{1}{x+y}}$

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Equilibrium

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:

Answer: (B)

For

A_{x} B_{y} ⇌ x A^{y +} + y B^{x -}

at concentration

c

and degree of dissociation

α

, the equilibrium concentrations are

[A^{y +}] = x c α

[B^{x -}] = y c α

and, for a weak electrolyte,

[A_{x} B_{y}] = c (1 - α) \approx c

. Hence

K = \frac{( x c α ) ^{x} ( y c α ) ^{y}}{c} = x^{x} y^{y} c^{x + y - 1} α^{x + y}

, and solving for the degree of dissociation gives

α = (\frac{K}{c ^{x + y - 1} x ^{x} y ^{y}})^{\frac{1}{x + y}}

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option