JEE Main 2026 Apr 6 Shift 2, Chemistry Q6: Ionization Of Acids And Bases

JEE Main2026Apr 6, Shift 2Chemistry

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:

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

Solution

$A_{x} B_{y} ⇌ x A^{y +} + y B^{x -}$

At equilibrium concentrations: $[A_{x} B_{y}] = c (1 - α)$ , $[A] = x c α$ , $[B] = y c α$ .

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

For weak electrolytes $α ≪ 1$ , so $1 - α \approx 1$ :

$K = x^{x} y^{y} c^{x + y - 1} α^{x + y}$ .

Solving for $α$ :

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

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JEE Main 2026 Apr 6 Shift 2, Chemistry Q6: Ionization Of Acids And Bases

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