Chemistry Q56 – JEE-MAIN 2026

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 :

Clue 1 of 3: Dissociation of a weak electrolyte

For a general weak electrolyte $A_{x} B_{y}$ , it dissociates into $x$ cations and $y$ anions.

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Step 1: Write the Dissociation Equilibrium and Concentrations✦ Active

The dissociation of the weak electrolyte $A_{x} B_{y}$ can be represented as:

A_{x} B_{y} ⇌ x A^{y +} + y B^{x -}

If the initial concentration of $A_{x} B_{y}$ is $c$ and the degree of dissociation is $α$ , then at equilibrium, the concentrations are:

[A_{x} B_{y}] = c (1 - α)

[A^{y +}] = c x α

[B^{x -}] = c y α

Step 2: Formulate the Dissociation Constant Expression○ Expand

The dissociation constant $K$ is given by the expression:

K = \frac{[A^{y +}]^{x} [B^{x -}]^{y}}{[A_{x} B_{y}]}

Substitute the equilibrium concentrations into the expression:

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

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

Step 3: Simplify for Weak Electrolyte and Solve for Degree of Dissociation○ Expand

For a weak electrolyte, the degree of dissociation $α$ is very small (i.e., $α ≪ 1$ ). Therefore, we can make the approximation $1 - α \approx 1$ . The expression for $K$ simplifies to:

K \approx c^{x + y - 1} x^{x} y^{y} α^{x + y}

Now, rearrange the equation to solve for $α$ :

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

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

This matches option 2.

💡 Teacher's Secret Hint

Remember to apply the weak electrolyte approximation ( $1 - α \approx 1$ ) to simplify the calculation for $α$ .

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Chemistry Question 56 – JEE-MAIN 2026

Clue 1 of 3: Dissociation of a weak electrolyte

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