JEE Main 2026 — Jan 24, Shift 2 — Chemistry solution

JEE Main2026Jan 24, Shift 2Chemistry

Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 1/30 of the molar conductivity of another weak acid HZ with concentration of 0.02 of M. If $λ_{Q^{-}}^{0}$ happened to be equal with $λ_{Z^{-}}^{0}$ , then the difference between the $pK_{a}$ values of the two weak acids $(pK_{a} (HQ) - pK_{a} (HZ))$ is ________ (Nearest integer).

[Given : degree of dissociation $(α) ≪ 1$ for both weak acids, $λ^{o}$ : limiting molar conductivity of ions]

Your answer

Solution

For a weak acid, $α = \frac{λ _{m}}{λ _{m}^{\infty}}$ and $K_{a} = C α^{2}$ when $α ≪ 1$ .

Since $λ_{Q^{-}}^{0} = λ_{Z^{-}}^{0}$ (and $λ_{H^{+}}^{0}$ is common), the limiting molar conductivities are equal: $λ_{m}^{\infty} (HQ) = λ_{m}^{\infty} (HZ)$ .

$\frac{K _{a} ( HQ )}{K _{a} ( HZ )} = \frac{C _{1} α _{1}^{2}}{C _{2} α _{2}^{2}} = \frac{C _{1}}{C _{2}} \cdot (\frac{λ _{m} ( HQ )}{λ _{m} ( HZ )})^{2} = \frac{0.18}{0.02} \cdot (\frac{1}{30})^{2} = 9 \cdot \frac{1}{900} = \frac{1}{100}$ .

$pK_{a} (HQ) - pK_{a} (HZ) = - lo g (\frac{1}{100}) = 2$ .

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