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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2026 · 24 January · Evening Shift · Chemistry · Electrochemistry

Molar conductivity of a weak acid HQ of concentration 0.18 M was...

Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 1/301 / 30 of the molar conductivity of another weak acid HZ with concentration of 0.02 M. If λ∘Q−\lambda^{\circ} \mathrm{Q}^{-}happened to be equal with λ∘Z−\lambda^{\circ} \mathrm{Z}^{-}, then the difference of the pKa\mathrm{pK}_{\mathrm{a}} values of the two weak acids (pKa(HQ)−pKa(HZ))\left(\mathrm{pK}_{\mathrm{a}}(\mathrm{HQ})-\mathrm{pK}_{\mathrm{a}}(\mathrm{HZ})\right) is ____\_\_\_\_ (Nearest integer).
[Given: degree of dissociation (α)≪1(\alpha) \ll 1 for both weak acids, λ∘\lambda^{\circ} : limiting molar conductivity of ions]

Correct answer

0

Step-by-step solution

For weak acids with λ∞Q−=λ∞Z−\lambda^{\infty}Q^- = \lambda^{\infty}Z^-, the molar conductivity is λm=αλ∞\lambda_m = \alpha \lambda^{\infty}.
From λmQ=130λmZ\lambda_m^Q = \frac{1}{30}\lambda_m^Z, we get αQ=αZ30\alpha_Q = \frac{\alpha_Z}{30}.
Using Ka=α2CK_a = \alpha^2 C for weak acids: Ka(HQ)=αQ2×0.18=αZ2×0.18900K_a(HQ) = \alpha_Q^2 \times 0.18 = \frac{\alpha_Z^2 \times 0.18}{900} and Ka(HZ)=αZ2×0.02K_a(HZ) = \alpha_Z^2 \times 0.02.
Therefore Ka(HZ)Ka(HQ)=0.02×9000.18=100\frac{K_a(HZ)}{K_a(HQ)} = \frac{0.02 \times 900}{0.18} = 100.
Thus pKa(HQ)−pKa(HZ)=log⁡(100)=2pK_a(HQ) - pK_a(HZ) = \log(100) = 2.
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