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JEE Main Chemistry Electrochemistry 2026 JEE Main 2026 (24 January Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

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 M. If ^ Q ^ - happened to be equal with ^ Z ^ - , then the difference of the pK _ a values of the two weak acids ( pK _ a ( HQ )- pK _ a ( HZ ) ) is \_\_\_\_ (Nearest integer). [Given: degree of dissociation ( ) 1 for both weak acids, ^ : limiting molar conductivity of ions]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

For weak acids with ^ Q^- = ^ Z^-, the molar conductivity is _m = ^ . From _m^Q = 1 30 _m^Z, we get _Q = _Z 30 . Using K_a = ^2 C for weak acids: K_a(HQ) = _Q^2 0.18 = _Z^2 0.18 900 and K_a(HZ) = _Z^2 0.02. Therefore K_a(HZ) K_a(HQ) = 0.02 900 0.18 = 100. Thus pK_a(HQ) - pK_a(HZ) = (100) = 2.

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