Q23 / 75 · JEE Main 2026 (24 January Shift 2) · Chemistry
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 $\lambda^{\circ} \mathrm{Q}^{-}$happened to be equal with $\lambda^{\circ} \mathrm{Z}^{-}$, then the difference of the $\mathrm{pK}_{\mathrm{a}}$ values of the two weak acids $\left(\mathrm{pK}_{\mathrm{a}}(\mathrm{HQ})-\mathrm{pK}_{\mathrm{a}}(\mathrm{HZ})\right)$ is $\_\_\_\_$ (Nearest integer).
[Given: degree of dissociation $(\alpha) \ll 1$ for both weak acids, $\lambda^{\circ}$ : limiting molar conductivity of ions]
[Given: degree of dissociation $(\alpha) \ll 1$ for both weak acids, $\lambda^{\circ}$ : limiting molar conductivity of ions]
Correct answer: 2
Solution
For weak acids with $\lambda^{\infty}Q^- = \lambda^{\infty}Z^-$, the molar conductivity is $\lambda_m = \alpha \lambda^{\infty}$.
From $\lambda_m^Q = \frac{1}{30}\lambda_m^Z$, we get $\alpha_Q = \frac{\alpha_Z}{30}$.
Using $K_a = \alpha^2 C$ for weak acids: $K_a(HQ) = \alpha_Q^2 \times 0.18 = \frac{\alpha_Z^2 \times 0.18}{900}$ and $K_a(HZ) = \alpha_Z^2 \times 0.02$.
Therefore $\frac{K_a(HZ)}{K_a(HQ)} = \frac{0.02 \times 900}{0.18} = 100$.
Thus $pK_a(HQ) - pK_a(HZ) = \log(100) = 2$.
From $\lambda_m^Q = \frac{1}{30}\lambda_m^Z$, we get $\alpha_Q = \frac{\alpha_Z}{30}$.
Using $K_a = \alpha^2 C$ for weak acids: $K_a(HQ) = \alpha_Q^2 \times 0.18 = \frac{\alpha_Z^2 \times 0.18}{900}$ and $K_a(HZ) = \alpha_Z^2 \times 0.02$.
Therefore $\frac{K_a(HZ)}{K_a(HQ)} = \frac{0.02 \times 900}{0.18} = 100$.
Thus $pK_a(HQ) - pK_a(HZ) = \log(100) = 2$.
Hence, the correct answer is 2.