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 of the pK a values of the two weak acids ( pK a ( HQ )- pK a ( HZ ) ) is \ \ \ \ (Nearest integer). [Given : degree of dissociation (α) ll 1 for both weak acids, λ ° : limiting molar conductivity of ions]

Chemistry · Electrochemistry

JEE Main 2026 — 24 January, Evening Shift — Question 70

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 $\lambda_{\mathrm{Q}^{-}}^{0}$ happened to be equal with $\lambda_{\mathrm{z}^{-}}^{0}$ , 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]

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{K}_{\mathrm{a}}(\mathrm{HQ})=\mathrm{C}_{1} \alpha_{1}^{2}$

\alpha_{1}=\frac{\lambda_{\mathrm{m}}(\mathrm{HQ})}{\lambda_{\mathrm{m}}^{\infty}(\mathrm{HQ})}

$\mathrm{K}_{\mathrm{a}}(\mathrm{HZ})=\mathrm{C}_{2} \alpha_{2}{ }^{2}$

\alpha_{2}=\frac{\lambda_{\mathrm{m}}(\mathrm{HZ})}{\lambda_{\mathrm{m}}^{\infty}(\mathrm{HZ})}

$\frac{\mathrm{K}_{\mathrm{a}}(\mathrm{HQ})}{\mathrm{K}_{\mathrm{a}}(\mathrm{HZ})}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}} \cdot\left(\frac{\alpha_{1}}{\alpha_{2}}\right)^{2}=\frac{0.18}{0.02} \cdot\left[\frac{\lambda_{\mathrm{m}}(\mathrm{HQ})}{\lambda_{\mathrm{m}}(\mathrm{HZ})}\right]^{2}$ $\frac{\mathrm{K}_{\mathrm{a}}(\mathrm{HQ})}{\mathrm{K}_{\mathrm{a}}(\mathrm{HQ})}=9 \times\left(\frac{1}{30}\right)^{2}=\frac{1}{100}$ $\mathrm{pK}_{\mathrm{a}}(\mathrm{HQ})-\mathrm{pK}_{\mathrm{a}}(\mathrm{HZ})=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Evening Shift
Subject: Chemistry
Chapter: Electrochemistry
Topic: Conductance of Solutions and Kohlrausch's Law