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Chemistry · Electrochemistry

JEE Main 2024 — 9 April, Shift 2 — Question 71

Which out of the following is a correct equation to show change in molar conductivity with respect to concentration for a weak electrolyte, if the symbols carry their usual meaning :

  1. Option A:

    Λm2C−KaΛm∘2+KaΛmΛm∘=0\Lambda_{\mathrm{m}}^{2} \mathrm{C}-\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ 2}+\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ}=0

  2. Option B:

    Λm−Λm∘+AC12=0\Lambda_{m}-\Lambda_{m}^{\circ}+A C^{\frac{1}{2}}=0

  3. Option C:

    Λm−Λm∘−AC12=0\Lambda_{m}-\Lambda_{m}^{\circ}-A C^{\frac{1}{2}}=0

  4. Option D:

    Λ2mC+KaΛm∘2−KaΛmΛm∘=0\Lambda^{2}{ }_{\mathrm{m}} \mathrm{C}+\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ 2}-\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ}=0

    Correct

Answer: D

Step-by-step solution

HA(aq)⇌H+(aq)+A−(aq)\mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})

Ka=α2C1−α\mathrm{K}_{\mathrm{a}}=\frac{\alpha^{2} \mathrm{C}}{1-\alpha}

α2C+Kaα−Ka=0\alpha^{2} \mathrm{C}+\mathrm{K}_{\mathrm{a}} \alpha-\mathrm{K}_{\mathrm{a}}=0

(λmλm∞)2C+Kaλmλm∞−Ka=0\left(\frac{\lambda_{m}}{\lambda_{m}^{\infty}}\right)^{2} \mathrm{C}+\mathrm{K}_{\mathrm{a}} \frac{\lambda_{m}}{\lambda_{m}^{\infty}}-\mathrm{K}_{\mathrm{a}}=0

λm2C+Kaλmλm∞−Ka(λm∞)2=0\lambda_{\mathrm{m}}^{2} \mathrm{C}+\mathrm{K}_{\mathrm{a}} \lambda_{\mathrm{m}} \lambda_{\mathrm{m}}^{\infty}-\mathrm{K}_{\mathrm{a}}\left(\lambda_{\mathrm{m}}^{\infty}\right)^{2}=0

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 2
Subject
Chemistry
Chapter
Electrochemistry
Topic
Conductance of Solutions and Kohlrausch's Law

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