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JEE Main 2016 · Chemistry · Electrochemistry · hard

At 298 K the limiting molar conductivities are Lambda^°m(CH3COONa)=91, Lambda^°m(HCl)=426 and Lambda^°m(NaCl)=126.5 (all in S cm^2 mol^-1). A 0.010 M…

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At 298 K298\ \mathrm{K} the limiting molar conductivities are Λm∘(CH3COONa)=91\Lambda^\circ_m(\mathrm{CH_3COONa})=91, Λm∘(HCl)=426\Lambda^\circ_m(\mathrm{HCl})=426 and Λm∘(NaCl)=126.5\Lambda^\circ_m(\mathrm{NaCl})=126.5 (all in S cm2 mol−1\mathrm{S\,cm^2\,mol^{-1}}). A 0.010 M0.010\ \mathrm{M} solution of acetic acid has conductivity 1.6×10−4 S cm−11.6\times10^{-4}\ \mathrm{S\,cm^{-1}}. The dissociation constant of acetic acid is nearly
  1. A. 4.1×10−44.1\times10^{-4}
  2. B. 1.75×10−51.75\times10^{-5}
  3. C. 1.7×10−31.7\times10^{-3}
  4. D. 1.75×10−71.75\times10^{-7}
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Answer: (B) 1.75×10−51.75\times10^{-5}

By Kohlrausch's law, Λm∘(CH3COOH)=91+426−126.5=390.5 S cm2 mol−1\Lambda^\circ_m(\mathrm{CH_3COOH})=91+426-126.5=390.5\ \mathrm{S\,cm^2\,mol^{-1}}. The molar conductivity of the solution is Λm=1000κ/c=1000(1.6×10−4)/0.010=16.0\Lambda_m=1000\kappa/c=1000(1.6\times10^{-4})/0.010=16.0. Degree of dissociation α=16.0/390.5=0.0410\alpha=16.0/390.5=0.0410. Ka=cα21−α=0.010(0.0410)20.959=1.75×10−5K_a=\dfrac{c\alpha^2}{1-\alpha}=\dfrac{0.010(0.0410)^2}{0.959}=1.75\times10^{-5}. Using Ka=αcK_a=\alpha c gives 4.1×10−44.1\times10^{-4} and Ka=α2K_a=\alpha^2 gives 1.7×10−31.7\times10^{-3}.
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Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake