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JEE Main 2020 · Chemistry · Electrochemistry · hard

A conductivity cell of cell constant 0.40 cm^-1 filled with 0.010 M acetic acid shows a resistance of 2500 Ω. Given Lambdam^°(CH3COOH)=390 S cm^2…

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A conductivity cell of cell constant 0.40 cm−10.40\ \mathrm{cm^{-1}} filled with 0.010 M0.010\ \mathrm{M} acetic acid shows a resistance of 2500 Ω2500\ \Omega. Given Λm∘(CH3COOH)=390 S cm2 mol−1\Lambda_m^\circ(\mathrm{CH_3COOH})=390\ \mathrm{S\,cm^2\,mol^{-1}}, the dissociation constant of acetic acid is closest to
  1. A. 1.76×10−51.76\times10^{-5}
  2. B. 4.1×10−44.1\times10^{-4}
  3. C. 3.6×10−53.6\times10^{-5}
  4. D. 1.6×10−41.6\times10^{-4}
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Answer: (A) 1.76×10−51.76\times10^{-5}

κ=cell constantR=0.402500=1.6×10−4 S cm−1\kappa=\dfrac{\text{cell constant}}{R}=\dfrac{0.40}{2500}=1.6\times10^{-4}\ \mathrm{S\,cm^{-1}}. Λm=1000κC=0.160.010=16 S cm2 mol−1\Lambda_m=\dfrac{1000\kappa}{C}=\dfrac{0.16}{0.010}=16\ \mathrm{S\,cm^2\,mol^{-1}}. α=Λm/Λm∘=16/390=0.0410\alpha=\Lambda_m/\Lambda_m^\circ=16/390=0.0410. Ka=Cα21−α=0.010×1.683×10−30.959=1.76×10−5K_a=\dfrac{C\alpha^2}{1-\alpha}=\dfrac{0.010\times1.683\times10^{-3}}{0.959}=1.76\times10^{-5}. The value Cα=4.1×10−4C\alpha=4.1\times10^{-4} is the ionic concentration, not KaK_a.
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