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JEE Main 2018 · Chemistry · Electrochemistry · medium

The conductivity of a 0.02 M solution of a weak monobasic acid HA at 298 K is 4.0×10^-4 S cm^-1, and the limiting molar conductivity of HA is 400 S…

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The conductivity of a 0.02 M0.02\ M solution of a weak monobasic acid HA at 298 K298\ \mathrm{K} is 4.0×10−4 S cm−14.0\times10^{-4}\ \mathrm{S\,cm^{-1}}, and the limiting molar conductivity of HA is 400 S cm2 mol−1400\ \mathrm{S\,cm^2\,mol^{-1}}. The acid dissociation constant KaK_a of HA is closest to
  1. A. 2.5×10−32.5\times10^{-3}
  2. B. 5.0×10−55.0\times10^{-5}
  3. C. 5.3×10−55.3\times10^{-5}
  4. D. 1.0×10−31.0\times10^{-3}
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Answer: (C) 5.3×10−55.3\times10^{-5}

Λm=κ×1000c=4.0×10−4×10000.02=20 S cm2 mol−1\Lambda_m=\dfrac{\kappa\times1000}{c}=\dfrac{4.0\times10^{-4}\times1000}{0.02}=20\ \mathrm{S\,cm^2\,mol^{-1}}. Degree of dissociation α=Λm/Λm∘=20/400=0.05\alpha=\Lambda_m/\Lambda_m^\circ=20/400=0.05. For HA, Ka=cα21−α=0.02×0.00250.95=5.26×10−5≈5.3×10−5K_a=\dfrac{c\alpha^2}{1-\alpha}=\dfrac{0.02\times0.0025}{0.95}=5.26\times10^{-5}\approx5.3\times10^{-5}. The simpler cα2=5.0×10−5c\alpha^2=5.0\times10^{-5} ignores the (1−α)(1-\alpha) term and is not the closest value when α=5%\alpha=5\%.
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