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JEE Main 2019 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

A 0.20 m aqueous solution of a weak monobasic acid HA freezes at -0.465 ^circC (Kf = 1.86 K kg mol^-1). Treating molarity as equal to molality, the…

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A 0.20 m0.20\ \mathrm{m} aqueous solution of a weak monobasic acid HA freezes at −0.465 ∘C-0.465\,^\circ\mathrm{C} (Kf=1.86 K kg mol−1K_f = 1.86\ \mathrm{K\,kg\,mol^{-1}}). Treating molarity as equal to molality, the acid dissociation constant of HA is nearest to
  1. A. 1.7×10−21.7\times10^{-2}
  2. B. 4.0×10−24.0\times10^{-2}
  3. C. 8.3×10−38.3\times10^{-3}
  4. D. 2.5×10−12.5\times10^{-1}
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Answer: (A) 1.7×10−21.7\times10^{-2}

ΔTf=iKfm\Delta T_f = iK_f m gives i=0.465/(1.86×0.20)=1.25i = 0.465/(1.86\times0.20) = 1.25. For HA⇌H++A−\mathrm{HA \rightleftharpoons H^+ + A^-}, i=1+αi = 1+\alpha, so α=0.25\alpha = 0.25. Then Ka=Cα21−α=0.20×0.06250.75=1.67×10−2K_a = \dfrac{C\alpha^2}{1-\alpha} = \dfrac{0.20\times0.0625}{0.75} = 1.67\times10^{-2}. The common error is Ka=Cα2=1.25×10−2K_a = C\alpha^2 = 1.25\times10^{-2} (ignoring 1−α1-\alpha), which is not adequate since α\alpha is large.
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