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JEE Advanced 2021 · Paper 2, Chemistry question 7

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Question Stem for Question Nos. 7 and 8

At 298 K, the limiting molar conductivity of a weak monobasic acid is 4×1024 \times 10^2 S cm2^2 mol−1^{-1}. At 298 K, for an aqueous solution of the acid the degree of dissociation is α\alpha and the molar conductivity is y×102\mathbf y \times 10^2 S cm2^2 mol−1^{-1}. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y×1023\mathbf y \times 10^2 S cm2^2 mol−1^{-1}.

The value of α\alpha is ____.

Answer and worked solution

Official answer: 0.2

Degree of dissociation from conductivity. Λ=αΛ∘\Lambda=\alpha\Lambda^\circ, so before dilution y×102=400αy\times10^2=400\alpha. After 20-fold dilution, Λ=3y×102\Lambda=3y\times10^2, so the new degree of dissociation is α′=3α\alpha'=3\alpha.

Ostwald's dilution law. Ka=cα21−αK_a=\dfrac{c\alpha^2}{1-\alpha} is unchanged by dilution:
cα21−α=(c/20)(3α)21−3α  ⟹  20(1−3α)=9(1−α)  ⟹  α=1151≈0.22.\frac{c\alpha^2}{1-\alpha}=\frac{(c/20)(3\alpha)^2}{1-3\alpha}\implies20(1-3\alpha)=9(1-\alpha)\implies\alpha=\frac{11}{51}\approx0.22.

Topics:Conductance and Kohlrausch's lawpH and ionisation
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