The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If λ⁰ₓ₋ ≈ λ⁰ᵧ₋, the difference in their pKₐ values, pKₐ(HX) - pKₐ(HY), is (consider degree of ionization of both acids to be <<1)
Answer: 3
Solution
Given λ⁰ₓ₋ ≈ λ⁰ᵧ₋, so λ⁰ₕₓ ≈ λ⁰ₕᵧ.
Molar conductivity λₘ = λ⁰ₘα, where α is the degree of dissociation.
Given: λₘ(HY) = 10λₘ(HX)
So: λ⁰α₂ = 10λ⁰α₁, which gives α₂ = 10α₁
For weak acids with α << 1: Kₐ = Cα²
Therefore:
Kₐ(HX)/Kₐ(HY) = (0.01α₁²)/(0.1α₂²) = (0.01/0.1) × (1/10)² = 1/1000
Taking logarithm:
log Kₐ(HX) - log Kₐ(HY) = -3
pKₐ(HX) - pKₐ(HY) = 3.
Molar conductivity λₘ = λ⁰ₘα, where α is the degree of dissociation.
Given: λₘ(HY) = 10λₘ(HX)
So: λ⁰α₂ = 10λ⁰α₁, which gives α₂ = 10α₁
For weak acids with α << 1: Kₐ = Cα²
Therefore:
Kₐ(HX)/Kₐ(HY) = (0.01α₁²)/(0.1α₂²) = (0.01/0.1) × (1/10)² = 1/1000
Taking logarithm:
log Kₐ(HX) - log Kₐ(HY) = -3
pKₐ(HX) - pKₐ(HY) = 3.
JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Integer answer