Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 1/30 of the molar conductivity of another weak acid HZ with concentration of 0.02 M. If λ°Q⁻ happened to be equal with λ°Z⁻, then the difference of the pKₐ values of the two weak acids (pKₐ(HQ) − pKₐ(HZ)) is ____ (Nearest integer). [Given: degree of dissociation (α) ≪ 1 for both weak acids, λ° : limiting molar conductivity of ions]

Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 1/30 of the molar conductivity of another weak acid HZ with concentration of 0.02 M

Q. Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 1/30 of the molar conductivity of another weak acid HZ with concentration of 0.02 M. If λ°Q⁻ happened to be equal with λ°Z⁻, then the difference of the pK_a values of the two weak acids (pK_a(HQ) − pK_a(HZ)) is ____ (Nearest integer).

[Given: degree of dissociation (α) ≪ 1 for both weak acids, λ° : limiting molar conductivity of ions]

Correct Answer: 2

Explanation

For a weak acid, molar conductivity is given by:

Λ_m = α Λ°

Given that λ°Q⁻ = λ°Z⁻, the limiting molar conductivities of HQ and HZ are equal. Hence:

Λ_HQ / Λ_HZ = α_HQ / α_HZ

According to the question:

Λ_HQ = (1/30) Λ_HZ

So,

α_HQ / α_HZ = 1/30

For weak acids (α ≪ 1),

K_a = C α²

Therefore,

K_a(HQ) / K_a(HZ) = [0.18 × α_HQ²] / [0.02 × α_HZ²]

Substitute α_HQ / α_HZ = 1/30:

K_a(HQ) / K_a(HZ) = (0.18 / 0.02) × (1/30)²

= 9 × 1/900 = 1/100

Taking negative logarithm:

pK_a(HQ) − pK_a(HZ) = −log(1/100) = 2

Hence, the required difference is 2.

Explanation

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