Question

0.18 M HQ solution has molar conductivity $\frac{1}{30}$ times the molar conductivity of 0.02 M HZ solution. Find the value of pK$_a$(HQ) − pK$_a$(HZ).
[Given that $\alpha$ is very less than 1]

Hint:

Lower molar conductivity implies lower degree of dissociation and hence weaker acid strength.

Answer 1

Correct Answer: 2

Step 1: Express dissociation constant using Ostwald’s dilution law

For a weak acid HA of concentration C:

K_a = C · (α / (1 − α))²

Since the acids are weak, α ≪ 1, hence:

K_a ≈ Cα²

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Step 2: Relate degree of dissociation to molar conductivity

Degree of dissociation:

α = λ_m / λ_m⁰

Given:

λ_m⁰(Q⁺) = λ_m⁰(Z⁺)

Hence, λ_m⁰ for HQ and HZ are equal.

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Step 3: Form ratio of dissociation constants

K_a(HQ) / K_a(HZ)

= [C_HQ α_HQ²] / [C_HZ α_HZ²]

= (C_HQ / C_HZ) · (α_HQ / α_HZ)²

= (C_HQ / C_HZ) · (λ_m(HQ) / λ_m(HZ))²

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Step 4: Substitute the given values

= (0.18 / 0.02) · (1 / 30)²

= 9 × 1/900

= 1/100

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Step 5: Convert ratio into pK difference

pK_a(HQ) − pK_a(HZ)

= log(100)

= 2

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Final Answer:

pK_a(HQ) − pK_a(HZ) = 2