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Question:

Λom for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm²mol⁻¹, respectively. If the conductivity of 0.001 M HA is 5 × 10⁻³ S cm⁻¹, degree of dissociation of HA is:

0.75

0.125

0.25

0.50

Solution:

Λ⁰m(HA) = Λ⁰m(HCl) + Λ⁰m(NaA) - Λ⁰m(NaCl) = 425.9 + 100.5 - 126.4 = 400 S cm²mol⁻¹
Λm = 1000κ/M = 1000 × 5 × 10⁻³/0.001 = 50 S cm²mol⁻¹
α = Λm/Λ⁰m = 50/400 = 0.125