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The molar conductivity of 0.001028 M acetic acid is 48.15 S cm² mol⁻¹, and Λ°_m(HAc) = 390.5 S cm² mol⁻¹. The dissociation constant K_a is approximately:
A1.78 × 10⁻⁵
B1.78 × 10⁻³
C0.123
D1.0 × 10⁻⁷
Answer & Solution
Correct answer: A. 1.78 × 10⁻⁵
α = 48.15/390.5 = 0.1233. K_a = cα²/(1−α) = (0.001028 × 0.1233²)/(1 − 0.1233) ≈ (0.001028 × 0.01520)/0.8767 ≈ **1.78 × 10⁻⁵** — the textbook value for acetic acid.
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