For 0.01 M acetic acid at 25 °C, $\Lambda_c = 16.5$ Ω⁻¹ cm² mol⁻¹ and $\Lambda_0 = 390.7$ Ω⁻¹ cm² mol⁻¹. The dissociation constant $K_a$ is approximately:
A$1.85 \times 10^{-7}$ M
B$1.85 \times 10^{-3}$ M
C$4.22 \times 10^{-2}$ M
D$1.85 \times 10^{-5}$ M
Answer & Solution
Correct answer: D. $1.85 \times 10^{-5}$ M
$\alpha = \Lambda_c/\Lambda_0 = 16.5/390.7 \approx 0.0422$. $K_a = \alpha^2 c/(1-\alpha) \approx (0.0422)^2 \times 0.01 / (1-0.0422) \approx 1.85 \times 10^{-5}$ M — the textbook value for acetic acid.
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