Calculate Ka of acetic acid if its 0.05 M solution has molar conductivity of 7.814 × 10−4 Ω−1 m2 mol

Question

Calculate Ka of acetic acid if its 0.05 M solution has molar conductivity of 7.814 × 10−4 Ω−1 m2 mol−1 at 25°C. Given: $\Lambda_{m}^{o}$ for CH3COOH = 3.907 × 10−2 Ω−1 m2 mol−1.

Accepted Answer

$\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{o}} = \frac{7.814 \times 10^{- 4}}{3.907 \times 10^{- 2}} = 0.02$

Now, $K_{a} = \frac{\alpha^{2}.C}{1 - \alpha} \simeq \alpha^{2}.C = (0.02)^{2} \times 0.05 = 2 \times 10^{- 5}$

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