Calculate of acetic acid if its 0.05 M solution has molar conductivity of at 25°C. Given: for CH3COO — Electrochemistry Chemistry Question

💡 Solution & Explanation

Degree of dissociation $\alpha = \Lambda_m / \Lambda_m^\circ = (7.814 \times 10^{–4}) / (3.907 \times 10^{–2}) = 2 \times 10^{–2} = 0.02$. Weak acid dissociation constant $K_a \approx C\alpha^2 = 0.05 \times (0.02)^2 = 0.05 \times 0.0004 = 2 \times 10^{–5}$. Therefore, correct answer is A.