The specific conductance of 0.0025 M acetic acid is 5 × 10–5 S cm–1 at a certain temperature. The di — JEE Mains Chemistry Past Papers Chemistry Question
Question
The specific conductance of 0.0025 M acetic acid is 5 × 10–5 S cm–1 at a certain temperature. The dissociation constant of acetic acid is ___________ × 10–7. (Nearest integer) Consider limiting molar conductivity of CH3COOH as 400 S cm2 mol–1
Answer: .
💡 Solution & Explanation
m = M C K x = . 10 x x 5 = 20 S cm2 mol–1 = 400 20 = 20 Ka = 2 C = 66 x 10–7
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