JEE Mains Chemistry Past PapershardNUMERICAL

The specific conductance of 0.0025 M acetic acid is 5 × 10–5 S cm–1 at a certain temperature. The diJEE 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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