See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

HCOOH HCOO H         1 2 a HCOO H H K HCOOH HCOOH                3 3 CH COOH CH COO H         2 2 3 a 3 3 CH COO H H K CH COOH CH COOH                As H    in both case are equal     1 2 a a 3 K HCOOH K CH COOH        1 2 4 a 3 3 5 a K