See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

2 2 H A OH HA H O      2 2 HA OH A H O       AITS-OT-II (Paper-1)-PCM(Sol.)-JEE(Advanced)/21 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 12 At first equivalent point, volume of base used is V1 and at second equivalence point, it is V2. So, V2 = 2V1. When, 1 3V V , 2  conc. of HAand 2 A will be equal and a buffer will be formed, so, 2 a pH pK  . Also, when 1 V V , 2  first buffer will be formed containing H2A and Ain equal concentration. So, 1 a pH pK  .