See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

pH = – In [In ] pk log [InH]  5 = In 75 pk log 25  pkIn = 4.523 pH = 4.523 + 10 log 90 = 4.523 – .954 = 3.56 AITS-FT-IX (Paper-2)-PCM(Sol.)-JEE(Advanced)/19 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 14 Mathematics PART – III SECTION – A