See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

2 2 Amphiprotic anionic salt H A NaOH NaHA H O    2 2 NaHA NaOH Na A H O    As from titration curve 2 V  volume of strong base required at 1st equivalence point i.e point B. And V4 = volume of strong base required at 2nd equivalence point i.e. point D. Hence at 2 2 1 3 V 3V V and V 2 2   indicating 1st half and 2nd half equivalence point.  Therefore at 1st half equivalence point     1 a 2 NaHA 1, pH pK H A   pH = 4 and similarly, at 2nd half equivalence point at C,     2 2 a Na A 1, pH pK NaHA   At point D, only salt Na2A will exist hence pH can be calculated by using salt hydrolysis phenomena. AITS-CRT-III (Paper-1)-PCM(Sol.)-JEE (Advanced)/2021 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 11 SECTION – C