JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

Given: Weak dibasic acid H_2A with K_ a1 = 8.1 10^ -8 and K_ a2 = 1.0 10^ -13 Initial concentration: [H_2A]_0 = 0.1 M The solution also contains 0.1 M HCl , which dissociates completely to give [H^+]_0 = 0.1 M . The main source of HA^- is the first dissociation: H_2A H^+ + HA^- Let x be the amount of H_2A that dissociates. Then at equilibrium: [H_2A] = 0.1 - x [H^+] = 0.1 + x [HA^-] = x Applying the equilibrium expression: K_ a1 = [H^+][HA^-] [H_2A] 8.1 10^ -8 = (0.1 + x)(x) 0.1 - x Since K_ a1 is very small and the common H^+ ion from HCl suppresses dissociation, x 0.1. Hence: 0.1 - x 0.1 and 0.1 + x 0.1 8.1 10^ -8 (0.1)(x) 0.1 = x Therefore: [HA^-] = x = 8.1 10^ -8 M The second dissociation (K_ a2 = 1.0 10^ -13 ) is negligible compared to the first, so the HA^- consumed in that step is insignificant. Hence, the correct option is (3)\ 8.1 10^ -8 M .