StemCET Logo

Chemistry Question 52 – JEE-MAIN 2026

← Back
The first and second ionization constants of a weak dibasic acid H2A are 8.1×10−8 and 1.0×10−13 respectively. 0.1 mol of H2A was dissolved in 1L of 0.1 M HCl solution. The concentration of HA− in the resultant solution is :

The presence of a strong acid (HCl) will significantly suppress the ionization of the weak dibasic acid (H2A) due to the common ion effect.

Step 1: Identify Initial Conditions and Dominant Species✦ Active

The solution contains 0.1 M H2A and 0.1 M HCl. Since HCl is a strong acid, it completely dissociates, providing an initial [H+] of 0.1 M. The ionization of the weak acid H2A will be suppressed by this high initial [H+] (common ion effect).

Step 2: Apply First Ionization Equilibrium○ Expand

Consider the first ionization of H2A: H2A⇌H++HA−. The equilibrium constant is Ka1=[H+][HA−][H2A]=8.1×10−8. Let x=[HA−] at equilibrium. Initial concentrations: [H2A]=0.1 M, [H+]=0.1 M (from HCl). At equilibrium, due to the common ion effect and small Ka1, we can approximate: [H2A]≈0.1 M and [H+]≈0.1 M. Substituting these into the Ka1 expression:

8.1×10−8=(0.1)(x)0.1

Solving for x:

x=[HA−]=8.1×10−8 M
Step 3: Verify Approximation and Consider Second Ionization○ Expand

The value of x=8.1×10−8 M is indeed much smaller than 0.1 M, validating the approximation. The second ionization, HA−⇌H++A2−, has Ka2=1.0×10−13, which is significantly smaller than Ka1. This means the dissociation of HA− into A2− will be even more negligible and will not significantly affect the concentration of HA−. Therefore, the concentration of HA− in the resultant solution is 8.1×10−8 M.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.