The first and second ionization constants of a weak dibasic acid H2A are 8.1 x 10^-8 and 1.0 x 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 correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q52 – JEE-MAIN 2026

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Home Chemistry EquilibriumQuestion #415

The first and second ionization constants of a weak dibasic acid

H_{2} A

are

8.1 \times 10^{- 8}

and

1.0 \times 10^{- 13}

respectively.

0.1

mol of

H_{2} A

was dissolved in

1 L

0.1

M HCl solution. The concentration of

H A^{-}

in the resultant solution is :

Clue 1 of 3: Strong Acid Effect

The presence of a strong acid (HCl) will significantly suppress the ionization of the weak dibasic acid ( $H_{2} A$ ) due to the common ion effect.

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Step 1: Identify Initial Conditions and Dominant Species✦ Active

The solution contains $0.1 M H_{2} A$ and $0.1 M H C l$ . Since $H C l$ is a strong acid, it completely dissociates, providing an initial $[H^{+}]$ of $0.1 M$ . The ionization of the weak acid $H_{2} A$ will be suppressed by this high initial $[H^{+}]$ (common ion effect).

Step 2: Apply First Ionization Equilibrium○ Expand

Consider the first ionization of $H_{2} A$ : $H_{2} A ⇌ H^{+} + H A^{-}$ . The equilibrium constant is $K_{a 1} = \frac{[H^{+}] [H A^{-}]}{[H_{2} A]} = 8.1 \times 10^{- 8}$ . Let $x = [H A^{-}]$ at equilibrium. Initial concentrations: $[H_{2} A] = 0.1 M$ , $[H^{+}] = 0.1 M$ (from $H C l$ ). At equilibrium, due to the common ion effect and small $K_{a 1}$ , we can approximate: $[H_{2} A] \approx 0.1 M$ and $[H^{+}] \approx 0.1 M$ . Substituting these into the $K_{a 1}$ expression:

8.1 \times 10^{- 8} = \frac{(0.1) (x)}{0.1}

Solving for $x$ :

x = [H A^{-}] = 8.1 \times 10^{- 8} M

Step 3: Verify Approximation and Consider Second Ionization○ Expand

The value of $x = 8.1 \times 10^{- 8} M$ is indeed much smaller than $0.1 M$ , validating the approximation. The second ionization, $H A^{-} ⇌ H^{+} + A^{2 -}$ , has $K_{a 2} = 1.0 \times 10^{- 13}$ , which is significantly smaller than $K_{a 1}$ . This means the dissociation of $H A^{-}$ into $A^{2 -}$ will be even more negligible and will not significantly affect the concentration of $H A^{-}$ . Therefore, the concentration of $H A^{-}$ in the resultant solution is $8.1 \times 10^{- 8} M$ .

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Chemistry Question 52 – JEE-MAIN 2026

Clue 1 of 3: Strong Acid Effect

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