The first and second ionization constants of a weak dibasic acid $\ce{H2A}$ are $8.1 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. 0.1 mol of $\ce{H2A}$ was dissolved in 1L of 0.1 M HCl solution. The concentration of $\ce{HA-}$ in the resultant solution is:

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Equilibrium

The first and second ionization constants of a weak dibasic acid

H X_{2} A

are

8.1 \times 1 0^{- 8}

and

1.0 \times 1 0^{- 13}

respectively. 0.1 mol of

H X_{2} A

was dissolved in 1L of 0.1 M HCl solution. The concentration of

HA X^{-}

in the resultant solution is:

Answer: (C)

Dissolving

H X_{2} A

0.1

M HCl supplies a large common-ion concentration of

H X^{+}

, so the first dissociation

H X_{2} A H X^{+} + HA X^{-}

is strongly suppressed and the amounts consumed are negligible, leaving

[H X_{2} A] \approx 0.1

M and

[H X^{+}] \approx 0.1

M. From the first ionization constant,

K_{a 1} = \frac{[ H X ^{+} ] [ HA X ^{-} ]}{[ H X _{2} A ]} = \frac{( 0.1 ) [ HA X ^{-} ]}{0.1} = [HA X^{-}] .

The second dissociation is far too weak to add appreciably to

HA X^{-}

. Hence the concentration of

HA X^{-}

equals

K_{a 1}

, which is

8.1 \times 1 0^{- 8}

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option