The first and second ionization constants of $\mathrm{H}_{2} \mathrm{X}$ are $2.5 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. The concentration of $\mathrm{X}^{2-}$ in $0.1 \mathrm{M}$ $\mathrm{H}_{2} \mathrm{X}$ solution is $\_\_\_\_$ $\times 10^{-15} \mathrm{M}$. (Nearest Integer)

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Equilibrium

The first and second ionization constants of

H_{2} X

are

2.5 \times 1 0^{- 8}

and

1.0 \times 1 0^{- 13}

respectively. The concentration of

X^{2 -}

0.1 M

H_{2} X

solution is

____

\times 1 0^{- 15} M

. (Nearest Integer)

Answer: 100

For the second ionisation,

K_{a_{2}} = \frac{[ H ^{+} ] [ X ^{2 -} ]}{[ HX ^{-} ]}

. The hydrogen ion and the

HX^{-}

ion are generated together in the first, dominant ionisation, so to a good approximation

[H^{+}] \approx [HX^{-}]

; these cancel and leave

[X^{2 -}] \approx K_{a_{2}}

, independent of the acid concentration. Hence

[X^{2 -}] = 1.0 \times 1 0^{- 13} M = 100 \times 1 0^{- 15} M

. The concentration of

X^{2 -}

is 100 in units of

1 0^{- 15} M

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Integer answer