JEE Main 2026 — Jan 21, Shift 2 — Chemistry solution

JEE Main2026Jan 21, Shift 2Chemistry

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 $X^{2 -}$ in 0.1 M $H_{2} X$ solution is _________ $\times 1 0^{- 15}$ M. (Nearest Integer)

Your answer

Solution

For a weak diprotic acid where $K_{a 1} ≫ K_{a 2}$ , a standard result is that $[X^{2 -}] \approx K_{a 2}$ — almost independent of the analytical concentration.

Quick derivation: from the first dissociation $[H^{+}] \approx [H X^{-}] = K_{a 1} C$ . From the second, $K_{a 2} = \frac{[ H ^{+} ] [ X ^{2 -} ]}{[ H X ^{-} ]}$ , and since $[H^{+}] \approx [H X^{-}]$ , $[X^{2 -}] \approx K_{a 2} = 1.0 \times 1 0^{- 13}$ M.

Expressing in the requested units: $[X^{2 -}] = 100 \times 1 0^{- 15}$ M, so the answer is 100.

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