JEE Advanced 2025 — Paper 1 — CHEM-III solution

JEE Advanced2025Paper 1CHEM-III

At 25°C, the concentration of H $^{+}$ ions in $1.00 \times 1 0^{- 3}$ M aqueous solution of a weak monobasic acid having acid dissociation constant ( $K_{a}$ ) $= 4.00 \times 1 0^{- 11}$ is $X \times 1 0^{- 7}$ M. The value of $X$ is _______.

Use: Ionic product of water ( $K_{w}$ ) $= 1.00 \times 1 0^{- 14}$ at 25°C

Your answer

Solution

Since $K_{a}$ is very small ( $1 0^{- 11}$ ), the H $^{+}$ contributed by the acid is comparable to that from water self-ionisation, so both equilibria must be solved together.

Let $x = [X^{-}]$ and $y = [OH^{-}]$ . Then $[H^{+}] = x + y$ from charge balance. With the approximation $[HX] \approx 1 0^{- 3}$ M (negligible dissociation):

$K_{a} = \frac{x ( x + y )}{1 0 ^{- 3}} = 4 \times 1 0^{- 11} \Rightarrow x (x + y) = 4 \times 1 0^{- 14}$ …(1)

$K_{w} = y (x + y) = 1 0^{- 14}$ …(2)

Adding (1) and (2):

$(x + y) (x + y) = 5 \times 1 0^{- 14} \Rightarrow [H^{+}] = 5 \times 1 0^{- 7}$ M

$5 \approx 2.236$ , so $X \approx 2.23$ (rounded to two decimals).

$∴ X = 2.23$ . (The official key also accepts 2.24.)

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