At 25°C, the concentration of $\mathrm{H}^{+}$ ions in $1.00 \times 10^{-3} \mathrm{M}$ aqueous solution of a weak monobasic acid having acid dissociation constant $\left(K_{a}\right)=4.00 \times 10^{-11}$ is $\boldsymbol{X} \times 10^{-7} \mathrm{M}$. The value of $\boldsymbol{X}$ is ____. Use: Ionic product of water $\left(K_{w}\right)=1.00 \times 10^{-14}$ at 25°C

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Equilibrium

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}

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

Answer: 2.24

Since

K_{a}

is very small, we must consider the autoionization of water.

Let the weak acid be HX:

HX (a q) ⇌ H^{+} (a q) + X^{-} (a q)

H_{2} O (l) ⇌ H^{+} (a q) + OH^{-} (a q)

Let

x

[X^{-}]

and

y

[OH^{-}]

, then

[H^{+}] = x + y

.

With approximation

(1 0^{- 3} - x) \approx 1 0^{- 3}

\frac{x ( x + y )}{1 0 ^{- 3}} = 4 \times 1 0^{- 11} ...(1)

y (x + y) = 1 0^{- 14} ...(2)

Adding (1) and (2):

(x + y)^{2} = 4 \times 1 0^{- 14} + 1 0^{- 14} = 5 \times 1 0^{- 14}

x + y = 5 \times 1 0^{- 7} = 2.236 \times 1 0^{- 7}

Therefore,

X = 2.24

(rounded to two decimal places).

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Numerical answer