At 25 $$^\circ$$C, the concentration of H$$^+$$ ions in $$1.00 \times 10^{-3}$$ M aqueous solution of a weak monobasic acid having acid dissociation constant ($$K_a$$) of $$4.00 \times 10^{-11}$$ is $$X \times 10^{-7}$$ M. The value of $$X$$ is ________.

Use: Ionic product of water ($$K_w$$) = $$1.00 \times 10^{-14}$$ at 25 $$^\circ$$C

The weak monobasic acid is represented as $$HA$$.

Dissociation in water:
$$HA \rightleftharpoons H^+ + A^-$$ with $$K_a = 4.00 \times 10^{-11}$$

Let the initial concentration of the acid be $$C = 1.00 \times 10^{-3}\;{\rm M}$$.
At equilibrium, suppose:

$$\begin{aligned} [H^+]_{\text{total}} &= x \;{\rm M}\\[2pt] \text{Degree of dissociation from the acid} &= y \;{\rm M}\\[2pt] \Rightarrow [A^-] &= y \;{\rm M}\\[2pt] [HA] &= C - y \end{aligned}$$

Water also dissociates: $$H_2O \rightleftharpoons H^+ + OH^-$$ with $$K_w = 1.00 \times 10^{-14}$$.
Hence $$[OH^-] = \dfrac{K_w}{x}$$.

The $$H^+$$ that comes only from water equals the $$OH^-$$ formed, so

$$[H^+]_{\text{from water}} = \dfrac{K_w}{x}$$

Therefore the $$H^+$$ contributed by the acid is

$$y = x - \dfrac{K_w}{x}\qquad -(1)$$

Write the $$K_a$$ expression:

$$K_a = \dfrac{[H^+][A^-]}{[HA]} = \dfrac{x \, y}{C - y}\qquad -(2)$$

Because the acid is very weak, $$y \ll C$$ (numerically, we will soon see $$y \approx 2 \times 10^{-7}\,{\rm M}$$ while $$C = 10^{-3}\,{\rm M}$$).
Hence we can put $$C - y \approx C$$ in $$(2)$$:

$$K_a \approx \dfrac{x\,y}{C}\qquad -(3)$$

Substituting $$y$$ from $$(1)$$ into $$(3)$$:

$$K_a \approx \dfrac{x\left(x - \dfrac{K_w}{x}\right)}{C} = \dfrac{x^2 - K_w}{C}$$

Re-arranging gives

$$x^2 = K_a C + K_w\qquad -(4)$$

Insert the numerical data:

$$\begin{aligned} K_a C &= (4.00 \times 10^{-11})(1.00 \times 10^{-3}) = 4.00 \times 10^{-14}\\[4pt] K_a C + K_w &= 4.00 \times 10^{-14} + 1.00 \times 10^{-14} = 5.00 \times 10^{-14}\\[4pt] x &= \sqrt{5.00 \times 10^{-14}} = (\sqrt{5}) \times 10^{-7}\,{\rm M}\\[4pt] \sqrt{5} &\approx 2.236 \end{aligned}$$

Thus

$$[H^+] = 2.236 \times 10^{-7}\;{\rm M}$$

Comparing with the required form $$X \times 10^{-7}\;{\rm M}$$, we have

$$X \approx 2.24$$

Any value in the range 2.2-2.3 is therefore correct.