At 25 ℃, the concentration of H⁺ ions in 1.00 × 10⁻³ M aqueous solution of a weak monobasic acid having acid dissociation constant (K_a) of 4.00 × 10⁻¹¹ is X × 10⁻⁷ M. The value of X is ______

Use: Ionic product of water (K_w) = 1.00 × 10⁻¹⁴ at 25 ℃

To find the value of $X$, we need to calculate the total hydrogen ion concentration, $\left[{\text{H}}^{+}\right]$, in the aqueous solution of the weak monobasic acid.

Given data:
Concentration of the weak acid ($C$) = $1.00\times {10}^{-3} \text{M}$
Acid dissociation constant ($K_a$) = $4.00\times {10}^{-11}$
Ionic product of water ($K_w$) = $1.00\times {10}^{-14}$ at 25 ℃

Since the acid dissociation constant ($K_a$) is extremely small, the concentration of ${\text{H}}^{+}$ ions produced solely by the acid is very low and comparable to the concentration of ${\text{H}}^{+}$ ions produced by the autoionization of water. Therefore, we must consider the contribution of water to the total ${\text{H}}^{+}$ ion concentration.

For a weak monobasic acid $\text{HA}$ in water, we have the following equilibria:
$\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$
${\text{H}}_2\text{O}\rightleftharpoons {\text{H}}^{+}+{\text{OH}}^{-}$

Using the charge balance equation for the solution:
$\left[{\text{H}}^{+}\right]=\left[{\text{A}}^{-}\right]+\left[{\text{OH}}^{-}\right]$

From the acid dissociation expression:
$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}\Rightarrow \left[{\text{A}}^{-}\right]=\frac{K_a\left[\text{HA}\right]}{\left[{\text{H}}^{+}\right]}$

Since the acid is very weak and its dissociation is negligible, we can approximate the equilibrium concentration of the undissociated acid as its initial concentration:
$\left[\text{HA}\right]\approx C=1.00\times {10}^{-3} \text{M}$

Thus, the expression for $\left[{\text{A}}^{-}\right]$ becomes:
$\left[{\text{A}}^{-}\right]\approx \frac{K_{a}C}{\left[{\text{H}}^{+}\right]}$

From the autoionization of water:
$\left[{\text{OH}}^{-}\right]=\frac{K_w}{\left[{\text{H}}^{+}\right]}$

Substituting these values into the charge balance equation:
$\left[{\text{H}}^{+}\right]=\frac{K_{a}C}{\left[{\text{H}}^{+}\right]}+\frac{K_w}{\left[{\text{H}}^{+}\right]}$

Multiplying both sides by $\left[{\text{H}}^{+}\right]$ gives:
${\left[{\text{H}}^{+}\right]}^2=K_{a}C+K_w$

Substitute the given values into the equation:
${\left[{\text{H}}^{+}\right]}^2=(4.00\times {10}^{-11})(1.00\times {10}^{-3})+1.00\times {10}^{-14}$
${\left[{\text{H}}^{+}\right]}^2=4.00\times {10}^{-14}+1.00\times {10}^{-14}$
${\left[{\text{H}}^{+}\right]}^2=5.00\times {10}^{-14}$

Taking the square root of both sides:
$\left[{\text{H}}^{+}\right]=\sqrt{5.00\times {10}^{-14}}$
$\left[{\text{H}}^{+}\right]=\sqrt{5.00}\times {10}^{-7} \text{M}$
$\left[{\text{H}}^{+}\right]\approx 2.236\times {10}^{-7} \text{M}\approx 2.24\times {10}^{-7} \text{M}$

Comparing this with the given concentration form $X\times {10}^{-7} \text{M}$, we find:
$X=2.24$