2,4,6-Trichlorophenol is a weak monoprotic acid, with $K_a = 2.51 \times 10^{-6} \ mol \ dm^{-3}$ at 298 K - AQA - A-Level Chemistry - Question 9 - 2017 - Paper 3

To find the hydrogen ion concentration from the weak acid, we will use the equilibrium expression for the dissociation of the weak acid:

$K_a = \frac{[H^+][A^-]}{[HA]}$

Assuming that $x$ is the concentration of $H^+$ and $A^-$ formed at equilibrium:

• The initial concentration of the acid, $HA$, is $[HA]_0 = 2.00 \times 10^{-3} \ mol \ dm^{-3}$.
• At equilibrium, $[HA] = [HA]_0 - x$.
• Therefore, $[H^+] = x$ and $[A^-] = x$.

This leads to:

$K_a = \frac{x^2}{[HA]_0 - x}$

Assuming $x$ is very small compared to $[HA]_0$, we simplify this to:

$K_a = \frac{x^2}{[HA]_0}$

Now we substitute the known values:

$2.51 \times 10^{-6} = \frac{x^2}{2.00 \times 10^{-3}}$

Rearranging gives:

$x^2 = 2.51 \times 10^{-6} \times 2.00 \times 10^{-3}$

Calculating gives:

$x^2 = 5.02 \times 10^{-9}$

Taking the square root:

$x = \sqrt{5.02 \times 10^{-9}} \approx 7.09 \times 10^{-5} \ mol \ dm^{-3}$

Thus, the concentration of hydrogen ions in the solution is $7.09 \times 10^{-5} \ mol \ dm^{-3}$, which corresponds to option B.