2,4,6-Trichlorophenol is a weak monoprotic acid, with K_a = 2.51 x 10^-5 mol dm^-3 at 298 K - AQA - A-Level Chemistry - Question 9 - 2017 - Paper 3

Given that 2,4,6-trichlorophenol is a weak acid, we use the formula for weak acid dissociation:

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

Assuming x is the concentration of H⁺ ions, we can express this as:

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

In this case,

• The initial concentration of the acid, [HA], is 2.00 x 10^-3 mol dm^-3,
• K_a is 2.51 x 10^-5 mol dm^-3,
• We can simplify this to:

$2.51 x 10^{-5} = \frac{x^2}{(2.00 x 10^{-3}) - x}$

Assuming x is very small, we simplify to:

$2.51 x 10^{-5} = \frac{x^2}{2.00 x 10^{-3}}$

Solving for x gives:

$x^2 = 2.51 x 10^{-5} \times 2.00 x 10^{-3}$

After calculating: $x^2 = 5.02 x 10^{-8}$ $x = \sqrt{5.02 x 10^{-8}} \approx 7.09 x 10^{-4}$

Thus, the concentration of hydrogen ions, [H⁺], in the solution is approximately 7.09 x 10^-4 mol dm^-3.