Final answer:
The acid ionization constant (Ka) for an acid with a pH of 2.11 and an equilibrium concentration of 0.30 M is approximately 2.0 × 10^-4.
Step-by-step explanation:
To determine the acid ionization constant, often represented as Ka, for a weak acid using its equilibrium concentrations, we apply the formula Ka = [H3O+][A-]/[HA], where [H3O+] is the concentration of hydronium ions, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
Given that the pH of the solution is 2.11, we first calculate the [H3O+] using the pH definition: [H3O+] = 10^-pH = 10^-2.11 ≈ 0.00775 M. As for the equilibrium concentration of the acid (HA) given as 0.30 M, this concentration does not change significantly for a weak acid, and so it remains the same at equilibrium in this context.
To find [A-], we need data on the concentration of the conjugate base at equilibrium, which is not provided directly in the question. However, the assumption often made is that [H3O+] equals the concentration of [A-] for a weak acid that dissociates slightly. Hence, [A-] is approximately equal to [H3O+], or 0.00775 M. The Ka can be calculated as follows: Ka = [H3O+][A-]/[HA] = (0.00775 M)(0.00775 M) / (0.30 M) ≈ 2.0 × 10^-4.