Final answer:
To calculate the pH of a 0.626 M anilinium hydrochloride solution, the dissociation constant (Ka) for the anilinium ion is calculated using the provided Kb for aniline and the ion product constant of water (Kw). The pH is then determined by taking the negative logarithm of the concentration of H3O+ ions, which gives a pH of approximately 0.20.
Step-by-step explanation:
The question is asking for the pH of a 0.626 M anilinium hydrochloride solution. Anilinium hydrochloride is a salt formed from the reaction of aniline, a weak base, with hydrochloric acid. To answer this, we need to first calculate the Ka for anilinium ion, using the given Kb of aniline and the relation Kw = Ka × Kb, where Kw is the ion product constant of water (1.0 × 10⁻¹⁴ at 25°C).
First, we find the Ka of the anilinium ion:
Ka = Kw / Kb = 1.0 × 10⁻¹⁴ / 3.83 × 10⁻¹⁴ = 2.61 × 10⁻¹⁻¹
Given that anilinium hydrochloride is an acid (the conjugate acid of aniline), the concentration of [H3O+] in the solution is approximately equal to its molarity because it fully dissociates. Consequently, the pH is calculated using the equation
pH = -log([H3O+])
Thus:
pH = -log(0.626) ≈ 0.20
Therefore, the pH of the 0.626 M anilinium hydrochloride solution is approximately 0.20.