Calculate the pH of 0.726 M anilinium hydrochloride (C₆H₅NH₃Cl)solution in water, given that K_b for aniline is 3.83 × 10^-4. A)1.78 B)12.22 C)5.36 D)8.64 E)12.42

Question

Quizplus · Accepted Answer

Anilinium ion is the conjugate acid of aniline. The equation for the dissociation of anilinium ion in water is:

C6H5NH3+ + H2O ⇌ C6H5NH2 + H3O+

The equilibrium constant for this reaction is:

Ka = [C6H5NH2][H3O+] / [C6H5NH3+]

Aniline is a weak base, and we can assume that [C6H5NH2] is negligible compared to [C6H5NH3+]. Therefore, we can simplify the equation to:

Ka = [H3O+] / [C6H5NH3+]

We know that Ka for aniline is 3.83 × 10-4. Therefore:

3.83 × 10-4 = [H3O+] / 0.726

[H3O+] = 2.78 × 10-4 M

pH = -log[H3O+] = 5.36

Calculate the PH of 0