Calculate the pH of 0.586 M anilinium hydrochloride (C₆H₅NH₃Cl)solution in water, given that K_b for aniline is 3.83 × 10^-4.

Question

Quizplus · Accepted Answer

Anilinium hydrochloride can be considered as an acid, which will ionize in water according to the following equation:

C6H5NH3+ Cl-   →   C6H5NH2 + H+ + Cl-

Kb for aniline is given, but we need Ka for anilinium ion:

Kb × Ka = Kw = 10^(-14)

Ka = Kw/Kb = 10^(-14)/3.83 × 10^(-4) = 2.61 × 10^(-11)

Now we can calculate the concentration of H+ ions in the solution as:

Ka = [H+][C6H5NH3+]/[C6H5NH2] = [H+]^2 / 0.586

[H+] = √(0.586 × 2.61 × 10^(-11)) = 1.82 × 10^(-6) M

pH = -log[H+] = 5.41

Calculate the PH of 0