Question

If the Ka of a monoprotic weak acid is 1.6 × 10-6, what is the pH of a 0.12 M solution of this acid?

Answer 1

Dear friend your answer should be by taking acetic acid because acetic acid is a weak monoprotic acid. only weak acids can have acid dissociation constant K_a

CH₃COOH -----> CH₃COO^- + H⁺
K_a = [H⁺][CH₃COO^-] / [CH₃COOH] = 1.6 x 10^-6

Let x = the amount of CH₃COOH that dissociates.
[H⁺] = x
[CH₃COO^-] = x
[CH₃COOH] = 0.12 - x = 0.12 (assuming x << 0.12)

1.6 x 10^-6 = (x) (x) / (0.12)
x = 0.00043817 = [H+]
pH = - log (0.00043817) = 3.3583

P^H of a given monoprotic acid is 3.3583

Thank you.