A 0.0310 M solution of an organic acid has an [H+] of 1.55×10-3 M .

Question

Using the values above, calculate the pH of the solution.

What is the percent ionization of the acid?

Calculate the K_a value of the acid.

Answer 1

pH = -log [H⁺]

pH = -log 1.55x10^-3

pH = 2.81

Let HA = weak acid

% ionization = [H⁺] / [HA] = 1.55x10^-3 / 0.0310 (x100%) = 5% ionized

HA ==> H⁺ + A^-

Ka = [H⁺][A^-] / [HA]

Ka = (1.55x10^-3)(1.55x10^-3) / 0.0310 - 1.55x10^-3)

Ka = 2.40x10^-6 / 0.02945

Ka = 8.15x10^-5