Calculate the concentration of H₃O⁺ ions in a mixture of 0.02 M acetic acid and 0.1 M sodium acetate.

K_a for acetic acid is 1.8 x 10^-5.
Acetic acid, CH₃COOH is a weak acid which ionizes partially and sodium acetate is a strong electrolyte which ionizes completely in the solution. Let x be the number of moles of acetic acid ionized.

Then, the concentrations of various species are:
CH₃COOH (aq) + H₂O <-----> CH₃COO^– (aq) + H₃O⁺ (aq)
CH₃COONa (aq) —————-> CH₃COO^– (aq) + Na⁺ (aq)
Concentration of acetate ions from acetic acid = x
Since sodium acetate is completely ionized,
The concentration of acetate ion from sodium acetate = 0.1 M
Therefore, total [CH₃COO^–] = 0.1 + x
Concentration of acetic acid left unionized = 0.02 – x
Now, x is very small in comparison to 0.1 so that
[CH₃COO^–] = 0.1 + x = 0.1 mol L^-1
[CH₃COOH] = 0.02 – x = 0.02 mol L^-1
Ionization constant, K_a is:
K_a = [H₃O⁺] [CH₃COO^–]
[CH₃COOH]
K_a for acetic acid is 1.8 x 10^-5 (given)
1.8 x 10^-5 = [H₃O⁺] x 0.1 / 0.02
[H₃O⁺] = 1.8 x 10^-5 x 0.02 / 0.1
[H₃O⁺] = 3.6 x 10^-6 mol L^-1