The 𝐾a of propanoic acid (C2H5COOH) is 1.34×10−5.

Question

Calculate the pH of the solution and the concentrations of 

C2H5COOH and C2H5COO- in a 0.489M propanoic acid solution at equilibrium.

Answer 1

C2H5COOH ==> C2H5COO^- + H⁺

Ka = [C2H5COO^-] [H⁺] / [C2H5COOH]

1.34x10^-5 = (x)(x) / 0.489 - x (and we can ignore -x assuming it is small relative to 0.489 M)

1.34x10^-5 = x2 / 0.489

x² = 6.55x10^-6

x = 2.56x10^-3 M (note that our assumption was correct since this is only about 0.5% of 0.489)

[H⁺] = 2.56x10^-3 M

pH = -log 2.56x10^-3

pH = 2.59