1) Equlibrium reaction
CH3COOH (aq) = CH3COO(-) (aq) + H(+) (aq)
2) Equilibrium constant
Keq = Ka = [CH3COO-] [H+] / [CH3COOH]
3) Equilibrium concentrations
CH3COOH CH3COO- H+
start 1.40 0 0
react x 0 0
produced 0 x x
equilibrium 1.40 - x x x
=> Ka = x * x / (1.40 - x)
Approximation: given that Ka is very small x <<< 1,40 and 1.40 - x ≈ 1.40
=> Ka ≈ x^2 / 1.40
=> x^2 ≈ 1.40Ka = 1.40 * 1.8 * 10^ - 5 = 2.52 * 10^-5
=> x ≈ √(2.52 * 10^-5) ≈ 5.02 * 10^ -3 M
4) pH = log 1 / [H+]
[H+] = x = 5.02 * 10^-3M
=> pH ≈ log (1 / 5.02 * 10^-3) ≈ 2.3
Answer: 2.3
