You need to use the Ka for the acetic acid and the equilibrium equation.
Ka = 1.85 * 10^ -5
Equilibrium reaction: CH3COOH (aq) ---> CH3COO(-) + H(+)
Ka = [CH3COO-][H+] / [CH3COOH]
Molar concentrations at equilibrium
CH3COOH CH3COO- H+
0.50 - x x x
Ka = x*x / (0.50 - x) = x^2 / (0.50 - x)
Given that Ka is << 1 => 0.50 >> x and 0.50 - x ≈ 0.50
=> Ka ≈ x^2 / 0.50
=> x^2 ≈ 0.50 * Ka = 0.50 * 1.85 * 10^ -5 = 0.925 * 10^ - 5 = 9.25 * 10 ^ - 6
=> x = √ [9.25 * 10^ -6] = 3.04 * 10^ -3 ≈ 0.0030
pH = - log [H+] = - log (x) = - log (0.0030) = 2.5
Answer: 2.5
