Chemical Equilibrium 2

When 10 cm³ of NaOH (Neglect dissociation of CH3COOH) :
CH3COOH(aq) + OH^-(aq) ≒ CH3COO^-(aq) + H2O
[CH3COO^-]o = 1 x (10/60) = 0.167 M
[CH3COOH]o = 1 x (50/60) - 1 x (10/60) = 0.667 M

Consider the dissociation of CH3COOH :
CH3COOH(aq) + H2O(l) ≒ CH3COO^-(aq) + H^+(aq)

At equilibrium :
Let [H^+] = y M
[CH3COO^-] = (0.167 + y) M ≈ 0.167 M (Assume 0.167 >> y)
[CH3COOH] = (0.667 + y) M ≈ 0.667 M (Assume 0.667 >> y)
Ka = [CH3COO^-][H^+]/[CH3COOH] = 1.8 x 10^-5
0.167y/0.667 = 1.8 x 10^-5
y = 7.2 x 10^-5
pH = -log(7.2 x 10^-5) = 4.14

Alternative method :
pH
= pKa - log([acid]o/[salt]o)
= -log(1.8 x 10^-5) - log(0.667/0.167)
= 4.14