Chemical Equilibrium

(i)
CH3COOH(aq) ≒ CH3COO^-(aq) + H^+(aq)
At equilibrium :
Let [CH3COO^-] = [H^+] = y M
[CH3COOH] = (1 - y) M

Ka = y^2/(1 - y) = 1.8 x 10^-5
y^2 + (1.8 x 10^-5)y - (1.8 x 10^-5) = 0
y = 0.00423
pH = -log(0.00423) = 2.37


(ii)
CH3COOH(aq) + OH^-(aq) → CH3COO^-(aq) + H2O(l)
OH^- is completely reacted.
Neglect the dissociation of CH3COOH:
[CH3COO^-]o = 1 x (25/75) = 0.333 M
[CH3COOH]o = 1 x (50/75) - 1 x (25/75) = 0.333 M

pH
= pKa - log([CH3COOH]o/[CH3COO^-]o)
= -log(1.8 x 10^-5) - log(0.333/0.333)
= 4.74


(iii)
CH3COOH(aq) + OH^-(aq) → CH3COO^-(aq) + H2O(l)
OH^‑ and CH3COOH are completely reacted.
Neglect the hydrolysis of CH3COO^-:
[CH3COO^-]o = 1 x (50/100) = 0.5 M

Consider the hydrolysis of CH3COO^- :
CH3COO^-(aq) + H2O(l) ≒ CH3COOH(aq) + OH^-(aq) .. (Kh)
At equilibrium :
Let [CH3COOH] = [OH^-] = z M
[CH3COO^-] = (1 - z) M

Kh = (1 x 10^-14)/(1.8 x 10^-5) = 5.56 x 10^-10 M
Kh = z^2/(1 - z) = 5.56 x 10^-10
z^2 + (5.56 x 10^-10)z - (5.56 x 10^-10) = 0
z = 2.36 x 10^-5
pOH = -log(2.36 x 10^-5) = 4.63
pH = 14 - 4.63 = 9.37

(iv)
CH3COOH(aq) + OH^-(aq) → CH3COO^-(aq) + H2O(l)
OH^- is in excess.
[CH3COO^-]o = 1 x (50/125) = 0.4 M
[OH^-]o = 1 x (75/125) - 1 x (50/125) = 0.2 M

OH^- is a strong base:
pOH = -log(0.2) = 0.7
pH = 14 - pOH = 14 - 0.7 = 13.3