F6 Acid-base eqm question

Given that the Ka for ethanoic acid are 1.80 x 10^-5 M and
the Kb for dimethylamine is 9.55 x 10^-4 M respectively at 298K.
a) Write down the expression of Ka for ethanoic acid and Kb of dimethylamine.

CH3COOH(aq) ⇌ CH3COO⁻(aq) + H⁺(aq)
Ka(CH3COOH) = [CH3COO⁻] [H⁺] / [CH3COOH]

(CH3)2NH(aq) + H2O(l) ⇌ (CH3)2NH2⁺(aq) + OH⁻(aq)
Kb((CH3)2NH) = [(CH3)2NH2⁺] [OH⁻] / [(CH3)2NH]


=====
b) 0.25M aqueous solution of dimethylammonium ethanoate at 298K is dissolved in water at 298K, which is known to exist the following eqm with eqm constant Kc.
CH3COO⁻(aq) + (CH3)2NH2⁺(aq) ⇌ CH3COOH(aq) + (CH3)2NH(aq) .. Kc
Determine Kc.

Kc = [CH3COOH] [(CH3)2NH] / [CH3COO⁻] [(CH3)2NH2⁺]
= ([H⁺] [OH⁻]) x ([CH3COOH] / [CH3COO⁻] [H⁺]) x ([(CH3)2NH] / [(CH3)2NH2⁺] [OH⁻])
= Kw / Ka Kb
= (1 x 10^-14) / (1.8 x 10^-5)(9.55 x 10^-4)
= 5.82 x 10^-7


=====
c) Calculate the pH of 0.25M aqueous solution of dimethylammonium ethanoate.

CH3COO⁻(aq) + (CH3)2NH2⁺(aq) ⇌ CH3COOH(aq) + (CH3)2NH(aq)

Initial concentrations:
[CH3COO⁻]o = [(CH3)2NH2⁺]o = 0.25 M
[CH3COOH]o = [(CH3­)2NH]o = 0 M

At eqm:
Since Kc is small, only small amounts of CH3COOH and (CH3)2NH are formed.
Let [CH3COOH] = [(CH3­)2NH] = y M
then [CH3COO⁻] = [(CH3)2NH2⁺] = (0.25 - y) M ≈ 0.25 M .. (Assume 0.25 >> y)

Kc = y^2/(0.25)^2 = 5.82 x 10^-7
y = 0.000191

Ka(CH3COOH) = (0.25) [H⁺] / (0.000191) = 1.8 x 10^-5
[H⁺] = 1.38 x 10^-8
pH = -log(2.74 x 10^-8) = 7.86