equilibrium

Each mol of (NH4)2SO4 contains 2 mol of NH4+ ions.
[NH4+]o = 0.2 x 2 = 0.4 M

start aaaNH4+(aq) + H2O(l) ≒ NH­3(aq) + H3O+(aq)
start aaa 0.4 M q) + H0 M) ≒ NH0 Mq) + H0 M
change a -y Maq) + H2O(l) ≒ NH+y M) + H+y M
eqm aa(0.4 - y) M+ H2O(l) ≒ NH y M) + HOy M

Hydrolysis constant Kh
= Kw / Kb
= (1 x 10-14) / (1.8 x 10-5)
= 5.56 x 10-10 mol dm-3

Kh = y2 / (0.4 - y) = 5.56 x 10-10
y2 + (5.56 x 10-10)y - (2.22 x 10-10) = 0
[H3O+] = y = 1.49 x 10-5 M
pH = -log(1.49 x 10-5) = 4.83