The pH of a 0.02 M NH4Cl solution will be: [given Kb(NH4OH)=10⁻⁵ and log2=0.301]

For the salt of strong acid and weak base
H+ = √Kw × CKb
[H+] = √10⁻¹⁴ × 0.02 × 10⁻⁵
[H+] = √4 × 10⁻¹¹
[H+] = 2 × 10⁻⁶
-log[H+] = 6 - log2
∴pH = 5.35