sebanyak 5,35gram NH4Cl dilarutkan dalam air sampai volume 250 mL jika Kb =10^-5maka pH larutan NH4Cl adalah

M NH4Cl = gr/Mr × 1000/v
= 5,35/53,5 × 1000/250
= 0,1 × 4
= 0,4

[H+] = √Kw/Kb . [B+]
= √10^-14/10^-5 . 4.10^-1
= √4.10^-10
= 2.10^-5
pH = - log [H+]
= - log 2.10^-5
= 5 - log 2