Sebanyak 100 mL NH4OH 0,1 M (kb=10^-5). Harga pH larutan tersebut adalah ...

100 × 0,1 M = 10
OH- = √kb.cb
OH- = √10^-5 . 10
OH- = √10^-4
OH- = 10^-2

POH = - log OH-
POH = - log 10^-2
POH = 2

PH = 14 - POH
PH = 14 - 2
PH = 12

100 mL NH4OH 0,1 M (Kb = 10^-5)

pH ... ?

NH4OH adalah basa lemah, jadi menggunakan rumus:
[OH^-] = √( Kb × Mb )
[OH^-] = √( 10^-5 × 0,1 )
[OH^-] = √( 10^-6 )
[OH^-] = 10^-3

pOH = - log [OH^-]
pOH = - log 10^-3
pOH = 3

pH = 14 - pOH
pH = 14 - 3
pH = 11