Jika harga Kb NH4OH= 10`5 ,maka pH larutan NH4OH 0,016 M adalah

[OH-] = √M×kb
= √16×10^-3 × 10^-5
= √16×10^-8
= 4×10-4
pOH = -log[OH-]
= -log(4×10^-4)
= 4-log4
pH = 14 - pOH
= 14-(4-log4)
= 10+log4