Tentukan pH larutan yang terbentuk dari 50 mL larutan NH3 0,1 M (Kb=10^-5) dicampurkan dengan 100 mL larutan NH4Cl 0,4 M.

50 mL NH3 0,1 M + 100 mL NH4Cl 0,4 M
Kb NH3 = 10^-5

pH .... ?

n NH3 = M × V
n NH3 = 0,1 × 50
n NH3 = 5 mmol

n NH4Cl = M × V
n NH4Cl = 0,4 × 100
n NH4Cl = 40 mmol

[OH^-] = Kb × n basa lemah/n asam konjugasi
[OH^-] = 10^-5 × 5/40
[OH^-] = 10^-5 × 0,125
[OH^-] = 1,25 × 10^-6

pOH = - log [OH^-]
pOH = - log 1,25 × 10^-6
pOH = 6 - log 1,25

pH = 14 - pOH
pH = 14 - (6 - log 1,25)
pH = 14 - 6 + log 1,25
pH = 8 + log 1,25