jika 100 ml larutan HCL 0,1 M dicampurkan dengan 50 ml larutan NH3 0,3 M (Kb=10(-5)) maka pH larutan yang terjadi adalah

100 mL HCl 0,1 M + 50 mL NH3 0,3 M
Kb = 10^-5

pH ... ?

*n HCl = M × V
n HCl = 0,1 × 100
n HCl = 10 mmol

*n NH3 = M × V
n NH3 = 0,3 × 50
n NH3 = 15 mmol

#Reaksi
..... HCl + NH3 => NH4Cl
M : 10 ...... 15
R : 10 ...... 10
__________---
S : --- ....... 5 ........... 10

[OH^-] = Kb × basa lemah/asam konjugasi
[OH^-] = 10^-5 × 5/10
[OH^-] = 10^-5 × 0,5
[OH^-] = 5 × 10^-6

pOH = - log [OH^-]
pOH = - log 5 × 10^-6
pOH = 6 - log 5

pH = 14 - pOH
pH = 14 - (6 - log 5)
pH = 14 - 6 + log 5
pH = 8 + log 5