Campuran dari 100 mL larutan Ba(OH)2 0,1 M dan 100 mL larutan HNO2 0,2 M (Ka=5 x 10^-4) memiliki pH sebesar

mol Ba(OH)2 = 0.1 x 100 = 10 mmol
mol HNO2 = 0.2 x 100 = 20 mmol

....Ba(OH)2 + 2HNO2 => Ba(NO2)2 + 2H2O
m 10.................20
r -10..................-20.................+10..............+20
s -.....................-......................10.................20

habis bereaksi, terjadi hidrolisis
[garam] = mol/v total = 10/200 = 0.05

[OH-] = akar Kw/ka x [garam]
[OH-] = akar 10^-14/5x10^-4 x 0.05 = 10^-6
pOH = 6
pH = 14-6 = 8