Hitung ph dari 100 ml NH3 0,2 m + 50 ml H2SO4 0,1M KB =10-5

100 mL NH3 0,2 M + 50 mL H2SO4 0,1 M
Kb = 10^-5

pH ... ?

n NH3 = M × V
n NH3 = 0,2 × 100 mL
n NH3 = 20 mmol

n H2SO4 = M × V
n H2SO4 = 0,1 × 50 mL
n H2SO4 = 5 mmol

#Reaksi
... 2 NH3 + H2SO4 => (NH4)2SO4
M : 20 ............ 5
R : 10 .............. 5
______________---
S : 10 ............. --- ................. 5

Campuran kedua larutan tersebut membentuk larutan penyangga yang bersifat basa.

[OH^-] = Kb × Basa lemah/Asam konjugasi
[OH^-] = 10^-5 × 10/5
[OH^-] = 10^-5 × 2

pOH = - log [OH^-]
pOH = - log 10^-5 × 2
pOH = 5 - log 2

pH = 14 - pOH
pH = 14 - (5 - log 2)
pH = 14 - 5 + log 2
pH = 9 + log 2