sebanyak 4 mL NH 3 0,5 M dicampur dengan 10 mL NH 4 Cl 0,1 M.jika Kb=2kali10 pangkat -5,maka pH campurannya adalah....

4 mL NH3 0,5 M + 10 mL NH4Cl 0,1 M
Kb = 2 × 10^-5

pH ... ?

n NH3 = M × V
n NH3 = 0,5 × 4 mL
n NH3 = 2 mmol

n NH4Cl = M × V
n NH4Cl = 0,1 × 10 mL
n NH4Cl = 1 mmol

[OH^-] = Kb × Basa lemah/A. konjugasi
[OH^-] = 2 × 10^-5 × 2/1
[OH^-] = 2 × 10^-5 × 2
[OH^-] = 4 × 10^-5

pOH = - log [OH^-]
pOH = - log 4 × 10^-5
pOH = 5 - log 4

pH = 14 - pOH
pH = 14 - (5 - log 4)
pH = 14 - 5 + log 4
pH = 9 + log 4