Rumus :

[ H+ ] = √ Kw / Kb × [ Garam ]

pH = -Log [ H+ ]

Jawab :

Diket :

V NH4OH = 100 mL

Molar NH4OH = 0,2 M

Kb NH4OH = 1 × 10^-5

V HCl = 100 mL

Molar HCl = 0,2 M

Ditan : pH ?

n mol NH4OH = 100 mL × 0,2 M = 20 mmol

n mol HCl = 100 mL × 0,1 M = 10 mmol

Hitung mol zat sisa hasil reaksi :

reaksi : 2NH4OH + HCl → (NH4)2Cl + 2H2O

mula2 : 20 mmol + 10 mmol

bereaksi : 20 mmol + 10 mmol → 10 mmol + 20 mmol

M (NH4)2Cl = mmol / V total = 10 mmol / 200 mL = 0,05 M

[H+] = √Kw/Kb × 2 × [ M (NH4)2Cl ]

[H+] = √10^-14/10^-5 × 2 × [ 5 × 10^-2 ]

[H+] = √10^-9 × 2 × 5 × 10^-2

[H+] = √10^-10

[H+] = 10^-5

pH = -Log [H+]

pH = 5-Log 1

pH = 5