2. Sebanyak 50 mL larutan CH3COOH 0,1 M (Ka = 10^-5) dicampur dgn 50 mL larutan NH4OH dgn konsentrasi yg sama. Bila Kb NH4OH = 10^-6, berapakah pH larutan yg te

CH3COOH + NH4OH → CHCOONH4 + H2O

→ CH3COOH
m = 50 x 0,1 = 5 mmol
r = 5 mmol
s = -

→ NH4OH
m = 50 x 0,1 = 5 mmol
r = 5 mmol
s = -

→ CHOONH4
m = -
r = 5 mmol
s = 5 mmol

M = 5 mmol / (50 + 50)ml
M = 5 / 100
M = 0,05 M

→ H2O
m = -
r = 5 mmol
s = 5 mmol

[H+] = √Kw . Ka / Kb
[H+] = √10^-14 . 10^-5 /10^-6
[H+] = √10^-13
[H+] = √10^-12 x 0,1
[H+] = √0,1 x 10^-6

pH = -log [H+]
pH = -log √0,1 x 10^-6
pH = 6 - log√0,1