Question

Hitunglah pH larutan penyangga yang dibuat dari campura 100 mL CH₃COOH 0,1 M dengan 200 mL larutan CH₃COONa 0,1 M ! (Ka CH₃COOH = 10⁻⁵)

Answer 1

n CH3COOH = 100 x 0,1 = 10 mmol
n CH3COONa = 200 x 0,1 = 20 mmol
[H+] = Ka mmol CH3COOH/mmol CH3COO-
[H+] = 10^-5 10/20
[H+] = 10^-5 (0,5)
[H+] = 5 x 10^-6
<--> pH = -log [H+]
pH= -log 5 x 10^-6
pH = 5,3

Answer 2

Diketahui : Ka CH3COOH = 10^-5
n CH3COOH = 0,1 M  X 0,1 L = 10^-2
n CH3COONa = 0,1 M  X 0,2 L = 2 X 10^-2
[H+] = Ka X n CH3COOH / n CH3COONa
       = 10^-5 X 10^-2 / 2 X 10^-2
       = 10^-5/2
       = 5 X 10^-6
PH = - Log [H+]
     = - Log 5 X 10^-6
     = 6 - Log 5
     = 6 - 0,7
     = 5,3