Question

Suatu buffer basa dengan pH=9 dibuat dengan melarutkan 100ml 1M NH3 dan 100ml H2SO4 0,25M. Jika larutan HCl 0.25M sebanyak 100ml ditambahkan ke dalam buffer, maka pH larutan akan berubah menjadi

Answer 1

CH₃COOH + Ba(CH₃COO)₂
mol CH₃COOH = 200 ml x 0,1 M = 20 mmol
mol Ba(CH₃COO)₂ = 100 ml x 0,1 = 10 mmol
mol OH⁻ sisa = 20-10 = 10 mmol
[OH⁻] = mol/V.total = 10 mmol / 300 ml = 10⁻²
pOH = - log [OH⁻] = - log [10⁻²] = 2 
pH = 14-pOH = 14 -2 = 12 

Answer 2

CH₃COOH + Ba(CH₃COO)₂ :

mol CH₃COOH
: 200 ml x 0,1 M = 20 mmol
mol Ba(CH₃COO)₂
: 100 ml x 0,1 = 10 mmol
mol OH⁻ sisa
: 20-10 = 10 mmol
[OH⁻] = mol/V.total = 10 mmol / 300 ml = 10⁻²
pOH = - log [OH⁻] = - log [10⁻²] = 2 
pH = 14-pOH = 14 -2 = 12