Jawabanmu

2014-02-26T17:28:26+07:00
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
2014-02-26T17:53:43+07:00
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