Jawabanmu

2014-02-26T18:58:23+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 
2014-02-26T19:09:37+07:00
M=G/Mr x 1000/V
M= 4,1/82 x 1000/500
M= 0,1
[H⁺] = √Ka.M.a
      = 
√10^⁻⁵.0,1.1
      = √10^⁻⁶
[H⁺]=10^⁻³
pH= - log [H⁺]
    = - log 10^⁻3
pH= 3
3 4 3