Jawabanmu

2014-04-18T17:08:41+07:00
Ph sebelum reaksi 
CH3COOH (asam lemah)
Ph = √Ka . konsentrasi
     =√2x10 ⁻⁵ .4
Ba (OH)2 (basa kuat )
OH- = n . konsentrasi
       = 2 .0,5
       = 1
Ph =13
2CH3COOH + Ba(OH)2  
⇒ (CH3COO)2Ba + 2H2O
  8 mmol          4mmol           4 mmol
 
PH 
(CH3COO)2Ba 
OH- =√Kw/ka . n . konsentrasi garam
      = √10⁻¹⁴ / 2x10 ⁻⁵ .2 . 4/10
      =√10 ⁻⁹ .0,4
      =2. 10⁻⁵
PH = 14 -(5 -log 2)
     = 9 +log 2