Jawabanmu

2014-06-14T19:29:12+07:00
PH = 5-log 2
H+ = 2 x 10^-5

H+ = Ka (asam lemah/Basa konjugasi )
2x10^-5 = 10^-5 (CH3COOH/CH3COO-)
2 CH3COO- = CH3COOH
2/1 = CH3COOH/CH3COONa


CH3COONa ⇒ CH3COO- + Na+    (CH3COO- = CH3COONa krn perbndingan koef 1/1)
    2                    2              2 

perbandinganya yaitu 2:1
1 5 1