Jawabanmu

2014-06-04T13:58:29+07:00
Mol CH3COOH = 40 x 0,15 = 6 mmol
mol NaOH = 20 x 0,3 = 6 mmol

[H+] = Ka x a/g
= 10^-5 x 6/6
= 10^-5
pH = - log 10^-5
= 5

2014-06-04T14:06:10+07:00
CH₃COOH + NaOH => CH₃COONa + H₂O
  6 mmol     6 mmol            -              -
  6 mmol     6 mmol         6 mmol      6 mmol
     -               -                6 mmol      6 mmol

reaktan habis bereaksi, maka memakai rumus hidrolisis. garam bersifat basa.
M(garam) =  \frac{n}{Va+Vb}  \\ = \frac{6}{60}  \\ =0,1 M
[OH⁻] = = \sqrt{ \frac{Kw}{Ka}[G] } \\ = \sqrt{ \frac{ 10^{-14} }{ 10^{-5}} 10^{-1}} \\ = \sqrt{ 10^{-10} }
[O H^{-}] = 10^{-5}
pOH = - log OH⁻
        = - log 10⁻⁵
        = 5
pH = 14 - pOH
      9

iya. mau nyari apa?
mau tnya penyelesaian jika BacrO4 ksp=1,2*10-10?
dicari kelarutannya
kelarutan = s.
Ksp = s^2. ; 1,2*10^-10 =s^2 ; maka, s = 2 akar 30 *10^-6