Question

Kedalam 1 liter larutan asam asetat 0,2 m ditambahkan padatan NaOH 8 gr . tentukan Ph campuran yang terjadi (Ka CH3COOH = 2,10^-3 , Ar H = 1 , C = 12, O = 16 , Na = 25 )

Answer 1

Diketahui :
Volume Asam asetat (CH3COOH) = 1 L
Molaritas CH3COOH = 0,2 M = 0,2 mol/L
m NaOH = 8 gram
Ka = 2 × 10^ -3
Ar H = 1 , C = 12 , Na = 23 , O = 16

Ditanya : pH campuran ?

Jawab :


n CH3COOH :
= M × V
= 0,2 mol/L × 1 L
= 0,2 mol

Mr NaOH :
= Ar Na + Ar O + Ar H
= 23 + 16 + 1
= 40 gram/mol

n NaOH :
= m / Mr
= 8 gram / 40 gram/mol
= 0,2 mol

Reaksi :
_______CH3COOH + NaOH --> CH3COONa + H2O
awal :0,2___0,2____________
reaksi:0,2__0,2____0,2__0,2
akhir :__-____-_____0,2__0,2

mol garam = CH3COONa = 0,2 mol

Molaritas garam :
= n / V larutan
= 0,2 mol / 1 L
= 0,2 mol/L


• Mencari pH larutan / campuran :
Kw = Tetapan Kesetimbangan Air = 10^ -14


[OH-] :
= √ Kw / Ka × M (Molaritas)
= √ 10^ -14 / 2 × 10^ -3 × 0,2
= 10^ -6

pOH :
= - log [OH-]
= - log 10^ -6
= 6

pH :
= 14 - pOH
= 14 - 6
= 8

Answer 2

Mol CH3COOH = 1 L x 0,2 M = 0,2 mol
mol NaOH = 8 gram / 40 gram/mol = 0,2 mol
   CH3COOH + NaOH --> CH3COONa + H2O
m 0,2 mol         0,2 mol          
b  0,2 mol        0,2 mol      0,2 mol          0,2 mol   
s      -                  -             0,2 mol          0,2 mol

M CH3COO- = 0,2 mol / 1 L = 0,2 M



pOH = 6
pH = 8

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