Question

Hitunglah pH dari larutan berikut :
50 mL larutan CH₃COOH 0,2 M + 50 mL larutan NaOH 0,05 M (Ka CH₃COOH = 10^-5)

Answer 1

             cari mol masing2 dlu
            Reaksi larutan penyangga
            CH3COOH + NaOH ---------> CH3COONa + H2O
            mol= Mx V    
            0.05x0,2=             0,05x0,05=          
  mula2    10^-2                2,5x10^-3           -                  -                       bereaksi  2,5x10^-3       2,5x10^-3         2,5x10^-3   2,5x10^-3
----------------------------------------------------------------------------------------- -
setimbang 7,5x10^-3            0                 2,5x10^-3   2,5x10^-3
lalu masukkan ke rumus
[H+]=ka.asam/garam
       = 10^-5 . 7,5x10^-3/2,5x10^-3
       = 10^-5 . 3
       = 3x10^-5
PH  = -log 3x10^-5
       = 5-log3
smoga mmbantu ya

Answer 2

Diketahui :
Volume CH3COOH = 50 ml = 0,05 L
Molaritas CH3COOH = 0,2 M = 0,2 mol/L
Volume NaOH = 50 ml = 0,05 L
Molaritas NaOH = 0,05 M = 0,05 mol/L
Ka CH3COOH = 10^ -5

Ditanya : pH larutan ?

Jawab :

• Mencari mol masing - masing zat :

n CH3COOH :
= M × V
= 0,2 mol/L × 0,05 L
= 0,01 mol

n NaOH :
= M × V
= 0,05 mol/L × 0,05 L
= 0,0025 mol

Reaksi :
_______NaOH + CH3COOH ---> CH3COONa + H2O
awal :0,0025_0,01____________
reaksi:0,0025_0,0025_0,0025_0,0025
akhir:__-______0,0075_0,0025_0,0025

mol yang dihasilkan :
n Asam CH3COOH = 0,0075 mol
n garam CH3COONa = 0,0025 mol

• Mencari pH larutan :

[H+] :
= Ka × n asam / n garam
= 10^ -5 × 0,0075 mol / 0,0025 mol
= 3 × 10^ -5

pH :
= - log [H+]
= - log 3 × 10^ -5
= - ( log 3 + log 10^ -5)
= - log 3 - log 10^ -5
= 5 - log 3