PH campuran yg terbentuk dari 100 ml KOH 0.1 M dan 100 ml asam format 0.2 M. (Ka 1.7 x 10^-4)
aryakusuma2000
Diketahui : Volume KOH = 100 ml = 0,1 L Molaritas KOH = 0,1 M = 0,1 mol/L Volume Asam format (HCOOH) = 100 ml = 0,1 L Molaritas HCOOH = 0,2 M = 0,2 mol/L Ka = 1,7 × 10^ -4
Ditanya : pH campuran ?
Jawab :
n KOH : = M × V = 0,1 mol/L × 0,1 L = 0,01 mol
n HCOOH : = M × V = 0,2 mol/L × 0,1 L = 0,02 mol
Reaksi : _______KOH + HCOOH --> HCOOK + H2O awal :0,01_0,02______________ reaksi:0,01_0,01__0,01_0,01 akhir :__-___0,01__0,01_0,01
Volume KOH = 100 ml = 0,1 L
Molaritas KOH = 0,1 M = 0,1 mol/L
Volume Asam format (HCOOH) = 100 ml = 0,1 L
Molaritas HCOOH = 0,2 M = 0,2 mol/L
Ka = 1,7 × 10^ -4
Ditanya : pH campuran ?
Jawab :
n KOH :
= M × V
= 0,1 mol/L × 0,1 L
= 0,01 mol
n HCOOH :
= M × V
= 0,2 mol/L × 0,1 L
= 0,02 mol
Reaksi :
_______KOH + HCOOH --> HCOOK + H2O
awal :0,01_0,02______________
reaksi:0,01_0,01__0,01_0,01
akhir :__-___0,01__0,01_0,01
n Asam : 0,01 mol
n garam : 0,01 mol
[H+] :
= Ka × mol asam / mol garam
= 1,7 × 10^ -4 × 0,01 mol / 0,01 mol
= 1,7 × 10^ -4
pH :
= - log [H+]
= - log 1,7 × 10^ -4
= - ( log 1,7 + log 10^ -4)
= - log 1,7 - log 10^ -4
= 4 - log 1,7