DATOS EQUIPO 1
TABLA 6.1 REGISTRO DE DATOS EXPERIMENTALES CNaOH = _0.0998M__ CAcético =__0.03M__ V Acético =_10mL__ |
V NaOH mL |
0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 2.6 | 2.7 | 2.8 | 2.9 | 3.0 | 3.1 | 3.2 | 3.3 | 3.4 | 3.5 | 4.0 | 4.5 | 5.0 |
pH |
2.90 | 3.70 | 3.98 | 4.12 | 4.59 | 4.97 | 5.06 | 5.09 | 5.14 | 5.19 | 5.30 | 5.61 | 5.75 | 5.79 | 5.95 | 9.71 |11.49 | 11.89 | 12.04 |
TABLA 6.2 CÁLCULO DE ?pH/?V |
V NaOH mL |
0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 2.6 | 2.7 | 2.8 | 2.9 | 3.0 | 3.1 | 3.2 | 3.3 | 3.4 | 3.5 | 4.0 | 4.5 | 5.0 |
?pH/?V |
? | 1.6 | 0.56 | 0.28 | 0.94 | 0.76 | 0.9 | 0.3 | 0.5 | 0.5 | 1.1 | 3.1 | 1.4 | 0.4 | 1.6 | 37.6 | 3.56 | 0.8 | 0.3 |
V promedio = (V inicial de NaOH + V final deNaOH)/2 |
? | 0.25 | 0.75 | 1.25 | 1.75 | 2.25 | 2.55 | 2.65 | 2.75 | 2.85 | 2.95 | 3.05 | 3.15 | 3.25 | 3.35 | 3.45 | 3.75 | 4.25 | 4.75 |
GRAFICA ?pH / ?V VS. V promedio
Vo = 3.35mL
a) Calcular la normalidad (concentración) del ácido N1
N1 V1 = No Vo
No = concentración molar de la base (M o)
N1=NoVoV1
N1=(0.0998)(3.35)10
N1=0.03343M
CALCULO DE Ka
Para realizar el cálculo de laconstante es necesario elaborar la curva de titulación
1.- En el punto inicial
pH = | 1 | log { K a [HA] } | ( 6.5 ) |
| 2 | | |
[HA] = molaridad del ácido débil ( M ).
Ka=10-pH2[HA]
Ka=10-2.902[0.03]=5.28 x10-5
2.- En la trayectoria 12.
pH = – log K a – log | [ (Vo – V) ] |
| V |
Ka = 10^(-pH + log Vo-VV)
Ka = 10^(-3.70 +log 3.35-0.50.5) =1.137×10-3
Ka = 10^(-3.98 + log 3.35-11) =2.46×10-4
3.-En la trayectoria 23.
pH = – log K a – log | [ (Vo – V) ] |
| V |
Ka = 10-(pH + log Vo-VV)
Ka = 10^(-5.4 + log 3.35-2.82.8) =7.81×10-7
Ka = 10^(-5.61 + log 3.35-3.13.1) =1.97×10-7
4.- En el punto medio
Ka = 10 – pH
Ka =10-5.95=1.12×10-6
Ka promedio =2.32×10-4
Calculo del error porcentual de Ka conrespecto al reportado en la literatura ( 1.85 x 10 – 5 )
erp=2.32×10-4-1.85×10-52.32×10-4×100 = 92.16%
Datos equipo 2
TABLA 6.1 REGISTRO DE DATOS EXPERIMENTALES CNaOH = __0.0998M__ CAcético =_0.03M__ V Acético =__10mL__ |
V NaOH mL |
0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 2.6 | 2.7 | 2.8 | 2.9 | 3.0 | 3.1 | 3.2 | 3.3 | 3.4 | 3.5 | 4.0 | 4.5 | 5.0 |
pH |
3 | 3.66| 4.10 | 4.46 | 4.73 | 5.06 | 5.15 | 5.18 | 5.30 | 5.39 | 5.46 | 5.65 | 5.77 | 6.0 | 6.36 | 7.38 | 11.47 | 11.88 | 12.03 |
TABLA 6.2 CÁLCULO DE ?pH/?V |
V NaOH mL |
0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 2.6 | 2.7 | 2.8 | 2.9 | 3.0 | 3.1 | 3.2 | 3.3 | 3.4 | 3.5 | 4.0 | 4.5 | 5.0 |
?pH/?V |
? | 0.66 | 0.44 | 0.36 | 0.27 | 0.33 | 0.09 | 0.03 | 0.12 | 0.09 |0.07 | 0.19 | 0.12 | 0.23 | 0.36 | 1.02 | 4.09 | 0.41 | 0.15 |
V promedio = (V inicial de NaOH + V final de NaOH)/2 |
? | 0.25 | 0.75 | 1.25 | 1.75 | 2.25 | 2.55 | 2.65 | 2.75 | 2.85 | 2.95 | 3.05 | 3.15 | 3.25 | 3.35 | 3.45 | 3.75 | 4.25 | 4.75 |
GRAFICA ?pH / ?V VS. V promedio
Vo= 3.75mL
b) Calcular la normalidad (concentración) del ácido N1
N1 V1 = No Vo
No = concentración molar de labase (M o)
N1=NoVoV1
N1=(0.0998)(3.75)10
N1=0.03742M
CALCULO DE Ka
Para realizar el cálculo de la constante es necesario elaborar la curva de titulación
1.- En el punto inicial
pH = | 1 | log { K a [HA] } | ( 6.5 ) |
| 2 | | |
[HA] = molaridad del ácido débil ( M ).
Ka=10-pH2[HA]
Ka=10-32[0.03]=3.33×10-5
2.- En latrayectoria 12.
pH = – log K a – log | [ (Vo – V) ] |
| V |
Ka = 10^(-pH + log Vo-VV)
Ka = 10^(-3.666 + log 3.75-0.50.5) =1.422×10-3
Ka = 10^(-4.10 + log 3.75-11) =2.18×10-4
3.-En la trayectoria 23.
pH = – log K a – log | [ (Vo – V) ] |
| V |
Ka = 10-(pH + log Vo-VV)
Ka = 10^(-5.3 + log 3.75-2.82.8) =1.70×10-6
Ka = 10^(-5.65 + log 3.75-3.13.1) =4.69×10-7
4.- En el punto…
