Pract

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) =1.137x10-3

Ka = 10^(-3.98 + log 3.35-11) =2.46x10-4

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) =7.81x10-7

Ka = 10^(-5.61 + log 3.35-3.13.1) =1.97x10-7

4.- En el punto medio

Ka = 10 – pH
Ka =10-5.95=1.12x10-6

Ka promedio =2.32x10-4

Calculo del error porcentual de Ka conrespecto al reportado en la literatura ( 1.85 x 10 – 5 )

erp=2.32x10-4-1.85x10-52.32x10-4x100 = 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]=3.33x10-5

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) =1.422x10-3

Ka = 10^(-4.10 + log 3.75-11) =2.18x10-4

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) =1.70x10-6

Ka = 10^(-5.65 + log 3.75-3.13.1) =4.69x10-7

4.- En el punto...