Solubilidad

TABLA DE DATOS EXPERIMENTALES
Porcentaje en vol. De acido en agua | ml de acido | ml de agua | Vol. De alcohol gastado |
10 % | 1 ml | 9 ml | 1.3 ml |
15% | 1.5 ml | 8.5 ml | 2.5 ml |
20% | 2 ml | 8 ml | 4.5 ml |
25% | 2.5 ml | 7.5 ml | 7.5 ml |
Porcentaje en vol. de acido en alcohol | ml de acido | ml de alcohol | Vol. de H20 gastado |
10% | 1 ml | 9 ml | 2.5 ml |
20% | 2 ml | 8ml | 5.8 ml |
30% | 3 ml | 7 ml | 9 ml |
40% | 4 ml | 6 ml | 13 ml |
TABLA # 1

TABLA # 2

TITULACION
M CH3COOH | 0.9526 g |
V NaOH | 15 ml |
TABLA #2

fase | Vol tomado | W muestra | NaOH gastado |
acuosa | 5 ml | 4.8931 | 6.2 ml |
organica | 5 ml | 5.1493 | 8.5 ml |
TABLA# 3

TABLA DE DATOS TEÓRICOS
Sustancia | agua | CH3COOH | N-BUTANOL |
densidad | 0.99829 g/ml |1.049 g/ml | 0.81 g/ml |

RESULTADOS
ACIDO EN AGUA
sustancia | Al 10% | Al 15% | Al 20% | Al 25% |
CH3COOH | 9.46% | 13.02% | 15.28% | 16.2% |
H2O | 81.04% | 70.22% | 58.17% | 46.26% |
n-butanol | 9.5% | 16.76% | 26.55% | 37.54% |

ACIDO EN ALCOHOL
sustancia | Al 10% | Al 20% | Al 30% | Al 40% |
CH3COOH | 9.68% | 14.6% | 17.68% | 19.04% |
H2O | 67.28% | 45.1% | 31.85% | 22.06%|
n-butanol | 23.04% | 40.3% | 50.47% | 58.9% |

* Titulo = 0.0635 g de CH3COOH/ ml de NaOH

* % en peso de la determinación de la línea de reparto
* CH3COOH = 11.42%
* N – butanol =39.68%
* H2O = 48.90%

* % en peso en la fase acuosa =8.05%
% en peso en la fase organica =10.48%

CÁLCULOS
a) Calculando los % en peso de cada componente:
ACIDO EN AGUAAl 10 % :
* CH3COOH
1ml de CH3COOH x 1.049gml=1.049g de CH3COOH
* H2O

9 ml De H2O x 0.99829 gml=8.984 g de H2O

* N – butanol

1.3 ml de n-butanol x 0.81gml=1.053 g de n-butanol

W total = W de CH3COOH + W de agua + W del n- butanol
W total = 1.049 + 8.984 + 1.053
W total = 11.086
* CH3COOH
1.049g de CH3COOH11.086 g total x100%=9.46%
* H2O
8.984 g den-butanol 11.086 g totalx100=81.04%
* N – butanol

1.053 g de H2O11.086 g total x 100%=9.50%
Al 15 % :

* CH3COOH
1.5 ml de CH3COOH x 1.049gml=1.5735g de CH3COOH
* H2O

8.5 ml De H2O x 0.99829 gml=8.486 g de H2O

* N – butanol

2.5 ml de n-butanol x 0.81gml=2.025 g de n-butanol

W total = W de CH3COOH + W de agua + W del n- butanol
W total = 1.5735 + 8.486 + 2.025W total = 12.0845
* CH3COOH
1.5735g de CH3COOH12.0845 g total x100%=13.02%
* H2O

8.486 g de n-butanol 12.0845 g totalx100%=70.22%
* N – butanol

2.025 g de H2O12.0845 g total x 100%=16.76%

Al 20 % :

* CH3COOH
2 ml de CH3COOH x 1.049gml=2.098 g de CH3COOH
* H2O

8 ml De H2O x 0.99829 gml=7.986 g de H2O

* N – butanol

4.5 ml de n-butanol x0.81gml=3.645 g de n-butanol

W total = W de CH3COOH + W de agua + W del n- butanol
W total = 2.098 + 7.986 + 3.645
W total = 13.729
* CH3COOH
2.098g de CH3COOH13.729 g total x100%=15.28%
* H2O

7.986 g de n-butanol 13.729 g totalx100%=58.17%
* N – butanol

3.645 g de H2O13.729 g total x 100%=26.55%

Al 25 % :

* CH3COOH
2.5 ml de CH3COOH x 1.049gml=2.6225 gde CH3COOH
* H2O

7.5 ml De H2O x 0.99829 gml=7.487 g de H2O

* N – butanol

7.5 ml de n-butanol x 0.81gml=6.075 g de n-butanol

W total = W de CH3COOH + W de agua + W del n- butanol
W total = 2.6225 + 7.487 + 6.075
W total = 16.1845
* CH3COOH
2.6225g de CH3COOH16.1845g total x100%=16.20%
* H2O

7.487 g de n-butanol 16.1845 g totalx100%=46.26%
* N –butanol

6.075g de H2O16.1845 g total x 100%=37.54%

ACIDO EN ALCOHOL

Al 10 % :
* CH3COOH
1ml de CH3COOH x 1.049gml=1.049g de CH3COOH
* N – butanol

9 ml de n-butanol x 0.81gml=7.29 g de n-butanol
* H2O

2.5 ml De H2O x 0.99829 gml=2.496 g de H2O

W total = W de CH3COOH + W del n- butanol + W de agua
W total = 1.049 + 7.29 + 2.496
W total = 10.835
*...