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TABLA DE DATOS EXPERIMENTALES
Practica #1
Tabla #1:
Gramo de NaAc(s) | 0.15 g ( ±0.01) |
Solucion de HAc(ac) | 100 ml |
Solucion de HCl (0.1M) | 0.5ml, 1 ml, 1.5 ml (±0.1) |
Agua Destilada | 20 ml (±0.1) |
Solucion de NaOH (0.1M) | 0.5 ml, 1 ml, 1.5 ml, 2 ml, 2.5 ml, 3 ml (±0.1) |

Practica #2 (Apreciación: Bureta (±0.1), ph-metro (±0.1))
Tabla #1:
Mililitros de NaOH (ml) (±0.1) | PH (±0.1) |
0.0 ml | 2.8 |
1.0 ml | 3.3 |
2.0 ml | 3.5 |
3.0 ml | 3.8 |
4.0 ml | 4.1 |
5.0 ml | 4.4 |
6.0 ml | 4.7 |
7.0 ml | 5.2 |
8.0 ml | 5.4 |
9.0 ml | 10.0 |
10.0 ml | 10.1 |
11.0 ml | 10.2 |
12 .0 ml | 10.9 |

Tabla #2:
Mililitros de NaOH (ml) (±0.1) | PH (±0.1) |
0.0 ml | 2.5 |
1.0 ml | 2.8 |
2.0 ml | 3.2 |
3.0 ml | 3.6 |
4.0 ml | 3.8 |
5.0ml | 4.1 |
6.0 ml | 4.1 |
7.0 ml | 4.2 |
8.0 ml | 4.4 |
8.2 ml | 4.6 |
8.4 ml | 4.7 |
8.6 ml | 4.8 |
8.8 ml | 4.9 |
9.0 ml | 5.0 |
9.2 ml | 5.1 |
9.4 ml | 5.2 |
9.6 ml | 5.4 |
9.8 ml | 5.5 |
10.0 ml | 5.8 |
10.2 ml | 7.5 |
10.4 ml | 10.0 |
10.6 ml | 10.5 |
10.8 ml | 10.6 |
11.0 ml | 10.8 |

Tabla #3:
Mililitros de NaOH (ml) (±0.1) | PH (±0.1) |
0.0 ml |2.6 |
1.0 ml | 2.7 |
2.0 ml | 2.9 |
3.0 ml | 3.3 |
4.0 ml | 3.6 |
5.0 ml | 3.8 |
6.0 ml | 4.2 |
7.0 ml | 4.4 |
8.0 ml | 4.6 |
8.1 ml | 4.7 |
8.2 ml | 4.7 |
8.3 ml | 4.8 |
8.4 ml | 4.9 |
8.5 ml | 4.9 |
8.6 ml | 5.0 |
8.7 ml | 5.1 |
8.8 ml | 5.2 |
8.9 ml | 5.3 |
9.0 ml | 5.4 |
9.1 ml | 5.4 |
9.2 ml | 5.5 |
9.3 ml | 5.5 |
9.4 ml | 5.6 |
9.5 ml | 5.7 |9.6 ml | 5.8 |
9.7 ml | 5.8 |
9.8 ml | 5.9 |
9.9 ml | 5.9 |
10.0 ml | 6.1 |
10.1 ml | 10.0 |
10.2 ml | 10.1 |
10.3 ml | 10.1 |
10.4 ml | 10.2 |
10.5 ml | 10.3 |
10.6 ml | 10.4 |
10.7 ml | 10.5 |

Tabla #4:
Militros de Vinagre (±0.1) | 1 ml |
Militros de Agua Destilada (±0.1) | 10 ml |

CALCULOS Y GRAFICAS

Practica #1
* Determinación del PH teórico del Buffermol NaAc=g NaAc*PM=0.15 g NaAc* 1 mol NaAc82.03 g NaAc=0.0018 mol NaAc
NaAcs → Na+ + Ac-
0.0018 mol 0.0018 mol 0.0018mol

HAc + H2O ↔ Acs + H3O+
i) (0.1M) 0.0018mol
r) - Mxmx Mx
e) (0.1 M - Mx) (0.0018 mol + mx) Mx
No se toma para el equilibrio
Ka=[H3O+][HAc]= [Mx][0.1M-Mx] , donde Ka=1.8*10-5
1.8*10-5= Mx0.1M- Mx
Mx=H3O+=0.000183 M
pH= -logH3O+= -log[0.000183]
pHteo=3.74

* pH obtenidos al mezclarle diferentes tipos de soluciones

* AguaDestilada
pHexp=3.6

* Solucion de HCl
0.5 ml | 3.2 |
1.0 ml | 3.1 |
1.5 ml | 3.1 |

* Solucion de NaOH
0.5 ml | 3.6 |
1.0 ml | 3.6 |
1.5 ml | 3.6 |
2.0 ml | 3.6 |
2.5 ml | 3.7 |
3.0 ml | 3.8 |

* Determinación de la desviación estándar
Para la solución de HCl:
x= 1N1Nxi
x= 13 i=13Xi= 13 3.2+3.1+3.1=3.133
σ= 1N-1i=1NXi-x2
σ= 12i=13Xi-3.1332
σ=123.2-3.1332+3.1-3.1332+3.1-3.1332
σ= 12*0.00667=0.0033
σ=0.057

Para la solución de NaOH:
x= 1N1Nxi
x= 16 i=13Xi= 16 3.6+3.6+3.6+3.6+3.7+3.8=3.65
σ= 1N-1i=1NXi-x2
σ= 15i=13Xi-3.652
σ= 153.6-3.652+3.6-3.652+3.6-3.652+3.6-3.652+3.7-3.652+3.8-3.652
σ= 15*[0.35]=0.007
σ=0.084

Practica #2
* Determinación de la concentración de vinagre

* Datos Teoricos
Mvinagre= 6 gr HAc100 ml vinagre×1 molHAc60 gr HAc×103ml1 Lts=1M

MaVa=MbVb
1M*(1ml)=0.1M*Vb
Vb=10 ml
, donde: Vb es el volumen de NaOH, Va es el volumen de vinagre, Ma es la molaridad del vinagre y Mb es la molaridad de NaOH.

#1:

Grafico #1: ph-metro variación de ph
Ecuacion de la recta:y= 0.7225x+1.6956 , Si y=7, x=7.34 ml
MaVa=MbVb
1M*Va=0.1M*(7.34ml)
Va=0.734 ml
%conct= 0.734 ml10 ml*100=7.34 %...
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