# Problemas

Solo disponible en BuenasTareas
• Páginas : 4 (777 palabras )
• Descarga(s) : 0
• Publicado : 29 de mayo de 2011

Vista previa del texto
EXPERIMENTO 1: Parte 1

Calcular cada punto para la curva de titulación de 15 ml de NaOH 0.05M con HCl 0.05M. Hasta haber agregando 30 ml , agregando del titulante ml por ml.

1) [NaOH]= 0.05M= [OH]
pOH= -log[0.05]= 1.3
pH= 12.7

2) V agregado HCl= 1 ml

[HCl] = 0.05M = nHCl / 1ml

nHCl = 0.05 m mol

[NaOH]= 0.05M = nNaOH / 15 ml

nNaOH = 0.75 m mol

Pero :nNaOH sin titular = (0.75 – 0.05) m mol

nNaOH sin titular = 0.7 m mol

[NaOH] = 0.7 m mol / (15 + 1) ml

[NaOH]= 0.044 M = [OH]
pOH= -log[0.044]= 1.36pH= 12.46

3) V agregado HCl= 1 ml

[HCl] = 0.05M = nHCl / 1ml

nHCl = 0.05 m mol

nNaOH = 0.7 m mol

Pero :

nNaOH sin titular = (0.7 – 0.05) m molnNaOH sin titular = 0.65 m mol

[NaOH] = 0.65 m mol / (16 + 1) ml

[NaOH]= 0.038 M = [OH]
pOH= -log[0.038]= 1.42
pH= 12.58

4) V agregado HCl= 1 ml

[HCl] = 0.05M = nHCl /1ml

nHCl = 0.05 m mol

nNaOH = 0.65 m mol

Pero :

nNaOH sin titular = (0.65 – 0.05) m mol

nNaOH sin titular = 0.6 m mol

[NaOH] = 0.6m mol / (17 + 1) ml

[NaOH]= 0.033 M = [OH]
pOH= -log[0.033]= 1.48
pH= 12.52

5) V agregado HCl= 1 ml

[HCl] = 0.05M = nHCl / 1ml

nHCl = 0.05 m mol

nNaOH = 0.6 m mol

Pero :nNaOH sin titular = (0.6 – 0.05) m mol

nNaOH sin titular = 0.55 m mol

[NaOH] = 0.55 m mol / (18 + 1) ml

[NaOH]= 0.029 M = [OH]
pOH= -log[0.029]=4.54
pH= 12.46

6) V agregado HCl= 1 ml

[HCl] = 0.05M = nHCl / 1ml

nHCl = 0.05 m mol

nNaOH = 0.55 m mol

Pero :

nNaOH sin titular = (0.55 – 0.05) mmol

nNaOH sin titular = 0.5 m mol

[NaOH] = 0.5 m mol / (19 + 1) ml

[NaOH]= 0.025 M = [OH]
pOH= -log[0.025]= 1.6
pH= 12.4

7) V agregado HCl= 1 ml

[HCl] = 0.05M =...