Titulacion

Laboratorio Titulaciones Polifuncionales
Mateo Múnera Manco
Lina Toro Ocampo
Análisis Químico-Biología

Titulación Acido Polifuncional
H2SO4 NaOH
0.100M + 0.200M 50.00mL X mL




H2SO4 + H2O Ka1 HSO4- + H3O+

HSO4- + H2O Ka2 SO42-+ H3O+

Ka2 = [SO42-] [H3O+] = 1.02 x 10-2 [HSO4-] |

PUNTO INICIAL

* VolNaOH= 0.00mL H2SO4 + H2O Ka1 HSO4- + H3O+

Balance de [H3O+]
[H3O+]TOTAL= [H3O+]H2SO4 +[H3O+]HSO4-

HSO4- + H2O Ka2 SO42- + H3O+

Ka2= [SO42-] [H3O+]
[HSO4-]
[H3O+]HSO4-= [SO42-]
[H3O+]HSO4-= 0.1 + [SO42-]
[SO42-]= [H3O+]HSO4- - 0.1

0.1M= [HSO4-] + [SO42-]
[HSO4-]= 0.1M - [SO42-]
[HSO4-]= 0.1M - ([H3O+]HSO4- - o.1)

[HSO4-]= 0.2M - [H3O+]HSO4-

Ka2= ([H3O+] -0.1M) [H3O+] = [H3O+]2 - 0.1[H3O+]
0.2M - [H3O+] 0.2M - [H3O+]


Primera Región Buffer
HSO4- + OH- SO42- + H2O
HSO4- + H2O SO42- + H30+
SO42- + H20 HS04- + OH-

* VolNaOH= 30.00mL

[HSO4-]= (50mL x 0.1molHSO4- - (30 – 25)mL x 0.2molNaOH x 1molHSO4-) x 1 x1000mL=0.05M
1000mL 1000mL 1molNaOH (50 + 30)mL 1L

[SO42-]= ((30 -25)mL x 0.2molOH- x 1molSO42-) x 1 x 1000mL= 0.0125M
1000mL 1molOH- (50+30)mL 1L

[HSO4-]= 0.05 – [H30+] + [OH-]= 0.05 – [H30+]
[SO42-]= 0.0125 – [0H-] + [H30+]=0.0125 + [H30+]

Ka2= [SO42-] [H30+] Ka2= (0.0125 + [H30+]) [H30+]
[HS04-] 0.05 – [H30+]

[H30+]2 + 0.0125[H3O+] = Ka2(0.05 – [H30+])
[H30+]2 + (Ka2 + 0.0125)[H30+] – Ka2(0.05)= 0
[H30+]2 + 0.0227 [H30+] – 5.1x10-4
[H30+]= 0.0139M

pH= -Log[H3O+]
pH= 1.86

* VolNaOH= 45.00mL

[HSO4-]= (50mL x 0.1molHSO4- - (45 –25)mL x 0.2molNaOH x 1molHSO4-) x 1 x 1000mL=0.011M
1000mL 1000mL 1molNaOH (50 + 45)mL 1L

[SO42-]= ((45 -25)mL x 0.2molOH- x 1molSO42-) x 1 x 1000mL= 0.042M
1000mL 1molOH- (50+45)mL 1L

[HSO4-]= 0.011 –[H30+] + [OH-]= 0.011 – [H30+]
[SO42-]= 0.042 – [0H-] + [H30+]= 0.042 + [H30+]

Ka2= [SO42-] [H30+] Ka2= (0.042 + [H30+]) [H30+]
[HS04-] 0.011 – [H30+]

[H30+]2 + 0.042[H3O+] = Ka2(0.011 – [H30+])
[H30+]2 + (Ka2 + 0.042)[H30+] – Ka2(0.011)= 0
[H30+]2 + 0.0522[H30+] – 1.122x10-4= 0
[H30+]= 0.00207M
pH=-Log[H3O+]
pH= 2.68

Segundo Punto de Equivalencia

* VolNaOH= 50.00mL

H3O+ + OH- 2H2O
HSO4- + OH- SO42- + H2O
SO42- + H2O HSO4- + OH-

Kb= [HSO4-][OH-] Kb= Kw = 10-14 = 9.80 x 10-13
[SO42-] Ka 1.02 x 10-2

[SO42-]= ((50 - 25)mL x 0.2molOH- x 1molSO42-) x 1 x 1000mL=0.05M
1000mL 1molOH- (50 +50)mL 1L
[HSO4-]= [OH-]
0.05M= [SO42-] [OH-]
Kb= [HSO4-][OH-] 9.80 x 10-13= [OH-][OH-]
[SO42-] 0.05M

[OH-]= √(9.80 x 10-13) (0.05M)
Ka2(0.2 - [H3O+]) = [H3O+]2 - 0.1[H3O+]
[H3O+]2 - 0.1[H3O+] = 0.2Ka2...