Transcript Titolazione base debole

curva di titolazione base debole-acido forte
100 mL di NH3 0.1 M con HCL 0.1 M
OH- + NH4+
/
/
x
x
H2O + NH3
i 0. 1 M
e 0.1 -x
[OH-] [NH4+]
Kb =
[NH3]
Kb = x x
0.1-x
5
x  1.8  10  0.1
x= 1.34x10-3
pOH= - log (1.34.10-3) = 2.87
pH = 14- pOH =11.13
V(ml) pH
0
[H+]
11.13 7.41x10-12
100 mL di NH3 0.1 M con 1 mL di HCL 0.1 M
nHCl = 1 x 10-3 x 0.1 = 10-4
nNH3 = 0.100 x 0.1 = 10-2
NH4+ +Cl/-4
10
10-4
NH3 +
HCl
-4
i 10-2
10
f 9.9x 10-3
/
OH- + NH4+
/
10-4
0.101
x x + 9.9.10-4
H2O + NH3
i
9.9x10-3
0.101
e 0.098 -x
1.8.10-5
x
.10-4x +x2
9.9
=
0.098
4
 9.9  10

.10-4 + x) x
(9.9
Kb =
0.098-x
x2 + 9.9.10-4x - 1.76.10-6 = 0
9.9  10 
pOH = - log (9.2.10-4)
[OH-] [NH4+]
Kb =
[NH3]
4 2
 4  1.76  10 6
2
pOH= 3.03
 9.2  10 4
pH= 14-pOH= 10.97
V(ml) pH
[H+]
0
11.13 7.41x10-12
1
10.97 1.07x10-11
100 mL di NH3 0.1 M con 5 mL di HCL 0.1 M
nHCl = 5 x 10-3 x 0.1 = 5x10-4
nNH3 = 0.100 x 0.1 = 10-2
NH4+ +Cl/-4
5x10
5x10-4
NH3 +
HCl
-4
i 10-2
5x10
f 9.5x 10-3
/
H2O + NH3
i
9.5x10-3
0.105
e 0.090 -x
1.8.10-5
x
OH- + NH4+
/
5x10-4
0.105
x x + 4.8.10-3
.10-3x +x2
4.8
=
0.090
3
 4.8  10 
.10-3 + x) x
(4.8
Kb =
0.090-x
x2 + 4.8.10-3x - 1.63.10-6 = 0
4.8  10 
3 2
6
 4  1.63  10
2
pOH = - log (3.2.10-4)
[OH-] [NH4+]
Kb =
[NH3]
pOH= 3.5
4
 3.2  10
pH=14-pOH= 10.5
V(ml)
0
1
5
10
pH
11.13
10.97
10.5
10.2
[H+]
7. 41x10-12
1.07 x10-11
3.1610-11
6.3x10-11
100 mL di NH3 0.1 M con 10 mL di HCL 0.1 M
nHCl = 10 x 10-3 x 0.1 = 10-3
nNH3 = 0.100 x 0.1 = 10-2
NH4+ +Cl/-3
10
10-3
NH3 +
HCl
i 10-2 -3
10-3
f 9x 10
/
OH- + NH4+
/
10-3
0.110
x x + 9.1. 10-3
H2O + NH3
i
9 x10-3
0.110
e 0.082 -x
1.8.10-5
x
.10-3x +x2
9.1
=
0.082
3
 9.1  10 
pOH = - log (1.6.10-4)
[OH-] [NH4+]
Kb =
[NH3]
.10-3 + x) x
(9.1
Kb =
0.082-x
x2 + 9.1.10-3x - 1.47.10-6 = 0
9.1  10 
3 2
 4  1.47  10 6
2
pOH= 3.8
 1.6  10 4
pH= 14-pOH= 10.2
V(ml)
0
1
5
10
pH
11.13
10.97
10.5
10.2
[H+]
7. 41x10-12
1.07 x10-11
3.1610-11
6.3x10-11
100 mL di NH3 0.1 M con 25 mL di HCL 0.1 M
nNH3= 0.100 x 0.1 = 10-2
NH3 +
HCl
i 10-2 -3 2.5x10-3
f 7.5x 10
/
H2O + NH3
i
7.5x10-3
0.125
e 0.06 -x
1.8.10-5
nHCl = 25 x 10-3 x 0.1 = 2.5x10-3
NH4+ + Cl/
-3
2.510 2.5x10-3
OH- + NH4+
/
2.5x10-3
0.125
x x + 2.10-2
.10-2 x
2
=
0.06
.10-5.0.06
1.8
.10-5
x=
=
5.4
2.10-2
[OH-] [NH4+]
Kb =
[NH3]
.10-2 + x) x
(2
Kb =
0.06-x
pOH = - log (5.4.10-5)
pOH= 4.27
pH=14-pOH= 9.73
V(ml)
0
1
5
10
25
pH
11.13
10.97
10.5
10.2
9.73
[H+]
7. 41x10-12
1.07 x10-11
3.1610-11
6.3x10-11
1.86x10-10
100 mL di NH3 0.1 M con 50 mL di HCL 0.1 M
nNH3 = 0.100 x 0.1 = 10-2
NH3 +
HCl
i 10-2 -3 5x10-3
f 5x 10
/
H2O + NH3
i
5x10-3
0.15
e 3.3.10-2 -x
1.8.10-5
nHCl= 50 x 10-3 x 0.1 = 5 x10-3
NH4+ + Cl/ -3
5x10 5x10-3
OH- + NH4+
/
5x10-3
0.15
x x + 3.3 . 10-2
[OH-] [NH4+]
Kb =
[NH3]
.10-2+x)x
(3.3
Kb = . -2
3.3 10 -x
.10-2 x
3.3
=
3.3.10-2
pOH = - log (1.8.10-5)
x = 1.8.10-5
pOH= 4.74
pH=14-pOH= 9.26
V(ml)
0
1
5
10
25
50
pH
11.13
10.97
10.5
10.2
9.73
9.26
[H+]
7. 41x10-12
1.07 x10-11
3.1610-11
6.3x10-11
1.86x10-10
5.49x10-10
100 mL di NH3 0.1 M con 75 mL di HCL 0.1 M
nNH3 = 0.100 x 0.1 = 10-2
nHCl= 75 x 10-3 x 0.1 = 7.5 x10-3
NH3 +
HCl
NH4+ + Cl-3
i 10-2
7.5x10
/
-3
-3
f 2.5x 10
/
7.5x10 7.5x10-3
H2O + NH3
i
2.5x10-3
0.175
e 1.4.10-2 -x
1.8.10-5
OH- + NH4+
/
7.5x10-3
0.175
x x + 4.3 . 10-2
.10-2 x
4.3
=
1.4.10-2
.10-2 .1.8.10-5
1.4
x=
4.3.10-2
[OH-] [NH4+]
Kb =
[NH3]
.10-2+x)x
(4.3
Kb = . -2
1.4 10 -x
pOH = - log (5.9.10-6)
x = 5.9.10-6
pH =14-pOH= 8.77
pOH= 5.23
V(ml)
0
1
5
10
25
50
75
pH
11.13
10.97
10.5
10.2
9.73
9.26
8.77
[H+]
1.34x10-3
9.2 x 10-4
3.1610-11
6.3x10-11
1.86x10-10
5.49x10-10
1.7x 10-9
100 mL di NH3 0.1 M con 100 mL di HCL 0.1 M
nHCl = 0.100 x 0.1 = 10-2
nNH3 = 0.100 x 0.1 = 10-2
NH3 +
i 10-2
f /
NH4+
i 10-2
0.2
e 5.10-2 -x
5.55.10-10
=
x = 5.27.10-6
NH4+ +Cl/ -2
10 10-2
HCl
10-2
/
H+ + NH3
/
/
x
x2
5.10-2
x
[H+] [NH3]
Ka =
[NH4+]
Ka = x. x-2
5 10 -x
x  5.55  1010  5  102
pH = - log (5.27.10-6)
pH= 5.28
punto equivalente
moli di acido = moli di base
V(ml)
0
1
5
10
25
50
75
100
pH
11.13
10.97
10.5
10.2
9.73
9.26
8.77
5.28
[H+]
7. 41x10-12
1.07 x10-11
3.1610-11
6.3x10-11
1.86x10-10
5.49x10-10
1.7x 10-9
5.27x10-6
100 mL di NH3 0.1 M con 110 mL di HCL 0.1 M
nHCl = 0.110 x 0.1 = 1.1x10-2
nNH3 = 0.100 x 0.1 = 10-2
NH3 +
i 10-2
f /
HCl
i 10-3
0.21
f /
HCl
-2
1.1x10
10-3
H+
/
NH4+ + Cl/
/ -2
-2
10
10
+ Cl/
4.76 x10-3 4.76 x10-3
pH = - log (4.76.10-3)
pH= 2.32
V(ml)
0
1
5
10
25
50
75
100
110
0
50
100
150
volume di acido aggiunto/mL
pH
11.13
10.97
10.5
10.2
9.73
9.26
8.77
5.28
2.32
[H+]
7. 41x10-12
1.07 x10-11
3.1610-11
6.3x10-11
1.86x10-10
5.49x10-10
1.7x 10-9
5.27x10-6
4.76x10-3