Transcript Titration Curves I. Strong Acid 0.1 M HCl

Titration Curves
I. Strong Acid +
0.1 M HCl
25.0 mL
2.5 x 10-3mol
1. Initial pH
HCl  H++ Cl0.1 M 0.1 M
[H+] = 0.1 M
pH = - log H+ = 1.00
Strong Base
0.1 M NaOH
25.0 mL
2.5 x 10-3mol
10.0
pH
5.0
.
0
10
20
30
volume NaOH added (mL)
40
50
Strong Acid + Strong Base
0.1 M HCl
0.1 M NaOH
25.0 mL
10.0 mL
2.5 x 10-3mol - 1.0x 10-3 mol = 1.5 x 10-3 mol
V = 25 + 10 mL
[H+] =1.5 x 10-3 mol
35 x 10-3 L
[H+] = 4.28 x 10-2 M
10.0
pH
5.0
.
pH =1.37
0
.
10
20
30
volume NaOH added (mL)
40
50
Strong Acid + Strong Base
0.1 M HCl
0.1 M NaOH
25.0 mL
20.0 mL
2.5 x 10-3mol - 2.0 x 10-3 mol = 0.5 x 10-3 mol
V = 25 + 20 mL
[H+] =0.5 x 10-3 mol
45 x 10-3 L
[H+] = 1.11 x 10-2 M
10.0
pH
5.0
.
pH =1.95
0
.
.
10
20
30
volume NaOH added (mL)
40
50
Strong Acid + Strong Base
0.1 M HCl
0.1 M NaOH
25.0 mL
25.0 mL
2.5 x 10-3mol - 2.5 x 10-3 mol = 0.0 mol
[H+] = 1.00 x 10-7 M
10.0
pH
pH =7.00
.
5.0
.
0
.
.
10
20
30
volume NaOH added (mL)
40
50
Strong Acid +
0.1 M HCl
25.0 mL
2.5 x 10-3mol
Strong Base
0.1 M NaOH
35.0 mL
3.5 x 10-3 mol OH-
V = 25 + 35 mL
[OH-] = 1.0 x 10-3 mol
60 x 10-3 L
[OH-] = 1.67 x 10-2 M
pOH =1.78
pH = 12.22
.
10.0
pH
.
5.0
.
0
.
.
10
20
30
volume NaOH added (mL)
40
50
Titration Curves
Weak Acid
+
0.1 M CH3COOH
25.0 mL
Initial
weak acid
Ka = 1.8 x
10-5
=
[H+]
Strong Base
0.1 M NaOH
25.0 mL
pH = pKa + log [CH3COO-]
[CH3COOH]
COO-]
[CH3
[CH3COOH]
half-way point pH = pKa = 4.74
10.0
equivalence point
pH
CH3COO + H2O  CH3COOH + OH5.0
Kb = 5.6 x 10-10 = [OH-] [CH3COOH]
[CH3COO-]
strong base
0
10
20
30
volume NaOH added (mL)
40
50
Titration Curves
Weak Base
+
Strong Acid
0.1 M NH3
0.1 M HCl
25.0 mL
25.0 mL
1. Initial pH
NH3 + H2O  NH4+ + OHKb= 1.8 x 10-5 = [OH-][NH4+]
[NH3]
10.0
pH
5.0
[OH-] = 1.34 x 10-3 M
pOH = 2.87 pH = 11.12
0
10
20
30
volume HCl added (mL)
40
50
Titration Curves
Weak Base
+ Strong Acid
0.1 M NH3
0.1 M HCl
25.0 mL
10.0 mL
2.5 x 10-3mol
- 1.0 x 10-3 mol
= 1.5 x 10-3 mol
V = 25 + 10 mL
x = 2.67 x 10-5 pOH = 4.57
NH3 + H2O  NH4+ + OHpH = 9.43
Kb= 1.8 x 10-5 = [NH4+] [OH-]
[NH3]
[NH3]
0.043
0.043 -x
[NH4+]
0.029
0.029 + x
[OH-]
0.0
x
1.8 x 10-5 = [x] [0.029 + x]
[0.043 - x]
.
10.0
pH
5.0
0
10
20
30
volume HCl added (mL)
40
50
Titration Curves
Weak Base
+ Strong Acid
0.1 M NH3
0.1 M HCl
25.0 mL
20.0 mL
2.5 x 10-3mol
- 2.0 x 10-3 mol
= 5.0 x 10-4 mol
V = 25 + 20 mL
x = 4.5 x 10-6 pOH = 5.35
NH3 + H2O  NH4+ + OHpH = 8.65
Kb= 1.8 x 10-5 = [NH4+] [OH-]
[NH3]
[NH3]
0.011
0.011 -x
[NH4+]
0.044
0.044 + x
[OH-]
0.0
x
1.8 x 10-5 = [x] [0.044 + x]
[0.011 - x]
.
10.0
pH
.
5.0
0
10
20
30
volume HCl added (mL)
40
50
Titration Curves
Weak Base
+ Strong Acid
0.1 M NH3
0.1 M HCl
25.0 mL
25.0 mL
2.5 x 10-3mol
- 2.5 x 10-3 mol
= 0.00
V = 25 + 25 mL
-6
x
=
5.9
x
10
NH4+  NH3 +H+
Ka= 5.6 x 10-10 = [NH3] [H+]
[NH4+]
[NH4]
0.05
0.05 -x
5.6 x 10-10 =
[NH3]
0.00
x
[x2]
[0.05 - x]
[H+]
0.0
x
.
10.0
10.0
pH = 5.27
.
pH
pH
.
5.0
5.0
0
0
10
10
20
20
30
30
volume
volume HCl
HCl added
added (mL)
(mL)
40
40
50
50
Titration Curves
Weak Base
0.1 M NH3
25.0 mL
2.5 x
+
Strong Acid
0.1 M HCl
20.0 mL
10-3mol
pH = 8.65
Ka = 1.8 x 10-5
pOH = pKb + log [NH4+]
[NH3]
5.0 x 10-4 mol NH3
10.0
pH
2.0 x 10-3 mol NH4+
5.0
V = 45 x 10-3 L
pOH = 4.74 + log (0.44) = 5.34
(0.11)
0
10
20
30
volume HCl added (mL)
40
50
Polyprotic Acid
H2SO3
 HSO3-
+ H+
HSO3-  SO32- + H+
2 equivalents of base
0.10 M H2SO3
40 mL
Initial pH
Ka1 = 1.4 x 10-2
Ka2 = 6.5 x 10-8
0.10 M NaOH
80 mL
1.4 x 10-2 = [HSO3-] [H+] = x2
[H2SO3] 0.1 - x
10.0
pH
x = 0.03
pH = 1.51
5.0
1
equivalents of base
2
Polyprotic Acid
H2SO3
 HSO3-
+ H+
HSO3-  SO32- + H+
2 equivalents of base
0.10 M H2SO3
half-way point
Ka1 = 1.4 x 10-2
Ka2 = 6.5 x 10-8
0.1 M NaOH
pH = pKa
buffering regions
- log 1.4 x 10-2 = 1.85
- log 6.5 x 10-8 = 7.19
1st equivalence point
1.84 + 7.19 = 4.52
2
2nd equivalence point
conjugate base, SO3-
.
10.0
10.0
pH
pH
.
.
5.0
5.0
.
2
1
1
equivalents of base
equivalents of base
2
Polyprotic Acid
H2SO3  HSO3- + H+ Ka1 = 1.4 x 10-2
HSO3-  SO32- + H+ Ka2 = 6.5 x 10-8
2 equivalents of base
0.10 M H2SO3
40 mL
0.1 M NaOH
80 mL
Initial pH
1.4 x 10-2 = [HSO3-][H+] = x2
[H2SO3]
0.1 - x
x = 0.03
pH = 1.51