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Titration Curves
I. 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 10-3mol
1. Initial pH HCl H++ Cl-0.1 M 0.1 M[H+] = 0.1 M
pH = - log H+ = 1.00
pH
volume NaOH added (mL)
5.0
10.0
010 20 30 40 50
.
pH
volume NaOH added (mL)
5.0
10.0
010 20 30 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 mol35 x 10-3 L
[H+] = 4.28 x 10-2 M
pH =1.37 .
pH
volume NaOH added (mL)
5.0
10.0
010 20 30 40 50
.
Strong Acid + Strong Base
0.1 M HCl 0.1 M NaOH
25.0 mL
2.5 x 10-3mol
.
20.0 mL
2.0 x 10-3 mol- = 0.5 x 10-3 mol
V = 25 + 20 mL
[H+] = 0.5 x 10-3 mol45 x 10-3 L
[H+] = 1.11 x 10-2 M
pH =1.95 .
pH
volume NaOH added (mL)
5.0
10.0
010 20 30 40 50
.
Strong Acid + Strong Base
0.1 M HCl 0.1 M NaOH
25.0 mL
2.5 x 10-3mol
.
25.0 mL
2.5 x 10-3 mol- = 0.0 mol
[H+] =
pH =7.00
.
1.00 x 10-7 M
.
pH
volume NaOH added (mL)
5.0
10.0
010 20 30 40 50
.
Strong Acid + Strong Base
0.1 M HCl 0.1 M NaOH
25.0 mL
2.5 x 10-3mol
.
35.0 mL
V = 25 + 35 mL
[OH-] = 1.0 x 10-3 mol60 x 10-3 L
[OH-] = 1.67 x 10-2 M
pOH =1.78 .
3.5 x 10-3 mol OH-
.
pH = 12.22
.
pH
volume NaOH added (mL)
5.0
10.0
010 20 30 40 50
Titration CurvesWeak Acid + Strong Base
0.1 M CH3COOH 0.1 M NaOH
25.0 mL 25.0 mL
pH
volume NaOH added (mL)
5.0
10.0
010 20 30 40 50
Initial weak acid
half-way point pH = pKa
Ka = 1.8 x 10-5
equivalence point
= [H+] [CH3COO-]
[CH3COOH]
= 4.74
CH3COO- + H2OCH3COOH + OH-
Kb = 5.6 x 10-10 = [OH-] [CH3COOH]
[CH3COO-]
strong base
pH = pKa + log [CH3COO-] [CH3COOH]
Titration Curves
Weak Base + Strong Acid
0.1 M NH3 0.1 M HCl
25.0 mL 25.0 mL
1. Initial pH
NH3 NH4+ + OH-
[OH-] = 1.34 x 10-3 M
pOH = 2.87
Kb= 1.8 x 10-5 = [OH-][NH4+]
[NH3]
+ H2O
pH = 11.12
pH
volume HCl added (mL)
5.0
10.0
010 20 30 40 50
Titration Curves
Weak Base + Strong Acid0.1 M NH3 0.1 M HCl
25.0 mL2.5 x 10-3mol
NH3 NH4+ + OH-
Kb= 1.8 x 10-5 = [NH4+] [OH-]
[NH3]
10.0 mL1.0 x 10-3 mol- = 1.5 x 10-3 mol
V = 25 + 10 mL
[NH3] [NH4+] [OH-]
0.043 0.029 0.00.043 -x 0.029 + x x
1.8 x 10-5 = [x] [0.029 + x]
[0.043 - x]
x = 2.67 x 10-5 pOH = 4.57+ H2O
pH = 9.43
pH
volume HCl added (mL)
5.0
10.0
010 20 30 40 50
.
Titration Curves
Weak Base + Strong Acid0.1 M NH3 0.1 M HCl
25.0 mL2.5 x 10-3mol
NH3 NH4+ + OH-
Kb= 1.8 x 10-5 = [NH4+] [OH-]
[NH3]
20.0 mL2.0 x 10-3 mol- = 5.0 x 10-4 mol
V = 25 + 20 mL
[NH3] [NH4+] [OH-]
0.011 0.044 0.00.011 -x 0.044 + x x
1.8 x 10-5 = [x] [0.044 + x]
[0.011 - x]
x = 4.5 x 10-6 pOH = 5.35+ H2O pH = 8.65
pH
volume HCl added (mL)
5.0
10.0
010 20 30 40 50
. .
Titration Curves
Weak Base + Strong Acid0.1 M NH3 0.1 M HCl
25.0 mL2.5 x 10-3mol
NH3NH4+ +H+
Ka= 5.6 x 10-10 = [NH3] [H+]
[NH4+]
25.0 mL2.5 x 10-3 mol- = 0.00
V = 25 + 25 mL
[NH4] [NH3] [H+]
0.05 0.00 0.00.05 -x x x
5.6 x 10-10 = [x2]
[0.05 - x]
x = 5.9 x 10-6 pH = 5.27
pH
volume HCl added (mL)
5.0
10.0
010 20 30 40 50
. .
.pH
volume HCl added (mL)
5.0
10.0
010 20 30 40 50
Titration Curves
Weak Base + Strong Acid0.1 M NH3 0.1 M HCl
25.0 mL
2.5 x 10-3mol
20.0 mL
pOH = pKb + log [NH4+]
[NH3]
5.0 x 10-4 mol NH3
2.0 x 10-3 mol NH4+
V = 45 x 10-3 L
pOH =
Ka = 1.8 x 10-5
4.74 + log (0.44) (0.11)
= 5.34
pH
volume HCl added (mL)
5.0
10.0
010 20 30 40 50
pH = 8.65
Polyprotic AcidH2SO3 HSO3
- + H+ Ka1 = 1.4 x 10-2
HSO3- SO3
2- + H+ Ka2 = 6.5 x 10-8
2 equivalents of base
0.10 M H2SO3 0.10 M NaOH
40 mL 80 mL
Initial pH
1.4 x 10-2 = [HSO3-] [H+]
[H2SO3]= x2
0.1 - x
x = 0.03 pH = 1.51
equivalents of base
1 2
pH
5.0
10.0
Polyprotic AcidH2SO3 HSO3
- + H+ Ka1 = 1.4 x 10-2
HSO3- SO3
2- + H+ Ka2 = 6.5 x 10-8
2 equivalents of base
0.10 M H2SO3 0.1 M NaOH
equivalents of base
1 2
pH
5.0
10.0
half-way point pH = pKa
- log 1.4 x 10-2 = 1.85
.
- log 6.5 x 10-8 = 7.19
.1st equivalence point
1.84 + 7.192
= 4.52 .
2nd equivalence point
conjugate base, SO3-
.
buffering regions
equivalents of base
1 2
pH
5.0
10.0
Polyprotic Acid
H2SO3 HSO3- + H+ Ka1 = 1.4 x 10-2
HSO3- SO3
2- + H+ Ka2 = 6.5 x 10-8
2 equivalents of base
0.10 M H2SO3 0.1 M NaOH40 mL 80 mL
Initial pH
1.4 x 10-2 = [HSO3-][H+]
[H2SO3]
= x2
0.1 - x
x = 0.03
pH = 1.51