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Titrations of acids and bases. Titrations of acids and bases. HA + H 2 O H 3 O + + A -. Titrations of acids and bases. HA + H 2 O H 3 O + + A -. B + H 2 O OH - + HB +. Titrations of acids and bases. HA + H 2 O H 3 O + + A -. - PowerPoint PPT Presentation
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Titrations of acids and bases
Titrations of acids and bases
HA + H2O H3O+ + A-
Titrations of acids and bases
HA + H2O H3O+ + A-
B + H2O OH- + HB+
Titrations of acids and bases
HA + H2O H3O+ + A-
B + H2O OH- + HB+
H3O+ + OH- 2 H2O
H3O+ + OH- 2 H2O
At equivalence point [H3O+] = [OH-]
For strong acid – strong base
H3O+ + OH- 2 H2O
At equivalence point [H3O+] = [OH-]
Kw = [ H3O +] = [OH-] = 10-14
H3O+ + OH- 2 H2O
At equivalence point [H3O+] = [OH-]
Kw = [ H3O +] = [OH-] = 10-14
pH = 7
Titrations of acids and bases
HA + H2O H3O+ + A-
B + H2O OH- + HB+
H3O+ + OH- 2 H2O
Volume of base added
Start with acid solution
pH
7
pH
4
10
100 ml 0.1 M HCl
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
H3O+ + OH- 2 H2O
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
H3O+ + OH- 2 H2O
Every part of a mole of NaOH added
reduces the moles of H3O+ by an equal amount.
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
H3O+ + OH- 2 H2O
As NaOH solution is added, the overall volume
increases. This further decreases [H3O+].
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Neutralization and dilution; both increase
the pH of the solution.
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 50 ml of NaOH solution.
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 50 ml of NaOH solution.
Start with 0.1 L x 0.1 M H3O+
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 50 ml of NaOH solution.
Start with 0.1 L x 0.1 M H3O+
0.01 moles H3O+
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 50 ml of NaOH solution.
0.01 moles H3O+
Add 0.05 L x 0.1 M OH-
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 50 ml of NaOH solution.
0.01 moles H3O+
Add 0.05 L x 0.1 M OH-
0.005 moles OH-
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 50 ml of NaOH solution.
0.01 moles H3O+
0.005 moles OH-
H3O+ is neutralized by OH- 1:1 ratio
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 50 ml of NaOH solution.
0.01 moles H3O+
0.005 moles OH-
Remaining H3O+ = 0.01 moles – 0.005 moles
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 50 ml of NaOH solution.
0.01 moles H3O+
0.005 moles OH-
Remaining H3O+ = 0.005 moles
Total volume = 150 ml = 0.15 L
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 50 ml of NaOH solution.
0.01 moles H3O+
0.005 moles OH-
[H3O+]= 0.005 moles0.15 L
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 50 ml of NaOH solution.
0.01 moles H3O+
0.005 moles OH-
[H3O+]= 0.005 moles0.15 L
= 0.033 M
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 50 ml of NaOH solution.
0.01 moles H3O+
0.005 moles OH-
[H3O+]= 0.005 moles0.15 L
= 0.033 M
pH = 1.48
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 99.9 ml of NaOH solution.
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 99.9 ml of NaOH solution.
0.01 moles H3O+
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 99.9 ml of NaOH solution.
0.01 moles H3O+
0.0999 L x 0.1 M = 0.00999 moles OH-
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 99.9 ml of NaOH solution.
0.01 moles H3O+
0.0999 L x 0.1 M = 0.00999 moles OH-
Moles H3O+ = 0.01 – 0.00999 = 1 x 10-5
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 99.9 ml of NaOH solution.
0.01 moles H3O+
0.0999 L x 0.1 M = 0.00999 moles OH-
Moles H3O+ = 0.01 – 0.00999 = 1 x 10-5
Total volume = 0.1 L + 0.0999 L = 0.1999 L
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 99.9 ml of NaOH solution.
0.01 moles H3O+
0.0999 L x 0.1 M = 0.00999 moles OH-
[H3O+]= 1 x 10-5 moles
0.1999 L
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 99.9 ml of NaOH solution.
0.01 moles H3O+
0.0999 L x 0.1 M = 0.00999 moles OH-
[H3O+]= 1 x 10-5 moles
0.1999 L = 5 x 10-5
100 ml 0.1 M HCl
Titrate with 0.1 M NaOH
Add 99.9 ml of NaOH solution.
0.01 moles H3O+
0.0999 L x 0.1 M = 0.00999 moles OH-
[H3O+]= 1 x 10-5 moles
0.1999 L = 5 x 10-5
pH = 4.3
4.3
4.3
+ 0.1 ml will give
pH = 7
If more 0.1 M NaOH solution is added
after the equivalence point, there is no
H3O+ to neutralize it.
At pH = 7 the volume is 0.2 L
At pH = 7 the volume is 0.2 L
Add 0.1 mL 0.1 M NaOH
At pH = 7 the volume is 0.2 L
Add 0.1 mL 0.1 M NaOH
Add 0.0001 L x 0.1 M = 10-5 moles OH-
[OH-] = 1 x 10-5 moles
0.201 L= 5 x 10-5 M
At pH = 7 the volume is 0.2 L
Add 0.1 mL 0.1 M NaOH
Add 0.0001 L x 0.1 M = 10-5 moles OH-
[OH-] = 1 x 10-5 moles
0.2001 L= 5 x 10-5 M
pOH = 4.3 pH = 9.7
4.3
pH = 7
V = 199.9 ml
V = 200 ml
pH = 9.7 V = 200.1 ml
Titrating a weak acid with a strong base
Titrating a weak acid with a strong base
HA + H2O H3O+ + A-
Ka < 1
Titrating a weak acid with a strong base
HA + H2O H3O+ + A-
Ka < 1
Much less than 100% dissociation.
Titrating a weak acid with a strong base
HA + H2O H3O+ + A-
Ka < 1
Much less than 100% dissociation.
Every OH- added neutralizes an H3O+
and shifts the equilibrium to the right.
Titrate 100 ml 0.10 M CH3COOH
With 0.10 M NaOH
Titrate 100 ml 0.10 M CH3COOH
With 0.10 M NaOH
Ka = 1.8 x 10-5
Titrate 100 ml 0.10 M CH3COOH
With 0.10 M NaOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
0 ml NaOH
Titrate 100 ml 0.10 M CH3COOH
With 0.10 M NaOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
0 ml NaOH
pH = 2.87
Titrate 100 ml 0.10 M CH3COOH
With 0.10 M NaOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
50 ml NaOH
.1 L x 0.1 M = 0.01 moles CH3COOH
0.05L x 0.1 M = 0.005 moles NaOH
CH3COOH + H2O H3O+ + CH3COO-
+OH-
2 H2O
CH3COOH + H2O H3O+ + CH3COO-
+OH-
2 H2O
Net result: for every OH- added, there is
one less CH3COOH and one more CH3COO-
Titrate 100 ml 0.10 M CH3COOH
With 0.10 M NaOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
50 ml NaOH
.1 L x 0.1 M = 0.01 moles CH3COOH
0.05L x 0.1 M = 0.005 moles NaOH
0.01 - 0.005 moles CH3COOH = 0.005 moles
Titrate 100 ml 0.10 M CH3COOH
With 0.10 M NaOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
50 ml NaOH
.1 L x 0.1 M = 0.01 moles CH3COOH
0.05L x 0.1 M = 0.005 moles NaOH
0.01 - 0.005 moles CH3COOH = 0.005 moles
0.005 moles CH3COO-
Titrate 100 ml 0.10 M CH3COOH
With 0.10 M NaOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
50 ml NaOH
0.01 - 0.005 moles CH3COOH = 0.005 moles
0.005 moles CH3COO-
0.15 L solution
[CH3COOH] = 0.005 moles
0.005 moles CH3COO-
0.15 L= 0.033 M
= 0.033 M
[CH3COOH] = 0.005 moles
0.005 moles CH3COO-
0.15 L= 0.033 M
= 0.033 M
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
Buffer solution: pKa = pH = 4.74
Titrate 100 ml 0.10 M CH3COOH
With 0.10 M NaOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
100 ml NaOH
0.01 moles CH3COO-
0.20 L solution
Titrate 100 ml 0.10 M CH3COOH
With 0.10 M NaOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
100 ml NaOH
0.01 moles CH3COO-
0.20 L solution
[CH3COO-] = 0.05 M
Titrate 100 ml 0.10 M CH3COOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
[CH3COO-] = 0.05 M
Kb = Kw
Ka
= 10-14
1.8 x 10-5
= 5.6 x 10-10
Kb =[CH3COOH][OH-]
[CH3COO-]
Titrate 100 ml 0.10 M CH3COOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
[CH3COO-] = 0.05 M
Kb = Kw
Ka
= 10-14
1.8 x 10-5
= 5.6 x 10-10
Kb =[CH3COOH][OH-]
[CH3COO-]=
y2
0.05
Titrate 100 ml 0.10 M CH3COOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
Kb = Kw
Ka
= 10-14
1.8 x 10-5
= 5.6 x 10-10
Kb =[CH3COOH][OH-]
[CH3COO-]=
y2
0.05
y2 = (5.6 x 10-10)(0.05) = 2.8 x 10-11
Titrate 100 ml 0.10 M CH3COOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
Kb = Kw
Ka
= 10-14
1.8 x 10-5
= 5.6 x 10-10
Kb =[CH3COOH][OH-]
[CH3COO-]=
y2
0.05
y2 = (5.6 x 10-10)(0.05) = 2.8 x 10-11
y = 5.3 x 10-6
Titrate 100 ml 0.10 M CH3COOH
Ka = 1.8 x 10-5 = [H3O+][CH3COO-]
[CH3COOH]
Kb = Kw
Ka
= 10-14
1.8 x 10-5
= 5.6 x 10-10
Kb =[CH3COOH][OH-]
[CH3COO-]=
y2
0.05
y2 = (5.6 x 10-10)(0.05) = 2.8 x 10-11
y = 5.3 x 10-6 pOH =5.3; pH = 8.7
After the equivalence point,
OH- is being added to a saturated
buffer system.
After the equivalence point,
OH- is being added to a saturated
buffer system.
pH increases rapidly
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