Petrucci • Harwood • Herring • Madura
GENERAL
CHEMISTRY
Ninth
Edition
Principles and Modern Applications
Chapter 16: Acids and Bases
Philip Dutton
University of Windsor, Canada
Prentice-Hall © 2007
Slide 1 of 52
General Chemistry: Chapter 16
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Contents
16-1
16-2
16-3
16-4
16-5
16-6
16-7
16-8
Slide 2 of 52
The Arrhenius Theory: A Brief Review
Brønsted-Lowry Theory of Acids and Bases
The Self-Ionization of Water and the pH Scale
Strong Acids and Strong Bases
Weak Acids and Weak Bases
Polyprotic Acids
Ions as Acids and Bases
Molecular Structure and Acid-Base Behaviour
General Chemistry: Chapter 16
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16-1 The Arrhenius Theory:
A Brief Review
H2O
HCl(g) → H+(aq) + Cl-(aq)
H2O
NaOH(s) → Na+(aq) + OH-(aq)
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq)
H+(aq) + OH-(aq) → H2O(l)
Arrhenius theory did not handle non OHbases such as ammonia very well.
Slide 3 of 52
General Chemistry: Chapter 16
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16-2 Brønsted-Lowry Theory of
Acids and Bases
 An acid is a proton donor.
 A base is a proton acceptor.
base
acid
NH3 + H2O
NH4+ + OHacid
Slide 4 of 52
base
NH4+ + OHNH3 + H2O
??
??
General Chemistry: Chapter 16
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The Solvated Proton
Slide 5 of 52
General Chemistry: Chapter 16
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Base Ionization Constant
base
acid
NH3 + H2O
Kc=
conjugate conjugate
base
acid
NH4+ + OH-
[NH4+][OH-]
[NH3][H2O]
[NH4+][OH-]
Kb= Kc[H2O] =
= 1.810-5
[NH3]
Slide 6 of 52
General Chemistry: Chapter 16
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Acid Ionization Constant
acid
base
CH3CO2H + H2O
conjugate
base
conjugate
acid
CH3CO2- + H3O+
[CH3CO2-][H3O+]
Kc=
[CH3CO2H][H2O]
[CH3CO2-][H3O+]
Ka= Kc[H2O] =
= 1.810-5
[CH3CO2H]
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General Chemistry: Chapter 16
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Slide 8 of 52
General Chemistry: Chapter 16
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A Weak Base
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General Chemistry: Chapter 16
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A Weak Acid
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General Chemistry: Chapter 16
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A Strong Acid
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General Chemistry: Chapter 16
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16-3 The Self-Ionization of Water and
the pH Scale
Slide 12 of 52
General Chemistry: Chapter 16
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Ion Product of Water
base
acid
H2O + H2O
conjugate conjugate
base
acid
H3O+ + OH-
[H3O+][OH-]
Kc=
[H2O][H2O]
KW= Kc[H2O][H2O] = [H3O+][OH-] = 1.010-14
Slide 13 of 52
General Chemistry: Chapter 16
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pH and pOH
 The potential of the hydrogen ion was defined in
1909 as the negative of the logarithm of [H+].
pH = -log[H3O+]
pOH = -log[OH-]
KW = [H3O+][OH-]= 1.010-14
-logKW = -log[H3O+]-log[OH-]= -log(1.010-14)
pKW = pH + pOH= -(-14)
pKW = pH + pOH = 14
Slide 14 of 52
General Chemistry: Chapter 16
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pH and pOH Scales
Slide 15 of 52
General Chemistry: Chapter 16
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16-4 Strong Acids and Bases
HCl
CH3CO2H
Thymol Blue Indicator
pH < 1.2 < pH < 2.8 < pH
Slide 16 of 52
General Chemistry: Chapter 16
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16-5 Weak Acids and Bases
Acetic Acid
Lactic Acid
Slide 17 of 52
General Chemistry: Chapter 16
Glycine
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2007
Weak Acids
Acetic Acid
[CH3CO2-][H3O+]
Ka=
= 1.810-5
[CH3CO2H]
pKa= -log(1.810-5) = 4.74
Slide 18 of 52
General Chemistry: Chapter 16
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2007
Weak Bases
[CH3NH3+][HO-]
Kb=
= 4.310-4
[CH3NH2]
pKb= -log(4.210-4) = 3.37
Slide 19 of 52
General Chemistry: Chapter 16
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Table 16.3 Ionization Constants of Weak Acids and Bases
Slide 20 of 52
General Chemistry: Chapter 16
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EXAMPLE 16-5
Determining a Value of KA from the pH of a Solution of a
Weak Acid. Butyric acid, HC4H7O2 (or CH3CH2CH2CO2H) is
used to make compounds employed in artificial flavorings and
syrups. A 0.250 M aqueous solution of HC4H7O2 is found to
have a pH of 2.72. Determine KA for butyric acid.
HC4H7O2 + H2O
C4H7O2 + H3O+
Ka = ?
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General Chemistry: Chapter 16
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EXAMPLE 16-5
Solution:
For HC4H7O2 KA is likely to be much larger than KW. Therefore
assume self-ionization of water is unimportant.
HC4H7O2 + H2O
C4H7O2 + H3O+
Initial conc.
0.250 M
0
0
Changes
-x M
+x M
+x M
Equilibrium
Concentration
(0.250-x) M
xM
xM
Slide 22 of 52
General Chemistry: Chapter 16
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EXAMPLE 16-5
HC4H7O2 + H2O
C4H7O2 + H3O+
Log[H3O+] = -pH = -2.72
[H3O+] = 10-2.72 = 1.910-3 = x
Ka=
[H3O+] [C4H7O2-]
[HC4H7O2]
Ka= 1.510-5
Slide 23 of 52
1.910-3 · 1.910-3
=
(0.250 – 1.910-3)
Check assumption: Ka >> KW.
General Chemistry: Chapter 16
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Percent Ionization
H3O+ + A-
HA + H2O
[H3O+] from HA
Degree of ionization =
[HA] originally
[H3O+] from HA
Percent ionization =
Slide 24 of 52
[HA] originally
General Chemistry: Chapter 16
 100%
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Percent Ionization
Ka =
[H3O+][A-]
[HA]
n H O nA 1
Ka =
nHA V
3
Slide 25 of 52
General Chemistry: Chapter 16
+
-
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16-6 Polyprotic Acids
Phosphoric acid:
A triprotic acid.
H3PO4 + H2O
H3O+ + H2PO4-
Ka = 7.110-3
H2PO4- + H2O
H3O+ + HPO42-
Ka = 6.310-8
HPO42- + H2O
H3O+ + PO43-
Ka = 4.210-13
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General Chemistry: Chapter 16
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Phosphoric Acid
 Ka1 >> Ka2
◦ All H3O+ is formed in the first ionization step.
 H2PO4- essentially does not ionize further.
◦ Assume [H2PO4-] = [H3O+].
 [HPO42-]  Ka2 regardless of solution molarity.
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General Chemistry: Chapter 16
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Slide 28 of 52
General Chemistry: Chapter 16
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EXAMPLE 16-9
Calculating Ion Concentrations in a Polyprotic Acid
Solution. For a 3.0 M H3PO4 solution, calculate:
(a) [H3O+];
(b) [H2PO4-];
(c) [HPO42-]
(d) [PO43-]
H3PO4 + H2O
H2PO4- + H3O+
Initial conc.
3.0 M
0
0
Changes
-x M
+x M
+x M
Equilibrium
Concentration
(3.0-x) M
xM
xM
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General Chemistry: Chapter 16
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EXAMPLE 16-9
H3PO4 +
Ka=
H2PO4- + H3O+
H2O
[H3O+] [H2PO4-]
[H3PO4]
=
x·x
(3.0 – x)
= 7.110-3
Assume that x << 3.0
x2 = (3.0)(7.110-3)
x = 0.14 M
[H2PO4-] = [H3O+] = 0.14 M
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General Chemistry: Chapter 16
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EXAMPLE 16-9
H2PO4- +
H2O
HPO42- + H3O+
Initial conc.
0.14 M
0
0.14 M
Changes
-y M
+y M
+y M
Equilibrium
Concentration
(0.14 - y) M
yM
(0.14 +y) M
Ka=
[H3O+] [HPO42-]
[H2PO4
y << 0.14 M
Slide 31 of 52
-]
=
y · (0.14 + y)
= 6.310-8
(0.14 - y)
y = [HPO42-] = 6.310-8
General Chemistry: Chapter 16
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EXAMPLE 16-9
HPO4- +
Ka=
[H3O+] [HPO42-]
[H2PO4-]
PO43- + H3O+
H2O
=
(0.14)[PO43-]
= 4.210-13 M
6.310-8
[PO43-] = 1.910-19 M
Slide 32 of 52
General Chemistry: Chapter 16
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Sulfuric Acid
Sulfuric acid:
A diprotic acid.
H2SO4 + H2O
H3O+ + HSO4-
Ka = very large
HSO4- + H2O
H3O+ + SO42-
Ka = 1.96
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General Chemistry: Chapter 16
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General Approach to Solution Equilibrium
Calculations
 Identify species present in any significant amounts
in solution (excluding H2O).
 Write equations that include these species.
 Number of equations = number of unknowns.
◦ Equilibrium constant expressions.
◦ Material balance equations.
◦ Electroneutrality condition.
 Solve the system of equations for the unknowns.
Slide 34 of 52
General Chemistry: Chapter 16
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16-7 Ions as Acids and Bases
CH3CO2- + H2O
base
acid
NH4+ + H2O
acid
Ka=
CH3CO2H + OH-
NH3 + H3O+
Ka=
base
[NH3] [H3O+] [OH-]
[NH4+] [OH-]
=
KW
Kb
[NH3] [H3O+]
[NH4
1.010-14
=
1.810-5
+]
=?
= 5.610-10
Ka Kb = Kw
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General Chemistry: Chapter 16
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Hydrolysis
 Water (hydro) causing cleavage (lysis) of a bond.
Slide 36 of 52
Na+ + H2O → Na+ + H2O
No reaction
Cl- + H2O → Cl- + H2O
No reaction
NH4+ + H2O → NH3 + H3O+
Hydrolysis
General Chemistry: Chapter 16
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16-8 Molecular Structure and
Acid-Base Behavior
 Why is HCl a strong acid, but HF is a weak one?
 Why is CH3CO2H a stronger acid than CH3CH2OH?
 There is a relationship between molecular structure
and acid strength.
 Bond dissociation energies are measured in the gas
phase and not in solution.
Slide 37 of 52
General Chemistry: Chapter 16
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Strengths of Binary Acids
HI
HBr
HCl
HF
Bond length
160.9 > 141.4 > 127.4 > 91.7
pm
Bond energy
297
< 368
kJ/mol
Acid strength
109
> 108 > 1.3106 >> 6.610-4
< 431
HF + H2O → [F-·····H3O+]
ion pair
H-bonding
Slide 38 of 52
General Chemistry: Chapter 16
< 569
F- + H3O+
free ions
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Strengths of Oxoacids
 Factors promoting electron withdrawal from the
OH bond to the oxygen atom:
 High electronegativity (EN) of the central atom.
 A large number of terminal O atoms in the molecule.
Slide 39 of 52
H-O-Cl
H-O-Br
ENCl = 3.0
ENBr= 2.8
Ka = 2.910-8
Ka = 2.110-9
General Chemistry: Chapter 16
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··
··
Strengths of Oxoacids
··
··
O
O
··
O
··
S
H
H
··
O
··
··
O
··
S
··
H
··
··
H
··
O
··
O
Ka 103
Ka =1.310-2
·· O
··
··
··
··
·· O
S
2+
··
O
··
H
H
··
O
··
S
··
+
··
O
··
H
-
··
··
H
··
O
··
O
··
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General Chemistry: Chapter 16
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H
H
··
··
Strengths of Organic Acids
O
C
H
··
O
··
C
C
H
H
H
acetic acid
ethanol
C
Ka = 1.810-5
Slide 41 of 52
H
H
H
··
O
··
H
Ka =1.310-16
General Chemistry: Chapter 16
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