1. Chapter 16: Acids and Bases
Chemistry 140 Fall 2002
CHEMISTRY
Ninth
Edition
GENERAL
Principles and Modern Applications
Petrucci • Harwood • Herring • Madura
Chapter 16: Acids and Bases
Juana Mendenhall, Ph.D.
Assistant Professor
Morehouse College
General Chemistry: Chapter 16
Prentice-Hall © 2007

2. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
Objectives
Calculate one of Ka, [H+] or [H3O+], or the molarity of a weak acid, given the other two (doing the same for a weak base).
Become familiar with percent ionization
Calculate the strength of acids and bases
General Chemistry: Chapter 16
Prentice-Hall © 2007

3. General Approach to Solution Equilibrium Calculations
Chemistry 140 Fall 2002
General Approach to Solution Equilibrium Calculations
Identify species present in any significant amounts in solution (excluding H2O).
Write equations that include these species.
Write the Ionization Equation
Arrange in tabular form and emphasize:.
Initial Conditions
Changes that occur, &
Equilibrium concentrations
Substitute equilibrium concentrations into the ionization constant expression.
Solve for the quantity of interest.
General Chemistry: Chapter 16
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4. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
Example
A M solution of HN3 has pH = What is Ka for this acid?
[H3O+][N3-]
HN3(aq) + H2O(aq)
H3O+(aq) + N3-(aq)
Ka =
[HN3]
1. Next,
[H3O+] = 10-pH = = 1.4 x 10-3 M
2. Make chart
HN3(aq) + H2O(aq)
H3O+(aq) + N3-(aq) I:
0.101 M M M C: M M M E:
0.100 M
[H3O+][N3-]
(0.0014)(0.0014)
Ka = = =
2.0 x M
[HN3]
0.100
General Chemistry: Chapter 16
Prentice-Hall © 2007

5. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
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-]
H3PO H2O H2PO H3O+
Initial conc. 3.0 M 0 0
Changes -x M +x M +x M
Equilibrium (3.0-x) M x M x M
Concentration
General Chemistry: Chapter 16
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6. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
EXAMPLE 16-9
H3PO H2O H2PO H3O+
[H3O+] [H2PO4-]
x · x
Ka= =
= 7.110-3
[H3PO4]
(3.0 – x)
Assume that x << 3.0
x2 = (3.0)(7.110-3) x = 0.14 M
[H2PO4-] = [H3O+] = 0.14 M
General Chemistry: Chapter 16
Prentice-Hall © 2007

7. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
EXAMPLE 16-9
H2PO H2O HPO H3O+
Initial conc M M
Changes -y M +y M +y M
Equilibrium ( y) M y M (0.14 +y) M
Concentration
[H3O+] [HPO42-]
[H2PO4-]
Ka=
y · ( y)
( y) =
= 6.310-8
y << 0.14 M y = [HPO42-] = 6.310-8
General Chemistry: Chapter 16
Prentice-Hall © 2007

8. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
EXAMPLE 16-9
HPO H2O PO H3O+
[H3O+] [HPO42-]
(0.14)[PO43-]
= 4.210-13 M
Ka= =
[H2PO4-]
6.310-8
[PO43-] = 1.910-19 M
General Chemistry: Chapter 16
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9. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
Percent Ionization
HA + H2O H3O+ + A-
Degree of ionization =
[H3O+] from HA
[HA] originally
Percent ionization =
[H3O+] from HA
[HA] originally
 100%
General Chemistry: Chapter 16
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10. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
% Ionization Example
A student prepared a 0.10 M solution of formic acid, HCHO2, and
measured its pH using a pH meter. The pH at 25 °C was found to be 2.38.
Calculate the Ka for formic acid at this temperature.
What percentage of the acid is ionized in this 0.10 M solution?
Answer
HCHO2(aq)
H+(aq) + CHO2-(aq)
General Chemistry: Chapter 16
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11. 16-7 Ions as Acids and Bases
Chemistry 140 Fall 2002
16-7 Ions as Acids and Bases
CH3CO2- + H2O CH3CO2H + OH-
base
acid
[NH3] [H3O+]
Ka=
[NH4+]
= ?
NH4+ + H2O NH3 + H3O+
acid
base
[NH3] [H3O+] [OH-]
Ka=
[NH4+] [OH-] = KW Kb =
1.010-14
1.810-5
= 5.610-10
Ka Kb = Kw
General Chemistry: Chapter 16
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12. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
Hydrolysis
Water (hydro) causing cleavage (lysis) of a bond.
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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13. 16-8 Molecular Structure and Acid-Base Behavior
Chemistry 140 Fall 2002
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.
Strength depends on properties of the solvent and temperature.
General Chemistry: Chapter 16
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14. Strengths of Binary Acids
Chemistry 140 Fall 2002
Strengths of Binary Acids
HI HBr HCl HF
Bond length
160.9 > > > pm
Bond energy
297 < 368 < 431 < kJ/mol
Acid strength
109 > 108 > 1.3106 >> 6.610-4
More energy is required to break the H—F bone, HF is the weak acid. HI has the lowest bond energy, so HI is the strongest acid of the group.
The polarity of the bond actually decreases from HF to HI b/c F is the most electonegative of the halogens.
HF + H2O → [F-·····H3O+] F- + H3O+
ion pair
H-bonding
free ions
General Chemistry: Chapter 16
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