2. UNIT 14: ACIDS AND BASES
• The Nature of Acids and Bases
• Acid Strength
• The pH Scale
• Calculating the pH of Strong Acid Solutions
• Calculating the pH of Weak Acid Solutions
• Bases
• Polyprotic Acids
• Acid-Base Properties of Salts
• The Effect of Structure on Acid-Base Properties
• Acid-Base Properties of Oxides
• The Lewis Acid-Base Model
• Strategy for Solving Acid-Base Problems

4. ARRHENIUS CONCEPT
The first person to recognize the essential
nature of acids and bases was Svante
Arrhenius. He stated that acids produce
hydrogen ions in aqueous solutions, while
bases produce hydroxide ion.
• Major step in quantifying acid-base
chemistry.
• Concept is very limited because it applies to
aqueous solutions only and it only allows
for only one kind of base – the hydroxide
ion.

5. BRONSTED-LOWRY MODEL
A more general definition of acids and
bases was suggested by the Danish
chemist, Bronsted, and the English
chemist Lowry. An acid is defined as a
proton (H+) donor, and a base is a proton
acceptor.
• includes reactions of gases and not just
aqueous reactions.

6. WATER AS A BASE
Water can act as a base because the water
molecule has two unshared electrons pairs
that allow a covalent bond with a proton
(H+). + -
HA(aq) + H 2O(l ) Û H 3O(aq) + A(aq)
• When water accepts a proton, H3O+,
hydronium is formed.
• HA is the acid (proton donor) and H2O is
the base (proton acceptor).

7. CONJUGATE ACID-BASE PAIR
A conjugate acid-base pair consists of two
substances that are related to each other
by the donation and acceptance of a
proton.
• H3O+is the conjugate acid. The result of
the accepted proton.
• A-is the conjugate base. The result of the
donated proton.
• These two components are considered
conjugate acid base pairs.
+ -
HA(aq) + H 2O(l ) Û H 3O (aq) +A (aq)

8. CONJUGATE ACID-BASE PAIR
H2O and A-, the base and conjugate base
are both competing for a proton.
• If H2O is a stronger base, the reaction
will lie to the right because water will
accept the proton before A- will.
+ -
HA(aq) + H 2O(l ) Þ H3O (aq) +A (aq)

9. CONJUGATE ACID-BASE PAIR
H2O and A-, the base and conjugate base
are both competing for a proton.
• If A- is a stronger base, the reaction will
lie to the left because A- will accept the
proton before H2O will.
+ -
HA(aq) + H 2O(l ) Ü H3O (aq) +A (aq)

10. ACID DISSOCIATION CONSTANT
The equilibrium expression for an acid-
base reaction gives Ka, the acid
dissociation constant:
+ - + -
[H 3O ][A ] [H ][A ]
Ka = =
[HA] [HA]
**Note that H2O is not in the expression
and H+ is commonly substituted for
hydronium.
+ -
HA(aq) + H 2O(l ) Û H 3O (aq) +A (aq)

11. ACID DISSOCIATION CONSTANT
Strong acids have large dissociation
constants.
[H 3O + ][A - ] [H + ][A - ]
Ka = =
[HA] [HA]
Weak acids have small dissociation
constants.
+ -
HA(aq) + H 2O(l ) Û H 3O (aq) +A (aq)

12. PRACTICE PROBLEM #1
Write the simple dissociation (ionization)
reaction (omitting water) for each of the
following acids.
a. hydrochloric acid
b. acetic acid
c. the ammonium ion
d. the anilinium ion (C6H5NH3+)
e. the hydrated aluminum (III) ion [Al(H2O)6]3+

14. ACID STRENGTH
The strength of an acid is defined by the
equilibrium position of its dissociation
reaction.
• A strong acid has an equilibrium that lies to
the right.
• Most HA has dissociated.
• A strong acid produces a weak conjugate
base.
• Water is the stronger base
+ -
HA(aq) + H 2O(l ) Þ H3O (aq) +A (aq)

15. ACID STRENGTH
The strength of an acid is defined by the
equilibrium position of its dissociation
reaction.
• A weak acid has an equilibrium that lies to
the left.
• Most HA is still present.
• A weak acid produces a strong conjugate
base.
• Water is the weaker base
+ -
HA(aq) + H 2O(l ) Ü H3O (aq) +A (aq)

16. ACID STRENGTH SUMMARY
Acid strength and conjugate base strength are
indirectly related.

17. COMMON STRONG ACIDS
1. HCl
2. HNO3
3. HClO4
4. H2SO4
Sulfuric acid is a diprotic acid, an acid with two
acidic protons. H2SO4 is a strong acid and
dissociates completely, but HSO4- is a weak
acid.

18. COMMON WEAK OXYACIDS
1.H3PO4
2.HNO2
3.HOCl
• Most acids are oxyacids, in which the
acidic proton is attached to an oxygen
atom.

19. ORGANIC ACIDS
Organic acids are those with a carbon
atom “backbone”, commonly contain the
carboxyl group:
Acids of this type are usually weak.
1. acetic acid (HC2H3O2 or CH3COOH)
2. benzoic acid (C6H5COOH)
*note: the rest of the hydrogens are not
acidic

20. COMMON MONOPROTIC ACIDS

21. PRACTICE PROBLEM #2
Using Table 14.2, arrange the following
species according to their strengths as
bases: H2O, F-, Cl-, NO2-, and CN-.
Cl-< H2O < F-< NO2-, < CN-

22. WATER AS AN ACID AND A BASE
A substance that can behave either as an
acid or as a base is called amphoteric.
• Water is the most common amphoteric
substance by autoionization.
+ -
H 2O(l) + H 2O(l) Û H3O (aq) + OH (aq)

23. WATER AS AN ACID AND A BASE
The equilibrium expression for this reaction is
the dissociation constant of water, also called
the ion-product constant.
+ - + -
Kw = [H3O ][OH ] = [H ][OH ]
• At 25°C, the ion-product constant, Kw, = 1.0 x
10-14
• [H+] = [OH-]= 1.0 x 10-7 M
+ -
H 2O(l) + H 2O(l) Û H3O (aq) + OH (aq)

24. USING THE ION-PRODUCT CONSTANT
It is important to recognize the meaning of Ks.
In any aqueous solution at 25°C, no matter
what it contains, the product of [H+] and [OH-]
must always equal 1.0 x 10-14.
• [H+] = [OH-] : neutral solution
• [H+] >[OH-] : acidic solution
• [H+] <[OH-] : basic solution

25. PRACTICE PROBLEM #3
Calculate [H+] or [OH-] as required for each
of the following solutions at 25°C, and
state whether the solution is neutral,
acidic, or basic.
a. 1.0 x 10-5 M OH-
b. 1.0 x 10-7 M OH-
c. 10.0 M H+
a. basic b. neutral c. acidic

26. PRACTICE PROBLEM #4
At 60°C, the value of Kw is 1 x 10-13.
+ -
2H 2O(l ) Û H 3O(aq) + OH (aq)
a. Using Le Chatelier’s principle, predict
whether the below is exothermic or
endothermic.
b.Calculate [H+] and [OH-] in a neutral solution
at 60°C.
Kw increases with temperature – endothermic
[H+] =[OH-] = 3 x 10-7 M

28. PH
The pH scale represents solution acidity with
a log based scale.
+
pH = -log[H ]
• [H+] = 1.0 x 10-7
• pH = 7.00
*The number of decimal places in the log is
equal to the number of sig figs in the original
number

29. PH
Because the pH scale is based on a scale
of 10, the pH changes by 1 for every
power of 10 change in [H+]
• example: a solution of pH = 3 has 10
times the concentration of that a solution
of pH 4 and 100 times the concentration
of a solution of pH 5.

30. POH
The pOH scale is similar to the pH scale
-
pOH = -log[OH ]
• [OH-] = 1.0 x 10-7
• pOH = 7.00
*Because the ion product constant of water,
pOH + pH always = 14.

31. PRACTICE PROBLEM #5
Calculate pH and POH for each of the
following solutions at 25°C.
a. 1.0 x 10-3 M OH-
b.1.0 M OH-
a. pH=11.00, pOH=3.00 b. pH=0.00,
pOH=14.00

32. PRACTICE PROBLEM #6
The pH of a sample of human blood was
measured to be 7.41 at 25°C. Calculate
pOH, [H+], and [OH-] for the sample.
pOH=6.59, [H+]= 3.9x10-8, [OH-]= 2.6x10-7M

34. SOLVING ACID-BASE PROBLEMS
Strategies:
1. Think chemistry: Focus on the solution
components and their reactions. Choose the
reaction that is most important.
2. Be systematic: Acid-base problems require a
step by step approach.
3. Be flexible: Treat each problem as a separate
entity. Look for both similarities and the
differences to other problems.

35. SOLVING ACID-BASE PROBLEMS
Strategies:
4. Be patient: The complete solution to a
complicated problem cannot be seen
immediately in all its detail. Pick the
problem apart into its workable steps.
5. Be confident: Do not rely on memorizing
solutions to problems. Understand and
think, do not memorize.

36. CALCULATING THE PH OF STRONG ACID SOLUTIONS
When calculating acid-base equilibria, focus
on the main solution components:
• example: 1.0 M HCl contains virtually no
HCl molecules. Because HCl is a strong
acid, it is completely dissociated. The main
solution components are H+, Cl-, and H2O.
• The first step of solving acid-base
problems is the writing of the major
species in the solution.

37. PRACTICE PROBLEM #7, PART 1
Calculate the pH of 0.10 M HNO3.
major species:
H+, NO-3, and H2O
concentration of species contributing:
H+ = 0.10 M
• H+ from autoionization of water is negligible
compared to H+ from HNO3.
pH
pH = 1.00

38. PRACTICE PROBLEM #7 PART 2
Calculate the pH of 1.0 x 10-10 M HCl.
major species:
H+, Cl-, and H2O
concentration of species contributing:
H+from HCl is negligible.
H+ from autoionization of water is most
important.
pH
pH = 7.00

40. CALCULATING THE PH OF A WEAK ACID
We will walk through a systematic approach to
solving pH for a weak acid:
• Look for major species.
• Narrow down contributing species and its
concentration. Write balanced equation
• Write equilibrium expression.
• ICE
• Use approximations when able
• Check validity

41. CALCULATING THE PH OF A WEAK ACID
What is the pH of a 1.00 M solution of HF, Ka=
7.2x10-4
Major Species:
HF and H2O because both have small
dissociation constants.
Concentration of species contributing:
H+ from HF is far more significant than H+ from
H2O
Ka = 7.2x10-4vs Kw= 1.0x10-14

42. CALCULATING THE PH OF A WEAK ACID
What is the pH of a 1.00 M solution of HF, Ka=
7.2x10-4
strategy:
pH  [H+]  Ka [H + ][F - ]
K a = 7.2x10 -4 =
[HF]
initial[HF] = 1.0 M Equilibrium [H+] = ?
ICE!

43. CALCULATING THE PH OF A WEAK ACID
What is the pH of a 1.00 M solution of HF, Ka=
7.2x10-4
ICE:
Initial Change Equilibriu
m
H+
F-
HF
[H + ][F - ]
K a = 7.2x10 -4 =
[HF]

44. CALCULATING THE PH OF A WEAK ACID
What is the pH of a 1.00 M solution of HF, Ka=
7.2x10-4
ICE:
Initial Change Equilibriu
m
H+ 0 x x
F- 0 x x
HF 1.0 -x 1.0-x
[H + ][F - ]
K a = 7.2x10 -4 =
[HF]

45. CALCULATING THE PH OF A WEAK ACID
What is the pH of a 1.00 M solution of HF, Ka=
7.2x10-4 [x][x]
K a = 7.2x10 -4 =
[1.0 - x]
1.0 - x @ 1.0
• Since Ka for HF is so small, HF will dissociate
only slightly, and x is expected to be small. If
x is very small compared to concentration (2-
3+ exponents), the denominator can be
simplified. + -
[H ][F ]
-4
K a = 7.2x10 =
[HF]

46. CALCULATING THE PH OF A WEAK ACID
What is the pH of a 1.00 M solution of HF, Ka= 7.2x10-
4
[x][x]
-4
K a = 7.2x10 =
[1.0]
-2
x @ 2.7x10
The validity of this answer needs to be checked due
to the approximation made in the calculations.
Typically Ka values are known to an accuracy of
±5%. We will use this same accuracy for answers
to be considered valid. + -
-4 [H ][F ]
K a = 7.2x10 =
[HF]

47. CALCULATING THE PH OF A WEAK ACID
What is the pH of a 1.00 M solution of HF, Ka=
7.2x10-4
Checking validity: x
x100%
x=2.7x10-2, [HF] = 1.00 M [HA]o
2.7x10 -2
x100% = 2.7%
1.0M
x is valid. [H+] = 2.7x10-2, pH=1.57
[H + ][F - ]
K a = 7.2x10 -4 =
[HF]

48. PRACTICE PROBLEM #8
The hypochlorite ion (OCl-) is a strong oxidizing
agent often found in household bleaches and
disinfectants. It is also the active ingredient
that forms when swimming pool water is treated
with chlorine. In addition to its oxidizing
abilities, the hypochlorite ion has a relatively
high affinity for protons (it is much stronger
base than Cl-, for example) and forms the
weakly acidic hypochlorous acid (HOCl,
Ka=3.5x10-8). Calculate the pH of a 0.100M
aqueous solution of hypochlorous acid.

49. PRACTICE PROBLEM #8
Major species:
HOCl and H2O
Narrow down contributing species and its
concentration. Write balanced equation
HOClKa=3.5x10-8 significant ≅ 0.100M
H2O Kw=1.0x10-14 negligible
+ -
HOCl(aq) Û H (aq) + OCl (aq)

50. PRACTICE PROBLEM #8
Write equilibrium expression.
[H + ][OCl - ]
Ka =
[HOCl]
ICE Initial Change Equilibriu
m
H+
OCl-
HOCl

51. PRACTICE PROBLEM #8
Use approximations when able:
(x)(x)
Ka =
(.100 - x)
x2
3.5x10 -8 =
(@ .100)
x≅5.9x10-5

52. PRACTICE PROBLEM #8
Check validity
x 5.7x10 -5
x100% = x100% = 0.059%
[HA] 0.100
pH:
[H+] = 5.7x10-8, pH=4.23

53. PRACTICE PROBLEM #9
Calculate the pH of a solution that contains
1.00M HCN (Ka=6.2 x 10-10) and 5.00 M
HNO2 (Ka=4.0 x 10-4). Also calculate the
concentration of cyanide ion (CN-) in this
solution at equilibrium.

54. PRACTICE PROBLEM #9
Major species:
Narrow down contributing species and its
concentration. Write balanced equation

55. PRACTICE PROBLEM #9
Write equilibrium expression.
ICE Initial Change Equilibriu
m
H+
NO2-
HNO2

56. PRACTICE PROBLEM #9
Use approximations when able:
Check validity.
pH= 1.35 But wait….what about
CN-?

57. PRACTICE PROBLEM #9
Calculate the pH of a solution that contains
1.00M HCN (Ka=6.2 x 10-10) and 5.00 M HNO2
(Ka=4.0 x 10-4). Also calculate the
concentration of cyanide ion (CN-) in this
solution at equilibrium.
We also want to calculate the CN-
concentration from the dissociation of HCN
HCN(a)<-> H+(aq)+ CN-(aq)Ka=6.2x10-10

58. PRACTICE PROBLEM #9
The molarity of the HCN is known and the
[H+]is equal to the prominent contribution of
HNO2. The source of the H+ ions is not
important in equilibrium.
[H + ][CN - ]
K a = 6.2x10 -10 =
[HCN ]
[CN-]=1.4x10-8M
[4.5x10 -2 ][CN - ]
6.2x10 -10 =
[1.00]

59. PERCENT DISSOCIATION
The percent dissociation is:
[amount dissociated]
%dissociation = x100%
[initial concentration]
For a given weak acid, the percent
dissociation increases as the acid
becomes more dilute.

60. PRACTICE PROBLEM #10
Calculate the percent dissociation of acetic acid (Ka
= 1.8 x 10-5) in each of the following solutions.
a. 1.00 M HC2H3O2
Major species:
HC2H3O2 and H2O
Narrow down contributing species and its
concentration. Write balanced equation
HC2H3O2 Ka = 1.8 x 10-5, H2O Kw = 1.0 x 10-14
HC2H3O2(aq) <-> H+(aq) + C2H3O2-(aq)
[H + ][C2 H3O2 ]
-
K a = 1.8x10 -5 =
[HC2 H3O2 ]

61. PRACTICE PROBLEM #10
Calculate the percent dissociation of acetic acid
(Ka = 1.8 x 10-5) in each of the following
solutions.
a. 1.00 M HC2H3O2
Initial Change Equilibriu
ICE
m
H+
C2H3O2-
HC2H3O
2
[H + ][C2 H3O2 ]
-
K a = 1.8x10 -5 =
[HC2 H3O2 ]

62. PRACTICE PROBLEM #10
Calculate the percent dissociation of acetic acid
(Ka = 1.8 x 10-5) in each of the following
solutions.
a. 1.00 M HC2H3O2
Use approximations when able:
4.2x10 -3
Apply Equation
x100% = 0.42%
1.00
[H + ][C2 H3O2 ]
-
K a = 1.8x10 -5 =
[HC2 H3O2 ]

63. PRACTICE PROBLEM #10
Calculate the percent dissociation of acetic acid
(Ka = 1.8 x 10-5) in each of the following
solutions.
b. 0.100 M HC2H3O2
Major species:
HC2H3O2 and H2O
Narrow down contributing species and its
concentration. Write balanced equation
HC2H3O2 Ka = 1.8 x 10-5, H2O Kw = 1.0 x 10-14
HC2H3O2(aq) <-> H+(aq) + C2H3O2-(aq)

64. PRACTICE PROBLEM #10
Calculate the percent dissociation of acetic acid
(Ka = 1.8 x 10-5) in each of the following
solutions.
b. 0.100 M HC2H3O2
Initial Change Equilibriu
ICE
m
H+
C2H3O2-
HC2H3O
2
[H + ][C2 H3O2 ]
-
K a = 1.8x10 -5 =
[HC2 H3O2 ]

65. PRACTICE PROBLEM #10
Calculate the percent dissociation of acetic acid
(Ka = 1.8 x 10-5) in each of the following
solutions.
b. 0.100 M HC2H3O2
Use approximations when able:
1.3x10 -3
Apply Equation
x100% = 1.3%
0.10
[H + ][C2 H3O2 ]
-
K a = 1.8x10 -5 =
[HC2 H3O2 ]

66. PERCENT DISSOCIATION
The previous example is a perfect example
of the percent dissociation relationship.
• For a given weak acid, the percent
dissociation increases as the acid
becomes more dilute.

67. PRACTICE PROBLEM #11
Lactic acid (HC3H5O3) is a waste product that
accumulates in muscle tissue during
exertion, leading to pain and a feeling of
fatigue. In a 0.100 M aqueous solution,
lactic acid is 3.7% dissociated. Calculate
the value of Ka for this acid.

68. PRACTICE PROBLEM #11
Major species:
HC3H5O3and H2O
Narrow down contributing species and its
concentration. Write balanced
equationHC3H5O3(aq)<-> H+(aq) + C3H5O3-(aq)
+ -
[H ][C3H 5O ]
Ka = 3
[HC3H 5O3 ]

69. PRACTICE PROBLEM #11
Percent Dissociation
x x -3
3.7% = 100% = 100%, x = 3.7x10
[HC3 H 5O3 ] (.10)
Substitute in the expression
[H + ][C3H 5O3 ] (3.7x10 -3 )(3.7x10 -3 )
-
Ka = =
[HC3H 5O3 ] 0.10
Ka=1.4x10-4
[H + ][C3H 5O3 ]
-
Ka =
[HC3H 5O3 ]