UNIT 14: ACIDS AND BASES
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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
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.
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.
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).
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)
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)
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)
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)
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)
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(H 2O)6]3+
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)
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)
ACID STRENGTH SUMMARY
Acid strength and conjugate base strength are
indirectly related.
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.
COMMON WEAK OXYACIDS
1.H3PO4
2.HNO2
3.HOCl
• Most acids are oxyacids, in which the acidic
proton is attached to an oxygen atom.
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
COMMON MONOPROTIC ACIDS
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-
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)
WATER AS AN ACID AND A BASE
The equilibrium expression for this reaction is the
dissociation constant of water, also called the ionproduct 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)
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
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 OHb. 1.0 x 10-7 M OHc. 10.0 M H+
a. basic b. neutral c. acidic
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
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
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.
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.
PRACTICE PROBLEM #5
Calculate pH and POH for each of the following
solutions at 25°C.
a. 1.0 x 10-3 M OHb. 1.0 M OH-
a. pH=11.00, pOH=3.00 b. pH=0.00, pOH=14.00
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
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.
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.
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.
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
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
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
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 = 7.2x10-4 vs Kw= 1.0x10-14
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
]
-4
K a = 7.2x10 =
[HF]
initial [HF] = 1.0 M
ICE!
Equilibrium [H+] = ?
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
Equilibrium
H+
FHF
+
[H
][F
]
-4
K a = 7.2x10 =
[HF]
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
H+
FHF
0
0
1.0
Change
x
x
-x
Equilibrium
x
x
1.0-x
+
[H
][F
]
-4
K a = 7.2x10 =
[HF]
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 =
[1.0 - x]
1.0 - x @ 1.0
-4
• 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]
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 =
[1.0]
-4
x @ 2.7x10 -2
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.
+
[H
][F
]
-4
K a = 7.2x10 =
[HF]
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
x=2.7x10-2, [HF] = 1.00 M
x100%
[HA]o
2.7x10 -2
x100% = 2.7%
1.0M
x is valid. [H+] = 2.7x10-2, pH=1.57
+
[H
][F
]
-4
K a = 7.2x10 =
[HF]
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.
PRACTICE PROBLEM #8
Major species:
HOCl and H2O
Narrow down contributing species and its
concentration. Write balanced equation
HOCl Ka=3.5x10-8
significant ≅ 0.100M
H2O Kw=1.0x10-14
negligible
HOCl(aq) Û H
+
(aq)
+ OCl
(aq)
PRACTICE PROBLEM #8
Write equilibrium expression.
[H + ][OCl - ]
Ka =
[HOCl]
ICE
Initial
H+
OClHOCl
Change
Equilibrium
PRACTICE PROBLEM #8
Use approximations when able:
(x)(x)
Ka =
(.100 - x)
2
x
3.5x10 -8 =
(@ .100)
x≅5.9x10-5
PRACTICE PROBLEM #8
Check validity
x
5.7x10 -5
x100% =
x100% = 0.059%
[HA]
0.100
pH:
[H+] = 5.7x10-8, pH=4.23
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.
PRACTICE PROBLEM #9
Major species:
Narrow down contributing species and its
concentration. Write balanced equation
PRACTICE PROBLEM #9
Write equilibrium expression.
ICE
Initial
H+
NO2HNO2
Change
Equilibrium
PRACTICE PROBLEM #9
Use approximations when able:
Check validity.
pH= 1.35
But wait….what about CN-?
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
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.
K a = 6.2x10 -10
[H + ][CN - ]
=
[HCN ]
-2
[4.5x10 -8M
][CN
-10-]=1.4x10
[CN
6.2x10
=
[1.00]
-
]
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.
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
][C
H
O
-5
2 3 2]
K a = 1.8x10 =
[HC2 H3O2 ]
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
ICE
Initial
Change Equilibrium
H+
C2H3O2HC2H3O2
+
[H
][C
H
O
-5
2 3 2]
K a = 1.8x10 =
[HC2 H3O2 ]
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:
Apply Equation
4.2x10 -3
x100% = 0.42%
1.00
+
[H
][C
H
O
-5
2 3 2]
K a = 1.8x10 =
[HC2 H3O2 ]
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)
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
ICE
Initial
Change Equilibrium
H+
C2H3O2HC2H3O2
+
[H
][C
H
O
-5
2 3 2]
K a = 1.8x10 =
[HC2 H3O2 ]
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:
Apply Equation
1.3x10 -3
x100% = 1.3%
0.10
+
[H
][C
H
O
-5
2 3 2]
K a = 1.8x10 =
[HC2 H3O2 ]
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.
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.
PRACTICE PROBLEM #11
Major species:
HC3H5O3 and H2O
Narrow down contributing species and its
concentration. Write balanced equation
HC3H5O3(aq) <-> H+(aq) + C3H5O3-(aq)
[H + ][C3H 5O3- ]
Ka =
[HC3H 5O3 ]
PRACTICE PROBLEM #11
Percent Dissociation
x
x
-3
3.7% =
100% =
100%, x = 3.7x10
[HC3 H 5O3 ]
(.10)
Substitute in the expression
+
3
-3
-3
[H ][C3H 5O ] (3.7x10 )(3.7x10 )
Ka =
=
[HC3H 5O3 ]
0.10
Ka=1.4x10-4
[H + ][C3H 5O3- ]
Ka =
[HC3H 5O3 ]