Chapter 14
Acids and Bases
QUESTION
Aniline, C6H5NH2, was isolated in the 1800s and began
immediate use in the dye industry. What is the formula of the
conjugate acid of this base?
1.
2.
3.
4.
C6H5NH2+
C6H5NH3+
C6H5NH–
C6H5NH+
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CRS Question, 14–2
ANSWER
Choice 2 correctly represents the result of aniline accepting a
H+ ion as bases typically do. The conjugate acid of a base is
represented as the base with the addition of a H+.
Section 14.1: The Nature of Acids and Bases
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CRS Question, 14–3
QUESTION
Nitric acid, HNO3, is considered to be a strong acid whereas
nitrous acid, HNO2, is considered to be a weak acid. Which of
the statements here is fully correct?
1. Nitric acid has an aqueous equilibrium that lies far to the
right and NO3– is considered a weak conjugate base.
2. Nitric acid has a stronger conjugate base than nitrous acid.
3. The dissociation of nitrous acid compared to an equal
concentration of nitric acid produces more H+.
4. The equilibrium of nitrous acid lies far to the left and the
conjugate base is weaker than the conjugate base of nitric
acid.
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CRS Question, 14–4
ANSWER
Choice 1 correctly compares equilibrium and conjugate base
characteristics. The conjugate base of a strong acid is
considered to be weak. The stronger the acid, the more
reaction in water. Therefore, a weak acid’s equilibrium is
favored to the left.
Section 14.2: Acid Strength
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CRS Question, 14–5
QUESTION
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CRS Question, 14–6
QUESTION (continued)
Use information on this table to determine which of the
following bases would have the weakest conjugate acid:
OC6H5–; C2H3O2–;
1.
2.
3.
4.
OCl–;
NH3
OC6H5–
C2H3O2–
OCl–
NH3
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CRS Question, 14–7
ANSWER
Choice 1 correctly identifies the base, among these four,
with the weakest conjugate acid. The Ka’s in the table can
be used to compare conjugate acid strength. The higher the
Ka value, the stronger the acid.
Section 14.2: Acid Strength
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CRS Question, 14–8
QUESTION
In a solution of water at a particular temperature the [H+]
may be 1.2  10–6 M. What is the [OH–] in the same solution?
Is the solution acidic, basic, or neutral?
1.
2.
3.
4.
1.2  10–20 M; acidic
1.2  10–20 M; basic
8.3  10–9 M; basic
8.3  10–9 M; acidic
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CRS Question, 14–9
ANSWER
Choice 4 correctly shows the OH– molarity and classifies the
solution as acidic. Kw = [H+][OH–] = 1.0  10–14 at 25°C. The
H+ molarity is approximately 1,000 times greater than the OH–
concentration. Solutions with higher H+ concentrations than
OH– are acidic.
Section 14.2: Acid Strength
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CRS Question, 14–10
QUESTION
An environmental chemist obtains a sample of rainwater near a
large industrial city. The [H+] was determined to be 3.5  10–6 M.
What is the pH, pOH, and [OH–] of the solution?
1.
2.
3.
4.
pH = 5.46 ; pOH = 8.54; [OH–] = 7.0  10–6 M
pH = 5.46 ; pOH = 8.54; [OH–] = 2.9  10–9 M
pH = 12.56 ; pOH =1.44 ; [OH–] = 3.6  10–2 M
pH = 8.54; pOH = 5.46; [OH–] = 2.9  10–9 M
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CRS Question, 14–11
ANSWER
Choice 2 provides all three correct responses. The
expression pH = –log[H+] can be used to find the pH then:
14.00 = pH + pOH can be used to obtain the pOH. Finally,
[OH–] = 10–pOH.
Section 14.3: The pH Scale
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CRS Question, 14–12
QUESTION
Which of the following correctly compares strength of acids,
pH, and concentrations?
1. A weak acid, at the same concentration of a strong acid,
will have a lower pH.
2. A weak acid, at the same concentration of a strong acid,
will have the same pH.
3. A weak acid, at a high enough concentration more than a
strong acid, could have a lower pH than the strong acid.
4. A weak acid, at a concentration below a strong acid,
could have a lower pH than a strong acid.
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CRS Question, 14–13
ANSWER
Choice 3 correctly predicts that it is possible to have a high
enough concentration of the weak acid compared to a strong
acid, and that the pH of the weaker acid would be lower (more
acidic) than the more dilute stronger acid. Strength of an acid
refers to its dissociation. The pH of a solution depends on the
concentration, regardless of source, of the H+ ion.
Section 14.4: Calculating the pH of Strong Acid Solutions
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CRS Question, 14–14
QUESTION
Butyric acid is a weak acid that can be found in spoiled butter.
The compound has many uses in synthesizing other flavors.
The Ka of HC4H7O2 at typical room temperatures is 1.5  10–5.
What is the pH of a 0.20 M solution of the acid?
1.
2.
3.
4.
5.52
4.82
2.76
–0.70
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CRS Question, 14–15
ANSWER
Choice 3 is correct assuming that the amount of dissociation
of this weak acid is negligible when compared to its molarity.
Ka = [H+][A–] / [HA – x] = x2/(0.20 M – x)
becomes 1.5  10–5 = x2/0.20; once x is found, representing the
[H+], taking –log of that yields the pH.
Section 14.5: Calculating the pH of Weak Acid Solutions
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CRS Question, 14–16
QUESTION
A 0.35 M solution of an unknown acid is brought into a lab.
The pH of the solution is found to be 2.67. From this data,
what is the Ka value of the acid?
1.
2.
3.
4.
6.1  10–3
1.3  10–5
7.5  10–4
2.1  10–3
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CRS Question, 14–17
ANSWER
Choice 2 shows the unknown acid’s Ka value. The pH could
be used to find the [H+] concentration , 10–2.67, then the Ka
expression only has one unknown:
Ka = [0.00214][0.00214]/(0.35–0.00214)
Section 14.5: Calculating the pH of Weak Acid Solutions
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CRS Question, 14–18
QUESTION
Two weak acids, HA and HB are placed in separate solutions so
that their molarities are the same. Which, if either, would have
the larger value for Ka if the pH of the HA solution were lower
than the pH of the HB solution?
1.
2.
3.
4.
HA has the larger Ka.
HB has the larger Ka.
The Ka of HA = Ka of HB.
My answer would actually just be a guess, does more
information need to be given?
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CRS Question, 14–19
ANSWER
Choice 1 correctly identifies the acid with the larger Ka
value. When equal concentrations are supplied there is an
equal opportunity to dissociate into H+. However, the lower
pH in the HA solution indicates a higher concentration of H+
present in the solution. Therefore, HA must have a higher
level of dissociation. This would be expected if the Ka
(product to reactant ratio) were larger.
Section 14.5: Calculating the pH of Weak Acid Solutions
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CRS Question, 14–20
QUESTION
Amine compounds often have pungent aromas. The compound
ethylamine can be associated with a fish-like aroma. The Kb,
at typical room conditions, of C2H5NH2 is 5.6  10–4. What is
the pH of a 0.15 M solution of this weak base?
1.
2.
3.
4.
9.92
2.04
11.96
10.75
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CRS Question, 14–21
ANSWER
Choice 3 provides the correct pH for the solution. The Kb
expression can be used to solve for [OH–]. This can be used to
either obtain the pOH (pH = 14.00 – pOH at 25°C) or [H+] =
(Kw/[OH–]); and then pH can be calculated.
Section 14.6: Bases
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CRS Question, 14–22
QUESTION
Pyridine is a very weak base with a strong, sharp odor. If a
0.57 M solution were measured to have a 0.0055%
dissociation, what is the value of the Kb of pyridine?
1.
2.
3.
4.
1.7  10–9
1.7  10–5
5.3  10–5
3.1  10–5
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CRS Question, 14–23
ANSWER
Choice 1 displays the correct Kb value. The initial molarity
(0.57 M) is relatively unchanged by the 0.0055% dissociation,
so in the Kb expression 0.57 M could be used for the
equilibrium concentration. Of the original concentration only
a 0.000055 fraction reacts. So 0.000031 is the molarity of
OH– and pyridine’s conjugate acid. Now, Kb can be calculated
as Kb = (0.000031)2/0.57
Section 14.6: Bases
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CRS Question, 14–24
QUESTION
Ascorbic acid, also known as vitamin C, has two hydrogen
atoms that ionize from the acid. Ka1 = 7.9  10–5; Ka2 = 1.6 
10–12. What is the pH, and C6H6O62– concentration of a 0.10 M
solution of H2C6H6O6?
1.
2.
3.
4.
2.55; [C6H6O62–] = 0.050 M
2.55; [C6H6O62–] = 1.6  10–12 M
1.00; [C6H6O62–] = 1.6  10–12 M
5.10; [C6H6O62–] = 0.050 M
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CRS Question, 14–25
ANSWER
Choice 2 shows both correct answers. In a diprotic acid with
two small, widely separated Ka values the pH of a solution
can be obtained from using the first Ka and the molarity. The
concentration of the dianion can be closely approximated by
assuming very little dissociation of the second acidic
hydrogen, so that Ka2 is very close to the molarity.
Section 14.7: Polyprotic Acids
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CRS Question, 14–26
QUESTION
The following salts were all placed in separate solutions at the
same temperature so that their concentrations were all equal.
Arrange them in order from lowest pH to highest pH.
NaCl;
NH4NO3;
Ca(C2H3O2)2;
AlCl3
Additional information: Kb for NH3 = 1.8  10–5; Ka for
HC2H3O2 = 1.8  10–5; Ka for Al(H2O)3+ = 1.4  10–5.
1.
2.
3.
4.
NaCl;
AlCl3;
AlCl3;
NH4NO3;
NH4NO3;
NaCl;
NH4NO3;
AlCl3;
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Ca(C2H3O2)2;
NH4NO3;
NaCl;
NaCl;
AlCl3
Ca(C2H3O2)2
Ca(C2H3O2)2
Ca(C2H3O2)2
CRS Question, 14–27
ANSWER
Choice 3 correctly ranks the salt solutions from lowest pH
(most acidic solution) to highest pH. The ranking is based
on production of H+ from the salt ions interacting with
water. Highly charged small metal ions such as Al3+ can
produce H+ as can NH4+. However, the Ka of the
aluminum’s reaction is larger than the Ka for NH4+. NaCl is
neutral and the acetate ion undergoes a reaction that
produces OH–, so it has a high pH.
Section 14.8: Acid–Base Properties of Salts
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CRS Question, 14–28
QUESTION
The Ka value for monochloroacetic acid is 1.35  10–3 at
25°C. Neutralizing the acid with KOH would produce the
salt potassium chloroacetate. What would be the pH of a
0.110 M solution of the salt at 25°C?
1.
2.
3.
4.
7.956
13.041
6.044
12.086
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CRS Question, 14–29
ANSWER
Choice 1 is the correct pH for this basic salt. The salt, in
water, will dissociate and provide the opportunity for the
cation and anion to interact with water. Since the anion is
from a weak acid (note the Ka) it can regain H+ from water
thus producing OH– ions. The Kb for this reaction is obtained
through Kw/Ka. Then the [OH–] can be obtained from the Kb
and molarity. This is then converted to pH after using the Kw
for water.
Section 14.8: Acid–Base Properties of Salts
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CRS Question, 14–30
QUESTION
To fit our traditional definition of an acid, a molecule must be
able to produce H+ ions in water. Some molecules can do this
while others cannot. For example, the C–H bond does not
allow H to easily leave. H–Cl however, does allow H to easily
leave the Cl. H–F also allows H+ to leave in water, but not as
well as HCl. Which statement here offers the best explanation
for these observations?
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CRS Question, 14–31
QUESTION (continued)
1. The C–H bond is strong, but not very polar. The H–Cl
bond is strong and polar. The H–F bond is unusually
strong and yet very polar.
2. The C–H bond is very strong, and polar. The H–Cl bond is
very strong and not polar. The H–F bond is weak and
polar.
3. The C–H bond is not strong, or very polar. The H–Cl bond
is very strong and very polar. The H–F bond is weak and
polar.
4. A weaker bond produces a more polar bond, hence a
stronger acid.
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CRS Question, 14–32
ANSWER
Choice 1 correctly relates bond strength and polarity to the
observations described in the question. Generally weaker
bonds such as O–H would be likely to release H+, but polarity
plays a key role in attracting H+ to water. C–H is also nonpolar so it does not release H+. H–F has a more polar bond
than H–Cl so it could be expected to release H+ to polar water,
but the H–F is unusually strong so only a small percentage of
HF molecules may behave this way.
Section 14.9: The Effect of Structure on Acid–Base Properties
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CRS Question, 14–33
QUESTION
Some CO2 gas is bubbled into water. Which of the
following compounds would be suitable to now neutralize
the CO2 infused solution?
1.
2.
3.
4.
K2O
NaCl
NH4Br
AlCl3
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CRS Question, 14–34
ANSWER
Choice 1 is the only compound that would begin to neutralize
an acid solution. Nonmetal oxides, such as CO2, produce acid
solutions in water. So a compound with basic properties
would be needed to neutralize the solution. Of the choices,
K2O (a metal oxide) is the only one with basic properties.
Section 14.10: Acid–Base Properties of Oxides
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CRS Question, 14–35
QUESTION
Hydrated metal ions are stabilized by Lewis acid-base
bonding. Which statement correctly identifies both the Lewis
classification and the rationale for the decision when a metal
ion bonds to a water molecule?
1. The metal ion is the Lewis base because it accepts a pair
of electrons from water.
2. The metal ion is the Lewis base because it donates a pair
of electrons to water.
3. Water is the Lewis base because it accepts a pair of
electrons from the metal ion.
4. Water is the Lewis base because it donates a pair of
electrons to the metal ion.
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CRS Question, 14–36
ANSWER
Choice 4 provides the correct classification for water as a
Lewis base and provides the correct rationale. Lewis bases
are electron donors. Metal ions have lost electrons; thus
water has the opportunity, through its lone electron pairs, to
donate electrons to the metal ion.
Section 14.11: The Lewis Acid–Base Model
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CRS Question, 14–37