2. Some Definitions
• Arrhenius
Acid: Substance that, when dissolved in
water, increases the concentration of
hydrogen ions.
Acids
and
Bases
3. Some Definitions
• Arrhenius
Acid: Substance that, when dissolved in
water, increases the concentration of
hydrogen ions.
Base: Substance that, when dissolved in
water, increases the concentration of
hydroxide ions.
Acids
and
Bases
7. A Brønsted–Lowry acid…
…must have a removable (acidic) proton.
A Brønsted–Lowry base…
…must have a pair of nonbonding electrons.
Acids
and
Bases
8. If it can be either…
...it is amphiprotic.
HCO3 −
HSO4 −
H 2O
Acids
and
Bases
9. What Happens When an Acid
Dissolves in Water?
• Water acts as a
Brønsted–Lowry base
and abstracts a proton
(H+) from the acid.
Acids
and
Bases
10. What Happens When an Acid
Dissolves in Water?
• Water acts as a
Brønsted–Lowry base
and abstracts a proton
(H+) from the acid.
• As a result, the
conjugate base of the
acid and a hydronium
ion are formed.
Acids
and
Bases
13. Is NH3 an acid, base, or could it be both?
1. Acid
2. Base
3. Both
Acids
and
13 Bases
14. Is NH3 an acid, base, or could it be both?
1. Acid
2. Base
3. Both
In this chapter we learned that a more
general definition for a base is a
substance that can accept another
proton, which is true for NH3 because of
Acids
and
the lone electron pair on the N atom.
14 Bases
15. Is H2O an acid, base, or could it
be both?
1. Acid
2. Base
3. Both
Acids
and
15 Bases
16. Is H2O an acid, base, or could it
be both?
As indicated in the
1. Acid equilibrium below, water is
an amphoteric substance
2. Base that can either accept
3. Both another proton or donate a
proton.
H H
+ H H
+ _
H O H H O H O Acids
+ and
16 Bases
17. Is the ion PO4 3 an
acid,
base, or could it be both?
- -
O O
1. Acid - + -
- -
O P O
2. Base O P O
-
3. Both O O
Acids
and
17 Bases
18. Is the ion PO4 3 an
acid,
base, or could it be both?
- -
O O
1. Acid - + -
- -
O P O
2. Base O P O
-
3. Both O O
Phosphate is a proton
acceptor, regardless of
which resonance structure Acids
and
is being considered. 18 Bases
19. Conjugate Acids and Bases:
• From the Latin word conjugare, meaning
“to join together.”
Acids
and
Bases
20. Conjugate Acids and Bases:
• From the Latin word conjugare, meaning
“to join together.”
• Reactions between acids and bases
always yield their conjugate bases and
acids.
Acids
and
Bases
21. SAMPLE EXERCISE 16.1 Identifying Conjugate Acids and Bases
(a) What is the conjugate base of each of the
following acids: HClO4, H2S, PH4+, HCO3– ?
(b) What is the conjugate acid of each of the
following bases: CN–, SO42–, H2O, HCO3– ?
22. SAMPLE EXERCISE 16.1 Identifying Conjugate Acids and Bases
(a) What is the conjugate base of each of the
following acids: HClO4, H2S, PH4+, HCO3– ?
(b) What is the conjugate acid of each of the
following bases: CN–, SO42–, H2O, HCO3– ?
conjugate base = parent substance minus one proton,
conjugate acid = parent substance plus one proton.
23. SAMPLE EXERCISE 16.1 Identifying Conjugate Acids and Bases
(a) What is the conjugate base of each of the
following acids: HClO4, H2S, PH4+, HCO3– ?
(b) What is the conjugate acid of each of the
following bases: CN–, SO42–, H2O, HCO3– ?
conjugate base = parent substance minus one proton,
conjugate acid = parent substance plus one proton.
(a)HClO4 less one proton (H+) is ClO4–. The other
conjugate bases are HS–, PH3, and CO32–.
24. SAMPLE EXERCISE 16.1 Identifying Conjugate Acids and Bases
(a) What is the conjugate base of each of the
following acids: HClO4, H2S, PH4+, HCO3– ?
(b) What is the conjugate acid of each of the
following bases: CN–, SO42–, H2O, HCO3– ?
conjugate base = parent substance minus one proton,
conjugate acid = parent substance plus one proton.
(a)HClO4 less one proton (H+) is ClO4–. The other
conjugate bases are HS–, PH3, and CO32–.
(b) CN– plus one proton (H+) is HCN. The other
conjugate acids are HSO4–, H3O+, and H2CO3.
25. SAMPLE EXERCISE 16.1 Identifying Conjugate Acids and Bases
(a) What is the conjugate base of each of the
following acids: HClO4, H2S, PH4+, HCO3– ?
(b) What is the conjugate acid of each of the
following bases: CN–, SO42–, H2O, HCO3– ?
conjugate base = parent substance minus one proton,
conjugate acid = parent substance plus one proton.
(a)HClO4 less one proton (H+) is ClO4–. The other
conjugate bases are HS–, PH3, and CO32–.
(b) CN– plus one proton (H+) is HCN. The other
conjugate acids are HSO4–, H3O+, and H2CO3.
Hydrogen carbonate ion (HCO3–) is amphiprotic: It can
act as either an acid or a base.
26. SAMPLE EXERCISE 16.2 Writing Equations for Proton-Transfer Reactions
The hydrogen sulfite ion (HSO3–) is amphiprotic.
(a) Write an equation for the reaction of HSO3– with
water, in which the ion acts as an acid.
(b) Write an equation for the reaction of HSO3– with
water, in which the ion acts as a base. In both cases
identify the conjugate acid-base pairs.
Acids
and
Bases
27. SAMPLE EXERCISE 16.2 Writing Equations for Proton-Transfer Reactions
The hydrogen sulfite ion (HSO3–) is amphiprotic.
(a) Write an equation for the reaction of HSO3– with
water, in which the ion acts as an acid.
(b) Write an equation for the reaction of HSO3– with
water, in which the ion acts as a base. In both cases
identify the conjugate acid-base pairs.
a)
Acids
and
Bases
28. SAMPLE EXERCISE 16.2 Writing Equations for Proton-Transfer Reactions
The hydrogen sulfite ion (HSO3–) is amphiprotic.
(a) Write an equation for the reaction of HSO3– with
water, in which the ion acts as an acid.
(b) Write an equation for the reaction of HSO3– with
water, in which the ion acts as a base. In both cases
identify the conjugate acid-base pairs.
a)
The conjugate pairs in this equation are HSO3– (acid) and SO32–
(conjugate base); and H2O (base) and H3O+ (conjugate acid).
(b)
Acids
and
Bases
29. SAMPLE EXERCISE 16.2 Writing Equations for Proton-Transfer Reactions
The hydrogen sulfite ion (HSO3–) is amphiprotic.
(a) Write an equation for the reaction of HSO3– with
water, in which the ion acts as an acid.
(b) Write an equation for the reaction of HSO3– with
water, in which the ion acts as a base. In both cases
identify the conjugate acid-base pairs.
a)
The conjugate pairs in this equation are HSO3– (acid) and SO32–
(conjugate base); and H2O (base) and H3O+ (conjugate acid).
(b)
Acids
and
Bases
30. SAMPLE EXERCISE 16.2 Writing Equations for Proton-Transfer Reactions
The hydrogen sulfite ion (HSO3–) is amphiprotic.
(a) Write an equation for the reaction of HSO3– with
water, in which the ion acts as an acid.
(b) Write an equation for the reaction of HSO3– with
water, in which the ion acts as a base. In both cases
identify the conjugate acid-base pairs.
a)
The conjugate pairs in this equation are HSO3– (acid) and SO32–
(conjugate base); and H2O (base) and H3O+ (conjugate acid).
(b)
The conjugate pairs in this equation are H2O (acid) and OH–
(conjugate base), and HSO3– (base) and H2SO3 (conjugateAcids
and
acid). Bases
31. Acid and Base Strength
• Substances with
negligible acidity
do not dissociate
in water.
Their conjugate
bases are
exceedingly strong.
Acids
and
Bases
32. Strong acid Weak acid Negligible acidity
Acids
and
Bases
33. Strong acid Weak acid Negligible acidity
HNO3 is a strong acid, meaning NO3– has
negligible base strength.
Acids
and
Bases
34. Acid and Base Strength
In any acid-base reaction, the
equilibrium will favor the reaction
that moves the proton to the
stronger base.
HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq)
Acids
and
Bases
35. Acid and Base Strength
In any acid-base reaction, the
equilibrium will favor the reaction
that moves the proton to the
stronger base.
HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq)
H2O is a much stronger base than Cl−,
so the equilibrium lies so far to the right Acids
and
K is not measured (K>>1). Bases
36. Acid and Base Strength
C2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
Acids
and
Bases
37. Acid and Base Strength
C2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
Acetate is a stronger base than
H2O, so the equilibrium favors the
left side (K<1).
Acids
and
Bases
38. SAMPLE EXERCISE 16.3 Predicting the Position of a Proton-Transfer Equilibrium
For the following proton-transfer reaction, predict whether
the equilibrium lies predominantly to the left (that is, Kc < 1)
or to the right (Kc > 1):
Acids
and
Bases
39. For the following proton-transfer reaction, predict whether
the equilibrium lies predominantly to the left (that is, Kc < 1)
or to the right (Kc > 1):
This is a proton-transfer reaction, and the position of
equilibrium will favor the proton going to the stronger
of two bases.
Solve: CO32– is a stronger base than SO42–. CO32–
will get the proton preferentially to become HCO3–,
while SO42– will remain mostly unprotonated. The
resulting equilibrium will lie to the right, favoring Acids
and
products (that is, Kc > 1). Bases
41. Autoionization of Water
• As we have seen, water is amphoteric.
• In pure water, a few molecules act as
bases and a few act as acids.
Acids
and
Bases
42. Autoionization of Water
• As we have seen, water is amphoteric.
• In pure water, a few molecules act as
bases and a few act as acids.
H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
Acids
and
Bases
43. Autoionization of Water
• As we have seen, water is amphoteric.
• In pure water, a few molecules act as
bases and a few act as acids.
H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
• This is referred to as autoionization.
Acids
and
Bases
45. Ion-Product Constant
• The equilibrium expression for this
process is
Kc = [H3O+] [OH−]
• This special equilibrium constant is
referred to as the ion-product constant
for water, Kw.
Acids
and
Bases
46. Ion-Product Constant
• The equilibrium expression for this
process is
Kc = [H3O+] [OH−]
• This special equilibrium constant is
referred to as the ion-product constant
for water, Kw.
• At 25°C, Kw = 1.0 × 10−14
Acids
and
Bases
47. SAMPLE EXERCISE 16.4 Calculating [H+] for Pure Water
Calculate the values of [H+] and
[OH–] in a neutral solution at 25°C.
Acids
and
Bases
48. SAMPLE EXERCISE 16.4 Calculating [H+] for Pure Water
Calculate the values of [H+] and
[OH–] in a neutral solution at 25°C.
By definition, [H+] = [OH–] in a neutral solution.
We will represent the concentration of [H+] and
[OH–] in neutral solution with x. This gives
In an acid solution [H+] is greater than
1.0 × 10–7 M ; in a basic solution [H+] is Acids
and
less than 1.0 × 10–7 M.
Bases
49. PRACTICE EXERCISE
Indicate whether solutions with each of the
following ion concentrations are neutral, acidic,
or basic:
(a) [H+] = 4 × 10–9 M;
(b) [OH–] = 1 × 10–7 M;
(c) [OH–] = 7 × 10–13M.
Acids
and
38 Bases
50. PRACTICE EXERCISE
Indicate whether solutions with each of the
following ion concentrations are neutral, acidic,
or basic:
(a) [H+] = 4 × 10–9 M;
(b) [OH–] = 1 × 10–7 M;
(c) [OH–] = 7 × 10–13M.
Answers: (a) basic, (b) neutral, (c) acidic
Acids
and
38 Bases
51. SAMPLE EXERCISE 16.5 Calculating [H+] from [OH–]
Calculate the concentration of H+ (aq) in (a) a
solution in which [OH–] is 0.010 M, (b) a solution
in which [OH–] is 1.8 × 10–9 M.
Acids
and
Bases
52. SAMPLE EXERCISE 16.5 Calculating [H+] from [OH–]
Calculate the concentration of H+ (aq) in (a) a
solution in which [OH–] is 0.010 M, (b) a solution
in which [OH–] is 1.8 × 10–9 M.
Solve: (a)
Acids
and
Bases
53. SAMPLE EXERCISE 16.5 Calculating [H+] from [OH–]
Calculate the concentration of H+ (aq) in (a) a
solution in which [OH–] is 0.010 M, (b) a solution
in which [OH–] is 1.8 × 10–9 M.
Solve: (a)
This solution is basic because
Acids
and
Bases
54. SAMPLE EXERCISE 16.5 Calculating [H+] from [OH–]
Calculate the concentration of H+ (aq) in (a) a
solution in which [OH–] is 0.010 M, (b) a solution
in which [OH–] is 1.8 × 10–9 M.
Solve: (a)
This solution is basic because
(b) In this instance
Acids
and
Bases
55. SAMPLE EXERCISE 16.5 Calculating [H+] from [OH–]
Calculate the concentration of H+ (aq) in (a) a
solution in which [OH–] is 0.010 M, (b) a solution
in which [OH–] is 1.8 × 10–9 M.
Solve: (a)
This solution is basic because
(b) In this instance
This solution is acidic because Acids
and
Bases
56. PRACTICE EXERCISE
Calculate the concentration of OH–
(aq) in a solution in which
(a) [H+] = 2 × 10–6 M;
(b) [H+] = [OH–];
(c) [H+] = 100 × [OH–].
Acids
and
Bases
57. PRACTICE EXERCISE
Calculate the concentration of OH–
(aq) in a solution in which
(a) [H+] = 2 × 10–6 M;
(b) [H+] = [OH–];
(c) [H+] = 100 × [OH–].
Answers: (a) 5 × 10–9 M, (b) 1.0 × 10–7 M, (c) 1.0 × 10–8 M
Acids
and
Bases
59. Bellwork
Predict the products of the acid-base reactions,
and also predict whether the equilibrium lies to
the left or the right of the equation.
a)HCO3-(aq) + F-(aq) ⇌
b)C2H3O2-(aq) + H3O+(aq) ⇌
c) NH4+(aq) + OH-(aq) ⇌
Acids
and
Bases
60. pH
pH is defined as the
negative base-10 logarithm
of the hydronium ion
concentration.
pH = −log [H3 O+] Acids
and
Bases
61. pH
• In pure water,
Kw = [H3O +] [OH−] = 1.0 × 10−14
Acids
and
Bases
62. pH
• In pure water,
Kw = [H3O +] [OH−] = 1.0 × 10−14
Because [H3O+] = [OH−] in pure
water,
[H3O+] = (1.0 × 10−14)1/2 = 1.0 × 10−7
Acids
and
Bases
63. pH
• Therefore, in pure water,
pH = −log (1.0 × 10−7) = 7.00
Acids
and
Bases
64. pH
• Therefore, in pure water,
pH = −log (1.0 × 10−7) = 7.00
• An acid has a higher [H3O+] than pure
water, so its pH is <7
Acids
and
Bases
65. pH
• Therefore, in pure water,
pH = −log (1.0 × 10−7) = 7.00
• An acid has a higher [H3O+] than pure
water, so its pH is <7
• A base has a lower [H3O+] than pure
water, so its pH is >7.
Acids
and
Bases
69. At pH = 7, [H+] = [OH–]
The pH increases as [OH–] increases.
Acids
and
Bases
70. What is [H+] in an aqueous
solution whose pH = 3.72?
1. 1.9 x 10-3 M
2. 5.1 x 10-3 M
3. 1.9 x 10-4 M
4. 5.1 x 10-4 M
5. 5.1 x 10-10 M
Acids
and
51 Bases
71. What is [H+] in an aqueous
solution whose pH = 3.72?
1. 1.9 x 10-3 M
2. 5.1 x 10-3 M
3. 1.9 x 10-4 M
4. 5.1 x 10-4 M
5. 5.1 x 10-10 M
3.72 = -log[H+]
-3.72 = -log[H +]
Acids
[H +] = 1.9 x 10- 4 M and
52 Bases
72. What is the pH of an aqueous
solution of [OH-] = 6.0 x 10-3 M?
2.22
1. 3.22
2. 7.00
3. 10.78
4. 11.78
Acids
and
53 Bases
73. What is the pH of an aqueous
solution of [OH-] = 6.0 x 10-3 M?
2.22
There is more than one method to
1. 3.22 solve this problem. One is presented
2. 7.00 below:
3. 10.78 pOH = -log(6.0 x 10-3)
4. 11.78 = 2.22
pH = 14.00 - pOH
= 14.00 - 2.22 Acids
= 11.78
and
54 Bases
74. What is the approximate pH of an
aqueous solution of 1x10-12 M HCl?
1. 2
2. 7
3. 12
4. 14
5. None of the above
Acids
and
55 Bases
75. What is the approximate pH of an
aqueous solution of 1x10-12 M HCl?
1. 2
2. 7
3. 12
4. 14
5. None of the above
This is an aqueous solution, therefore
water will autoionize. The concentration
of protons due to autoionization of water
Acids
and
will dominate that of such a dilute acid.
56 Bases
76. Other “p” Scales
• The “p” in pH tells us to take the
negative log of the quantity (in this
case, hydrogen ions).
Acids
and
Bases
77. Other “p” Scales
• The “p” in pH tells us to take the
negative log of the quantity (in this
case, hydrogen ions).
• Some similar examples are
pOH = −log [OH−]
pKw = −log Kw
Acids
and
Bases
78. Watch This!
Because
[H3O+] [OH−] = Kw = 1.0 × 10−14,
we know that
−log [H3O+] + −log [OH−] = −log Kw = 14.0
Acids
and
Bases
79. Watch This!
Because
[H3O+] [OH−] = Kw = 1.0 × 10−14,
we know that
−log [H3O+] + −log [OH−] = −log Kw = 14.0
Acids
and
Bases
80. Watch This!
Because
[H3O+] [OH−] = Kw = 1.0 × 10−14,
we know that
−log [H3O+] + −log [OH−] = −log Kw = 14.0
or, in other words,
pH + pOH = pKw = 14.00 Acids
and
Bases
83. How Do We Measure pH?
• For less accurate
measurements, one
can use
Litmus paper
• “Red” paper turns
blue above ~pH = 8
• “Blue” paper turns
red below ~pH = 5
An indicator
84. How Do We Measure pH?
• For less accurate
measurements, one
can use
Litmus paper
• “Red” paper turns
blue above ~pH = 8
• “Blue” paper turns
red below ~pH = 5
An indicator
85. How Do We Measure pH?
• For less accurate
measurements, one
can use
Litmus paper
• “Red” paper turns
blue above ~pH = 8
• “Blue” paper turns
red below ~pH = 5
An indicator
86. How Do We Measure pH?
For more accurate
measurements,
one uses a pH
meter, which
measures the
voltage in the
solution.
Acids
and
Bases
87. Strong Acids
• You will recall that the seven strong
acids are HCl, HBr, HI, HNO3, H2SO4,
HClO3, and HClO4.
Acids
and
Bases
88. Strong Acids
• You will recall that the seven strong
acids are HCl, HBr, HI, HNO3, H2SO4,
HClO3, and HClO4.
• These are, by definition, strong
electrolytes and exist totally as ions in
aqueous solution.
Acids
and
Bases
89. Strong Acids
• You will recall that the seven strong
acids are HCl, HBr, HI, HNO3, H2SO4,
HClO3, and HClO4.
• These are, by definition, strong
electrolytes and exist totally as ions in
aqueous solution.
• For the monoprotic strong acids,
[H3O+] = [acid]. Acids
and
Bases
90. SAMPLE EXERCISE 16.8 Calculating the pH of a Strong Acid
What is the pH of a 0.040 M
solution of HClO4?
Acids
and
Bases
91. SAMPLE EXERCISE 16.8 Calculating the pH of a Strong Acid
What is the pH of a 0.040 M
solution of HClO4?
HClO4 is a strong acid, so it is
completely ionized.
[H+] = [ClO4–] = 0.040 M.
Because [H+] lies between 1 × 10–2 and 1 × 10–1 the pH will
be between 2.0 and 1.0.
Acids
and
Bases
92. SAMPLE EXERCISE 16.8 Calculating the pH of a Strong Acid
What is the pH of a 0.040 M
solution of HClO4?
HClO4 is a strong acid, so it is
completely ionized.
[H+] = [ClO4–] = 0.040 M.
Because [H+] lies between 1 × 10–2 and 1 × 10–1 the pH will
be between 2.0 and 1.0.
Acids
pH = –log(0.040) = 1.40. and
Bases
93. PRACTICE EXERCISE
An aqueous solution of HNO3
has a pH of 2.34. What is the
concentration of the acid?
Acids
and
71 Bases
94. PRACTICE EXERCISE
An aqueous solution of HNO3
has a pH of 2.34. What is the
concentration of the acid?
Answer: 0.0046 M
Acids
and
71 Bases
95. Strong Bases
• Strong bases are the soluble
hydroxides, which are the alkali metal
and heavier alkaline earth metal
hydroxides (Ca2+, Sr2+, and Ba2+).
Acids
and
Bases
96. Strong Bases
• Strong bases are the soluble
hydroxides, which are the alkali metal
and heavier alkaline earth metal
hydroxides (Ca2+, Sr2+, and Ba2+).
• Again, these substances dissociate
completely in aqueous solution. Acids
and
Bases
97. SAMPLE EXERCISE 16.9 Calculating the pH of a Strong Base
What is the pH of (a) a 0.028 M
solution of NaOH, (b) a 0.0011 M
solution of Ca(OH)2?
Acids
and
Bases
98. SAMPLE EXERCISE 16.9 Calculating the pH of a Strong Base
What is the pH of (a) a 0.028 M
solution of NaOH, (b) a 0.0011 M
solution of Ca(OH)2?
(a) NaOH dissociates in water to give
Solve:
one OH– ion per formula unit. Therefore, the
OH– concentration is 0.028 M.
Acids
and
Bases
99. SAMPLE EXERCISE 16.9 Calculating the pH of a Strong Base
What is the pH of (a) a 0.028 M
solution of NaOH, (b) a 0.0011 M
solution of Ca(OH)2?
Acids
and
Bases
100. SAMPLE EXERCISE 16.9 Calculating the pH of a Strong Base
What is the pH of (a) a 0.028 M
solution of NaOH, (b) a 0.0011 M
solution of Ca(OH)2?
(b) Ca(OH)2
is a strong base that dissociates in water
to give two OH– ions per formula unit. Thus, the
concentration of OH–(aq) is 2 × (0.0011M) = 0.0022 M.
Acids
and
Bases
101. PRACTICE EXERCISE
What is the concentration of a
solution of (a) KOH for which the
pH is 11.89; (b) Ca(OH)2 for which
the pH is 11.68?
Acids
and
Bases
102. PRACTICE EXERCISE
What is the concentration of a
solution of (a) KOH for which the
pH is 11.89; (b) Ca(OH)2 for which
the pH is 11.68?
Answers: (a) 7.8 × 10–3 M, (b) 2.4 × 10–13 M
Acids
and
Bases
107. Dissociation Constants
• For a generalized acid dissociation,
HA(aq) + H2O(l) A−(aq) + H3O+(aq)
the equilibrium expression would be
[H3O+] [A−]
Kc =
[HA]
Acids
and
Bases
108. Dissociation Constants
• For a generalized acid dissociation,
HA(aq) + H2O(l) A−(aq) + H3O+(aq)
the equilibrium expression would be
[H3O+] [A−]
Kc =
[HA]
• This equilibrium constant is called the
acid-dissociation constant, Ka.
Acids
and
Bases
110. The [H+] in an 0.020 M solution of
HNO2 is 3.0 x 10 -3 M. What is the
N
Ka of HNO2?
O OH
nitrous acid
1. 4.5 x 10-4
2. 6.0 x 10-5
3. 9.0 x 10-6
4. 1.5 x 10-1
5. None of Acids
and
Bases
the above
111. The [H+] in an 0.020 M solution of
HNO2 is 3.0 103 M. What is the Ka
N
of HNO2?
O OH
[H + ][A − ]
Ka = nitrous acid
[HA]
[3.0 × 10−3 ]2 9.0 × 10−6 1. 4.5 x 10-4
Ka = =
0.020 0.020 2. 6.0 x 10-5
= 4.5 × 10−4 3. 9.0 x 10-6
Ka
4. 1.5 x 10-1
5. None of Acids
and
Bases
the above
112. Calculating Ka from the pH
• The pH of a 0.10 M solution of formic acid,
HCOOH, at 25°C is 2.38. Calculate Ka for
formic acid at this temperature.
Acids
and
Bases
113. Calculating Ka from the pH
• The pH of a 0.10 M solution of formic acid,
HCOOH, at 25°C is 2.38. Calculate Ka for
formic acid at this temperature.
• We know that
[H3O+] [COO−]
Ka =
[HCOOH]
Acids
and
Bases
114. Calculating Ka from the pH
• The pH of a 0.10 M solution of formic acid,
HCOOH, at 25°C is 2.38. Calculate Ka for
formic acid at this temperature.
Acids
and
Bases
115. Calculating Ka from the pH
• The pH of a 0.10 M solution of formic acid,
HCOOH, at 25°C is 2.38. Calculate Ka for
formic acid at this temperature.
• To calculate Ka, we need the equilibrium
concentrations of all three things.
Acids
and
Bases
116. Calculating Ka from the pH
• The pH of a 0.10 M solution of formic acid,
HCOOH, at 25°C is 2.38. Calculate Ka for
formic acid at this temperature.
• To calculate Ka, we need the equilibrium
concentrations of all three things.
• We can find [H3O+], which is the same as
[HCOO−], from the pH.
Acids
and
Bases
117. Calculating Ka from the pH
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = 10log [H3O+] = [H3O+]
4.2 × 10−3 = [H3O+] = [HCOO−]
Acids
and
Bases
118. Calculating Ka from pH
Now we can set up a table…
[HCOOH], M [H3O+], M [HCOO−], M
Initially 0.10 0 0
Change −4.2 × 10-3 +4.2 × 10-3 +4.2 × 10−3
At 0.10 − 4.2 × 10−3 4.2 × 10−3 4.2 × 10−3
Equilibrium = 0.0958 = 0.10
Acids
and
Bases
119. Calculating Ka from pH
[4.2 × 10−3] [4.2 × 10−3]
Ka =
[0.10]
Acids
and
Bases
120. Calculating Ka from pH
[4.2 × 10−3] [4.2 × 10−3]
Ka =
[0.10]
= 1.8 × 10−4
Acids
and
Bases
121. PRACTICE EXERCISE
Niacin, one of the B vitamins, has the
following molecular structure:
A 0.020 M solution of niacin has a pH of 3.26.
(a) What percentage of the acid is ionized in
this solution?
(b) What is the acid-dissociation constant, Ka,
for niacin? Acids
and
Bases
122. PRACTICE EXERCISE
Niacin, one of the B vitamins, has the
following molecular structure:
A 0.020 M solution of niacin has a pH of 3.26.
(a) What percentage of the acid is ionized in
this solution?
(b) What is the acid-dissociation constant, Ka,
for niacin? Acids
and
Bases
Answers: (a) 2.7%, (b) 1.5 × 10–5
124. Calculating Percent
Ionization
[H3O+]eq
• Percent Ionization = × 100
[HA]initial
• In this example
[H3O+]eq = 4.2 × 10−3 M
[HCOOH]initial = 0.10 M
Acids
and
Bases
127. The pH of a 0.050 M weak acid is 3.00.
What is the percentage ionization?
1. 0.10%
2. 0.20%
3. 1.0%
4. 2.0%
5. 3.0%
Acids
and
97 Bases
128. The pH of a 0.050 M weak acid is 3.00.
What is the percentage ionization?
1. 0.10% Since pH = 3.00,
2. 0.20% [H+] = 1.0 × 10−3 M, so
3. 1.0% +
[H ] eq
% ionization = × 100%
4. 2.0% [HA]o
5. 3.0% % ionization =
[1.0 ×10 -3 ]
×100%
[0.050]
% ionization = 2.0%
Acids
and
98 Bases
129. Calculating pH from Ka
Calculate the pH of a 0.30 M solution of
acetic acid, HC2H3O2, at 25°C.
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
Ka for acetic acid at 25°C is 1.8 × 10−5.
Acids
and
Bases
130. Calculating pH from Ka
The equilibrium constant expression
is
Acids
and
Bases
131. Calculating pH from Ka
The equilibrium constant expression
is
[H3O+] [C2H3O2−]
Ka =
[HC2H3O2]
Acids
and
Bases
132. Calculating pH from Ka
We next set up a table…
[C2H3O2], M [H3O+], M [C2H3O2−], M
Initially 0.30 0 0
Change −x +x +x
At 0.30 − x ≈ 0.30 x x
Equilibrium
Acids
and
Bases
133. Calculating pH from Ka
We next set up a table…
[C2H3O2], M [H3O+], M [C2H3O2−], M
Initially 0.30 0 0
Change −x +x +x
At 0.30 − x ≈ 0.30 x x
Equilibrium
We are assuming that x will be very small
Acids
compared to 0.30 and can, therefore, beBases
and
ignored.
135. Calculating pH from Ka
Now,
(x)2
1.8 × 10−5 =
(0.30)
(1.8 × 10−5) (0.30) = x2
5.4 × 10−6 = x2
2.3 × 10−3 = x
Acids
(Check to ensure that ionization is < 5% or the “simplifying
and
Bases
assumption” is not valid
136. Calculating pH from Ka
pH = −log [H3O+]
pH = −log (2.3 × 10−3)
pH = 2.64
Acids
and
Bases
138. because weak acids typically
undergo very little ionization,
often less than 1% in solution.
Acids
and
Bases
139. PRACTICE EXERCISE
The Ka for niacin (Practice Exercise 16.10) is
1.5 × 10–5. What is the pH of a 0.010 M
solution of niacin?
Acids
and
Bases
140. PRACTICE EXERCISE
The Ka for niacin (Practice Exercise 16.10) is
1.5 × 10–5. What is the pH of a 0.010 M
solution of niacin?
Answer: 3.42 Acids
and
Bases
141. SAMPLE EXERCISE 16.12 Using Ka to Calculate Percent Ionization
Calculate the percentage of HF molecules
ionized in (a) a 0.10 M HF solution, (b) a
0.010 M HF solution.
Acids
and
Bases
142. SAMPLE EXERCISE 16.12 Using Ka to Calculate Percent Ionization
Calculate the percentage of HF molecules
ionized in (a) a 0.10 M HF solution, (b) a
0.010 M HF solution.
Acids
and
Bases
143. SAMPLE EXERCISE 16.12 Using Ka to Calculate Percent Ionization
Calculate the percentage of HF molecules
ionized in (a) a 0.10 M HF solution, (b) a
0.010 M HF solution.
The equilibrium-constant expression is
Acids
and
Bases
144. SAMPLE EXERCISE 16.12 Using Ka to Calculate Percent Ionization
Calculate the percentage of HF molecules
ionized in (a) a 0.10 M HF solution, (b) a
0.010 M HF solution.
The equilibrium-constant expression is
When we try solving this equation using the
approximation 0.10 – x = 0.10 (that is, by neglecting the
Acids
concentration of acid that ionizes in comparison with the and
initial concentration), we obtain Bases
145. SAMPLE EXERCISE 16.12 continued
Because this value is greater than 5% of 0.10 M, we should work the
problem without the approximation, using an equation-solving calculator or
the quadratic formula. Rearranging our equation and writing it in standard
quadratic form, we have
Acids
and
Bases
146. SAMPLE EXERCISE 16.12 continued
Because this value is greater than 5% of 0.10 M, we should work the
problem without the approximation, using an equation-solving calculator or
the quadratic formula. Rearranging our equation and writing it in standard
quadratic form, we have
This equation can be solved using the standard quadratic
formula.
Acids
and
Bases
147. SAMPLE EXERCISE 16.12 continued
Because this value is greater than 5% of 0.10 M, we should work the
problem without the approximation, using an equation-solving calculator or
the quadratic formula. Rearranging our equation and writing it in standard
quadratic form, we have
This equation can be solved using the standard quadratic
formula.
Substituting the appropriate numbers gives
Acids
and
Bases
148. SAMPLE EXERCISE 16.12 continued
Because this value is greater than 5% of 0.10 M, we should work the
problem without the approximation, using an equation-solving calculator or
the quadratic formula. Rearranging our equation and writing it in standard
quadratic form, we have
This equation can be solved using the standard quadratic
formula.
Substituting the appropriate numbers gives
Of the two solutions, only the one that gives a positive value for x is Acids
and
chemically reasonable. Thus, Bases
149. SAMPLE EXERCISE 16.12 continued
From our result, we can calculate the percent of molecules
ionized:
Acids
and
Bases
150. SAMPLE EXERCISE 16.12 continued
From our result, we can calculate the percent of molecules
ionized:
(b) Proceeding similarly for the 0.010 M solution, we have
Acids
and
Bases
151. SAMPLE EXERCISE 16.12 continued
From our result, we can calculate the percent of molecules
ionized:
(b) Proceeding similarly for the 0.010 M solution, we have
Solving the resultant quadratic expression, we obtain
Acids
and
Bases
152. SAMPLE EXERCISE 16.12 continued
From our result, we can calculate the percent of molecules
ionized:
(b) Proceeding similarly for the 0.010 M solution, we have
Solving the resultant quadratic expression, we obtain
The percentage of molecules ionized is
Acids
and
Bases
153. SAMPLE EXERCISE 16.12 continued
From our result, we can calculate the percent of molecules
ionized:
(b) Proceeding similarly for the 0.010 M solution, we have
Solving the resultant quadratic expression, we obtain
The percentage of molecules ionized is
Comment: Notice that if we do not use the quadratic formula to solve the problem properly, we
calculate 8.2% ionization for (a) and 26% ionization for (b). Notice also that in diluting the solution by a
factor of 10, the percentage of molecules ionized increases by a factor of 3. This result is in accord
with what we see in Figure 16.9. It is also what we would expect from Le Châtelier’s principle. • Acids
(Section 15.6) There are more “particles” or reaction components on the right side of the equation than and
on the left. Dilution causes the reaction to shift in the direction of the larger number of particles Bases
because this counters the effect of the decreasing concentration of particles.
154. PRACTICE EXERCISE
In Practice Exercise 16.10, we found that
the percent ionization of niacin (Ka = 1.5 ×
10–5) in a 0.020 M solution is 2.7%.
Calculate the percentage of niacin
molecules ionized in a solution that is
(a) 0.010 M,
(b) 1.0 × 10–3 M.
Acids
and
Bases
155. PRACTICE EXERCISE
In Practice Exercise 16.10, we found that
the percent ionization of niacin (Ka = 1.5 ×
10–5) in a 0.020 M solution is 2.7%.
Calculate the percentage of niacin
molecules ionized in a solution that is
(a) 0.010 M,
(b) 1.0 × 10–3 M.
Answers: (a) 3.8%, (b) 12%
Acids
and
Bases
157. Polyprotic Acids
• Have more than one acidic proton.
• If the difference between the Ka for the
first dissociation and subsequent Ka
values is 103 or more, the pH generally
depends only on the first dissociation.
Acids
and
Bases
159. This is the acid dissociation constant for
the 3rd and final proton from H3PO4.
Acids
and
Bases
160. SAMPLE EXERCISE 16.13 Calculating the pH of a Polyprotic Acid Solution
The solubility of CO2 in pure water at 25°C and 0.1 atm
pressure is 0.0037 M. The common practice is to
assume that all of the dissolved CO2 is in the form of
carbonic acid (H2CO3), which is produced by reaction
between the CO2 and H2O:
What is the pH of a 0.0037 M solution of H2CO3?
Acids
and
Bases
161. SAMPLE EXERCISE 16.13 Calculating the pH of a Polyprotic Acid Solution
The solubility of CO2 in pure water at 25°C and 0.1 atm
pressure is 0.0037 M. The common practice is to
assume that all of the dissolved CO2 is in the form of
carbonic acid (H2CO3), which is produced by reaction
between the CO2 and H2O:
What is the pH of a 0.0037 M solution of H2CO3?
H2CO3 is a diprotic acid; the two acid-dissociation
constants, Ka1 and Ka2 (Table 16.3), differ by more
than a factor of 103. Consequently, the pH can be
determined by considering only Ka1, thereby Acids
and
treating the acid as if it were a monoprotic acid. Bases
162. The solubility of CO2 in pure water at 25°C and 0.1 atm
pressure is 0.0037 M. The common practice is to
assume that all of the dissolved CO2 is in the form of
carbonic acid (H2CO3), which is produced by reaction
between the CO2 and H2O:
What is the pH of a 0.0037 M solution of H2CO3?
Acids
and
Bases
163. The solubility of CO2 in pure water at 25°C and 0.1 atm
pressure is 0.0037 M. The common practice is to
assume that all of the dissolved CO2 is in the form of
carbonic acid (H2CO3), which is produced by reaction
between the CO2 and H2O:
What is the pH of a 0.0037 M solution of H2CO3?
Acids
and
Bases
164. The solubility of CO2 in pure water at 25°C and 0.1 atm
pressure is 0.0037 M. The common practice is to
assume that all of the dissolved CO2 is in the form of
carbonic acid (H2CO3), which is produced by reaction
between the CO2 and H2O:
What is the pH of a 0.0037 M solution of H2CO3?
Acids
and
Bases
166. Solving this equation using an equation-solving calculator,
we get
Because Ka1 is small, we can make the simplifying
approximation that x is small, so that
Acids
and
Bases
167. Solving this equation using an equation-solving calculator,
we get
Because Ka1 is small, we can make the simplifying
approximation that x is small, so that
Acids
and
Bases
168. Solving this equation using an equation-solving calculator,
we get
Because Ka1 is small, we can make the simplifying
approximation that x is small, so that
Acids
and
Bases
169. Solving this equation using an equation-solving calculator,
we get
Because Ka1 is small, we can make the simplifying
approximation that x is small, so that
The small value of x indicates that our simplifying
assumption was justified.
Acids
and
Bases
170. Comment: If we were asked to solve for [CO32–], we would need
to use Ka2. Using the values of [HCO3–] and [H+] calculated
above, and setting [CO32–] = y
171. Comment: If we were asked to solve for [CO32–], we would need
to use Ka2. Using the values of [HCO3–] and [H+] calculated
above, and setting [CO32–] = y
Assuming that y is small compared to 4.0 × 10–5, we have
172. Comment: If we were asked to solve for [CO32–], we would need
to use Ka2. Using the values of [HCO3–] and [H+] calculated
above, and setting [CO32–] = y
Assuming that y is small compared to 4.0 × 10–5, we have
The value calculated for y is indeed very small compared to 4.0 × 10–5, showing that our assumption
was justified. It also shows that the ionization of HCO3– is negligible compared to that of H2CO3, as far
as production of H+ is concerned. However, it is the only source of CO32–, which has a very low
concentration in the solution. Our calculations thus tell us that in a solution of carbon dioxide in water,
most of the CO2 is in the form of CO2 or H2CO3, a small fraction ionizes to form H+ and HCO3–, and an
even smaller fraction ionizes to give CO32–. Notice also that [CO32–] is numerically equal to Ka2.
173. PRACTICE EXERCISE
(a) Calculate the pH of a 0.020 M
solution of oxalic acid (H2C2O4).
(See Table 16.3 for Ka1 and Ka2.)
(b) Calculate the concentration of
oxalate ion, [C2O42–], in this
solution.
Acids
and
Bases
174. PRACTICE EXERCISE
(a) Calculate the pH of a 0.020 M
solution of oxalic acid (H2C2O4).
(See Table 16.3 for Ka1 and Ka2.)
(b) Calculate the concentration of
oxalate ion, [C2O42–], in this
solution.
Answers: (a) pH = 1.80, (b) [C2O42–] = 6.4 × 10–5 M Acids
and
Bases
176. Weak Bases
The equilibrium constant expression
for this reaction is
[HB] [OH−]
Kb =
[B−]
where Kb is the base-dissociation constant.
Acids
and
Bases
177. Weak Bases
Kb can be used to find [OH−] and, through it, pH.
Acids
and
Bases
178. pH of Basic Solutions
What is the pH of a 0.15 M solution of
NH3?
NH3(aq) + H2O(l) NH4+(aq) + OH−(aq)
Acids
and
Bases
179. pH of Basic Solutions
What is the pH of a 0.15 M solution of
NH3?
NH3(aq) + H2O(l) NH4+(aq) + OH−(aq)
[NH4+] [OH−]
Kb = = 1.8 × 10−5
[NH3]
Acids
and
Bases
180. pH of Basic Solutions
Tabulate the data.
[NH3], M [NH4+], M [OH−], M
Initially 0.15 0 0
At Equilibrium 0.15 - x ≈ 0.15 x x
Acids
and
Bases
182. pH of Basic Solutions
Therefore,
[OH−] = 1.6 × 10−3 M
pOH = −log (1.6 × 10−3)
pOH = 2.80
pH = 14.00 − 2.80
pH = 11.20
The value obtained for x is only about 1% of the
NH3 concentration, 0.15 M. Therefore, neglecting Acids
and
x relative to 0.15 was justified. Bases
183. SAMPLE EXERCISE 16.15 Using pH to Determine the Concentration of a Salt
A solution made by adding solid sodium
hypochlorite (NaClO) to enough water to make
2.00 L of solution has a pH of 10.50. Calculate
the number of moles of NaClO that were added
to the water.
ClO– ion is a weak base with Kb = 3.33 × 10–7
Acids
and
Bases
184. SAMPLE EXERCISE 16.15 Using pH to Determine the Concentration of a Salt
A solution made by adding solid sodium
hypochlorite (NaClO) to enough water to make
2.00 L of solution has a pH of 10.50. Calculate
the number of moles of NaClO that were added
to the water.
ClO– ion is a weak base with Kb = 3.33 × 10–7
Acids
and
Bases
185. SAMPLE EXERCISE 16.15 Using pH to Determine the Concentration of a Salt
Acids
and
Bases
186. SAMPLE EXERCISE 16.15 Using pH to Determine the Concentration of a Salt
This concentration is high enough that we can
neglect any OH– produced by the
autoionization of H2O.
Acids
and
Bases
189. SAMPLE EXERCISE 16.15 continued
We say that the solution is 0.30 M in NaClO,
even though some of the ClO– ions have
reacted with water.
Because the solution is 0.30 M in NaClO and
the total volume of solution is 2.00 L, 0.60 mol
Acids
of NaClO is the amount of the salt that wasBases
and
added to the water.
190. PRACTICE EXERCISE
A solution of NH3 in water has a
pH of 11.17. What is the molarity
of the solution?
Acids
and
Bases
191. PRACTICE EXERCISE
A solution of NH3 in water has a
pH of 11.17. What is the molarity
of the solution?
Answer: 0.12 M
Acids
and
Bases
192. Chapter 16
Acid/Base Equilibrium
Sections 16.8 - 16.9
Ka from Kb
pH of Salt solutions
Read pages 696-702
HOMEWORK
Pg 712 #5, 7, 79, 81, 83, 85, 87, 89
Acids
and
Bases
194. Ka and Kb
Ka and Kb are related in this way:
Ka × Kb = Kw
Therefore, if you know one of them, you can
calculate the other. Acids
and
Bases
195. SAMPLE EXERCISE 16.16 Calculating Ka or Kb for a Conjugate Acid-Base Pair
Calculate (a) the base-dissociation constant, Kb, for the
fluoride ion (F–); (b) the acid-dissociation constant, Ka, for
the ammonium ion (NH4+).
Acids
and
Bases
196. SAMPLE EXERCISE 16.16 Calculating Ka or Kb for a Conjugate Acid-Base Pair
Calculate (a) the base-dissociation constant, Kb, for the
fluoride ion (F–); (b) the acid-dissociation constant, Ka, for
the ammonium ion (NH4+).
(a) Ka for the weak acid, HF, is given in Table 16.2 and
Solve:
Appendix D as Ka = 6.8 × 10–4 . We can use Equation
16.40 to calculate Kb for the conjugate base, F–:
Acids
and
Bases
197. SAMPLE EXERCISE 16.16 Calculating Ka or Kb for a Conjugate Acid-Base Pair
Calculate (a) the base-dissociation constant, Kb, for the
fluoride ion (F–); (b) the acid-dissociation constant, Ka, for
the ammonium ion (NH4+).
(a) Ka for the weak acid, HF, is given in Table 16.2 and
Solve:
Appendix D as Ka = 6.8 × 10–4 . We can use Equation
16.40 to calculate Kb for the conjugate base, F–:
Kb for NH3 is listed in Table 16.4 and in Appendix D as
(b)
Kb = 1.8 × 10–5. Using Equation 16.40, we can calculate Ka
for the conjugate acid, NH4+:
Acids
and
Bases
198. SAMPLE EXERCISE 16.16 continued
PRACTICE EXERCISE
(a)Which of the following anions has the largest base-
dissociation constant: NO2–, PO43– , or N3– ?
(b) The base quinoline has the following structure:
Acids
and
Bases
199. SAMPLE EXERCISE 16.16 continued
PRACTICE EXERCISE
(a)Which of the following anions has the largest base-
dissociation constant: NO2–, PO43– , or N3– ?
(b) The base quinoline has the following structure:
Its conjugate acid is listed in handbooks as having a pKa of
4.90. What is the base-dissociation constant for quinoline?
Acids
and
Bases
200. SAMPLE EXERCISE 16.16 continued
PRACTICE EXERCISE
(a)Which of the following anions has the largest base-
dissociation constant: NO2–, PO43– , or N3– ?
(b) The base quinoline has the following structure:
Its conjugate acid is listed in handbooks as having a pKa of
4.90. What is the base-dissociation constant for quinoline?
Answers: (a) PO43–(Kb = 2.4 × 10–2), (b) 7.9 × 10–10 Acids
and
Bases
203. Reactions of Anions with Water
• Anions are bases.
• As such, they can react with water in a
hydrolysis reaction to form OH− and the
conjugate acid:
X−(aq) + H2O(l) HX(aq) + OH−(aq)
Acids
and
Bases
205. Reactions of Cations with Water
• Cations with acidic protons
(like NH4+) will lower the pH
of a solution.
Acids
and
Bases
206. Reactions of Cations with Water
• Cations with acidic protons
(like NH4+) will lower the pH
of a solution.
• Most metal cations that are
hydrated in solution also
lower the pH of the solution.
Acids
and
Bases
208. Effect of Cations and Anions
1. An anion that is the
conjugate base of a strong
acid will not affect the pH.
Acids
and
Bases
209. Effect of Cations and Anions
1. An anion that is the
conjugate base of a strong
acid will not affect the pH.
2. An anion that is the
conjugate base of a weak
acid will increase the pH.
Acids
and
Bases
210. Effect of Cations and Anions
1. An anion that is the
conjugate base of a strong
acid will not affect the pH.
2. An anion that is the
conjugate base of a weak
acid will increase the pH.
3. A cation that is the
conjugate acid of a weak
base will decrease the pH.
Acids
and
Bases
212. Effect of Cations and Anions
4. Cations of the strong
Arrhenius bases will not
affect the pH.
Acids
and
Bases
213. Effect of Cations and Anions
4. Cations of the strong
Arrhenius bases will not
affect the pH.
5. Other metal ions will
cause a decrease in pH.
Acids
and
Bases
214. Effect of Cations and Anions
4. Cations of the strong
Arrhenius bases will not
affect the pH.
5. Other metal ions will
cause a decrease in pH.
6. When a solution contains
both the conjugate base
of a weak acid and the
conjugate acid of a weak
base, the affect on pH
depends on the Ka and Kb
values. Acids
and
Bases
220. SAMPLE EXERCISE 16.17 Predicting the Relative Acidity of Salt Solutions
List the following solutions in order of increasing
pH: (i) 0.1 M Ba(C2H3O2)2, (ii) 0.1 M NH4Cl,
(iii) 0.1 M NH3CH3Br, (iv) 0.1 M KNO3.
Acids
and
Bases
221. SAMPLE EXERCISE 16.17 Predicting the Relative Acidity of Salt Solutions
List the following solutions in order of increasing
pH: (i) 0.1 M Ba(C2H3O2)2, (ii) 0.1 M NH4Cl,
(iii) 0.1 M NH3CH3Br, (iv) 0.1 M KNO3.
Solve: Solution (i) contains barium ions and acetate ions. Ba2+ is an ion of one of
the heavy alkaline earth metals and will therefore not affect the pH (summary
point 4). The anion, C2H3O2–, is the conjugate base of the weak acid HC2H3O2
and will hydrolyze to produce OH– ions, thereby making the solution basic
(summary point 2). Solutions (ii) and (iii) both contain cations that are conjugate
acids of weak bases and anions that are conjugate bases of strong acids. Both
solutions will therefore be acidic. Solution (i) contains NH4+, which is the
conjugate acid of NH3 (Kb = 1.8 × 10–5). Solution (iii) contains NH3CH3+, which
is the conjugate acid of NH2CH3 (Kb = 4.4 × 10–4). Because NH3 has the smaller
Kb and is the weaker of the two bases, NH4+ will be the stronger of the two
conjugate acids. Solution (ii) will therefore be the more acidic of the two.
Solution (iv) contains the K+ ion, which is the cation of the strong base KOH,
and the NO3– ion, which is the conjugate base of the strong acid HNO3. Neither Acids
and
of the ions in solution (iv) will react with water to any appreciable extent, making
Bases
the solution neutral. Thus, the order of pH is 0.1 M NH4Cl < 0.1 M NH3CH3Br <
223. PRACTICE EXERCISE
In each of the following, indicate which salt
will form the more acidic (or less basic)
0.010 M solution:
(a) NaNO3, Fe(NO3)3; (b) KBr, KBrO; (c)
CH3NH3Cl, BaCl2, (d) NH4NO2, NH4NO3.
Acids
and
Bases
224. PRACTICE EXERCISE
In each of the following, indicate which salt
will form the more acidic (or less basic)
0.010 M solution:
(a) NaNO3, Fe(NO3)3; (b) KBr, KBrO; (c)
CH3NH3Cl, BaCl2, (d) NH4NO2, NH4NO3.
Acids
and
Answers: (a) Fe(NO3)3, (b) KBr, (c) CH3NH3Cl, (d) NH4NO3 Bases
225. SAMPLE EXERCISE 16.18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic
Predict whether the salt Na2HPO4 will form an acidic
solution or a basic solution on dissolving in water.
Acids
and
Bases
226. SAMPLE EXERCISE 16.18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic
Predict whether the salt Na2HPO4 will form an acidic
solution or a basic solution on dissolving in water.
Na+ ion will not affect pH, what will HPO4- do?
Acids
and
Bases
227. SAMPLE EXERCISE 16.18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic
Predict whether the salt Na2HPO4 will form an acidic
solution or a basic solution on dissolving in water.
Na+ ion will not affect pH, what will HPO4- do?
The reaction with the larger equilibrium constant will dominate
and determine whether the solution is acidic or basic.
Acids
and
Bases
228. SAMPLE EXERCISE 16.18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic
Predict whether the salt Na2HPO4 will form an acidic
solution or a basic solution on dissolving in water.
Na+ ion will not affect pH, what will HPO4- do?
The reaction with the larger equilibrium constant will dominate
and determine whether the solution is acidic or basic.
From appendix, the value of Ka for the first equation is 4.2 × 10–13.
Calculate the value of Kb for Equation #2 from the value of Ka for
its conjugate acid, H2PO4– , using Ka × Kb = Kw
Acids
and
Bases
229. SAMPLE EXERCISE 16.18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic
Predict whether the salt Na2HPO4 will form an acidic
solution or a basic solution on dissolving in water.
Na+ ion will not affect pH, what will HPO4- do?
The reaction with the larger equilibrium constant will dominate
and determine whether the solution is acidic or basic.
From appendix, the value of Ka for the first equation is 4.2 × 10–13.
Calculate the value of Kb for Equation #2 from the value of Ka for
its conjugate acid, H2PO4– , using Ka × Kb = Kw
Ka for H2PO4– is 6.2 × 10–8, so Kb = 1.6x10-7.
The second equation has a bigger K so it will dominate and
Acids
and
solution is basic. Bases
230. Chapter 16
Acid/Base Equilibrium
Sections 16.10 - 16.11
Acid/Base structure vs. properties
Lewis acids and bases
Read pages 702-712
HOMEWORK
Pg 716 # 8, 9, 10, 91, 93, 95, 99, 101,103
Acids
and
Bases
231. Bellwork
Predict whether the dipotassium salt of citric
acid (K2HC6H5O7) will form an acidic or basic
solution in water
232. Bellwork
Predict whether the dipotassium salt of citric
acid (K2HC6H5O7) will form an acidic or basic
solution in water
Answer: acidic
233. Acid-Base Behavior &
Chemical Structure
FOR ANY H-X BOND
• Polar bonds where H has a ∂+ are
acidic. (X= non-metal)
• Polar bonds where H has a ∂- are basic.
(X= metal)
• Non-polar bonds with H have no acid-
base properties. (C-H bond)
Acids
and
144 Bases
235. Factors Affecting Acid Strength
• The weaker the H-X bond, the more acidic
the compound.
Acids
and
Bases
236. Factors Affecting Acid Strength
• The weaker the H-X bond, the more acidic
the compound.
• Acidity increases from left to right across a
row (↑ polarity) and from top to bottom down
Acids
and
a group (↑atom size =↓ bond strength). Bases
238. Factors Affecting Acid Strength
• The stability of the conjugate base will also
affect the strength of an acid.
Acids
and
146 Bases
239. Factors Affecting Acid Strength
• The stability of the conjugate base will also
affect the strength of an acid.
• Strong acids have very stable conjugate
bases.
Acids
and
146 Bases
240. Factors Affecting Acid Strength
• The stability of the conjugate base will also
affect the strength of an acid.
• Strong acids have very stable conjugate
bases.
Acids
and
146 Bases
241. Factors Affecting Acid Strength
• The stability of the conjugate base will also
affect the strength of an acid.
• Strong acids have very stable conjugate
bases.
• Acid strength is determined by all three
factors: bond polarity, bond strength, and
conjugate stability. Acids
and
146 Bases
243. Moving down a column, H-X
bond strength decreases;
Acids
and
Bases
244. Moving down a column, H-X
bond strength decreases;
Moving left-to-right across a
period, electronegativity
increases.
Acids
and
Bases
245. Compound that have -OH bound to an
atom of low electronegativity (a metal)
are bases that produce OH-.
Other compounds have -OH groups
where only the H+ comes off. They are
oxyacids and they involve non-metals.
Acids
Sulfuric acid and
148 Bases
246. Factors Affecting Acid Strength
In oxyacids, in
which an OH is
bonded to
another atom,
Y, the more
electronegative
Y is, the more
acidic the acid. Acids
and
Bases
247. Factors Affecting Acid Strength
For a series of oxyacids, acidity
increases with the number of
oxygens. More EN means more polar Acids
and
Bases
HX bond and more stable anion.
248. SAMPLE EXERCISE 16.19 Predicting Relative Acidities from Composition
and Structure
Arrange the compounds in each of the following
series in order of increasing acid strength: (a)
AsH3, HI, NaH, H2O; (b) H2SeO3, H2SeO4, H2O.
249. SAMPLE EXERCISE 16.19 Predicting Relative Acidities from Composition
and Structure
Arrange the compounds in each of the following
series in order of increasing acid strength: (a)
AsH3, HI, NaH, H2O; (b) H2SeO3, H2SeO4, H2O.
Solve: (a) The elements from the left side of the periodic table form the most basic
binary hydrogen compounds because the hydrogen in these compounds carries a
negative charge. Thus NaH should be the most basic compound on the list.
Because arsenic is less electronegative than oxygen, we might expect that AsH3
would be a weak base toward water. That is also what we would predict by an
extension of the trends shown in Figure 16.13. Further, we expect that the binary
hydrogen compounds of the halogens, as the most electronegative element in
each period, will be acidic relative to water. In fact, HI is one of the strong acids in
water. Thus the order of increasing acidity is NaH < AsH3 < H2O < HI.
250. SAMPLE EXERCISE 16.19 Predicting Relative Acidities from Composition
and Structure
Arrange the compounds in each of the following
series in order of increasing acid strength: (a)
AsH3, HI, NaH, H2O; (b) H2SeO3, H2SeO4, H2O.
Solve: (a) The elements from the left side of the periodic table form the most basic
binary hydrogen compounds because the hydrogen in these compounds carries a
negative charge. Thus NaH should be the most basic compound on the list.
Because arsenic is less electronegative than oxygen, we might expect that AsH3
would be a weak base toward water. That is also what we would predict by an
extension of the trends shown in Figure 16.13. Further, we expect that the binary
hydrogen compounds of the halogens, as the most electronegative element in
each period, will be acidic relative to water. In fact, HI is one of the strong acids in
water. Thus the order of increasing acidity is NaH < AsH3 < H2O < HI.
(b) The acidity of oxyacids increases as the number of oxygen atoms bonded
to the central atom increases. Thus, H2SeO4 will be a stronger acid than H2SeO3;
in fact, the Se atom in H2SeO4 is in its maximum positive oxidation state, and so
we expect it to be a comparatively strong acid, much like H2SO4. H2SeO3 is an
oxyacid of a nonmetal that is similar to H2SO3. As such, we expect that H2SeO3 is
able to donate a proton to H2O, indicating that H2SeO3 is a stronger acid than H2O.
Thus, the order of increasing acidity is H2O < H2SeO3 < H2SeO4.
251.
252. PRACTICE EXERCISE
In each of the following pairs choose the compound
that leads to the more acidic (or less basic) solution:
(a) HBr, HF; (b) PH3, H2S; (c) HNO2, HNO3; (d)
H2SO3, H2SeO3.
253. PRACTICE EXERCISE
In each of the following pairs choose the compound
that leads to the more acidic (or less basic) solution:
(a) HBr, HF; (b) PH3, H2S; (c) HNO2, HNO3; (d)
H2SO3, H2SeO3.
Answers: (a) HBr, (b) H2S, (c) HNO3, (d) H2SO3
254. Factors Affecting Acid Strength
Resonance in the conjugate bases of
carboxylic acids stabilizes the base
and makes the conjugate acid more
acidic. -COOH = carboxyl group
Acids
and
Bases
261. Lewis Bases
• Lewis bases are defined as electron-pair donors.
Acids
and
Bases
262. Lewis Bases
• Lewis bases are defined as electron-pair donors.
• Anything that could be a Brønsted–Lowry base is
a Lewis base.
Acids
and
Bases
263. Lewis Bases
• Lewis bases are defined as electron-pair donors.
• Anything that could be a Brønsted–Lowry base is
a Lewis base.
• Lewis bases can interact with things other than
protons, however. Acids
and
Bases