Chapter 8 Acids and Bases
1
Acids
Arrhenius acids
• produce H+ ions in water
H2O(l)
HCl(g)  H+(aq) + Cl(aq)
•
•
•
•
are electrolytes
have a sour taste
turn litmus red
neutralize bases
2
Naming Acids
• Acids with H and a nonmetal are named with the prefix
hydro and end with ic acid.
HCl
hydrochloric acid
• Acids with H and a polyatomic ion are named by changing
the end of the name of the polyatomic ion from
ate to ic acid or
ite to ous acid
ClO3 chlorate
ClO2 chlorite
HClO3 chloric acid
HClO2 chlorous acid
3
Names of Common Acids
4
Bases
Arrhenius bases
• produce OH ions in water
• taste bitter or chalky
• are electrolytes
• feel soapy and slippery
• neutralize acids
5
Naming Bases
Bases with OH ions are named as the hydroxide of the metal
in the formula.
NaOH
sodium hydroxide
KOH
potassium hydroxide
Ba(OH)2
barium hydroxide
Al(OH)3
aluminum hydroxide
Fe(OH)3
iron(III) hydroxide
6
Brønsted–Lowry Acids and Bases
According to the Brønsted–Lowry theory,
• acids donate a proton (H+)
• bases accept a proton (H+)
7
NH3, a Brønsted–Lowry Base
In the reaction of ammonia and water,
• NH3 is the base that accepts H+
• H2O is the acid that donates H+
H
H
N
H
+ H2O
H
N
+
H
+ OH-
H
ammonia
+ water
H
ammonium ion + hydroxide ion
8
Characteristics of Acids and Bases
9
Learning Check
In each of the following identify the Brønsted–Lowry acid
and base:
A. HNO3(aq) + H2O(l)  H3O+(aq) + NO3(aq)
B. HF(aq) + H2O(l)
H3O+ (aq) + F(aq)
10
Solution
In each of the following identify the Brønsted–Lowry acid
and base:
A. HNO3(aq) + H2O(l)  H3O+(aq) + NO3(aq)
acid
base
B. HF(aq) + H2O(l)
acid
H3O+ (aq) + F(aq)
base
11
Conjugate Acid–Base Pairs
In any acid–base reaction, there are two conjugate
acid–base pairs.
• Each pair is related by the loss and gain of H+.
• One pair occurs in the forward direction.
• One pair occurs in the reverse direction.
conjugate acid–base pair 1
HA + B
A + BH+
conjugate acid–base pair 2
12
Conjugate Acid–Base Pairs
In this acid–base reaction,
• an acid, HF, donates H+ to form its conjugate base, F
• a base, H2O, accepts H+ to form its conjugate acid, H3O+
• there are two conjugate acid–base pairs
13
Conjugate Acid–Base Pairs
In this acid–base reaction,
• the first conjugate acid–base pair is HF, which donates H+ to
form its conjugate base, F
• the other conjugate acid–base pair is H2O, which accepts H+
to form its conjugate acid, H3O+
• each pair is related by a loss and gain of H+
14
Conjugate Acid–Base Pairs
In the reaction of NH3 and H2O,
• one conjugate acid–base pair is NH3/NH4+
• the other conjugate acid–base is H2O/H3O+
15
Learning Check
1. Write the conjugate base of each of the following:
A. HBr
B. H2S
C. H2CO3
2. Write the conjugate acid of each of the following:
A. NO2
B. NH3
C. OH
16
Solution
1. Write the conjugate base of each of the following:
Subtract an H+ from each acid to get the conjugate base:
A. HBr
– H+
 Br
B. H2S
– H+
 HS
C. H2CO3 – H+
 HCO3
17
Solution
2. Write the conjugate acid of each of the following:
Add an H+ to each base to get the conjugate acid.
A. NO2
+ H+
B. NH3
+ H+  NH4+
C. OH
+ H+
 HNO2
 H2O
18
Learning Check
1. The conjugate base of HCO3 is
A. CO32
B. HCO3 C. H2CO3
2. The conjugate acid of HCO3 is
A. CO32
B. HCO3
C. H2CO3
3. The conjugate base of H2O is
A. OH
B. H2O C. H3O+
4. The conjugate acid of H2O is
A. OH
B. H2O C. H3O+
19
Solutions
1. The conjugate base of HCO3 is
A. CO32
B. HCO3
C. H2CO3
2. The conjugate acid of HCO3 is
A. CO32
B. HCO3
C. H2CO3
3. The conjugate base of H2O is
A. OH
B. H2O C. H3O+
4. The conjugate acid of H2O is
A. OH
B. H2O C. H3O+
20
Acids
• A strong acid completely ionizes (100%) in aqueous
solutions.
HCl(g) + H2O(l)  H3O+(aq) + Cl(aq)
• A weak acid dissociates only slightly in water to form a
few ions in aqueous solutions.
H2CO3(aq) + H2O(l)
H3O+(aq) + HCO3(aq)
21
Strong Acids
In water, the dissolved molecules
of HA, a strong acid,
• dissociate into ions 100%
• give large concentrations of H3O+
and the anion (A)
HCl(g) + H2O(l) 
H3O+(aq) + Cl(aq)
22
Strong Acids
Strong acids
• make up six of all the acids
• have weak conjugate bases
23
Weak Acids
In weak acids,
• only a few molecules
dissociate
• most of the weak acid
remains as the undissociated
(molecular) form of the acid
• the concentrations of the
H3O+ and the anion (A) are
small
H2CO3(aq) + H2O(l)
H3O+(aq) + HCO3(aq)
24
Weak Acids
Weak acids
• make up most of the acids
• have strong conjugate bases
25
Strong Bases
Strong bases
• are formed from metals of
Groups 1A (1) and 2A (2)
• include LiOH, NaOH, KOH,
and Ca(OH)2
• dissociate completely in water
KOH(s)  K+(aq) + OH(aq)
26
Ionization of Water
In water,
• H+ is transferred from one H2O molecule to another
• one water molecule acts as an acid, while another acts as
a base
H2O + H2O
O H + O H
H
water
H
water
H3O+
+ OH
+
H O H +
O H
H
hydronium
ion (+)
hydroxide
ion ()
27
Pure Water Is Neutral
In pure water,
• the ionization of water molecules
produces small but equal
quantities of H3O+ and OH ions
• molar concentrations are
indicated in brackets, as [H3O+]
and [OH]
[H3O+]
[OH−]
= 1.0  107 M
= 1.0  107 M
28
Acidic Solutions
Adding an acid to pure water
• increases the [H3O+]
• causes the [H3O+] to exceed
1.0  107 M
• decreases the [OH−]
29
Basic Solutions
Adding a base to pure water
• increases the [OH−]
• causes the [OH−] to exceed
1.0  107M
• decreases the [H3O+]
30
Comparison of [H3O+] and [OH−]
31
Ion Product of Water, Kw
The ion product constant, Kw, for water
• is the product of the concentrations of the hydronium and
hydroxide ions
Kw = [H3O+] [OH]
• can be obtained from the concentrations in pure water
Kw = [H3O+] [OH]
Kw = [1.0  107 M]  [1.0  107 M]
= 1.0  1014
32
[H3O+] and [OH−] in Solutions
In neutral, acidic, or basic solutions, the Kw is always
1.0 x 1014.
33
Guide to Calculating [H3O+]
34
Calculating [H3O+]
What is the [H3O+] of a solution if [OH] is 5.0  108 M?
Step 1 Write the Kw for water.
Kw = [H3O+][OH] = 1.0  1014
Step 2 Solve the Kw expression for unknown [H3O+].
[H3O+] = 1.0  1014
[OH]
35
Calculating [H3O+]
What is the [H3O+] of a solution if [OH] is 5.0  108 M?
Step 3 Substitute in the known [H3O+] or [OH] and calculate.
[H3O+] = 1.0  1014 = 2.0  107 M
5.0  10−8
36
Learning Check
If lemon juice has [H3O+] of 2  103 M, what is the [OH] of
the solution?
A.
2  1011 M
B.
5  1011 M
C.
5  1012 M
37
Solution
If lemon juice has [H3O+] of 2  103 M, what is the [OH] of
the solution?
Step 1 Write the Kw for water.
Kw = [H3O+ ][OH] = 1.0  1014
Step 2 Solve the Kw expression for unknown [OH].
[OH] = 1.0  1014
[H3O+]
38
Solution
If lemon juice has [H3O+] of 2  103 M, what is the [OH] of
the solution?
Step 3 Substitute in the known [H3O+] or [OH]
and calculate.
[OH] = 1.0  1014 = 5  1012 M
2  103
The answer is C, 5  1012 M.
39
The pH Scale
The pH of a solution
• is used to indicate the acidity of a solution
• has values that usually range from 0 to 14
• is acidic when the values are less than 7
• is neutral with a pH of 7
• is basic when the values are greater than 7
40
pH of Everyday Substances
On the pH scale, values below
7.0 are acidic, a value
of 7.0 is neutral, and values
above 7.0 are basic.
41
Calculating pH of Solutions
pH is the negative log of the hydronium ion concentration.
pH = log [H3O+]
Example: For a solution with [H3O+] = 1  10 4
pH = log [1  10 4 ]
pH = [4.0]
pH = 4.0
Note: The number of decimal places in the pH equals
the significant figures in the coefficient of [H3O+].
4.0
1 SF: 1  104
42
Significant Figures in pH Calculations
When expressing log values, the number of decimal
places in the pH is equal to the number of significant
figures in the coefficient of [H3O+].
[H3O+] = 1  10 4 pH = 4.0
[H3O+] = 8.0  106
pH = 5.10
[H3O+] = 2.4  108
pH = 7.62
43
Guide to Calculating pH of Solutions
44
Calculating pH
Find the pH of a solution with a [H3O+] of 1.0  103:
Step 1 Enter the [H3O+].
Enter 1  103 by pressing 1 (EE) 3.
The EE key gives an exponent of 10.
Enter change sign (+/ key or – key).
Step 2 Press the log key and change the sign.
log (1  103) = [3]
45
Calculating pH
Find the pH of a solution with a [H3O+] of 1.0  103:
Step 3 Adjust the number of significant figures on the right
of the decimal point to equal the SF in the coefficient.
The answer is 3 with 2 decimal places: 3.00.
(Two significant figures in 1.0  103)
46
Learning Check
The [H3O+] of tomato juice is 2  104 M.
What is the pH of the solution?
A. 4.0
B. 3.7
C. 10.3
47
Solution
1. The [H3O+] of tomato juice is 2  104 M.
Step 1 Enter the [H3O+].
Enter 2  104 by pressing 2 (EE) 4.
The EE key gives an exponent of 10.
Enter change sign (+/ key or – key).
Step 2 Press the log key and change the sign.
log (2  104) = [3.7]
Step 3 Adjust the number of significant figures on the
right of the decimal point to equal the SF in the
coefficient.
pH = log [ 2  104] = 3.7 The answer is B.
48
Calculating [H3O+] from pH
The [H3O+] can be expressed by using the pH as the negative
power of 10.
[H3O+] = 1  10pH
For pH = 3.0, the [H3O+] = 1  103
On a calculator:
1. Enter the pH value
3.0
2. Change sign
3.0
3. Use the inverse log key (or 10x) to obtain
the [H3O+].
= 1  103 M
49
Learning Check
1. What is the [H3O+] of a solution with a pH of 10.0?
A. 1  104 M
B. 1  1010 M
C. 1  1010 M
2. What is the [OH] of a solution with a pH of 2.00?
A. 1.0  102 M
B. 1.0  1012 M
C. 2.0 M
50
Solution
1. What is the [H3O+] of a solution with a pH of 10.0?
Using equation: 1  10pH = 1  1010 M
The answer is C, 1  1010 M.
2. What is the [OH] of a solution with a pH of 2.00?
[H3O+] = 1.0  102 M 1  10pH
[OH] = 1.0  1014 = 1.0  1012 M
1.0  102
The answer is B, 1.0  1012 M.
51
Acids and Metals
Acids react with metals
• such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn
• to produce hydrogen gas and the salt of the metal
Molecular equations:
2K(s) + 2HCl(aq)  2KCl(aq) + H2(g)
metal
acid
salt
hydrogen gas
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
metal
acid
salt
hydrogen gas
52
Acids and Carbonates
Acids react
• with carbonates and hydrogen carbonates
• to produce carbon dioxide gas, a salt, and water
2HCl(aq) + CaCO3(s)  CO2(g) + CaCl2(aq) + H2O(l)
acid
carbonate
carbon
salt
water
dioxide
HCl(aq) + NaHCO3(s)  CO2(g) + NaCl (aq) + H2O(l)
acid
bicarbonate
carbon
salt
water
dioxide
53
Neutralization Reactions
In a neutralization reaction,
• an acid reacts with a base to produce salt and water
• the acid HCl reacts with NaOH to produce salt and water
• the salt formed is the anion from the acid and cation of
the base
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
acid
base
salt
water
54
Neutralization Reactions
In neutralization reactions,
• if we write the strong acid and strong base as ions, we
see that H+ reacts with OH− to form water, leaving the ions
Na+ and Cl in solution:
H+(aq) + Cl(aq) + Na+(aq) + OH(aq) 
Na+(aq) + Cl(aq) + H2O(l)
• the overall reaction is H3O+ from the acid and OH from
the base form water:
H+(aq) + OH(aq)  H2O(l)
55
Guide for Balancing Neutralization Reactions
56
Balancing Neutralization Reactions
Write the balanced equation for the neutralization of
magnesium hydroxide and nitric acid.
Step 1 Write the reactants and products.
Mg(OH)2 + HNO3
Step 2 Balance the H+ in the acid with the OH in the base.
Mg(OH)2 + 2HNO3
Step 3 Balance the H2O with H+ and the OH.
Mg(OH)2 + 2HNO3  salt + 2H2O
Step 4 Write the salt from the remaining ions.
Mg(OH)2 + 2HNO3  Mg(NO3)2 + 2H2O
57
Basic Compounds in Antacids
58
Learning Check
Write the neutralization reactions for stomach acid, HCl,
and the ingredients in Mylanta.
Mylanta: Al(OH)3 and Mg(OH)2
59
Solution
Write the neutralization reactions for stomach acid, HCl, and
the ingredients in Mylanta.
Mylanta: For Al(OH)3:
Step 1 Write the reactants and products.
Al(OH)3 + HCl
Step 2 Balance the H+ in the acid with the OH in the base.
Al(OH)3 + 3HCl
Step 3 Balance the H2O with H+ and the OH.
Al(OH)3 + 3HCl  salt + 3H2O
Step 4 Write the salt from the remaining ions.
Al(OH)3(aq) + 3HCl(aq)  AlCl3(aq) + 3H2O(l)
60
Solution
Write the neutralization reactions for stomach acid, HCl, and
the ingredients in Mylanta.
Mylanta: For Mg(OH)2:
Step 1 Write the reactants and products.
Mg(OH)2 + HCl
Step 2 Balance the H+ in the acid with the OH in the base.
Mg(OH)2 + 2HCl
Step 3 Balance the H2O with H+ and the OH.
Mg(OH)2 + 2HCl  salt + 2H2O
Step 4 Write the salt from the remaining ions.
2HCl(aq) + Mg(OH)2(aq)  MgCl2(aq) + 2H2O(l)
61
Acid–Base Titration
Titration
• is a laboratory procedure
used to determine the
molarity of an acid
Base 
NaOH
• uses a base such as NaOH
to neutralize a measured
volume of an acid
Acid 
Solution
62
Indicator
An indicator
• is added to the acid in the
flask
• causes the solution to change
color when the acid is
neutralized
63
Endpoint of Titration
At the endpoint,
• the indicator gives the
solution a permanent
pink color
• the volume of the base
used to reach the endpoint
is measured
• the molarity of the acid
is calculated using the
neutralization equation for
the reaction
64
Guide to Calculations for
Acid–Base Titrations
65
Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of
0.225 M NaOH are required to neutralize 10.0 mL of HCl?
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Step 1 State given and needed quantities.
Given: 18.5 mL of 0.225 M NaOH
10.0 mL of HCl
Needed: Molarity of HCl
66
Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of
0.225 M NaOH are required to neutralize 10.0 mL of HCl?
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Step 2 Write a plan to calculate molarity or volume.
molarity
NaOH
equation
coefficients
molarity
HCl
18.5 mL  L  moles NaOH  moles HCl  L HCl
67
Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of
0.225 M NaOH are required to neutralize 10.0 mL of HCl?
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Step 3 State equalities and conversion factors including
concentration.
1 L = 1000 mL
0.225 mole NaOH
1 L NaOH
and
1 mole HCl x
1 mole NaOH
68
Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of
0.225 M NaOH are required to neutralize 10.0 mL of HCl?
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Step 4 Set up the problem to calculate the needed quantity.
18.5 mL NaOH  1 L NaOH  0.225 mole NaOH
1000 mL NaOH
1 L NaOH
 1 mole HCl = 0.00416 mole of HCl
1 mole NaOH
MHCl = 0.00416 mole HCl = 0.416 M HCl
0.0100 L HCl
69
Buffers
When an acid or base is
added to water, the pH
changes drastically.
In a buffer solution, the
pH is maintained; pH does
not change when acid or
base is added.
70
How Buffers Work
Buffers work because they
• resist changes in pH from the addition of acid or base
• in the body, absorb H3O+ or OH from foods and cellular
processes to maintain pH
• are important in the proper functioning of cells and blood
• maintain a pH close to 7.4 in blood. A change in the
pH of the blood affects the uptake of oxygen and
cellular processes
71
Components of a Buffer
A buffer solution
• contains a combination of acid–base conjugate pairs
For example: HA and NaA
• may contain a weak acid and a salt of its conjugate base
• typically has equal concentrations of a weak acid and
its salt
• may also contain a weak base and a salt of the
conjugate acid
72
How Buffers Work
In the acetic acid/acetate buffer with acetic acid (HC2H3O2)
and sodium acetate (NaC2H3O2),
• the salt produces acetate ions and sodium ions
NaC2H3O2(aq)  C2H3O2(aq) + Na+(aq)
• the salt is added to provide a higher concentration of
the conjugate base C2H3O2 than the weak acid alone
HC2H3O2(aq) + H2O(l)
C2H3O2(aq) + H3O+(aq)
Large amount
Large amount
73
Function of a Weak Acid in a Buffer
The function of the weak acid in a buffer is to neutralize a base.
The acetate ion produced in the neutralization reaction adds to
the concentration of acetate already in solution from the salt.
HC2H3O2 + OH
acetic acid
base
 C2H3O2 + H2O
acetate ion
water
74
Function of Conjugate Base in a Buffer
The function of the acetate ion, C2H3O2, is to neutralize
H3O+ from acids. The acetic acid produced contributes to
the available weak acid.
C2H3O2 + H3O+  HC2H3O2 + H2O
acetate ion
acid
acetic acid
water
75
Working Buffers
Buffers work because
• the weak acid in a buffer neutralizes bases and the
conjugate base in the buffer neutralizes acids
• the pH of the solution is maintained
The buffer described here consists of about equal concentrations of acetic acid (HC 2H3O2) and its
conjugate base, acetate ion (C2H3O2). Adding H3O+ to the buffer reacts with C2H3O2 whereas
adding OH neutralizes HC2H3O2. The pH of the solution is maintained as long as the added
amounts of acid or base are small compared to the concentrations of the buffer components.
76
Learning Check
Which of the following combinations make a buffer solution?
A. HCl and KCl
B. H2CO3 and NaHCO3
C. H3PO4 and NaCl
D. HC2H3O2 and KC2H3O2
77
Solution
Which of the following combinations make a buffer solution?
A buffer consists of a weak acid and salt of its conjugate base.
A.
B.
C.
D.
HCl and KCl
H2CO3 and NaHCO3
H3PO4 and NaCl
HC2H3O2 and KC2H3O2
Not a buffer.
Buffer, a weak acid and its salt.
Not a buffer.
Buffer, a weak acid and its salt.
78