Chapter 10
Acids and Bases
10.1 Acids
10.1 Acids and Bases
10.2 Conjugate Acid-Base Pairs
10.3 Strengths of Acids and Bases
10.4 Dissociation Constants
10.5 Ionization of Water
10.6 The pH Scale
10.7 Reactions of Acids and Bases
(just neutralization reactions)
10.9 Buffers
 Arrhenius acids
 Produce H+ ions in water.
H2O
HCl
H+(aq) + Cl– (aq)
 Are electrolytes.
 Have a sour taste.
 Corrode metals.
 React with bases to form salts
and water.
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Bases
 Arrhenius bases
 Produce OH– ions in
water.
 Taste bitter or chalky.
 Are electrolytes.
 Feel soapy and slippery.
 React with acids to form
salts and water.
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Examples:
 Vinegar (5% acetic acid)
 Citric acid (fruits)
 Ascorbic acid (vitamin C)
 Lactic acid (in muscles)
 Phosphoric acid (Coke)
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Comparing Acids and Bases
Examples:
 Draino (NaOH)
 Baking Soda (NaHCO3)
 Bleach (NaOCl)
 Ammonia (NH3)
 Antacid
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1
Learning Check
Solution
Identify each as a characteristic of an
A) acid
or
B) base
____1. Has a sour taste.
____2. Produces OH- in aqueous solutions.
____3. Has a chalky taste.
____4. Is an electrolyte.
____5. Produces H+ in aqueous solutions.
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Identify each as a characteristic of an
A) acid or B) base
A
1. Has a sour taste.
B
2. Produces OH– in aqueous solutions.
B
3. Has a chalky taste.
A, B 4. Is an electrolyte.
A
5. Produces H+ in aqueous solutions.
5
Names of Acids


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Some Acids and Their Anions
Acids with H and one 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 an –ate ion
to -ic acid and an –ite ion to -ous acid.
HClO3 chloric acid
HClO2 chlorous acid
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2
Learning Check
Solution
A. HBr
3. hydrobromic acid
The name of an acid with H and one nonmetal
begins with the prefix hydro- and ends with -ic
acid.
Name each of the following as acids:
A. HBr
1. bromic acid
2. bromous acid
3. hydrobromic acid
B. H2CO3
B. H2CO3
1. carbonic acid
An acid with H and a polyatomic ion is named
by changing the end of an –ate ion to -ic acid.
1. carbonic acid
2. hydrocarbonic acid
3. carbonous acid
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Some Common Bases
Match the formulas with the names:
sodium hydroxide
potassium hydroxide
barium hydroxide
aluminum hydroxide
iron (III) hydroxide
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Learning Check
 Bases with OH- ions are named as the
hydroxide of the metal in the formula.
NaOH
KOH
Ba(OH)2
Al(OH)3
Fe(OH)3
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A. ___ HNO2
1) hydrochloric acid
B. ___Ca(OH)2
2) sulfuric acid
C. ___H2SO4
3) sodium hydroxide
D. ___HCl
4) nitrous acid
E. ___NaOH
5) calcium hydroxide
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3
Solution
Johannes BrØnsted and Thomas Lowry (1923)
According to the Brønsted-Lowry theory,
 Acids are hydrogen ion (H +) donors.
 Bases are hydrogen ion (H+) acceptors.
donor
acceptor
hydronium ion
HCl
+ H 2O
H3O+ + Cl+
-
Match the formulas with the names:
A. 4 HNO2
4) nitrous acid
B. 5 Ca(OH)2
5) calcium hydroxide
C. 2 H2SO4
2) sulfuric acid
D. 1 HCl
1) hydrochloric acid
E. 3 NaOH
3) sodium hydroxide
+
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10.2 Conjugate Acids and Bases
When NH3 dissolves in water, a few NH3
molecules react with water to form ammonium
ion NH4+ and a hydroxide ion.
NH3 + H2O



NH4+(aq) + OH- (aq)
An acid that donates H+ forms a conjugate base.
A base that accepts a H+ forms a conjugate acid.
In an acid-base reaction, there are two conjugate
acid-base pairs.
acid 1
acceptor donor
+
+
+
-
HF
conjugate base 1
+ HO
2
base 2
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NH3, A Bronsted-Lowry Base

+
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H3O+
F+
conjugate acid 2
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4
Conjugate Acid-Base Pairs

Learning Check
A conjugate acid-base pair is two substances
related by a loss or gain of H+.
HF +
H 2O
H3O+ +
acid 1– conjugate base 1
base 2– conjugate acid 2
A. Write the conjugate base of the following:
1. HBr
2. H2S
3. H2CO3
B. Write the conjugate acid of the following:
1. NO2–
2. NH3
3. OH–
F-
HF, FH2O, H3O+
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Solution
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Learning Check
A. Write the conjugate base of the following:
1. HBr
Br2. H2S
HS–
3. H2CO3
HCO3–
B. Write the conjugate acid of the following:
1. NO2–
HNO2
2. NH3
NH4+
3. OH–
H2O
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Identify the following that are acid-base
conjugate pairs.
1. HNO2, NO2–
2. H2CO3, CO32 –
3. HCl, ClO4 –
4. HS–, H2S
5. NH3, NH4+
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5
Solution
Learning Check
A. The conjugate base of HCO3– is
1. CO32–
2. HCO3–
3.
–
B. The conjugate acid of HCO3 is
1. CO32–
2. HCO3–
3.
C. The conjugate base of H2O is
1. OH–
2. H2O
3.
D. The conjugate acid of H2O is
1. OH–
2. H2O
3.
Identify the following that are acid-base
conjugate pairs.
1. HNO2, NO2–
4. HS–, H2S
5. NH3, NH4+
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H2CO3
H3O+
H3O+
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Solution
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10.3 Strengths of Acids and Bases
A. The conjugate base of HCO3 – is
1. CO32–
B. The conjugate acid of HCO3– is
3. H2CO3
C. The conjugate base of H2O is
1. OH–
D. The conjugate acid of H2O is
3. H3O+
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H2CO3
Strong acids completely ionize (100%) in
aqueous solutions.
HCl + H2O
H3O+ (aq) + Cl– (aq)
(100 % ions)
 Strong bases completely (100%) dissociate
into ions in aqueous solutions.
H2O
NaOH
Na+ (aq) + OH–(aq)
(100 % ions)

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6
Strong and Weak Acids


Strong Acids
In an HCl solution,
the strong acid HCl
dissociates 100%.
A solution of the
weak acid
CH3COOH contains
mostly molecules
and a few ions.
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

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Only a few acids are strong acids.
The conjugate bases of strong acids are weak
bases.
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Weak Acids

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Strong Bases
Most acids are weak acids. The conjugate
bases of weak acids are strong bases.

Most bases in Groups 1A and 2A are
strong bases. They include
LiOH, NaOH, KOH, and Ca(OH)2

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Most other bases are weak bases.
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7
Learning Check
Solution
Identify each of the following as a strong or
weak acid or base.
A. HBr
strong acid
B. HNO2
weak acid
C. NaOH
strong base
D. H2SO4
strong acid
E. Cu(OH)2 weak base
Identify each of the following as a strong or
weak acid or base.
A. HBr
B. HNO2
C. NaOH
D. H2SO4
E. Cu(OH)2
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Learning Check
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Solution
A. Identify the stronger acid in each pair.
1. HNO2 or H2S
2. HCO3– or HBr
3. H3PO4 or H3O+
A. Identify the stronger acid in each pair.
1. HNO2
2. HBr
3. H3O+
B. Identify the strong base in each pair.
1. NO3– or F2. CO32– or NO2–
3. OH– or H2O
B. Identify the stronger base in each pair.
1. F2. CO32–
3. OH–
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8
Equilibrium
10.4 Strong Acids

H2SO4 +
H2O
H3O+ + HSO4-
stronger
acid
stronger
base
weaker
acid
CO3- +
weaker
base
H2O
HCO3-
weaker
acid
stronger
acid
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weaker
base
+
OH-

stronger
base
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In water, the
dissolved molecules
of a strong acid are
essentially all
separated into ions.
The concentrations
of H3O+ and the
anion (A–) are
large.
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Weak Acids
Dissociation Constants
In weak acids,
 The equilibrium
favors the
undissociated
(molecular) form of
the acid.
 The concentrations
of the H3O+ and the
anion (A–) are
small.
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

In a weak acid, the rate of the dissociation of
the acid is equal to the rate of the association.
HA + H2O
H3O+ + A–
The equilibrium expression for a weak acid is
Keq = [H3O+][A-]
[HA][H2O]
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9
Acid Dissociation Constants (Ka)

Writing Ka for a Weak Acid
The [H2O] is large and remains unchanged.
It is combined with the Keq to write the acid
dissociation constant, Ka.
Ka = Keq[H2O]
=
Write the Ka for H2S.
1. Write the equation for the dissociation of H2S.
H2S + H2O
H3O+ + HS–
[H3O+][A-]
[HA]
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2. Set up the Ka expression
Ka = [H3O+][HS-]
[H2S]
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Learning Check
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Solution
Write the Ka for HCN.
Write the equation for the dissociation of HCN.
HCN + H2O
H3O+ + CN–
Write the Ka for HCN.
Set up the Ka expression
Ka =
[H3O+][CN–]
[HCN]
Note: Ka = Keq [H2O]
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10
Acid dissociation constants
10.5 Ionization of Water
Larger Ka (stronger acid)
Larger Kb (stronger base)
 In water, H+ is transferred from one H2O molecule
to another.
 One water molecule acts as an acid, while another
acts as a base (amphiprotic).
H2O + H2O
H3O+
+ OH –
Smaller Ka (weaker acid)
Smaller Kb (weaker base)
..
..
:O: H + :O:H
..
..
H
H
water molecules
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Pure Water is Neutral
..
H
hydronium
ion (+)
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..
+ :O:H–
..
hydroxide
ion (–)
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Acidic Solutions
 In pure water, the ionization of
molecules produces small, but
equal quantities of H3O+ and OHions.
H2O + H2O
H3O+ + OH-
 Adding an acid to pure
water increases the
[H3O+].
 In acids, [H3O+] exceeds
1.0 x 10–7 M.
 As [H3O+] increases,
[OH–] decreases.
 Molar concentrations are
indicated as [H3O+] and [OH-].
[H3O+] = 1.0 x 10-7 M
[OH-]
= 1.0 x 10-7 M
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..
H:O:H +
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11
Comparison of [H3O+] and [OH–]
Basic Solutions
 Adding a base to pure
water increases the
[OH–].
 In bases, [OH–]
exceeds 1.0 x 10–7M.
 As [OH–] increases,
[H3O+] decreases.
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Ion Product of Water, Kw


Kw in Acids and Bases
The ion product constant, Kw, for water is
the product of the concentrations of the
hydronium and hydroxide ions.
Kw = [H3O+][OH–]
We can obtain the value of Kw using the
concentrations in pure water.
Kw = [1.0 x 10–7][1.0 x 10–7]
= 1.0 x 10–14
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
47
In neutral, acidic, and basic solutions, the
Kw is equal to 1.0 x 10–14.
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12
Calculating [H3O+]
Learning Check
What is the [H3O+] of a solution if [OH–] is
1.0 x 10-8 M?
The [H3O+] of lemon juice is 1.0 x 10–3 M.
What is the [OH–] of the solution?
Rearrange the Kw expression for [H3O+ ].
Kw
=
[H3O+][OH–] = 1.0 x 10-14
[H3O+]
=
1.0 x 10-14
[OH–]
[H3O+]
= 1.0 x 10-14 = 1.0 x 10-6 M
1.0 x 10- 8
1) 1.0 x 103 M
2) 1.0 x 10–11 M
3) 1.0 x 1011 M
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Solution
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Learning Check
The [H3O+] of lemon juice is 1.0 x 10–3 M.
What is the [OH–] of the solution?
The [OH–] of a solution is 5 x 10–2 M.
What is the [H3O+ ] of the solution?
1) 2 x 10–2 M
2) 1.0 x 10–11 M
Rearrange the Kw to solve for [OH–]
Kw
= [H3O+ ][OH–]
[OH- ]
= 1.0 x 10 -14
1.0 x 10 - 3
=
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2) 1 x 102 M
= 1.0 x 10–14
3) 2 x 10–13 M
1.0 x 10!–11 M
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Solution
Learning Check
The [OH–] of a water solution is 5 x 10–2 M.
What is the [H3O+] of the solution?
3) 2 x 10–13 M
A. What is the [OH–] if the [H3O+ ] is 1 x 10–4 M?
Identify the solution as acidic, basic, or neutral.
1) 1 x 10–6 M 2) 1 x 10–8 M 3) 1 x 10–10 M
[ H3O+] = 1.0 x 10 –14
5 x 10–2
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B. What is [H3O+] if the [OH–] is 5 x 10–9 M?
Identify the solution as acidic, basic, or neutral.
1) 1 x 10–6 M 2) 2 x 10–6 M 3) 2 x 10–7 M
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Solution
=
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10.6 pH Scale
The pH scale:
 Is used to indicate the acidity of a solution.
 Has values that usually range from 0 to 14.
 Indicates an acidic solution when the values
are less than 7.
 Indicates a neutral solution with a pH of 7.
 Indicates a basic solution when the values
are greater than 7.
Kw = [H3O+][OH–] = 1.0 x 10 14
A. 3) [OH–]
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1.0 x 10–14 = 1.0 x 10–10
1.0 x 10–4
acidic
B. 2) [H3O+] = 1.0 x 10–14 = 2 x 10–6
5 x 10–9
acidic
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14
pH Range
0
1
2
3
4
5
6
7
8 9 10 11 12 13 14
Basic
Acidic
O+]>[OH-]
[H3
Neutral
[H3O+] = [OH-]
[H3O+]<[OH-]
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Learning Check
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Solution
Identify each solution as
1. acidic
2. basic
3. neutral
A. 1 HCl with a pH = 1.5
B. 2 Pancreatic fluid [H3O+] = 1 x 10–8 M
C. 1 Sprite soft drink pH = 3.0
D. 3 pH = 7.0
E. 1 [OH–] = 3 x 10–10 M
F. 2 [H3O+] = 5 x 10–12
Identify each solution as
1. acidic
2. basic
3. neutral
A. ___ HCl with a pH = 1.5
B. ___ Pancreatic fluid [H3O+] = 1 x 10–8 M
C. ___ Sprite soft drink pH = 3.0
D. ___ pH = 7.0
E. ___ [OH– ] = 3 x 10–10 M
F. ___ [H3O+ ] = 5 x 10–12
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Testing the pH of Solutions

Logarithms (log)
mx = n, then logm n = x, m = 10
The pH of solutions can be determined using a
a) pH meter, b) pH paper, and c) indicators
that have specific colors at different pHs.
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Ordinarily log10 (log in calculator)
log10 100 = 2.000
(102 = 100)
log10 1 = 0.000
(100 = 1)
log10 0.001 = -3.000
(10-3 = 0.001)
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Logarithms (log) Sig Figs
pH is:
 Defined as the negative log of the hydrogen
ion concentration.
pH = - log [H3O+]
 The value of the negative exponent of [H3O+]
for concentrations with a coefficient of 1.
[H3O+] = 1 x 10-4
pH = 4.0
+
-11
[H3O ] = 1 x 10
pH = 11.0
log10 2.00
= 0.301
log10 2.0
= 0.30
log10 2
= 0.3
3
log10 2.0 x 10 = 3.30
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pH Scale
1. Count the significant digits in the number (n)
(log n = x )
2. Add that same number of decimal points to the
log of the number (x).
Example:
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16
Calculations of pH
Significant Figures in pH
A calculator can be used to find the pH of a
solution with a [H3O+] of 1 x 10-3 as follows.
1. Enter 1 x 10-3 by pressing 1 (EE) 3 (+/–).
The EE key gives an exponent of 10 and the +/– key
changes the sign.
2. Press the log key to obtain log 1 x 10-3.
log (1 x 10-3) = –3
3. Press the +/– key to multiply by –1.
(-3) +/– = 3

[H3O+] = 1 x 10-4
pH = 4.0
+
-6
[H3O ] = 1.0 x 10 pH = 6.00
[H3O+] = 2.4 x 10-8 pH = 7.62
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Learning Check
2) 3.7
66
A. 2) 3.7
pH = – log [ 2 x 10-4] = 3.7
2 (EE) 4 (+/–) log (+/–)
3) 10.3
B. 2) 11.00
Use the Kw to obtain [H3O+] = 1.0 x 10-11
pH = – log [1.0 x 10- 11]
= – 11.00 (+/–) = 11.00
B. The [OH-] of a solution is 1.0 x 10-3 M.
What is the pH of the solution?
1) 3.00
2) 11.00 3) –11.00
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Solution
A. The [H3O+] of tomato juice is 2 x 10-4 M.
What is the pH of the solution?
1) 4.0
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+].
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17
Calculating [H3O+] from pH

Learning Check
A. What is the [H3O+] of a solution with a
pH of 10.0?
1) 1 x 10 -4 M
2) 1 x 1010 M
3) 1 x 10 -10 M
B. What is the [H3O+] of a solution with a
pH of 5.7?
1) 2 x 10- 6 M
2) 5 x 105 M
-5
3) 1 x 10 M
The [H3O+] can be expressed with the pH as
the negative power of 10.
[H3O+] = 1 x 10-pH
If the pH is 3.0, the [H3O+] = 1 x 10-3
On a calculator
1. Enter the pH value
3.0
2. Use (+/–) to change sign 3.0 (+/–) = –3.0
3. Use the inverse log key (or 10x) to obtain
the [H30+]. =
1 x 10-3 M
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Solution
70
pOH
A. What is the [H3O+] of a solution with a pH
of 10.0?
3) 1 x 10-10 M
1 x 10-pH
The pOH of a solution:
 Is related to the [OH-].
 Is similar to pH.
pOH = – log [OH-]
 Added to the pH value is equal to 14.0.
pH + pOH = 14.0
B. What is the [H3O+] of a solution with a pH
of 5.7?
1) 2 x 10- 6 M
5.7 (+/–) inv log (or 10x) = 2 x 10-6
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18
pH and pOH

Learning Check
The sum of the
pH and pOH
values is equal to
14.0 for any
aqueous
solution.
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What is the pH and the pOH of coffee
if the [H3O+] is 1 x 10-5 M?
1) pH = 5.0
pOH =7.0
2) pH = 7.0
pOH = 9.0
3) pH = 5.0
pOH = 9.0
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Solution
10.7 Neutralization Reactions
What is the pH and the pOH of coffee if the [H3O+]
is 1 x 10-5 M?
3) pH = 5.0
pOH = 9.0
pH = – log [1 x 10-5] = –(– 5.0) = 5.0
pH + pOH = 14.0
pOH = 14.0 – pH = 14.0 – 5.0 = 9.0
or [OH-] = 1 x 10-9
pOH = – log [1 x 10-9] = –(– 9.0) = 9.0
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75



An acid such as HCl produces H3O+ and a base
such as NaOH provides OH-.
HCl + H2O
H3O+ + ClNaOH
Na+ + OHIn a neutralization reaction, the H3O+ and the OHcombine to form water.
H3O+ + OH2 H 2O
The positive ion (metal) and the negative ion
(nonmetal) are the ions of a salt.
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19
Balancing Neutralization
Reactions
Neutralization Equations

Write the balanced equation for the neutralization
of magnesium hydroxide and nitric acid.
1. Write the formulas of the acid and base.
Mg(OH)2 + HNO3
2. Balance to give equal OH- and H+.
Mg(OH)2 + 2 HNO3
3. Write the products (a salt and water) and balance:
Mg(OH)2 + 2HNO3
Mg(NO3)2 + 2H2O
In the equation for neutralization, an acid
and a base produce a salt and water.
NaOH + HCl
NaCl + H2O
base
acid
salt
water
Ca(OH)2 + 2 HCl
base
acid
CaCl2 + 2H2O
salt
water
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Ionic Equations for Neutralization
Learning Check
Select the correct group of coefficients for the
following neutralization equations.
Write strong acids, bases, and salts as ions.
H+ + Cl- + Na+ + OH-
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Na+ + Cl- + H2O
A. __ HCl + __ Al(OH)3
Cross out matched ions.
H+ + Cl- + Na+ + OH-
1) 1, 3, 3, 1
Na+ + Cl- + H2O
H+ +
OH-
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2) 3, 1, 1, 1
B. __ Ba(OH)2 + __H3PO4
Write a net ionic reaction.
1) 3, 2, 2, 2
H2O
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__AlCl3 + __ H2O
__Ba3(PO4)2 + __ H2O
2) 3, 2, 1, 6
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3) 3, 1, 1 3
3) 2, 3, 1,
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Solution
Antacids
A. 3) 3, 1, 1 3
3HCl + 1Al(OH)3
 Antacids neutralize stomach acid (HCl).
 Alka-Seltzer
NaHCO3, citric acid, aspirin
 Chooz, Tums
CaCO3
 Di-gel
CaCO3 and Mg(OH)2
 Gelusil, Maalox,
Al(OH)3 and Mg(OH)2
 Mylanta
 Milk of Magnesia Mg(OH)2
 Rolaids
AlNa(OH)2CO3
1AlCl3 + 3H2O
B. 2) 3, 2, 1, 6
3Ba(OH)2 + 2H3PO4
1Ba3(PO4)2 + 6H2O
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Learning Check
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Solution
Write the neutralization reactions for stomach
acid HCl and Mylanta.
Mylanta: Al(OH)3 and Mg(OH)2
Write the neutralization reactions for
stomach acid HCl and Mylanta.
3HCl + Al(OH)3
2HCl + Mg(OH)2
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AlCl3 + 3H2O
MgCl2 + 2H2O
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10.9 Buffers


Buffers
Buffers
 Absorb H 3O+ or OH- from foods and
cellular processes to maintain pH.
 Are important in the proper functioning of
cells and blood.
 In blood maintain a pH close to 7.4. A
change in the pH of the blood affects the
uptake of oxygen and cellular processes.
When an acid or
base is added to
water, the pH
changes
drastically.
A buffer solution
resists a change in
pH when an acid
or base is added.
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Components of a Buffer
Buffer Action
A buffer solution
 Contains a combination of acid-base conjugate
pairs.
 Contains a weak acid and a salt of the
conjugate base of that acid.
 Typically has equal concentrations of a weak
acid and its salt.
 May also contain a weak base and a salt with
the conjugate acid.
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


The acetic acid/acetate buffer contains acetic
acid (CH3COOH) and sodium acetate
(CH3COONa).
The salt produces sodium and acetate ions.
CH3COONa
CH3COO- + Na+
The salt provides a higher concentration of the
conjugate base CH3COO- than the weak acid.
CH3COOH + H2O
CH3COO- + H3O+
Large amount
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Large amount
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Function of the Weak Acid

Function of the Conjugate Base
The function of the weak acid is to neutralize a
base. The acetate ion in the product adds to the
available acetate.
CH3COOH + OH–
CH3COO– + H2O
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
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Summary of Buffer Action



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Learning Check
The weak acid in a buffer neutralizes base.
The conjugate base in the buffer neutralizes acid.
The pH of the solution is maintained.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
The function of the acetate ion CH3COO(conjugate base) is to neutralize H3O+ from acids.
The weak acid product adds to the weak acid
available.
CH3COO- + H3O+
CH3COOH + H2O
Does each combination make a buffer solution?
1) yes
2) no
Explain.
A. HCl and KCl
B. H2CO3 and NaHCO3
C. H3PO4 and NaCl
D. CH3COOH and CH3COOK
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Solution
pH of a Buffer
Does each combination make a buffer solution?
1) yes
2) no
Explain.
A. HCl + KCl
no; HCl is a strong acid.
B. H2CO3 + NaHCO3 yes; weak acid and its salt.
C. H3PO4 + NaCl
no; NaCl does not contain a
conjugate base of H3PO4.
D. CH3COOH + CH3COOK
yes; weak acid and its salt.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.

The [H3O+] in the Ka expression is used to
determine the pH of a buffer.
Weak acid + H2O
H3O+ + Conjugate base
[H3O+][conjugate base]
[weak acid]
[H3O+] = Ka x
[weak acid]
[conjugate base]
pH = –log [H3O+]
Ka =
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Calculation of Buffer pH
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Learning Check
The weak acid H2PO4- in a blood buffer H2PO4/HPO42- has Ka = 6.2 x 10-8. What is the pH of the
buffer if it is 0.20 M in both H2PO4- and HPO42-?
What is the pH of a H2CO3 buffer that is 0.20 M
H2CO3 and 0.10 M HCO3-? Ka(H2CO3) = 4.3 x 10-7
[H3O+] = Ka x [H2PO4-]
[HPO42-]
[H3O+] = 6.2 x 10-8 x [0.20 M] = 6.2 x 10-8
[0.20 M]
pH = –log [6.2 x 10-8] = 7.17
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Solution
What is the pH of a H2CO3 buffer that is 0.20 M
H2CO3 and 0.10 M HCO3- ?
Ka (H2CO3) = 4.3 x 10-7
[H3O+] = Ka x [H2CO3]
[HCO3-]
[H3O+] = 4.3 x 10-7 x [0.20 M] = 8.6 x 10-7
[0.10 M]
-7
pH = –log [8.6 x 10 ] = 6.07
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