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Chapter 8
Acids and Bases and
Oxidation-Reduction
Denniston
Topping
Caret
5th Edition
8.1 Acids and Bases
•
•
•
Acids: Taste sour, dissolve some metals,
cause plant dye to change color
Bases: Taste bitter, are slippery, are
corrosive
Two theories that help us to understand
the chemistry of acids and bases
1. Arrhenius Theory
2. Brønsted-Lowry Theory
8.1 Acids and Bases
Arrhenius Theory of Acids
and Bases
• Acid - a substance, when dissolved in
water, dissociates to produce hydrogen
ions
– Hydrogen ion: H+ also called “protons”
HCl is an acid:
HCl(aq)  H+(aq) + Cl-(aq)
8.1 Acids and Bases
Arrhenius Theory of Acids
and Bases
• Base - a substance, when dissolved in
water, dissociates to produce hydroxide
ions
NaOH is a base
NaOH(aq)  Na+(aq) + OH-(aq)
8.1 Acids and Bases
Arrhenius Theory of Acids
and Bases
• Where does NH3 fit?
• When it dissolves in water it has basic
properties but it does not have OH- ions
in it
• The next acid-base theory gives us a
broader view of acids and bases
8.1 Acids and Bases
Brønsted-Lowry Theory of
Acids and Bases
• Acid - proton donor
• Base - proton acceptor
– Notice that acid and base are not defined
using water
– When writing the reactions, both accepting
and donation are evident
8.1 Acids and Bases
Brønsted-Lowry Theory of
Acids and Bases
HCl(aq) + H2O(l)  Cl-(aq) + H3O+(aq)
acid
base
What donated the proton?
Is it an acid or base?
HCl
Acid
What accepted the proton? H2O
Is it an acid or base?
Base
8.1 Acids and Bases
Brønsted-Lowry
Theory
of
.
Acids and Bases
NH3(aq) + H2O(l)
base
acid
NH4+(aq) + OH-(aq)
Now, let us look at NH3 and see why it is a
base.
Did NH3 donate or accept a proton? Accept
Is it an acid or base? Base
What is water in this reaction? Acid
8.1 Acids and Bases
Acid-Base Properties of Water
• Water possesses both acid and base
properties
– Amphiprotic - a substance possessing both acid
and base properties
– Water is the most commonly used solvent for
both acids and bases
– Solute-solvent interactions between water and
both acids and bases promote solubility and
dissociation
8.1 Acids and Bases
Acid and Base Strength
• Acid and base strength – degree of
dissociation
– Not a measure of concentration
– Strong acids and bases – reaction with water is
virtually 100% (Strong electrolytes)
8.1 Acids and Bases
Strong Acids and Bases
• Strong Acids:
–
–
–
–
HCl, HBr, HI
HNO3
H2SO4
HClO4
Hydrochloric Acid, etc.
Nitric Acid
Sulfuric Acid
Perchloric Acid
• Strong Bases:
– NaOH, KOH, Ba(OH)2
– All metal hydroxides
8.1 Acids and Bases
Weak Acids
• Weak acids and bases – only a small
percent dissociates (Weak electrolytes)
• Weak acid examples:
– Acetic acid:
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
– Carbonic Acid:
H2CO3(aq) + H2O(l)
HCO3-(aq) + H3O+(aq)
8.1 Acids and Bases
Weak Bases
• Weak base examples:
– Ammonia:
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
– Pyridine:
C5H5NH2(aq) + H2O(l)
C5H5NH3+(aq) + OH-(aq)
– Aniline:
C6H5NH2(aq) + H2O(l)
C6H5NH3+(aq) + OH-(aq)
8.1 Acids and Bases
Conjugate Acids and Bases
• The acid base reaction can be written in
the general form:
HA + B
A– + HB+
acid
base
• Notice the reversible arrows
• The products are also an acid and base
called the conjugate acid and base
8.1 Acids and Bases
HA + B
acid
base
A- + HB+
base acid
• Conjugate acid - what the base becomes
after it accepts a proton
• Conjugate base - what the acid becomes
after it donates its proton
• Conjugate acid-base pair - the acid and
base on the opposite sides of the
equation
8.1 Acids and Bases
Acid-Base Dissociation
HA + B
A– + HB+
• The reversible arrow isn’t always written
– Some acids or bases essentially dissociate 100%
– One way arrow is used
• HCl + H2O  Cl- + H3O+
– All of the HCl is converted to Cl– HCl is called a strong acid – an acid that
dissociates 100%
• Weak acid - one which does not dissociate
100%
8.1 Acids and Bases
Conjugate Acid-Base Pairs
• Which acid is
stronger:
HF or HCN? HF
• Which base is
stronger:
CN- or H2O? CN -
8.1 Acids and Bases
Acid-Base Practice
• Write the chemical reaction for the following acids
or bases in water
• Identify the conjugate acid-base pairs
1. HF (a weak acid)
2. H2S (a weak acid)
3. HNO3 (a strong acid)
4. CH3NH2 (a weak base)
Note: The degree of dissociation also defines weak
and strong bases
8.1 Acids and Bases
The Dissociation of Water
• Pure water is virtually 100% molecular
• Very small number of molecules dissociate
– Dissociation of acids and bases is often called
ionization
H2O(l) + H2O(l)
• Called autoionization
• Very weak electrolyte
H3O+(aq) + OH-(aq)
8.1 Acids and Bases
Hydronium Ion
• H3O+ is called the hydronium ion
• In pure water at room temperature:
– [H3O+] = 1 x 10-7 M
– [OH-] = 1 x 10-7 M
• What is the equilibrium expression for:
H2O(l) + H2O(l)
H3O+(aq) + OH-(aq)

Keq  [H3O ][OH ]
-
Remember, liquids are not included in equilibrium
expressions
8.1 Acids and Bases
Ion Product of Water
• This constant is called the ion product for
water and has the symbol Kw

K w  [H 3O ][OH ]
-
• Since [H3O+] = [OH-] = 1.0 x 10-7 M, what
is the value for Kw?
– 1.0 x 10-14
– It is unitless
8.2 pH: A Measurement Scale
for Acids and Bases
• pH scale - a scale that indicates the acidity
or basicity of a solution
– Ranges from 0 (very acidic) to 14 (very basic)
• The pH scale is rather similar to the
temperature scale assigning relative values
of hot and cold
• The pH of a solution is defined as:
pH = -log[H3O+]
8.2 pH: A Measurement
Scale for Acids and Bases
A Definition of pH
• Use these observations to develop a concept
of pH
– if know one concentration, can calculate the
other
– if add an acid, [H3O+]  and [OH-] 
– if add a base, [OH-]  and [H3O+] 
– [H3O+] = [OH-] when equal amounts of acid
and base are present
• In each of these cases 1 x 10-14 = [H3O+][OH-]
8.2 pH: A Measurement
Scale for Acids and Bases
Measuring pH
• pH of a solution can be:
– Calculated if the concentration of either is
known
• [H3O+]
• [OH-]
– Approximated using indicator / pH paper
that develops a color related to the solution
pH
– Measured using a pH meter whose sensor
measures an electrical property of the
solution that is proportional to pH
8.2 pH: A Measurement
Scale for Acids and Bases
Calculating pH
• How do we calculate the pH of a solution
when either the hydronium or hydroxide
ion concentration is known?
• How do we calculate the hydronium or
hydroxide ion concentration when the pH
is known?
• Use two facts:
pH = -log[H3O+]
1 x 10-14 = [H3O+][OH-]
8.2 pH: A Measurement
Scale for Acids and Bases
Calculating pH from
Acid Molarity
What is the pH of a 1.0 x 10-4 M HCl solution?
– HCl is a strong acid and dissociates in water
– If 1 mol HCl is placed in 1 L of aqueous
solution it produces 1 mol [H3O+]
– 1.0 x 10-4 M HCl solution has [H3O+]=1.0x10-4M
pH = -log[H3O+]
= -log [H3O+]
= -log [1.0 x 10-4]
= -[-4.00] = 4.00
8.2 pH: A Measurement
Scale for Acids and Bases
Calculating [H3O+] From pH
What is the [H3O+] of a solution with pH = 6.00?
pH = -log[H3O+]
• 4.00 = -log [H3O+]
• Multiply both sides of equation by –1
• -4.00 = log [H3O+]
• Take the antilog of both sides
• Antilog -4.00 = [H3O+]
• Antilog is the exponent of 10
• 1.0 x 10-4 M = [H3O+]
8.2 pH: A Measurement
Scale for Acids and Bases
Calculating the pH of a Base
What is the pH of a 1.0 x 10-3 M KOH solution?
• KOH is a strong base (as are any metal hydroxides)
• 1 mol KOH dissolved and dissociated in aqueous
solution produces 1 mol OH• 1.0 x 10-3 M KOH solution has [OH-] = 1.0 x 10-3 M
1 x 10-14 = [H3O+][OH-]
• Solve equation for [H3O+] = 1 x 10-14 / [OH-]
• [H3O+] = 1 x 10-14 / 1.0 x 10-3 = 1 x 10-11
• pH = -log [1 x 10-11]
= 11.00
pH = -log[H3O+]
8.2 pH: A Measurement
Scale for Acids and Bases
Calculating pH from Acid
Molarity
What is the pH of a 2.5 x 10-4 M HNO3 solution?
• We know that as a strong acid HNO3 dissociates
to produce 2.5 x 10-4 M [H3O+]
pH = -log[H3O+]
• pH = -log [2.5 x 10-4]
•
= 3.60
8.2 pH: A Measurement
Scale for Acids and Bases
Calculating [OH-] From pH
What is the [OH-] of a solution with pH = 4.95?
• First find [H3O+]
• 4.95 = -log [H3O+]
•
•
•
•
pH = -log[H3O+]
[H3O+] = 10-4.95
[H3O+] = 1.12 x 10-5
-14 = [H O+][OH-]
1
x
10
3
Now solve for [OH-]
[OH-] = 1 x 10-14 / 1.12 x 10-5
= 1.0 x 10-9
8.2 pH: A Measurement
Scale for Acids and Bases
The pH Scale
More basic
More Acidic
8.2 pH: A Measurement
Scale for Acids and Bases
For a strong acid
HCl molarity pH
1.0 x 100
0.00
1.0 x 10-1
1.00
1.0 x 10-2
2.00
1.0 x 10-3
3.00
1.0 x 10-4
4.00
1.0 x 10-5
5.00
1.0 x 10-6
6.00
1.0 x 10-7
7.00
For a strong base
NaOH molarity pH
1.0 x 100
14.00
1.0 x 10-1
13.00
1.0 x 10-2
12.00
1.0 x 10-3
11.00
1.0 x 10-4
10.00
1.0 x 10-5
9.00
1.0 x 10-6
8.00
1.0 x 10-7
7.00
Each 10 fold change in concentration
changes the pH by one unit
8.3 Reactions Between Acids
and Bases
• Neutralization reaction - the reaction of an acid
with a base to produce a salt and water
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Acid
Base
Salt
Water
• Break apart into ions:
H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O
• Net ionic equation
– Show only the changed components
– Omit any ions appearing the same on both sides of
equation = Spectator ions
H+ + OH-  H2O
8.3 Reactions Between
Acids and Bases
Net Ionic Neutralization Reaction
• The net ionic neutralization reaction is more
accurately written:
H3O+(aq) + OH-(aq)  2H2O(l)
• This equation applies to any strong acid / strong
base neutralization reaction
• An analytical technique to determine the
concentration of an acid or base is titration
• Titration involves the addition of measured
amount of a standard solution to neutralize the
second, unknown solution
• Standard solution - solution of known
concentration
8.3 Reactions Between
Acids and Bases
Acid – Base Titration
Standard solution
is slowly added
until the color
changes
The equivalence
point is when the
moles of H3O+
and OH- are equal
Buret – long glass
tube calibrated in mL
which contains the
standard solution
Indicator – a
substance which
changes color as
pH changes
Flask contains a
solution of unknown
concentration plus
indicator
8.3 Reactions Between
Acids and Bases
Relationship Between pH and
Color in Acid-Base Indicators
8.3 Reactions Between
Acids and Bases
Determine the Concentration of a
Solution of Hydrochloric Acid
• Place a known volume of
acid whose concentration
is not known into a flask
• Add an indicator,
experience guides
selection, here phenol red
is good
• Known concentration of
NaOH is placed in a buret
• Drip NaOH into the flask
until the indicator changes
color
8.3 Reactions Between
Acids and Bases
Determine the Concentration of a
Solution of Hydrochloric Acid
• Indicator changes color
– Equivalence point is reached
– Mol OH- = Mol H3O+ present in the unknown acid
• Volume dispensed from buret is determined
• Calculate acid concentration:
–
–
–
–
Volume of Hydrochloric Acid: 25.00 mL
Volume of NaOH added:
35.00 mL
Concentration of NaOH:
0.1000 M
Balanced reaction shows that HCl and NaOH react 1:1
8.3 Reactions Between
Acids and Bases
Determine the Concentration of a
Solution of Hydrochloric Acid
• 35.00 mL NaOH x 1L NaOH x 0.1000 mol NaOH
103 mL NaOH
1L NaOH
= 3.500 x 10-3 mol NaOH
• 3.500 x 10-3 mol NaOH x 1 mol HCl
1 mol NaOH
= 3.500 x 10-3 mol HCl
– this amount of HCl is contained in 25.00 mL
• 3.500 x 10-3 mol HCl x 103 mL HCl
25.00 mL HCl
1 L HCl
• = 1.400 x 10-1 mol HCl / L HCl = 0.1400 M HCl
8.3 Reactions Between
Acids and Bases
Determine the Concentration of a
Solution of Hydrochloric Acid
• Alternative strategy to solve the acid concentration
(Macid)(Vacid) = (Mbase)(Vbase)
• (Macid) = (Mbase)(Vbase)
(Vacid)
• (Macid) = (0.1000 M) (35.00 mL)
(25.00 mL)
• = 0.1400 M HCl
8.3 Reactions Between
Acids and Bases
Calculating the Concentration of
Sodium Hydroxide
Calculate [NaOH] if 25.00 mL of this solution were
required to neutralized 45.00 mL of 0.3000 M HCl
•Alternative strategy to solve the acid concentration
(Macid)(Vacid) = (Mbase)(Vbase)
•(Mbase) = (Macid)(Vacid)
(Vbase)
•(Mbase) = (0.3000 M) (45.00 mL)
(25.00 mL)
= 0.5400 M NaOH
8.3 Reactions Between
Acids and Bases
Polyprotic Substances
• The previous examples have the acid and base
at a 1:1 combining ratio
– Not all acid-base pairs do this
• Polyprotic substance - donates or accepts more
than one proton per formula unit
– Hydrochloric acid is monoprotic, producing one H+
ion for each unit of HCl
– Sulfuric acid is diprotic, each unit of H2SO4
produces 2 H+ ions
H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
8.3 Reactions Between
Acids and Bases
Dissociation of Polyprotic
Substances
Step 1.
H2SO4(aq) + H2O(l) HSO4-(aq) + H3O+(aq)
Step 2.
HSO4-(aq) + H2O(l)
SO42-(aq) + H3O+(aq)
• In Step 1, H2SO4 behaves as a strong acid –
dissociating completely
• In Step 2, HSO4-( behaves as a weak acid –
reversibly dissociating, note the double arrow
8.4 Acid-Base Buffers
• Buffer solution - solution which resists large
changes in pH when either acids or bases are
added
• These solutions are frequently prepared in
laboratories to maintain optimum conditions
for chemical reactions
• Buffers are also used routinely in
commercial products to maintain optimum
conditions for product behavior
8.4 Acid-Base Buffers
The Buffer Process
• Buffers act to establish an equilibrium between a
conjugate acid – base pair
• Buffers consist of either
– a weak acid and its salt (conjugate base)
– a weak base and its salt (conjugate acid)
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
– Acetic acid (CH3COOH) with sodium acetate
(CH3COONa)
• An equilibrium is established in solution
between the acid and the salt anion
• A buffer is Le Chatelier’s principle in action
8.4 Acid-Base Buffers
(OH )
Addition of Base
to a
Buffer Solution
• Adding a basic substance to a buffer causes
changes
– The OH- will react with the H3O+ producing water
– Acid in the buffer system dissociates to replace
the H3O+ consumed by the added base
– Net result is to maintain the pH close to the initial
level
• The loss of H3O+ (the stress) is compensated
by the dissociation of the acid to produce
more H3O+
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
8.4 Acid-Base Buffers
+
O)
Addition of Acid (H3
Buffer Solution
to a
• Adding an acidic substance to a buffer causes
changes
– The H3O+ from the acid will increase the overall
H3O+
– Conjugate base in the buffer system reacts with
the H3O+ to form more acid
– Net result is to maintain the H3O+ concentration
and the pH close to the initial level
• The gain of H3O+ (the stress) is compensated
by the reaction of the conjugate base to
produce more acid
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
8.4 Acid-Base Buffers
Buffer Capacity
• Buffer capacity - a measure of the ability
of a solution to resist large changes in
pH when a strong acid or strong base is
added
• Also described as the amount of strong
acid or strong base that a buffer can
neutralize without significantly changing
pH
8.4 Acid-Base Buffers
Preparation of a Buffer Solution
• Buffering process is an equilibrium reaction
described by an equilibrium-constant
expression
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
– In acids, this constant is Ka

-
[H 3O ][CH3COO ]
Ka 
[CH3COOH]
• If you want to know the pH of the buffer,
solve for [H3O+], then calculate pH
8.4 Acid-Base Buffers
Calculating the pH of a
Buffer Solution
Calculate the pH of a buffer solution in which
– Both the acetic acid (acid) and sodium acetate (salt)
concentrations are 2.0 x 10-2 M
– Sodium acetate, the salt, is also the conjugate base
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
– Ka = 1.75 x 10-5
[H 3O  ][CH3COO- ]
Ka 
– [H3O+] = [acid]Ka
[CH3COOH]
[conjugate base]
= {(2.0 x 10-2 M) x (1.75 x 10-5)} / 2.0 x 10-2 M
= 1.75 x 10-5
– pH = -log 1.75 x 10-5 = 4.76
8.4 Acid-Base Buffers
Henderson-Hasselbalch Equation
• Solution of equilibrium-constant expression
and pH can be combined into one operation
• Henderson-Hasselbalch equation is this
combined expression
[H 3O  ][CH3COO- ]
• Using these two equations: K a  [CH COOH]
3
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
– pKa = -log Ka just as pH = -log[H3O+]
– pKa = pH – log ( [CH3COO-] / [CH3COOH] )
– Henderson-Hasselbalch –
– pH = pKa + log( [CH3COO-] / [CH3COOH] )
8.4 Acid-Base Buffers
Henderson-Hasselbalch Equation
• pH = pKa + log ([CH3COO-] / [CH3COOH])
can be rewritten
– pH = pKa + log ([conjugate base] / [weak acid])
8.5 Oxidation-Reduction Processes
• Oxidation-reduction processes are responsible
for many types of chemical change
• Oxidation - defined by one of the following
– loss of electrons
– loss of hydrogen atoms
– gain of oxygen atoms
• Example: NaNa+ + e– Oxidation half reaction
8.5 Oxidation-Reduction
Processes
Oxidation-Reduction Processes
• Reduction - defined by one of the
following:
– gain of electrons
– gain of hydrogen
– loss of oxygen
• Example: Cl + e-  Cl– Reduction half reaction
• Cannot have oxidation without reduction
8.5 Oxidation-Reduction
Processes
Oxidation and Reduction as
Complementary Processes
Na
 Na+ + eCl + e-  ClNa + Cl Na+ + ClOxidizing Agent
Reducing Agent
• Is reduced
• Gains electrons
• Causes oxidation
• Is oxidized
• Loses electrons
• Causes reduction
8.5 Oxidation-Reduction
Processes
Applications of Oxidation
and Reduction
• Corrosion - the deterioration of metals
caused by an oxidation-reduction
process
– Example: rust (oxidation of iron)
4Fe(s) + 3O2(g)  2Fe2O3(s)
• Combustion of Fossil Fuels
– Example: natural gas furnaces
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
8.5 Oxidation-Reduction
Processes
Applications of Oxidation and
Reduction
• Bleaching
• Most bleaching agents are oxidizing
agents
• The oxidation of the stains produces
compounds that do not have color
– Example: Chlorine bleach - sodium
hypochlorite (NaOCl)
8.5 Oxidation-Reduction
Processes
Biological Processes
• Respiration
– Electron-transport chain of aerobic
respiration uses reversible oxidation and
reduction of iron atoms in cytochrome c
• Metabolism
– Break down of molecules into smaller pieces
by enzymes
8.5 Oxidation-Reduction
Processes
Voltaic Cells
• Voltaic cell - electrochemical cell that
converts stored chemical energy into
electrical energy
• Let’s consider the following reaction:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
• Is Zn oxidized or reduced?
• Oxidized
• Copper is reduced
8.5 Oxidation-Reduction
Processes
Voltaic Cells
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
• If the two reactants are placed in the same
flask they cannot produce electrical
current
• A voltaic cell separates the two half
reactions
• This makes the electrons flow through a
wire to allow the oxidation and reduction
to occur
8.5 Oxidation-Reduction
Processes
Voltaic Cell Generating Electrical
Current
Zn  Zn2+ + 2eOxidation
Cu2+ + 2e-  Cu
Reduction
anode – electrode cathode – electrode where
reduction occurs
where oxidation occurs
8.5 Oxidation-Reduction
Processes
Voltaic Cell Generating Electrical
Current
8.5 Oxidation-Reduction
Processes
Silver Battery
• Batteries use the concept of the voltaic cell
• Modern batteries are:
– Smaller
– Safer
– More dependable
8.5 Oxidation-Reduction
Processes
Electrolysis
• Electrolysis reactions - use electrical
energy to cause nonspontaneous
oxidation-reduction reactions to occur
• These reactions are the reverse of a
voltaic cell
– Rechargeable battery
•
•
•
•
When powering a device behaves as a voltaic cell
With time, the chemical reaction nears completion
Battery appears to “run down”
Cell reaction is reversible when battery attached to
charger
8.5 Oxidation-Reduction
Processes
Comparison of Voltaic and
Electrolytic Cells