Chapter 14
Acids and
Bases
Properties of Acids and Bases
• Generally, an acid is a compound that releases hydrogen
ions, H+, into water.
– Blue litmus is used to test for acids. Blue litmus paper turns
red in the presence of hydrogen ions.
– Acids are generally sour in taste and can be corrosive.
• Generally, a base is a compound that releases hydroxide
ions, OH –, into water.
– Red litmus is used to test for bases. Red litmus paper turns
blue in the presence of hydroxide ions.
– Bases have a bitter taste and a slippery, soapy feel.
Chapter 14
2
Classification of Compounds
Chapter 14
3
Binary Acids
• A binary acid is an aqueous solution of a
compound containing hydrogen and a
nonmetal.
• The formula of an acid always begins with
H, followed by the second element:
HF (aq)
• Binary acids are named by using the prefix
hydro- before the non-metal element stem
and adding the suffix -ic acid.
HF (aq) is hydrofluoric acid
H2S (aq) is hydrosulfuric acid
Chapter 14
4
Ternary Oxyacids
• Ternary oxyacids are aqueous solutions of a
compound containing hydrogen and an
oxyanion (a polyatomic ion!).
• If the acid is derived from an oxyanion ending
in -ate, the suffix is changed to -ic acid.
HNO3 (aq) is nitric acid
• If the acid is derived from an oxyanion ending
in -ite, the suffix is changed to -ous acid.
HNO2 (aq) is nitrous acid
Chapter 14
You must know the polyatomic ions
previously assigned in chapter 5
5
Naming Acids
Formula
Name
H2CO3 (aq)
Perchloric Acid
HBr (aq)
Hydroselenic Acid
HSO3 (aq)
Phosphoric Acid
H3N (aq)
Hydroiodic Acid
H2CrO4 (aq)
Acetic Acid
Chapter 14
6
Naming Bases
• Typical bases are a combination of a metal and a
hydroxide ion.
– These compounds are named as you would name an ionic
compound.
Formula
Name
NaOH
NH4OH
Lithium hydroxide
KOH
Al(OH)3
Barium hydroxide
Chapter 14
7
Definitions of an Acid and Base
• There are three definitions used to describe an
acid or a base:
Arrhenius Acids and Bases
Brønsted-Lowry Acids and Bases
Lewis Acids and Bases
Chapter 14
8
Arrhenius Acids and Bases
• Svante Arrhenius proposed the following definitions for
acids and bases in 1884:
– An Arrhenius acid is a substance that ionizes in water to
produce hydrogen ions (H+).
– An Arrhenius base is a substance that ionizes in water to
release hydroxide ions (OH-).
• For example, HCl is an Arrhenius acid and NaOH is an
Arrhenius base.
• To determine whether you have an Arrhenius acid or
base, write the dissociation reaction of your compound.
Chapter 14
9
Brønsted-Lowry Acids & Bases
• The Brønsted-Lowry definitions of acids and bases are broader
than the Arrhenius definitions.
• A Brønsted-Lowry acid is a substance that donates a hydrogen ion
(H+) to any other substance.
– It is a proton donor.
– All Arrhenius acids are classified as acids by the B-L definition
Chapter 14
10
Brønsted-Lowry Acids & Bases
• A Brønsted-Lowry base is a substance that accepts a
hydrogen ion (H+).
– It is a proton acceptor (So no OH- required!).
– All Arrhenius bases are classified as bases by the B-L definition
Chapter 14
11
Acids in Solution
• All Arrhenius and B-L acids have
a hydrogen atom attached to the
rest of the molecule by a polar
bond.
• This bond is broken when the acid
ionizes.
• Polar water molecules help ionize
the acid by pulling the hydrogen
atom away:
• The hydronium ion, H3O+, is
formed when the aqueous
hydrogen ion (H+) attaches to a
water molecule.
Chapter 14
12
Bases in Solution
• When we dissolve Arrhenius bases in solution,
they dissociate completely to give a cation and a
hydroxide anion in solution.
• Strong bases dissociate almost fully and weak
bases dissociate very little:
NaOH(aq) → Na+(aq) + OH–(aq)
NH4OH(aq) ⇄ NH4+(aq) + OH–(aq)
(~100%)
(~1%)
• The hydroxide ion, OH-, is formed when the base
dissociates in solution.
Chapter 14
13
Lewis Acids and Bases
• The Lewis definition of acids and bases is the
broadest of the three.
– A Lewis Acid is a substance that can accept (and share) an
electron pair
– A Lewis Base is a substance that can donate (and share) an
electron pairLewis
Acid Lewis
H
H H
H
Base
H
B
Lewis H
AcidLewis
Chapter 14
Base
N
H
H
H
B
N
H
H
H
14
Summary of Acids & Bases
Arrhenius Acids/Bases
Brønsted-Lowry Acids/Bases
Lewis Acids/Bases
Chapter 14
15
Conjugate Acid-Base Pairs
• When a pair of molecules are related by the loss or
gain of one H+, they are called a conjugate pair
• A conjugate acid-base pair is
the proton donor and the ion
formed by the loss of the
proton.
• A conjugate base-acid pair is
the proton acceptor and the ion
formed by the gain of the
proton.
Chapter 14
16
Conjugate Acid-Base Pairs
HF (g) + H2O (l) ⇄ F- (aq) + H3O+ (aq)
NH3 (g) + H2O (l) ⇄ NH4+ (aq) + OH- (aq)
Chapter 14
17
Conjugate Acid-Base Pairs
• Let’s practice writing and identifying conjugate pairs
What is the conjugate acid or base for the following:
Acid
HCl
H2O
HNO2
HSO4-
Conjugate Base
Base
HCO3H2O
NH3
CN-
Conjugate Acid
Identify the acid and base and their conjugates:
HCN + H2O ⇄ CN- + H3O+
NO2- + H2Se ⇄ HSe- + HNO2
Chapter 14
18
Strengths of Acids and Bases
• Acids and bases have varying strengths.
• The strength of an acid is measured by the amount of H+ that are
produced for each mole of acid that dissolves
– Ionization is the process where polar compounds separate into cations and
anions in solution.
– The acid HCl ionizes into H+ and Cl– ions in solution.
• The strength of a base is measured by the degree of dissociation in
solution.
– Dissociation is the process where cations and anions in an ionic compound
separate in solution.
– A formula unit of NaOH dissociates into Na+ and OH– ions in solution.
Chapter 14
19
Strong Acids and Bases
Strong Acids
Sulfuric Acid
H2SO4
Hydroiodic Acid
HI
Hydrobromic Acid
HBr
Hydrochloric Acid
HCl
Nitric Acid
HNO3
• Only a few acids are considered
strong acids.
• The conjugate bases of strong
acids are weak bases.
• You need to know these!!
• Bases made from Groups IA and IIA metals and hydroxide are
strong bases:
LiOH, NaOH, KOH, Ca(OH)2, etc.
• Most other bases are weak bases
NH3, NH4OH, Al(OH)3, etc.
Chapter 14
20
Chapter 14
21
Autoionization of Water
• Water undergoes an autoionization reaction.
H2O(l) + H2O(l) ⇄ H3O+(aq) + OH-(aq)
OR
H2O(l) ⇄ H+(aq) + OH-(aq)
• Only about 1 in 5 million water molecules is
present as ions so water is a weak electrolyte.
• The concentration of hydrogen ions, [H+], in
pure water is about 1 × 10-7 mol/L at 25°C.
Chapter 14
22
Autoionization of Water
• The molar ratio of H+ to OH- in the reaction is 1 to
1, so if the [H+] = 1 × 10-7 mol/L at 25°C, then the
[OH-] must also be 1 × 10-7 mol/L at 25°C:
H2O (l) ⇄ H+ (aq) + OH- (aq)
[H+] • [OH-] = (1 × 10-7)(1 × 10-7) = 1.0 × 10-14
• This value is called the ionization constant of
water, Kw.
• We can use this value to calculate [H+] or [OH-]
Chapter 14
23
Calculating [H+] and [OH-]
What is the [H+] of an ammonia cleaning solution with a
[OH-] = 2.61 x 10-5 M?
KW = [H+] • [OH-]
[H+] = KW / [OH-]
[H+] = 1.00 x 10-14
2.61 x 10-5 M
Chapter 14
[H+] = 3.83 x 10-10 M
24
[H+] and [OH-] Relationship
• What is the [OH-] if [H+] = 0.100 M ?
– Step 1: Determine what you have: [H+] = 0.100 M.
– Step 2: Determine what you want: ??? M [OH-]
– Step 3: Use the relationship between [H+] and [OH-] to
solve for [OH-].
[H+] • [OH-] = KW
[OH-] = 1.0 × 10-14 / 0.100 M
[OH-] = 1.00 × 10-13 M
Chapter 14
25
The pH Scale
• A pH value expresses the
acidity or basicity of a solution.
• Most solutions have a pH
between 0 and 14.
• Acidic solutions have a pH
less than 7.
– As a solution becomes more
acidic, the pH decreases.
• Basic solutions have a pH
greater than 7.
– As a solution becomes more
basic, the pH increases.
Chapter 14
26
The pH Concept
• pH is a measure of the acidity of a solution.
• The pH scale uses powers of ten to express the hydrogen
ion concentration.
• Mathematically:
pH = – log [H+]
OR
pH = – log [H3O+]
Chapter 14
27
Calculating pH
• What is the pH if the hydrogen ion concentration in a
vinegar solution is 0.001 M?
– Step 1: Determine what you have: [H+] = 0.001 M
– Step 2: Determine what you want: pH = ???
– Step 3: Use formula for pH and solve.
pH = – log [H+]
pH = – log (0.001)
pH = – ( – 3) = 3
Is the vinegar is acidic, basic or neutral?
Chapter 14
28
Calculating [H+] from pH
• Milk has a pH of 6.0. What is the concentration of
hydrogen ion in milk?
– Step 1: Determine what you have: pH = 6.0
– Step 2: Determine what you want: [H+] = ??? M
– Step 3: Use formula for pH but rearrange to solve for [H+] .
• If we rearrange the pH equation for [H+], we get:
[H+] = 10–pH
[H+] = 10–pH = 10–6.0 = 0.000001 M
[H+] = 1.0 × 10–6 M
Chapter 14
29
The pOH Concept
• pOH is similar to pH except it is a measure of basicity
of a solution.
– The lower the value, the more basic the solution
pOH = - log [OH-]
• There is also a relationship between pH and pOH:
pH + pOH = 14
• So, if you know the [H+] you can determine the pH,
[OH-] and pOH
Chapter 14
30
pH, pOH, [H+] and [OH-] Calculations
[H+]
[OH-]
pH
pOH
Acidic, Basic or Neutral?
6.15 x 10-4 M
2.61
5.28 x 10-8 M
3.45
pH = - log [H+]
pOH = - log [OH-]
[H+] = 10–pH
[OH-] = 10–pOH
[H+] • [OH-] = KW
pH + pOH = 14.00
Chapter 14
31
Acid/Base Neutralization
• When a strong acid and a strong base react, the
products are always water and a salt.
– The formula of the salt depends on the anion of the
acid and the cation of the base.
• When nitric acid reacts with sodium hydroxide,
water and a salt are produced. What is the
identity of the salt?
HNO3(aq) + NaOH(aq) →
Chapter 14
32
Balancing Neutralization Reactions
Write the balanced reaction of HCl(aq) and Be(OH)2(aq):
HCl(aq) + Be(OH)2(aq) → H2O(l) + SALT
There are two OH- ions in the base so we need
two H+ ions in the acid:
2HCl(aq) + Be(OH)2 (aq) → H2O(l) + SALT
There are four Hs and two Os on the reactant side
so we need two waters:
2HCl(aq) + Be(OH)2 (aq) → 2H2O(l) + SALT
The remaining ions are Be2+ and Cl- so the resulting
salt is BeCl2:
2HCl(aq) + Be(OH)2 (aq) → 2H2O(l) + BeCl2(aq)
Chapter 14
33
Predicting Neutralization Reactions
• We can identify the strong acid and base that react in a
neutralization reaction to produce a given salt.
• For example, what strong acid and strong base react to
give the salt, calcium sulfate?
Strong Acid + Strong Base → CaSO4(aq) + H2O(l)
– The calcium must be from a strong base, calcium hydroxide.
– The sulfate must be from a strong acid, sulfuric acid
H2SO4 (aq) + Ca(OH)2 (s) → CaSO4(aq) + 2 H2O(l)
Chapter 14
34
Acid-Base Titrations
• A titration is used to analyze an acid solution using a solution of a
base (with a known concentration) or vice versa.
– A measured volume of base of known concentration is added to an acid
solution of unknown concentration.
– When the acid is completely neutralized by the added base, the pH of the
solution is is 7.
– This point is called the endpoint
• We use a substance called
an indicator (a pH sensitive
compound) to show us when
the solution reaches the
endpoint.
Chapter 14
35
Titration Problem
• A 10.0 mL sample of acetic acid requires 37.55 mL of 0.223 M NaOH to
reach the endpoint. What is the concentration of the acetic acid?
– Step 0: Write the balanced neutralization reaction.
HC2H3O2(aq) + NaOH(aq) → NaC2H3O2(aq) + H2O(l)
– Step 1: Determine what you have: 37.55 mL of 0.223 M NaOH and 10.0 mL of
Acetic Acid
– Step 2: Determine want you want: ??? M acetic acid (AA).
– Step 3: Write out your plan to convert from mL and M NaOH to M Acetic Acid
– Step 4: Select your conversion factor(s) to complete the plan
Vol of SH
Chapter 14
Molarity
of SH
Moles of SH
Mole
Ratio
Moles of AA
Volume
of AA
Molarity
of AA
36
Another Titration Problem
• A 10.0 mL sample of 0.555 M H2SO4 is titrated with 0.233 M NaOH. What
volume of NaOH is required for the titration?
– Step 0: Write the balanced neutralization reaction.
H2SO4(aq) + 2NaOH(aq) → Na2SO4 (aq) + 2H2O(l)
– Step 1: Determine what you have: 10.0 mL of 0.555 M H2SO4 and 0.233 M NaOH
– Step 2: Determine want you want: ??? L NaOH.
– Step 3: Write out your plan to convert from mL and M H2SO4 to L NaOH
– Step 4: Select your conversion factor(s) to complete the plan
Vol. of SA
Molarity
of SA
Moles of SA
Mole
Ratio
Moles of SH
Molarity
of SH
Vol. of SH
Chapter 14
37
Acid-Base Properties of Salt Solutions
• When a salt dissolves in solution, it dissociates into
cations and anions.
• Solutions of salts can be acidic, basic or neutral.
• Anions and cations from strong acids and bases do not
affect pH.
• Anions and cations from weak acids and bases do affect
the pH.
• So, to determine whether your salt solution will be acidic,
basic or neutral, you have to figure out whether the anion
and cation would come from a strong or weak acid/base.
Chapter 14
38
Salts that Form Neutral Solutions
• A solution of a salt containing the cation from a
strong base and the anion from a strong acid will
be neutral (pH = 7)
NaCl (s) → Na+ (aq) + Cl- (aq)
Cation of
Strong Base
(NaOH)
Chapter 14
Anion of
Strong Acid
(HCl)
Both from strong
Acid/Base so this solution
will be neutral
39
Salts that Form Basic Solutions
• A solution of a salt containing the cation from a
strong base and the anion from a weak acid will
be basic (pH > 7)
KF (s) → K+ (aq) + F- (aq)
Cation of
Strong Base
(KOH)
Chapter 14
Anion of
Weak Acid
(HF)
The anion is the conjugate of a
weak acid so it is a strong base,
making the solution basic
40
Salts that Form Acidic Solutions
• A solution of a salt containing the cation from a
weak base and the anion from a strong acid will
be acidic (pH < 7)
NH4Br (s) → NH4+ (aq) + Br- (aq)
Cation of
Weak Base
(NH4OH)
Chapter 14
Anion of
Strong Acid
(HBr)
The cation is the conjugate of a
weak base so it is a strong acid,
making the solution acidic
41
Acid-Base Properties of Salt Solutions
Chapter 14
42
Buffers
• A buffer is a solution that
resists changes in pH when an
acid or a base is added.
• A buffer always contains a
combination of an acid-base
conjugate pair.
• For example, a buffer can be
made up from equal
concentrations of a weak acid
and a salt of the conjugate base
of that acid.
• May also contain a weak base
and a salt with the conjugate
acid.
Chapter 14
43
Identifying Buffer Solutions
• When trying to determine whether a solution is a buffer, you
need to determine: 1) whether a conjugate acid-base pair is
present; and 2) whether that pair consists of weak acids/bases.
• Will a mixture of NaCl and Na2CO3 make a buffer solution?
– NO. There is no conjugate pair here!!
• Will a mixture of NaCl and HCl make a buffer solution?
– NO. A solution of a strong acid and salt is completely
ionized.
• Will a mixture of NaF and HF make a buffer solution?
– YES!!! HF is a weak acid and the F- (from the NaF) is its
conjugate base!
Chapter 14
44
Calculating the pH of a Buffer
• We cannot use the pH formula to determine the pH of a
buffer because there is an acid and a base present.
• We use the Henderson-Hasslebach Equation for this
calculation.
pH = pKa + log
[Base]
[Acid]
Chapter 14
45
Calculating the pH of a Buffer
• Acetic acid has a pKa of 4.74. What is the pH of a buffer
solution containing 0.250 M Acetic acid (CH3COOH) and 0.125
M NaCH3COO?
– Step 1: Identify the acid and the base.
– Step 2: Write out the Henderson-Hasselback equation and make sure you
have all the information.
– Step 3: Plug your values into the H-H equation and solve for pH.
pH = pKa + log
[Base]
[Acid]
pH = 4.74 + log (0.125 M / 0.250 M)
pH = 4.44
Chapter 14
46