Transcript x - Highline Community College

Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Chapter 15
Acids and
Bases
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright  2011 Pearson Education, Inc.
Stomach Acid & Heartburn
• The cells that line your stomach produce
hydrochloric acid
 to kill unwanted bacteria
 to help break down food
 to activate enzymes that break down food
• If the stomach acid backs up into your
esophagus, it irritates those tissues, resulting in
heartburn
 acid reflux
Tro, Chemistry: A Molecular Approach, 2/e
2
Copyright  2011 Pearson Education, Inc.
Curing Heartburn
• Mild cases of heartburn can be cured by
neutralizing the acid in the esophagus
swallowing saliva, which contains bicarbonate
ion
taking antacids that contain hydroxide ions
and/or carbonate ions
Tro, Chemistry: A Molecular Approach, 2/e
3
Copyright  2011 Pearson Education, Inc.
GERD
• Chronic heartburn is a problem for some people
• GERD (gastroesophageal reflux disease) is
•
•
chronic leaking of stomach acid into the
esophagus
In people with GERD, the muscles separating
the stomach from the esophagus do not close
tightly, allowing stomach acid to leak into the
esophagus
Physicians diagnose GERD by attaching a pH
sensor to the esophagus to measure the acidity
levels of the fluids over time
Tro, Chemistry: A Molecular Approach, 2/e
4
Copyright  2011 Pearson Education, Inc.
Properties of Acids
• Sour taste
• React with “active” metals
 i.e., Al, Zn, Fe, but not Cu, Ag, or Au
2 Al + 6 HCl AlCl3 + 3 H2
 corrosive
• React with carbonates, producing CO2
 marble, baking soda, chalk, limestone
CaCO3 + 2 HCl CaCl2 + CO2 + H2O
• Change color of vegetable dyes
 blue litmus turns red
• React with bases to form ionic salts
Tro, Chemistry: A Molecular Approach, 2/e
5
Copyright  2011 Pearson Education, Inc.
Common Acids
Tro, Chemistry: A Molecular Approach, 2/e
6
Copyright  2011 Pearson Education, Inc.
Structures of Acids
• Binary acids have acid hydrogens attached
to a nonmetal atom
HCl, HF
Tro, Chemistry: A Molecular Approach, 2/e
7
Copyright  2011 Pearson Education, Inc.
Structure of Acids
• Oxy acids have acid hydrogens attached to
an oxygen atom
H2SO4, HNO3
Tro, Chemistry: A Molecular Approach, 2/e
8
Copyright  2011 Pearson Education, Inc.
Structure of Acids
• Carboxylic acids
have COOH group
 HC2H3O2, H3C6H5O7
• Only the first H in the
formula is acidic
 the H is on the
COOH
Tro, Chemistry: A Molecular Approach, 2/e
9
Copyright  2011 Pearson Education, Inc.
Properties of Bases
• Also known as alkalis
• Taste bitter
 alkaloids = plant product that is alkaline
 often poisonous
• Solutions feel slippery
• Change color of vegetable dyes
 different color than acid
 red litmus turns blue
• React with acids to form ionic salts
 neutralization
Tro, Chemistry: A Molecular Approach, 2/e
10
Copyright  2011 Pearson Education, Inc.
Common Bases
Tro, Chemistry: A Molecular Approach, 2/e
11
Copyright  2011 Pearson Education, Inc.
Structure of Bases
• Most ionic bases contain OH− ions
NaOH, Ca(OH)2
• Some contain CO32− ions
CaCO3 NaHCO3
• Molecular bases contain structures
that react with H+
mostly amine groups
Tro, Chemistry: A Molecular Approach, 2/e
12
Copyright  2011 Pearson Education, Inc.
Indicators
• Chemicals that change color depending on
•
the solution’s acidity or basicity
Many vegetable dyes are indicators
 anthocyanins
• Litmus
 from Spanish moss
 red in acid, blue in base
• Phenolphthalein
 found in laxatives
 red in base, colorless in acid
Tro, Chemistry: A Molecular Approach, 2/e
13
Copyright  2011 Pearson Education, Inc.
Arrhenius Theory
• Bases dissociate in water to produce OH− ions
and cations
ionic substances dissociate in water
NaOH(aq) → Na+(aq) + OH−(aq)
• Acids ionize in water to produce H+ ions and
anions
because molecular acids are not made of ions, they
cannot dissociate
they must be pulled apart, or ionized, by the water
HCl(aq) → H+(aq) + Cl−(aq)
in formula, ionizable H written in front
HC2H3O2(aq) → H+(aq) + C2H3O2−(aq)
Tro, Chemistry: A Molecular Approach, 2/e
14
Copyright  2011 Pearson Education, Inc.
Arrhenius Theory
NaOH dissociates in water,
producing Na+ and OH– ions
HCl ionizes in water,
producing H+ and Cl– ions
Tro, Chemistry: A Molecular Approach, 2/e
15
Copyright  2011 Pearson Education, Inc.
Hydronium Ion
• The H+ ions produced by the acid are so reactive
they cannot exist in water
 H+ ions are protons!!
• Instead, they react with water molecules to produce
complex ions, mainly hydronium ion, H3O+
H+ + H2O  H3O+
 there are also minor amounts of H+ with multiple water
molecules, H(H2O)n+
Tro, Chemistry: A Molecular Approach, 2/e
16
Copyright  2011 Pearson Education, Inc.
Arrhenius Acid–Base Reactions
• The H+ from the acid combines with the OH−
from the base to make a molecule of H2O
it is often helpful to think of H2O as H-OH
• The cation from the base combines with the
anion from the acid to make a salt
acid + base → salt + water
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Tro, Chemistry: A Molecular Approach, 2/e
17
Copyright  2011 Pearson Education, Inc.
Problems with Arrhenius Theory
• Does not explain why molecular substances, such
as NH3, dissolve in water to form basic solutions –
even though they do not contain OH– ions
• Does not explain how some ionic compounds,
such as Na2CO3 or Na2O, dissolve in water to
form basic solutions – even though they do not
contain OH– ions
• Does not explain why molecular substances, such
as CO2, dissolve in water to form acidic solutions
– even though they do not contain H+ ions
• Does not explain acid–base reactions that take
place outside aqueous solution
Tro, Chemistry: A Molecular Approach, 2/e
18
Copyright  2011 Pearson Education, Inc.
Another Definition:
Brønsted-Lowry Acid-Base Theory
• Brønsted and Lowry redefined acids and bases
•
based on what happens in a reaction.
Any reaction that involves H+ being transferred
from one molecule to another is an acid–base
reaction
regardless of whether it occurs in aqueous solution,
or if there is OH− present
• All reactions that fit the Arrhenius definition
also fit the Brønsted-Lowry definition, but many
more do as well
Tro, Chemistry: A Molecular Approach, 2/e
19
Copyright  2011 Pearson Education, Inc.
Brønsted-Lowry Theory
• In a Brønsted-Lowry acid–base reaction, an
•
•
H+ is transferred
The acid is an H donor
The base is an H acceptor
 base structure must contain an atom with an
unshared pair of electrons
• In a Brønsted-Lowry acid-base reaction, the
acid molecule gives an H+ to the base
molecule
H–A + :B  :A– + H–B+
Tro, Chemistry: A Molecular Approach, 2/e
20
Copyright  2011 Pearson Education, Inc.
Brønsted-Lowry Acids
• Brønsted-Lowry acids are H+ donors
any material that has H can potentially be a
Brønsted-Lowry acid
because of the molecular structure, often one H in
the molecule is easier to transfer than others
• When HCl dissolves in water, the HCl is the
acid because HCl transfers an H+ to H2O,
forming H3O+ ions
water acts as base, accepting H+
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
acid
base
Tro, Chemistry: A Molecular Approach, 2/e
21
Copyright  2011 Pearson Education, Inc.
Brønsted-Lowry Bases
• Brønsted-Lowry bases are H+ acceptors
any material that has atoms with lone pairs can
potentially be a Brønsted-Lowry base
because of the molecular structure, often one atom
in the molecule is more willing to accept H+ transfer
than others
• When NH3 dissolves in water, the NH3(aq) is
the base because NH3 accepts an H+ from
H2O, forming OH–(aq)
water acts as acid, donating H+
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
base
acid
Tro, Chemistry: A Molecular Approach, 2/e
22
Copyright  2011 Pearson Education, Inc.
A Warning!
• Because chemists know common bonding
patterns, we often do not draw lone pair
electrons on our structures. You need to be
able to recognize when an atom in a molecule
has lone pair electrons and when it doesn’t!
Tro, Chemistry: A Molecular Approach, 2/e
23
Copyright  2011 Pearson Education, Inc.
Practice – Draw structures of the
following that include lone pairs of
electrons
HClO
HCO3−
Tro, Chemistry: A Molecular Approach, 2/e
24
Copyright  2011 Pearson Education, Inc.
Amphoteric Substances
• Amphoteric substances can act as either an
acid or a base
because they have both a transferable H and an
atom with lone pair electrons
• Water acts as base, accepting H+ from HCl
•
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
Water acts as acid, donating H+ to NH3
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
Tro, Chemistry: A Molecular Approach, 2/e
25
Copyright  2011 Pearson Education, Inc.
Brønsted-Lowry
Acid-Base Reactions
• One of the advantages of Brønsted-Lowry
theory is that it allows reactions to be
reversible
H–A + :B  :A– + H–B+
• The original base has an extra H+ after the
reaction, so it will act as an acid in the reverse
process
• And the original acid has a lone pair of
electrons after the reaction – so it will act as a
base in the reverse process
:A– + H–B+  H–A + :B
Tro, Chemistry: A Molecular Approach, 2/e
26
Copyright  2011 Pearson Education, Inc.
Conjugate Pairs
• In a Brønsted-Lowry acid–base reaction, the
original base becomes an acid in the reverse
reaction, and the original acid becomes a base
in the reverse process
• Each reactant and the product it becomes is
called a conjugate pair
• The original base becomes its conjugate acid;
and the original acid becomes its conjugate
base
Tro, Chemistry: A Molecular Approach, 2/e
27
Copyright  2011 Pearson Education, Inc.
Brønsted-Lowry
Acid–Base Reactions
H–A
acid
+
HCHO2
acid
H2O
acid
+ H2O
base
+

:B
base

NH3
base
Tro, Chemistry: A Molecular Approach, 2/e
:A–
+
H–B+
conjugate
conjugate
base
acid
CHO2–
+
H3O+
conjugate
conjugate
base
acid

28
HO–
+
conjugate
base
NH4+
conjugate
acid
Copyright  2011 Pearson Education, Inc.
Conjugate Pairs
In the reaction H2O + NH3  HO– + NH4+
H2O and HO– constitute an
acid/conjugate base pair
NH3 and NH4+ constitute
a base/conjugate acid
pair
Tro, Chemistry: A Molecular Approach, 2/e
29
Copyright  2011 Pearson Education, Inc.
Practice – Write the formula for the
conjugate acid of the following
H2O
H3O+
NH3
NH4+
CO32−
HCO3−
H2PO41−
H3PO4
Tro, Chemistry: A Molecular Approach, 2/e
30
Copyright  2011 Pearson Education, Inc.
Practice – Write the formula for the
conjugate base of the following
H2O
HO−
NH3
NH2−
CO32−
because CO32− does not have an H, it
cannot be an acid
H2PO41−
H2PO41−
HPO42−
Tro, Chemistry: A Molecular Approach, 2/e
31
Copyright  2011 Pearson Education, Inc.
Example 15.1a: Identify the Brønsted-Lowry acids
and bases, and their conjugates, in the reaction
H2SO4 + H2O  HSO4–
+
H3O+
When the H2SO4 becomes HSO4, it loses an H+ so
H2SO4 must be the acid and HSO4 its conjugate base
When the H2O becomes H3O+, it accepts an H+ so
H2O must be the base and H3O+ its conjugate acid
H2SO4
acid
+ H2 O
base
Tro, Chemistry: A Molecular Approach, 2/e

32
HSO4–
+
conjugate
base
H3O+
conjugate
acid
Copyright  2011 Pearson Education, Inc.
Example 15.1b: Identify the Brønsted-Lowry acids
and bases and their conjugates in the reaction
HCO3– +
H2O

H2CO3
+
HO–
When the HCO3 becomes H2CO3, it accepts an H+ so
HCO3 must be the base and H2CO3 its conjugate acid
When the H2O becomes OH, it donats an H+ so H2O
must be the acid and OH its conjugate base
HCO3– +
base
H2O
acid

Tro, Chemistry: A Molecular Approach, 2/e
H2CO3
+
conjugate
acid
33
HO–
conjugate
base
Copyright  2011 Pearson Education, Inc.
Practice – Identify the Brønsted-Lowry acid,
base, conjugate acid, and conjugate base in
the following reaction
HSO4−(aq) +
Acid
HCO3−(aq)

Conjugate
Base
Base
Tro, Chemistry: A Molecular Approach, 2/e
SO42−(aq) +
34
H2CO3(aq)
Conjugate
Acid
Copyright  2011 Pearson Education, Inc.
Practice—Write the equations for the following
reacting with water and acting as a monoprotic
acid & label the conjugate acid and base
HBr
HBr + H2O  Br− + H3O+
Acid
HSO4−
Base
Conj.
base
Conj.
acid
HSO4− + H2O  SO42− + H3O+
Acid
Tro, Chemistry: A Molecular Approach, 2/e
Base
35
Conj.
base
Conj.
acid
Copyright  2011 Pearson Education, Inc.
Practice—Write the equations for the following
reacting with water and acting as a
monoprotic-accepting base and label the
conjugate acid and base
I−
I− +
Base
CO32−
H 2O 
Acid
HI +
Conj.
acid
OH−
Conj.
base
CO32− + H2O  HCO3− + OH−
Base
Tro, Chemistry: A Molecular Approach, 2/e
Acid
36
Conj.
acid
Conj.
base
Copyright  2011 Pearson Education, Inc.
Comparing Arrhenius Theory and
Brønsted-Lowry Theory
• Arrhenius theory
•
HCl(aq) 
H+(aq) + Cl−(aq)
HF(aq) 
H+(aq) + F−(aq)
NaOH(aq) 
Na+(aq) + OH−(aq)
NH4OH(aq) 
NH4+(aq) + OH−(aq)
Tro, Chemistry: A Molecular Approach, 2/e
37
Brønsted–Lowry theory
HCl(aq) + H2O(l) 
Cl−(aq) + H3O+(aq)
HF(aq) + H2O(l) 
F−(aq) + H3O+(aq)
NaOH(aq) 
Na+(aq) + OH−(aq)
NH3(aq) + H2O(l) 
NH4+(aq) + OH−(aq)
Copyright  2011 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two kinds
•
•
of arrows in reactions to indicate the
degree of completion of the
reactions
A single arrow indicates all the
reactant molecules are converted to
product molecules at the end
A double arrow indicates the
reaction stops when only some of
the reactant molecules have been
converted into products
 in these notes
Tro, Chemistry: A Molecular Approach, 2/e
38
Copyright  2011 Pearson Education, Inc.
Strong or Weak
• A strong acid is a strong electrolyte
practically all the acid molecules ionize, →
• A strong base is a strong electrolyte
practically all the base molecules form OH– ions,
either through dissociation or reaction with water, →
• A weak acid is a weak electrolyte
only a small percentage of the molecules ionize, 
• A weak base is a weak electrolyte
only a small percentage of the base molecules form
OH– ions, either through dissociation or reaction with
water, 
Tro, Chemistry: A Molecular Approach, 2/e
39
Copyright  2011 Pearson Education, Inc.
Strong Acids
• The stronger the acid, the
+ + Cl−
HCl

H
more willing it is to donate H
+ + Cl−
HCl
+
H
O

H
O
2
3
 we use water as the standard
base to donate H to
• Strong acids donate
practically all their H’s
 100% ionized in water
 strong electrolyte
• [H3O+] = [strong acid]
 [X] means the molarity of X 0.10 M HCl = 0.10 M H3O+
Tro, Chemistry: A Molecular Approach, 2/e
40
Copyright  2011 Pearson Education, Inc.
Weak Acids
• Weak acids donate a small
fraction of their H’s
most of the weak acid
molecules do not donate H
to water
much less than 1% ionized
in water
HF  H+ + F−
HF + H2O  H3O+ + F−
• [H3O+] << [weak acid]
0.10 M HF ≠ 0.10 M H3O+
Tro, Chemistry: A Molecular Approach, 2/e
41
Copyright  2011 Pearson Education, Inc.
Strong & Weak Acids
Tro, Chemistry: A Molecular Approach, 2/e
42
Copyright  2011 Pearson Education, Inc.
Tro, Chemistry: A Molecular Approach, 2/e
Copyright  2011 Pearson Education, Inc.
Strengths of Acids & Bases
• Commonly, acid or base strength is measured by
•
•
determining the equilibrium constant of a
substance’s reaction with water
HAcid + H2O  Acid− + H3O+
Base: + H2O  HBase+ + OH−
The farther the equilibrium position lies toward the
products, the stronger the acid or base
The position of equilibrium depends on the strength
of attraction between the base form and the H+
 stronger attraction means stronger base or weaker acid
Tro, Chemistry: A Molecular Approach, 2/e
44
Copyright  2011 Pearson Education, Inc.
General Trends in Acidity
• The stronger an acid is at donating H, the
weaker the conjugate base is at accepting H
• Higher oxidation number = stronger oxyacid
H2SO4 > H2SO3; HNO3 > HNO2
• Cation stronger acid than neutral molecule;
neutral stronger acid than anion
H3O+ > H2O > OH−; NH4+ > NH3 > NH2−
trend in base strength opposite
Tro, Chemistry: A Molecular Approach, 2/e
45
Copyright  2011 Pearson Education, Inc.
Acid Ionization Constant, Ka
• Acid strength measured by the size of the
equilibrium constant when reacts with H2O
HAcid + H2O  Acid− + H3O+
• The equilibrium constant for this reaction is
called the acid ionization constant, Ka
larger Ka = stronger acid
Ka 
Tro, Chemistry: A Molecular Approach, 2/e
[Acid ]  [H3O ]
[HAcid]
46
Copyright  2011 Pearson Education, Inc.
Tro, Chemistry: A Molecular Approach, 2/e
47
Copyright  2011 Pearson Education, Inc.
Autoionization of Water
• Water is actually an extremely weak electrolyte
therefore there must be a few ions present
• About 2 out of every 1 billion water molecules
•
form ions through a process called
autoionization
H2O  H+ + OH–
H2O + H2O  H3O+ + OH–
All aqueous solutions contain both H3O+ and OH–
the concentration of H3O+ and OH– are equal in water
[H3O+] = [OH–] = 10−7M @ 25 °C
Tro, Chemistry: A Molecular Approach, 2/e
48
Copyright  2011 Pearson Education, Inc.
Ion Product of Water
• The product of the H3O+ and OH–
concentrations is always the same number
• The number is called the Ion Product of
Water and has the symbol Kw
aka the Dissociation Constant of Water
• [H3O+] x [OH–] = Kw = 1.00 x 10−14 @ 25 °C
if you measure one of the concentrations, you
can calculate the other
• As [H3O+] increases the [OH–] must decrease
so the product stays constant
inversely proportional
Tro, Chemistry: A Molecular Approach, 2/e
49
Copyright  2011 Pearson Education, Inc.
Acidic and Basic Solutions
• All aqueous solutions contain both H3O+ and OH–
ions
• Neutral solutions have equal [H3O+] and [OH–]
[H3O+] = [OH–] = 1.00 x 10−7
• Acidic solutions have a larger [H3O+] than [OH–]
[H3O+] > 1.00 x 10−7; [OH–] < 1.00 x 10−7
• Basic solutions have a larger [OH–] than [H3O+]
[H3O+] < 1.00 x 10−7; [OH–] > 1.00 x 10−7
Tro, Chemistry: A Molecular Approach, 2/e
50
Copyright  2011 Pearson Education, Inc.
Practice – Complete the table
[H+] vs. [OH−]
[H+] 100 10−1
+
H
OH−
10−3
10−5
+
H
OH−
10−7
+
H
10−9
10−11
H+
−
−
OH OH
10−13 10−14
H+
−
OH
[OH−]
Tro, Chemistry: A Molecular Approach, 2/e
51
Copyright  2011 Pearson Education, Inc.
Practice – Complete the table
[H+] vs. [OH−]
Acid
[H+] 100 10−1
+
H
OH−
10−3
Base
10−5
+
H
OH−
[OH−]10−14 10−13 10−11
10−9
10−7
10−9
−
OH
OH
10−5
10−13 10−14
H+
H+
+
H
10−7
10−11
−
− OH
10−3
10−1 100
Even though it may look like it, neither H+ nor OH− will ever be 0
The sizes of the H+ and OH− are not to scale
because the divisions are powers of 10 rather than units
Tro, Chemistry: A Molecular Approach, 2/e
52
Copyright  2011 Pearson Education, Inc.
Example 15.2b: Calculate the [OH] at 25 °C when
the [H3O+] = 1.5 x 10−9 M, and determine if the
solution is acidic, basic, or neutral
Given:
Find:
Conceptual
Plan:
[H3O+] = 1.5 x 10−9 M
[OH]
[OH]
[H3O+]
Relationships:
Solution:
Check: the units are correct; the fact that the
[H3O+] < [OH] means the solution is basic
Tro, Chemistry: A Molecular Approach, 2/e
53
Copyright  2011 Pearson Education, Inc.
Practice – Determine the [H3O+] when the
[OH−] = 2.5 x 10−9 M
Tro, Chemistry: A Molecular Approach, 2/e
54
Copyright  2011 Pearson Education, Inc.
Practice – Determine the [H3O+] when the
[OH−] = 2.5 x 10−9 M
Given:
Find:
Conceptual
Plan:
[OH] = 2.5 x 10−9 M
[H3O+]
[OH]
[H3O+]
Relationships:
Solution:
Check: the units are correct; the fact that the
[H3O+] > [OH] means the solution is acidic
Tro, Chemistry: A Molecular Approach, 2/e
55
Copyright  2011 Pearson Education, Inc.
Measuring Acidity:
pH
• The acidity or basicity of a
solution is often expressed as pH
• pH = −log[H3O+]
exponent on 10 with a positive sign
pHwater = −log[10−7] = 7
need to know the [H3O+]
concentration to find pH
• pH < 7 is acidic; pH > 7 is basic,
•
pH = 7 is neutral
[H3O+] = 10−pH
Tro, Chemistry: A Molecular Approach, 2/e
56
Copyright  2011 Pearson Education, Inc.
Sig. Figs. & Logs
• When you take the log of a number written in
•
scientific notation, the digit(s) before the decimal
point come from the exponent on 10, and the
digits after the decimal point come from the
decimal part of the number
log(2.0 x 106) = log(106) + log(2.0)
= 6 + 0.30303… = 6.30303...
Because the part of the scientific notation number
that determines the significant figures is the
decimal part, the sig figs are the digits after the
decimal point in the log
log(2.0 x 106) = 6.30
Tro, Chemistry: A Molecular Approach, 2/e
57
Copyright  2011 Pearson Education, Inc.
What Does the pH Number Imply?
• The lower the pH, the more acidic the solution;
the higher the pH, the more basic the solution
1 pH unit corresponds to a factor of 10 difference
in acidity
• Normal range of pH is 0 to 14
pH 0 is [H3O+] = 1 M, pH 14 is [OH–] = 1 M
pH can be negative (very acidic) or larger than 14
(very alkaline)
Tro, Chemistry: A Molecular Approach, 2/e
58
Copyright  2011 Pearson Education, Inc.
Example 15.3b: Calculate the pH at 25 °C when the
[OH] = 1.3 x 10−2 M, and determine if the solution is
acidic, basic, or neutral
Given:
Find:
Conceptual
Plan:
[OH] = 1.3 x 10−2 M
pH
[OH]
[H3O+]
pH
Relationships:
Solution:
Check: pH is unitless; the fact that the pH > 7 means
the solution is basic
Tro, Chemistry: A Molecular Approach, 2/e
59
Copyright  2011 Pearson Education, Inc.
Practice – Determine the pH @ 25 ºC of a
solution that has [OH−] = 2.5 x 10−9 M
Tro, Chemistry: A Molecular Approach, 2/e
60
Copyright  2011 Pearson Education, Inc.
Practice – Determine the pH @ 25 ºC of a solution
that has [OH−] = 2.5 x 10−9 M
Given:
Find:
Conceptual
Plan:
[OH] = 2.5 x 10−9 M
pH
[OH]
[H3O+]
pH
Relationships:
Solution:
Check: pH is unitless; the fact that the pH < 7 means
the solution is acidic
Tro, Chemistry: A Molecular Approach, 2/e
61
Copyright  2011 Pearson Education, Inc.
Practice – Determine the [OH−] of a solution
with a pH of 5.40
Tro, Chemistry: A Molecular Approach, 2/e
62
Copyright  2011 Pearson Education, Inc.
Practice – Determine the [OH−] of a solution with a
pH of 5.40
Given:
Find:
Conceptual
Plan:
pH = 5.40
[OH−], M
[H3O+]
pH
[OH]
Relationships:
Solution:
Check: because the pH < 7, [OH−] should be less
than 1 x 10−7; and it is
Tro, Chemistry: A Molecular Approach, 2/e
63
Copyright  2011 Pearson Education, Inc.
pOH
• Another way of expressing the acidity/basicity of
•
a solution is pOH
pOH = −log[OH], [OH] = 10−pOH
pOHwater = −log[10−7] = 7
need to know the [OH] concentration to find pOH
• pOH < 7 is basic; pOH > 7 is acidic, pOH = 7 is
•
neutral
pH + pOH = 14.0
Tro, Chemistry: A Molecular Approach, 2/e
64
Copyright  2011 Pearson Education, Inc.
pH and pOH
Practice – Complete the table
pH
[H+] 100 10−1
+
H
OH−
10−3
10−5
+
H
OH−
[OH−]10−14 10−13 10−11
10−9
10−7
10−9
−
OH
10−13 10−14
H+
H+
+
H
10−7
10−11
−
− OH
OH
10−5
10−3
10−1 100
pOH
Tro, Chemistry: A Molecular Approach, 2/e
65
Copyright  2011 Pearson Education, Inc.
pH and pOH
Practice – Complete the table
Acid
pH
0
1
[H+] 100 10−1
+
H
OH−
3
5
7
9
10−3
10−5
10−7
10−9
+
H
OH−
[OH−]10−14 10−13 10−11
pOH 14
13
Base
11
+
H
−
OH
OH
10−5
9
7
5
66
H+
−
− OH
10−7
Tro, Chemistry: A Molecular Approach, 2/e
14
10−13 10−14
H+
10−9
11
10−11
13
10−3
3
10−1 100
1
0
Copyright  2011 Pearson Education, Inc.
Relationship between pH and pOH
• pH + pOH = 14.00
at 25 °C
you can use pOH to find pH of a solution
[H3O ][OH ]  K w  1.0  10 14



 
 ]  14.00

log
[H
O
]


log
[OH
     
 log [H3O ][OH ]   log 1.0  10 14

3
pH  pOH  14.00
Tro, Chemistry: A Molecular Approach, 2/e
67
Copyright  2011 Pearson Education, Inc.
Example: Calculate the pH at 25 °C when the [OH]
= 1.3 x 10−2 M, and determine if the solution is
acidic, basic, or neutral
Given:
Find:
Conceptual
Plan:
[OH] = 1.3 x 10−2 M
pH
[OH]
pOH
pH
Relationships:
Solution:
Check: pH is unitless; the fact that the pH > 7 means
the solution is basic
Tro, Chemistry: A Molecular Approach, 2/e
68
Copyright  2011 Pearson Education, Inc.
Practice – Determine the pOH @ 25 ºC of a
solution that has [H3O+] = 2.5 x 10−9 M
Tro, Chemistry: A Molecular Approach, 2/e
69
Copyright  2011 Pearson Education, Inc.
Practice – Determine the pOH @ 25 ºC of a solution
that has [H3O+] = 2.5 x 10−9 M
Given:
Find:
Conceptual
Plan:
[H3O+] = 2.5 x 10−9 M
pOH
[H3O+]
pH
pOH
Relationships:
Solution:
Check: pH is unitless; the fact that the pH < 7 means
the solution is acidic
Tro, Chemistry: A Molecular Approach, 2/e
70
Copyright  2011 Pearson Education, Inc.
pK
• A way of expressing the strength of an acid or
•
•
•
base is pK
pKa = −log(Ka), Ka = 10−pKa
pKb = −log(Kb), Kb = 10−pKb
The stronger the acid, the smaller the pKa
larger Ka = smaller pKa
because it is the –log
• The stronger the base, the smaller the pKb
larger Kb = smaller pKb
Tro, Chemistry: A Molecular Approach, 2/e
71
Copyright  2011 Pearson Education, Inc.
[H3O+] and [OH−] in a
Strong Acid or Strong Base Solution
• There are two sources of H3O+ in an aqueous
•
•
solution of a strong acid – the acid and the water
There are two sources of OH− in an aqueous
solution of a strong acid – the base and the water
For a strong acid or base, the contribution of the
water to the total [H3O+] or [OH−] is negligible
the [H3O+]acid shifts the Kw equilibrium so far that
[H3O+]water is too small to be significant
except in very dilute solutions, generally < 1 x 10−4 M
Tro, Chemistry: A Molecular Approach, 2/e
72
Copyright  2011 Pearson Education, Inc.
Finding pH of a Strong Acid or
Strong Base Solution
• For a monoprotic strong acid [H3O+] = [HAcid]
 for polyprotic acids, the other ionizations can generally be
ignored
for H2SO4, the second ionization cannot be ignored
 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00
• For a strong ionic base, [OH−] = (number
OH−)x[Base]
 for molecular bases with multiple lone pairs available, only
one lone pair accepts an H, the other reactions can
generally be ignored
 0.10 M Ca(OH)2 has [OH−] = 0.20 M and pH = 13.30
Tro, Chemistry: A Molecular Approach, 2/e
73
Copyright  2011 Pearson Education, Inc.
Finding the pH of a Weak Acid
• There are also two sources of H3O+ in an
aqueous solution of a weak acid – the acid and
the water
• However, finding the [H3O+] is complicated by
the fact that the acid only undergoes partial
ionization
• Calculating the [H3O+] requires solving an
equilibrium problem for the reaction that
defines the acidity of the acid
HAcid + H2O  Acid + H3O+
Tro, Chemistry: A Molecular Approach, 2/e
74
Copyright  2011 Pearson Education, Inc.
Example 15.6: Find the pH of 0.200 M
HNO2(aq) solution @ 25 °C
write the reaction for
the acid with water
HNO2 + H2O  NO2 + H3O+
[HNO2] [NO2−] [H3O+]
0.200
0
≈0
construct an ICE
table for the reaction initial
change
enter the initial
concentrations –
equilibrium
assuming the [H3O+]
from water is ≈ 0
because no products initially, Qc = 0, and the reaction is proceeding
forward
Tro, Chemistry: A Molecular Approach, 2/e
75
Copyright  2011 Pearson Education, Inc.
Example 15.6: Find the pH of 0.200 M
HNO2(aq) solution @ 25 °C
represent the
change in the
concentrations in
terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
[HNO2] [NO2−] [H3O+]
0.200
initial
0
0
x
+x
+x
change
x
x
equilibrium 0.200 x
HNO2 + H2O(l)  NO2−(aq) + H3O+(aq)
substitute into the
equilibrium constant
expression
Tro, Chemistry: A Molecular Approach, 2/e
76
Copyright  2011 Pearson Education, Inc.
Example 15.6: Find the pH of 0.200 M
HNO2(aq) solution @ 25 °C
Ka for HNO2 = 4.6 x 10−4
determine the value
of Ka from Table 15.5
because Ka is very
small, approximate
the [HNO2]eq =
[HNO2]init and solve
for x
[HNO2] [NO2−] [H3O+]
0.200
initial
0
≈0
change
−x
+x
+x
equilibrium 0.200
0.200x
x
x
Tro, Chemistry: A Molecular Approach, 2/e
77
Copyright  2011 Pearson Education, Inc.
Example 15.6: Find the pH of 0.200 M
HNO2(aq) solution @ 25 °C
Ka for HNO2 = 4.6 x 10−4
check if the
approximation is
valid by seeing if
x < 5% of [HNO2]init
[HNO2] [NO2−] [H3O+]
0.200
initial
0
≈0
change
−x
+x
+x
equilibrium 0.200
x
x
x = 9.6 x 10−3
the approximation is valid
Tro, Chemistry: A Molecular Approach, 2/e
78
Copyright  2011 Pearson Education, Inc.
Example 15.6: Find the pH of 0.200 M
HNO2(aq) solution @ 25 °C
Ka for HNO2 = 4.6 x 10−4
substitute x into
the equilibrium
concentration
definitions and
solve
initial
change
equilibrium
[HNO2] [NO2−] [H3O+]
0.200
0
≈0
−x
+x
+x
0.200−x
0.190 0.0096
0.0096
x
x
x = 9.6 x 10−3
Tro, Chemistry: A Molecular Approach, 2/e
79
Copyright  2011 Pearson Education, Inc.
Example 15.6: Find the pH of 0.200 M
HNO2(aq) solution @ 25 °C
Ka for HNO2 = 4.6 x 10−4
substitute [H3O+]
into the formula for
pH and solve
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
80
[HNO2] [NO2−] [H3O+]
0.200
0
≈0
−x
+x
+x
0.190
0.0096 0.0096
Copyright  2011 Pearson Education, Inc.
Example 15.6: Find the pH of 0.200 M
HNO2(aq) solution @ 25 °C
Ka for HNO2 = 4.6 x 10−4
check by substituting
[HNO2] [NO2−] [H3O+]
the equilibrium
0.200
initial
0
≈0
concentrations back
change
−x
+x
+x
into the equilibrium
constant expression equilibrium
0.190 0.0096 0.0096
and comparing the
calculated Ka to the
given Ka
though not exact,
the answer is
reasonably close
Tro, Chemistry: A Molecular Approach, 2/e
81
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.012 M
solution of nicotinic acid, HC6H4NO2?
(Ka = 1.4 x 10−5 @ 25 °C)
Tro, Chemistry: A Molecular Approach, 2/e
82
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.012 M
solution of nicotinic acid, HC6H4NO2?
write the reaction
for the acid with
water
construct an ICE
table for the
reaction
enter the initial
concentrations –
assuming the
[H3O+] from water
is ≈ 0
HC6H4NO2 + H2O  C6H4NO2 + H3O+
[HA]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
83
0.012
[A−]
0
[H3O+]
≈0
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.012 M
solution of nicotinic acid, HC6H4NO2?
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
HC6H4NO2 + H2O  C6H4NO2 + H3O+
[A−]
0.012
0
x
+x
x
0.012 x
[HA]
initial
change
equilibrium
[H3O+]
0
+x
x
substitute into the
equilibrium
constant
expression
Tro, Chemistry: A Molecular Approach, 2/e
84
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5 @ 25 °C
determine the
value of Ka
because Ka is very
small, approximate
the [HA]eq = [HA]init
and solve for x
HC6H4NO2 + H2O  C6H4NO2 + H3O+
[HA]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
0.012
−x
0.012
x
0.012
[A−]
0
+x
x
[H3O+]
≈0
+x
x
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5 @ 25 °C
Ka for HC6H4NO2 = 1.4 x 10−5
check if the
approximation is
valid by seeing if
x < 5% of
[HC6H4NO2]init
[HA]
initial
change
equilibrium
0.012
−x
0.012
[A−]
0
+x
x
[H3O+]
≈0
+x
x
x = 4.1 x 10−4
the approximation is valid
Tro, Chemistry: A Molecular Approach, 2/e
86
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5 @ 25 °C
substitute x into the
equilibrium
concentration
definitions and
solve
[HA]
initial
change
equilibrium
0.012
−x
0.012−x
[A−]
0
+x
x
[H3O+]
≈0
+x
x
x = 4.1 x 10−4
Tro, Chemistry: A Molecular Approach, 2/e
87
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5 @ 25 °C
substitute [H3O+]
into the formula for
pH and solve
[HA]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
88
0.012
−x
[A−]
0
+x
[H3O+]
≈0
+x
0.012 0.00041 0.00041
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.012 M solution of
nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5 @ 25 °C
check by substituting
the equilibrium
concentrations back
into the equilibrium
constant expression
and comparing the
calculated Ka to the
given Ka
[HA]
initial
change
equilibrium
0.012
−x
[A−]
0
+x
[H3O+]
≈0
+x
0.012 0.00041 0.00041
the values match
Tro, Chemistry: A Molecular Approach, 2/e
89
Copyright  2011 Pearson Education, Inc.
Example 15.7: Find the pH of 0.100 M
HClO2(aq) solution @ 25 °C
write the reaction
for the acid with
water
construct an ICE
table for the
reaction
enter the initial
concentrations –
assuming the
[H3O+] from water
is ≈ 0
HClO2 + H2O  ClO2 + H3O+
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
90
[HClO2] [ClO2−] [H3O+]
0.100
0
≈0
Copyright  2011 Pearson Education, Inc.
Example 15.7: Find the pH of 0.100 M
HClO2(aq) solution @ 25 °C
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
initial
change
equilibrium
[HClO2] [ClO2−] [H3O+]
0.100
0
≈0
−x
+x
+x
0.100−x
x
x
substitute into the
equilibrium
constant
expression
Tro, Chemistry: A Molecular Approach, 2/e
91
Copyright  2011 Pearson Education, Inc.
Example 15.7: Find the pH of 0.100 M
HClO2(aq) solution @ 25 °C
Ka for HClO2 = 1.1 x 10−2
determine the value
of Ka from Table 15.5
because Ka is very
small, approximate
the [HClO2]eq =
[HClO2]init and solve
for x
[HClO2] [ClO2−] [H3O+]
0.100
0
≈0
−x
+x
+x
initial
change
equilibrium 0.100−x
Tro, Chemistry: A Molecular Approach, 2/e
92
x
x
Copyright  2011 Pearson Education, Inc.
Example 15.7: Find the pH of 0.100 M
HClO2(aq) solution @ 25 °C
Ka for HClO2 = 1.1 x 10−2
check if the
approximation is
valid by seeing if x
< 5% of [HNO2]init
initial
change
equilibrium
[HClO2] [ClO2−] [H3O+]
0.100
0
≈0
−x
+x
+x
0.100−x
x
x
x = 3.3 x 10−2
the approximation is invalid
Tro, Chemistry: A Molecular Approach, 2/e
93
Copyright  2011 Pearson Education, Inc.
Example 15.7: Find the pH of 0.100 M
HClO2(aq) solution @ 25 °C
Ka for HClO2 = 1.1 x 10−2
if the
approximation is
invalid, solve for x
using the quadratic
formula
Tro, Chemistry: A Molecular Approach, 2/e
94
Copyright  2011 Pearson Education, Inc.
Example 15.7: Find the pH of 0.100 M
HClO2(aq) solution @ 25 °C
Ka for HClO2 = 1.1 x 10−2
substitute x into
the equilibrium
concentration
definitions and
solve
initial
change
equilibrium
[HClO2] [ClO2−] [H3O+]
0.100
0
≈0
−x
+x
+x
0.100−x
0.072
0.028
x
0.028
x
x = 0.028
Tro, Chemistry: A Molecular Approach, 2/e
95
Copyright  2011 Pearson Education, Inc.
Example 15.7: Find the pH of 0.100 M
HClO2(aq) solution @ 25 °C
Ka for HClO2 = 1.1 x 10−2
substitute [H3O+]
into the formula
for pH and solve
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
96
[HClO2] [ClO2−] [H3O+]
0.100
0
≈0
−x
+x
+x
0.072
0.028
0.028
Copyright  2011 Pearson Education, Inc.
Example 15.7: Find the pH of 0.100 M
HClO2(aq) solution @ 25°C
check by substituting
the equilibrium
concentrations back
into the equilibrium
constant expression
and comparing the
calculated Ka to the
given Ka
Ka for HClO2 = 1.1 x 10−2
initial
change
equilibrium
[HClO2] [ClO2-] [H3O+]
0.100
0
≈0
−x
+x
+x
0.072
0.028
0.028
the answer matches
Tro, Chemistry: A Molecular Approach, 2/e
97
Copyright  2011 Pearson Education, Inc.
Example 15.8: What is the Ka of a weak acid if
a 0.100 M solution has a pH of 4.25?
use the pH to find
the equilibrium
[H3O+]
write the reaction
for the acid with
water
construct an ICE
table for the
reaction
HA + H2O  A + H3O+
[HA]
initial
change
equilibrium
0.100
[A−]
0
[H3O+]
≈0
5.6E-05
enter the initial
concentrations and
[H3O+]equil
Tro, Chemistry: A Molecular Approach, 2/e
98
Copyright  2011 Pearson Education, Inc.
Example 15.8: What is the Ka of a weak acid if
a 0.100 M solution has a pH of 4.25?
fill in the rest of the
table using the
[H3O+] as a guide
if the difference is
insignificant,
[HA]equil = [HA]initial
substitute into the
Ka expression and
compute Ka
HA + H2O  A + H3O+
[HA]
[A−]
0
[H3O+]
0
initial
change
−5.6E-05 +5.6E-05 +5.6E-05
equilibrium
0.100 
0.100 5.6E-05 5.6E-05
5.6E-05
Tro, Chemistry: A Molecular Approach, 2/e
99
0.100
Copyright  2011 Pearson Education, Inc.
Practice – What is the Ka of nicotinic
acid, HC6H4NO2, if a 0.012 M solution
of nicotinic acid has a pH of 3.40?
Tro, Chemistry: A Molecular Approach, 2/e
100
Copyright  2011 Pearson Education, Inc.
Practice – What is the Ka of nicotinic acid,
HC6H4NO2, if a 0.012 M solution of nicotinic
acid has a pH of 3.40?
use the pH to find the
equilibrium [H3O+]
HA + H2O  A + H3O+
write the reaction for
the acid with water
construct an ICE
table for the reaction
enter the initial
concentrations and
[H3O+]equil
[HA]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
101
0.012
[A−]
0
[H3O+]
≈0
4.0E-04
Copyright  2011 Pearson Education, Inc.
Practice – What is the Ka of nicotinic acid,
HC6H4NO2, if a 0.012 M solution of nicotinic
acid has a pH of 3.40?
HA + H2O  A + H3O+
fill in the rest of the
table using the
[H3O+] as a guide
if the difference is
insignificant,
[HA]equil = [HA]initial
substitute into the
Ka expression and
compute Ka
[HA]
initial
change
0.012
[A−]
0
[H3O+]
0
−4.0E-04 +4.0E-04 +4.0E-04
0.012  4.0E-04 4.0E-04
equilibrium 4.0E-04
0.012
Tro, Chemistry: A Molecular Approach, 2/e
102
Copyright  2011 Pearson Education, Inc.
Polyprotic Acids
• Acid molecules often have more than one ionizable
H – these are called polyprotic acids
 the ionizable H’s may have different acid strengths or be
equal
 1 H = monoprotic, 2 H = diprotic, 3 H = triprotic
 HCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic
• Polyprotic acids ionize in steps
 each ionizable H is removed sequentially
• Removing of the first H automatically makes
removal of the second H harder
 H2SO4 is a stronger acid than HSO4
Tro, Chemistry: A Molecular Approach, 2/e
103
Copyright  2011 Pearson Education, Inc.
Percent Ionization
• Another way to measure the strength of an acid
is to determine the percentage of acid molecules
that ionize when dissolved in water – this is
called the percent ionization
the higher the percent ionization, the stronger the acid
• Because [ionized acid]equil = [H3O+]equil
Tro, Chemistry: A Molecular Approach, 2/e
104
Copyright  2011 Pearson Education, Inc.
Example 15.9: What is the percent ionization
of a 2.5 M HNO2 solution?
HNO2 + H2O  NO2 + H3O+
write the reaction for
the acid with water
construct an ICE
table for the reaction
enter the Initial
Concentrations
define the change in
concentration in
terms of x
[HNO2] [NO2−] [H3O+]
initial
change
equilibrium
2.5
0
≈≈00
x
+x
x
+x
x
2.5  x
sum the columns to
define the
equilibrium
concentrations
Tro, Chemistry: A Molecular Approach, 2/e
105
Copyright  2011 Pearson Education, Inc.
Example 15.9: What is the percent ionization
of a 2.5 M HNO2 solution?
Ka for HNO2 = 4.6 x 10−4
determine the value
of Ka from Table 15.5
because Ka is very
small, approximate
the [HNO2]eq =
[HNO2]init and solve
for x
[HNO2] [NO2−] [H3O+]
2.5
0
≈0
−x
+x
+x
initial
change
equilibrium 2.5−x ≈2.5
Tro, Chemistry: A Molecular Approach, 2/e
106
x
x
Copyright  2011 Pearson Education, Inc.
Example 15.9: What is the percent ionization
of a 2.5 M HNO2 solution?
HNO2 + H2O  NO2 + H3O+
substitute x into
the equilibrium
concentration
definitions and
solve
x = 3.4 x 10−2
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
107
[HNO2] [NO2−] [H3O+]
2.5
0
≈0
−x
+x
+x
2.5 x
2.5
0.034
x
0.034
x
Copyright  2011 Pearson Education, Inc.
Example 15.9: What is the percent ionization
of a 2.5 M HNO2 solution?
HNO2 + H2O  NO2 + H3O+
apply the definition
and compute the
percent ionization
initial
change
equilibrium
[HNO2] [NO2−] [H3O+]
2.5
0
≈0
−x
+x
+x
2.5
0.034
0.034
because the percent
ionization is < 5%,
the “x is small”
approximation is
valid
Tro, Chemistry: A Molecular Approach, 2/e
108
Copyright  2011 Pearson Education, Inc.
Practice – What is the percent ionization of a
0.012 M solution of nicotinic acid, HC6H4NO2?
(Ka = 1.4 x 10−5 @ 25 °C)
Tro, Chemistry: A Molecular Approach, 2/e
109
Copyright  2011 Pearson Education, Inc.
Practice – What is the percent ionization of a
0.012 M solution of nicotinic acid, HC6H4NO2?
write the reaction for HC6H4NO2 + H2O  C6H4NO2 + H3O+
the acid with water
construct an ICE
table for the reaction
enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
[HA]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
110
0.012
[A−]
0
[H3O+]
≈0
Copyright  2011 Pearson Education, Inc.
Practice – What is the percent ionization of a
0.012 M solution of nicotinic acid, HC6H4NO2?
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
HC6H4NO2 + H2O  C6H4NO2 + H3O+
[HA]
initial
change
0.012
x
equilibrium 0.012 x
[A−]
0
+x
x
[H3O+]
0
+x
x
substitute into the
equilibrium
constant
expression
Tro, Chemistry: A Molecular Approach, 2/e
111
Copyright  2011 Pearson Education, Inc.
Practice – What is the percent ionization of a 0.012 M
solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5
determine the value
of Ka
because Ka is very
small, approximate
the [HA]eq = [HA]init
and solve for x
HC6H4NO2 + H2O  C6H4NO2 + H3O+
[HA]
0.012
initial
change
−x
x
equilibrium 0.012
0.012
Tro, Chemistry: A Molecular Approach, 2/e
112
[A−]
0
+x
x
[H3O+]
≈0
+x
x
Copyright  2011 Pearson Education, Inc.
Practice – What is the percent ionization of a 0.012 M
solution of nicotinic acid, HC6H4NO2?
substitute x into
the equilibrium
concentration
definitions and
solve
[HA]
initial
change
equilibrium
0.012
−x
0.012−x
[A−]
0
+x
x
[H3O+]
≈0
+x
x
x = 4.1 x 10−4
Tro, Chemistry: A Molecular Approach, 2/e
113
Copyright  2011 Pearson Education, Inc.
Practice – What is the percent ionization of a
0.012 M solution of nicotinic acid, HC6H4NO2?
apply the definition
and compute the
percent ionization
because the
percent ionization
is < 5%, the
“x is small”
approximation is
valid
[HA]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
114
0.012
−x
0.012
[A−]
0
+x
[H3O+]
≈0
+x
4.1E-04 4.1E-04
Copyright  2011 Pearson Education, Inc.
Relationship Between
[H3O+]equilibrium & [HA]initial
• Increasing the initial
•
•
concentration of acid results
in increased [H3O+] at
equilibrium
Increasing the initial
concentration of acid results
in decreased percent
ionization
This means that the increase
in [H3O+] concentration is
slower than the increase in
acid concentration
Tro, Chemistry: A Molecular Approach, 2/e
115
Copyright  2011 Pearson Education, Inc.
Why doesn’t the increase in H3O+ keep
up with the increase in HA?
• The reaction for ionization of a weak acid is
HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
• According to Le Châtelier’s Principle, if we reduce the
concentrations of all the (aq) components, the
equilibrium should shift to the right to increase the total
number of dissolved particles
 we can reduce the (aq) concentrations by using a more
dilute initial acid concentration
• The result will be a larger [H3O+] in the dilute solution
compared to the initial acid concentration
• This will result in a larger percent ionization
Tro, Chemistry: A Molecular Approach, 2/e
116
Copyright  2011 Pearson Education, Inc.
Finding the pH of Mixtures of Acids
• Generally, you can ignore the contribution of the
•
•
weaker acid to the [H3O+]equil
For a mixture of a strong acid with a weak acid,
the complete ionization of the strong acid provides
more than enough [H3O+] to shift the weak acid
equilibrium to the left so far that the weak acid’s
added [H3O+] is negligible
For mixtures of weak acids, generally only need to
consider the stronger for the same reasons
 as long as one is significantly stronger than the other,
and their concentrations are similar
Tro, Chemistry: A Molecular Approach, 2/e
117
Copyright  2011 Pearson Education, Inc.
Example 15.10: Find the pH of a mixture of
0.150 M HF(aq) solution and 0.100 M HClO2(aq)
write the reactions for
the acids with water
and determine their Kas
HF + H2O  F + H3O+
if the Kas are
sufficiently different,
use the strongest acid
to construct an ICE
table for the reaction
H2O + H2O  OH + H3O+
enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
Tro, Chemistry: A Molecular Approach, 2/e
Ka = 3.5 x 10−4
HClO + H2O  ClO + H3O+ Ka = 2.9 x 10−8
[HF]
initial
change
equilibrium
118
0.150
Kw = 1.0 x 10−14
[F−]
0
[H3O+]
≈0
Copyright  2011 Pearson Education, Inc.
Example 15.10: Find the pH of a mixture of 0.150
M HF(aq) solution and 0.100 M HClO2(aq)
represent the
change in the
concentrations in
term of x
sum the columns to
find the equilibrium
concentrations in
terms of x
0.150
[F-]
0
[H3O+]
0
x
+x
+x
x
x
[HF]
initial
change
equilibrium 0.150 x
substitute into the
equilibrium constant
expression
Tro, Chemistry: A Molecular Approach, 2/e
119
Copyright  2011 Pearson Education, Inc.
Example 15.10: Find the pH of a mixture of 0.150
M HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10−4
because Ka is very
small, approximate
the [HF]eq = [HF]init
and solve for x
Tro, Chemistry: A Molecular Approach, 2/e
[HF]
initial
change
equilibrium
120
0.150
−x
0.150
0.150x
[F-]
0
+x
x
[H3O+]
≈0
+x
x
Copyright  2011 Pearson Education, Inc.
Example 15.10: Find the pH of a mixture of 0.150
M HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10−4
check if the
approximation is
valid by seeing if
x < 5% of [HF]init
[HF]
initial
change
equilibrium
0.150
−x
0.150
[F-]
0
+x
x
[H3O+]
≈0
+x
x
x = 7.2 x 10−3
the approximation is valid
Tro, Chemistry: A Molecular Approach, 2/e
121
Copyright  2011 Pearson Education, Inc.
Example 15.10: Find the pH of a mixture of 0.150
M HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10−4
substitute x into
the equilibrium
concentration
definitions and
solve
[F-] [H3O+]
0.150
0
≈0
−x
+x
+x
0.150-x
0.143 0.0072
0.0072
x
x
[HF]
initial
change
equilibrium
x = 7.2 x 10−3
Tro, Chemistry: A Molecular Approach, 2/e
122
Copyright  2011 Pearson Education, Inc.
Example 15.10: Find the pH of a mixture of 0.150
M HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10−4
substitute [H3O+]
into the formula
for pH and solve
Tro, Chemistry: A Molecular Approach, 2/e
[HF]
initial
change
equilibrium
123
0.150
−x
0.143
[F-]
0
+x
[H3O+]
≈0
+x
0.0072 0.0072
Copyright  2011 Pearson Education, Inc.
Example 15.10: Find the pH of a mixture of 0.150
M HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10−4
check by substituting
the equilibrium
concentrations back
into the equilibrium
constant expression
and comparing the
calculated Ka to the
given Ka
[HF]
initial
change
equilibrium
0.150
−x
0.143
[F-]
0
+x
[H3O+]
≈0
+x
0.0072 0.0072
though not exact,
the answer is
reasonably close
Tro, Chemistry: A Molecular Approach, 2/e
124
Copyright  2011 Pearson Education, Inc.
Practice – Determine the pH @ 25 ºC of a
solution that is a mixture of 0.045 M HCl and
0.15 M HF
Tro, Chemistry: A Molecular Approach, 2/e
125
Copyright  2011 Pearson Education, Inc.
Practice – Determine the pH @ 25 ºC of a solution
that is a mixture of 0.045 M HCl and 0.15 M HF
Given:
Find:
[HCl] = 4.5 x 10−2 M, [HF] = 0.15 M
pH
Conceptual Because HCl is a strong acid and HF is a weak acid,
Plan: [H3O+] = [HCl]
[H3O+]
[HCl]
pH
Relationships:
Solution:
Check: pH is unitless; the fact that the pH < 7 means
the solution is acidic
Tro, Chemistry: A Molecular Approach, 2/e
126
Copyright  2011 Pearson Education, Inc.
Strong Bases
• The stronger the base, the
more willing it is to accept H
NaOH  Na+ + OH−
 use water as the standard acid
• For ionic bases, practically all
units are dissociated into OH–
or accept H’s
 strong electrolyte
 multi-OH strong bases
completely dissociated
• [HO–] = [strong base] x (# OH)
Tro, Chemistry: A Molecular Approach, 2/e
127
Copyright  2011 Pearson Education, Inc.
Example 15.11: Calculate the pH at 25 °C of a 0.0015
M Sr(OH)2 solution and determine if the solution is
acidic, basic, or neutral
Given:
Find:
[Sr(OH)2] = 1.5 x 10−3 M
pH
Conceptual [Sr(OH) ]
2
Plan:
Relationships:
[OH]
[H3O+]
pH
[OH]=2[Sr(OH)2]
Solution:
[OH]
= 2(0.0015)
= 0.0030 M
Check:
pH is unitless; the fact that the pH > 7 means
the solution is basic
Tro, Chemistry: A Molecular Approach, 2/e
128
Copyright  2011 Pearson Education, Inc.
Practice – Calculate the pH of the
following strong acid or base solutions
0.0020 M HCl
[H3O+] = [HCl] = 2.0 x 10−3 M
pH = −log(2.0 x 10−3) = 2.70
0.0015 M Ca(OH)2
[OH−] = 2 x [Ca(OH)2] = 3.0 x 10−3 M
pOH = −log(3.0 x 10−3) = 2.52
pH = 14.00 − pOH = 14.00 − 2.52
pH = 11.48
Tro, Chemistry: A Molecular Approach, 2/e
130
Copyright  2011 Pearson Education, Inc.
Weak Bases
• In weak bases, only a small
fraction of molecules accept H’s
 weak electrolyte
NH3 + H2O  NH4+ + OH−
 most of the weak base molecules
do not take H from water
 much less than 1% ionization in
water
• [HO–] << [weak base]
• Finding the pH of a weak base
solution is similar to finding the
pH of a weak acid
Tro, Chemistry: A Molecular Approach, 2/e
131
Copyright  2011 Pearson Education, Inc.
Base Ionization Constant, Kb
• Base strength measured by the size of the
•
equilibrium constant when react with H2O
:Base + H2O  OH− + H:Base+
The equilibrium constant is called the base
ionization constant, Kb
larger Kb = stronger base
Tro, Chemistry: A Molecular Approach, 2/e
132
Copyright  2011 Pearson Education, Inc.
Tro, Chemistry: A Molecular Approach, 2/e
133
Copyright  2011 Pearson Education, Inc.
Structure of Amines
Tro, Chemistry: A Molecular Approach, 2/e
134
Copyright  2011 Pearson Education, Inc.
Example 15.12:Find the pH of 0.100 M NH3(aq)
write the reaction
for the base with
water
construct an ICE
table for the
reaction
enter the initial
concentrations –
assuming the
[OH] from water
is ≈ 0
NH3 + H2O  NH4+ + OH
initial
change
equilibrium
[NH3] [NH4+] [OH]
0.100
0
≈0
because no products initially, Qc = 0,
and the reaction is proceeding forward
Tro, Chemistry: A Molecular Approach, 2/e
135
Copyright  2011 Pearson Education, Inc.
Example 15.12: Find the pH of 0.100 M NH3(aq)
represent the
change in the
concentrations in
terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
[NH4+] [OH]
0.100
initial
0
0
x
+x
+x
change
x
x
equilibrium 0.100 x
[NH3]
substitute into the
equilibrium
constant
expression
Tro, Chemistry: A Molecular Approach, 2/e
136
Copyright  2011 Pearson Education, Inc.
Example 15.12: Find the pH of 0.100 M NH3(aq)
determine the value
of Kb from Table
15.8
because Kb is very
small, approximate
the [NH3]eq =
[NH3]init and solve
for x
Kb for NH3 = 1.76 x 10−5
[NH4+] [OH]
0.100
initial
0
≈0
change
−x
+x
+x
equilibrium 0.100
0.100x
x
x
Tro, Chemistry: A Molecular Approach, 2/e
[NH3]
137
Copyright  2011 Pearson Education, Inc.
Example 15.12: Find the pH of 0.100 M NH3(aq)
Kb for NH3 = 1.76 x 10−5
check if the
approximation is
valid by seeing if
x < 5% of [NH3]init
initial
change
equilibrium
[NH3] [NH4+] [OH]
0.100
0
≈0
−x
+x
+x
0.100
x
x
x = 1.33 x 10−3
the approximation is valid
Tro, Chemistry: A Molecular Approach, 2/e
138
Copyright  2011 Pearson Education, Inc.
Example 15.12: Find the pH of 0.100 M NH3(aq)
Kb for NH3 = 1.76 x 10−5
substitute x into
the equilibrium
concentration
definitions and
solve
[NH4+] [OH]
0.100
initial
0
≈0
change
-x
+x
+x
0.099x 1.33E-3
1.33E-3
equilibrium 0.100
x
x
[NH3]
x = 1.33 x 10−3
Tro, Chemistry: A Molecular Approach, 2/e
139
Copyright  2011 Pearson Education, Inc.
Example 15.12: Find the pH of 0.100 M NH3(aq)
Kb for NH3 = 1.76 x 10−5
use the [OH−] to
find the [H3O+]
using Kw
substitute [H3O+]
into the formula
for pH and solve
[NH3]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
140
0.100
−x
0.099
[NH4+]
0
+x
[OH]
≈0
+x
1.33E−3 1.33E−3
Copyright  2011 Pearson Education, Inc.
Example 15.12: Find the pH of 0.100 M NH3(aq)
check by
substituting the
equilibrium
concentrations back
into the equilibrium
constant expression
and comparing the
calculated Kb to the
given Kb
Kb for NH3 = 1.76 x 10−5
[NH3]
initial
change
equilibrium
0.100
−x
0.099
[NH4+]
0
+x
[OH]
≈0
+x
1.33E−3 1.33E−3
though not exact,
the answer is
reasonably close
Tro, Chemistry: A Molecular Approach, 2/e
142
Copyright  2011 Pearson Education, Inc.
Practice – Find the pH of a 0.0015 M
morphine solution, Kb = 1.6 x 10−6
Tro, Chemistry: A Molecular Approach, 2/e
143
Copyright  2011 Pearson Education, Inc.
Practice – Find the pH of a 0.0015 M morphine solution
write the reaction
for the base with
water
construct an ICE
table for the
reaction
enter the initial
concentrations –
assuming the
[OH] from water
is ≈ 0
B + H2O  BH+ + OH
[B]
initial
change
equilibrium
0.0015
[BH+]
0
[OH]
≈0
because no products initially, Qc = 0,
and the reaction is proceeding forward
Tro, Chemistry: A Molecular Approach, 2/e
144
Copyright  2011 Pearson Education, Inc.
Practice – Find the pH of a 0.0015 M morphine solution
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
substitute into
the equilibrium
constant
expression
B + H2O  BH+ + OH
[BH+]
0.0015
initial
0
x
+x
change
x
equilibrium 0.0015 x
Tro, Chemistry: A Molecular Approach, 2/e
[B]
145
[OH]
0
+x
x
Copyright  2011 Pearson Education, Inc.
Practice – Find the pH of a 0.0015 M morphine solution
Kb for morphine = 1.6 x 10−6
determine the
value of Kb
because Kb is very
small,
approximate the
[B]eq = [B]init and
solve for x
[BH+]
0.0015
0
−x
+x
0.0015
x
0.0015
x
[B]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
146
[OH]
≈0
+x
x
Copyright  2011 Pearson Education, Inc.
Practice – Find the pH of a 0.0015 M morphine solution
Kb for morphine = 1.6 x 10−6
check if the
approximation is
valid by seeing if
x < 5% of [B]init
[B]
initial
change
equilibrium
0.0015
−x
0.0015
[BH+]
0
+x
x
[OH]
≈0
+x
x
x = 4.9 x 10−5
the approximation is valid
Tro, Chemistry: A Molecular Approach, 2/e
147
Copyright  2011 Pearson Education, Inc.
Practice – Find the pH of a 0.0015 M morphine solution
Kb for morphine = 1.6 x 10−6
substitute x into
the equilibrium
concentration
definitions and
solve
[BH++] [OH
[OH]]
0.0015
initial
0
≈≈ 00
change
−x
+x
+x
+x
0.0015x 4.9E−5
4.9E−5
equilibrium 0.0015
x
x
[B]
x = 4.9 x 10−5
Tro, Chemistry: A Molecular Approach, 2/e
148
Copyright  2011 Pearson Education, Inc.
Practice – Find the pH of a 0.0015 M morphine solution
Kb for morphine = 1.6 x 10−6
use the [OH−] to
find the [H3O+]
using Kw
substitute [H3O+]
into the formula
for pH and solve
[B]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
149
0.0015
−x
0.0015
[BH+]
0
+x
[OH]
≈0
+x
4.9E−5 4.9E−5
Copyright  2011 Pearson Education, Inc.
Practice – Find the pH of a 0.0015 M morphine solution
check by
substituting the
equilibrium
concentrations back
into the equilibrium
constant expression
and comparing the
calculated Kb to the
given Kb
Kb for morphine = 1.6 x 10−6
[B]
initial
change
equilibrium
0.0015
−x
0.0015
[BH+]
0
+x
[OH]
≈0
+x
4.9E−5 4.9E−5
the answer
matches the
given Kb
Tro, Chemistry: A Molecular Approach, 2/e
151
Copyright  2011 Pearson Education, Inc.
Acid–Base Properties of Salts
• Salts are water-soluble ionic compounds
• Salts that contain the cation of a strong base and
an anion that is the conjugate base of a weak acid
are basic
 NaHCO3 solutions are basic
 Na+ is the cation of the strong base NaOH
 HCO3− is the conjugate base of the weak acid H2CO3
• Salts that contain cations that are the conjugate
acid of a weak base and an anion of a strong acid
are acidic
 NH4Cl solutions are acidic
 NH4+ is the conjugate acid of the weak base NH3
 Cl− is the anion of the strong acid HCl
Tro, Chemistry: A Molecular Approach, 2/e
152
Copyright  2011 Pearson Education, Inc.
Anions as Weak Bases
• Every anion can be thought of as the conjugate
•
base of an acid
Therefore, every anion can potentially be a
base
 A−(aq) + H2O(l)  HA(aq) + OH−(aq)
• The stronger the acid is, the weaker the
•
conjugate base is
An anion that is the conjugate base of a strong
acid is pH neutral
Cl−(aq) + H2O(l)  HCl(aq) + OH−(aq)
• An anion that is the conjugate base of a weak
acid is basic
F−(aq) + H2O(l)  HF(aq) + OH−(aq)
Tro, Chemistry: A Molecular Approach, 2/e
153
Copyright  2011 Pearson Education, Inc.
Example 15.13: Use the table to determine
if the given anion is basic or neutral
a) NO3−
b)
the conjugate base of
a strong acid,
therefore neutral
NO2−
the conjugate base of
a weak acid, therefore
basic
Tro, Chemistry: A Molecular Approach, 2/e
154
Copyright  2011 Pearson Education, Inc.
Relationship between Ka of an Acid and
Kb of Its Conjugate Base
• Many reference books only give tables of Ka
values because Kb values can be found from them
when you add
equations, you
multiply the
K’s
Tro, Chemistry: A Molecular Approach, 2/e
155
Copyright  2011 Pearson Education, Inc.
Na+ is the cation
of a strong base –
pH neutral. The
CHO2− is the
anion of a weak
acid – pH basic
write the reaction
for the anion with
water
construct an ICE
table for the
reaction
enter the initial
concentrations –
assuming the
[OH] from water
is ≈ 0
Example 15.14: Find the pH of
0.100 M NaCHO2(aq) solution
CHO2− + H2O  HCHO2 + OH
[CHO2−] [HCHO2] [OH]
initial
change
equilibrium
0.100
0
≈0
Copyright  2011 Pearson Education, Inc.
represent the
change in the
concentrations in
terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
Example 15.14: Find the pH of 0.100
M NaCHO2(aq) solution
[CHO2−] [HCHO2] [OH]
0.100
initial
x
change
equilibrium 0.100 x
0
≈0
+x
+x
x
x
Calculate the value
of Kb from the value
of Ka of the weak
acid from Table 15.5
substitute into the
equilibrium constant
expression
Tro, Chemistry: A Molecular Approach, 2/e
157
Copyright  2011 Pearson Education, Inc.
Example 15.14: Find the pH of 0.100 M NaCHO2(aq)
Kb for CHO2− = 5.6 x 10−11
because Kb is very
small, approximate
the [CHO2−]eq =
[CHO2−]init and
solve for x
[CHO2−] [HCHO2] [OH]
0.100
initial
change
−x
equilibrium 0.100
0.100x
Tro, Chemistry: A Molecular Approach, 2/e
158
0
≈0
+x
x
+x
x
Copyright  2011 Pearson Education, Inc.
Example 15.14: Find the pH of 0.100 M NaCHO2(aq)
Kb for CHO2− = 5.6 x 10−11
check if the
approximation is
valid by seeing if
x < 5% of
[CHO2−]init
[CHO2−] [HCHO2] [OH]
initial
change
equilibrium
0.100
0
≈0
−x
+x
x
+x
x
0.100
x = 2.4 x 10−6
the approximation is valid
Tro, Chemistry: A Molecular Approach, 2/e
159
Copyright  2011 Pearson Education, Inc.
Example 15.14: Find the pH of 0.100 M NaCHO2(aq)
Kb for CHO2− = 5.6 x 10−11
substitute x into
the equilibrium
concentration
definitions and
solve
[CHO2−] [HCHO2] [OH]
initial
change
equilibrium
0.100
0
≈0
−x
+x
2.4E-6
x
+x
2.4E-6
x
0.100
0.100−x
x = 2.4 x 10−6
Tro, Chemistry: A Molecular Approach, 2/e
160
Copyright  2011 Pearson Education, Inc.
Example 15.14: Find the pH of 0.100 M NaCHO2(aq)
use the [OH−] to
find the [H3O+]
using Kw
substitute [H3O+]
into the formula
for pH and solve
Kb for CHO2− = 5.6 x 10−11
[CHO2−] [HCHO2]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
161
[OH]
0.100
0
≈0
−x
+x
+x
0.100
2.4E-6
2.4E-6
Copyright  2011 Pearson Education, Inc.
Example 15.14: Find the pH of 0.100 M NaCHO2(aq)
check by
substituting the
equilibrium
concentrations
back into the
equilibrium
constant
expression and
comparing the
calculated Kb to
the given Kb
Kb for CHO2− = 5.6 x 10−11
[CHO2−] [HCHO2]
initial
change
equilibrium
[OH]
0.100
0
≈0
−x
+x
+x
0.100
2.4E−6
2.4E−6
though not exact,
the answer is
reasonably close
Tro, Chemistry: A Molecular Approach, 2/e
163
Copyright  2011 Pearson Education, Inc.
Practice – If a 0.15 M NaA solution has a pOH
of 5.45, what is the Ka of HA?
Tro, Chemistry: A Molecular Approach, 2/e
164
Copyright  2011 Pearson Education, Inc.
Na+ is the cation
of a strong base – If a 0.0015 M NaA solution has a pOH
of 5.45, what is the Ka of HA?
pOH neutral.
Because pOH is <
− + H O  HA + OH
7, the solutinon is
A
2
−
basic. A is basic.
[A−]
[HA]
[OH]
write the reaction
for the anion with
0.100
0
≈0
initial
water
construct an ICE
table for the
reaction
change
equilibrium
enter the initial
concentrations –
assuming the
[OH] from water
is ≈ 0
165
Copyright  2011 Pearson Education, Inc.
Practice – If a 0.15 M NaA solution has a
pOH of 5.45, what is the Ka of HA?
use the pOH to
find the [OH−]
use [OH−] to fill
in other items
[A−]
[HA]
[OH]
0.15
0
≈0
initial
−3.6E−6 +3.6E−6 +3.6E−6
change
0.15
3.6E-6
3.6E-6
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
166
Copyright  2011 Pearson Education, Inc.
Practice – If a 0.15 M NaA solution has a
pOH of 5.45, what is the Ka of HA?
calculate the
value of Kb of A−
[A−]
[HA]
[OH]
0.15
0
≈0
initial
change
−3.6E−6 +3.6E−6 +3.6E−6
0.15
3.6E-6
3.6E-6
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
167
Copyright  2011 Pearson Education, Inc.
Practice – If a 0.15 M NaA solution has a
pOH of 5.45, what is the Ka of HA?
use Kb of A− to
find Ka of HA
[A−]
[HA]
[OH]
0.15
0
≈0
initial
change
−3.6E−6 +3.6E−6 +3.6E−6
0.15
3.6E-6
3.6E-6
equilibrium
11
Kb  8.39 10
Tro, Chemistry: A Molecular Approach, 2/e
168
Copyright  2011 Pearson Education, Inc.
Polyatomic Cations as
Weak Acids
• Some cations can be thought of as the conjugate
acid of a weak base
 others are the counter-ions of a strong base
• Therefore, some cations can potentially be acidic
 MH+(aq) + H2O(l)  MOH(aq) + H3O+(aq)
• The stronger the base is, the weaker the
conjugate acid is
 a cation that is the counter-ion of a strong base is pH
neutral
 a cation that is the conjugate acid of a weak base is acidic
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
 because NH3 is a weak base, the position of this equilibrium favors
the right
Tro, Chemistry: A Molecular Approach, 2/e
169
Copyright  2011 Pearson Education, Inc.
Metal Cations as Weak Acids
• Cations of small, highly charged metals are
weakly acidic
 alkali metal cations and alkali earth metal cations are
pH neutral
 cations are hydrated
Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+ (aq) + H3O+(aq)
Tro, Chemistry: A Molecular Approach, 2/e
170
Copyright  2011 Pearson Education, Inc.
Example 15.15: Determine if the given
cation Is acidic or neutral
a) C5N5NH2+
b)
c)
the conjugate acid of the weak base pyridine,
therefore acidic
Ca2+
the counter-ion of the strong base Ca(OH)2,
therefore neutral
Cr3+
a highly charged metal ion, therefore acidic
Tro, Chemistry: A Molecular Approach, 2/e
171
Copyright  2011 Pearson Education, Inc.
Classifying Salt Solutions as
Acidic, Basic, or Neutral
• If the salt cation is the counter-ion of a strong
base and the anion is the conjugate base of a
strong acid, it will form a neutral solution
NaCl
Ca(NO3)2
KBr
• If the salt cation is the counter-ion of a strong
base and the anion is the conjugate base of a
weak acid, it will form a basic solution
NaF
Ca(C2H3O2)2
Tro, Chemistry: A Molecular Approach, 2/e
KNO2
172
Copyright  2011 Pearson Education, Inc.
Classifying Salt Solutions as
Acidic, Basic, or Neutral
• If the salt cation is the conjugate acid of a weak
base and the anion is the conjugate base of a
strong acid, it will form an acidic solution
NH4Cl
• If the salt cation is a highly charged metal ion
and the anion is the conjugate base of a strong
acid, it will form an acidic solution
Al(NO3)3
Tro, Chemistry: A Molecular Approach, 2/e
173
Copyright  2011 Pearson Education, Inc.
Classifying Salt Solutions as
Acidic, Basic, or Neutral
• If the salt cation is the conjugate acid of a weak
base and the anion is the conjugate base of a
weak acid, the pH of the solution depends on
the relative strengths of the acid and base
NH4F because HF is a stronger acid than NH4+, Ka
of NH4+ is larger than Kb of the F−; therefore the
solution will be acidic
Tro, Chemistry: A Molecular Approach, 2/e
174
Copyright  2011 Pearson Education, Inc.
Example 15.16: Determine whether a solution
of the following salts is acidic, basic, or neutral
a) SrCl2
Sr2+ is the counter-ion of a strong base, pH neutral
Cl− is the conjugate base of a strong acid, pH neutral
solution will be pH neutral
b) AlBr3
Al3+ is a small, highly charged metal ion, weak acid
Cl− is the conjugate base of a strong acid, pH neutral
solution will be acidic
c) CH3NH3NO3
CH3NH3+ is the conjugate acid of a weak base, acidic
NO3− is the conjugate base of a strong acid, pH neutral
solution will be acidic
Tro, Chemistry: A Molecular Approach, 2/e
175
Copyright  2011 Pearson Education, Inc.
Example 15.16: Determine whether a solution
of the following salts is acidic, basic, or neutral
d) NaCHO2
e)
Na+ is the counter-ion of a strong base, pH neutral
CHO2− is the conjugate base of a weak acid, basic
solution will be basic
NH4F
NH4+ is the conjugate acid of a weak base, acidic
F− is the conjugate base of a weak acid, basic
Ka(NH4+) > Kb(F−); solution will be acidic
Tro, Chemistry: A Molecular Approach, 2/e
176
Copyright  2011 Pearson Education, Inc.
Practice – Determine whether a solution of the
following salts is acidic, basic, or neutral
•
•
•
K+ is the counter-ion of a strong base, pH neutral
KNO3
NO3− is the counter-ion of a strong acid, pH neutral
the
is pH
neutralcation, pH acidic
Co3+solution
is a highly
charged
CoCl3 Cl− is the counter-ion of a strong acid, pH neutral
the solution is pH acidic
Ba2+ is the counter-ion of a strong base, pH neutral
Ba(HCO3)2 the solution is pH basic
− is the conjugate of a weak acid, pH basic
HCO
CH NH3 + is the conjugate of a weak base, pH acidic
3
3
• CH3NH3NO3
the solution is pH acidic
NO3− is the counter-ion of a strong acid, pH neutral
Tro, Chemistry: A Molecular Approach, 2/e
177
Copyright  2011 Pearson Education, Inc.
Ionization in Polyprotic Acids
• Because polyprotic acids ionize in steps, each H
•
•
has a separate Ka
Ka1 > Ka2 > Ka3
Generally, the difference in Ka values is great
enough so that the second ionization does not
happen to a large enough extent to affect the pH
 most pH problems just do first ionization
 except H2SO4  use [H2SO4] as the [H3O+] for the
second ionization
• [A2−] = Ka2 as long as the second ionization is
negligible
Tro, Chemistry: A Molecular Approach, 2/e
178
Copyright  2011 Pearson Education, Inc.
Tro, Chemistry: A Molecular Approach, 2/e
179
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.12 M solution
of carbonic acid, H2CO3?
(Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11)
Tro, Chemistry: A Molecular Approach, 2/e
180
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.12 M solution
of carbonic acid, H2CO3?
write the reactions
for the acid with
water one H at a
time
construct an ICE
table for the
reaction
enter the initial
concentrations –
assuming the
second ionization
is negligible
H2CO3 + H2O  HCO3 + H3O+
HCO3− + H2O  CO32− + H3O+
[HA]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
181
0.12
[A−]
0
[H3O+]
≈0
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.12 M solution
of carbonic acid, H2CO3?
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
H2CO3 + H2O  HCO3 + H3O+
[HA]
initial
change
equilibrium
0.12
x
0.12 x
[A−]
0
+x
x
[H3O+]
0
+x
x
substitute into the
equilibrium
constant
expression
Tro, Chemistry: A Molecular Approach, 2/e
182
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.12 M solution of
carbonic acid, H2CO3?
determine the
value of Ka1
because Ka1 is very
small, approximate
the [HA]eq = [HA]init
and solve for x
Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11
[HA]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
183
0.12
−x
0.12
x
0.012
[A−]
0
+x
x
[H3O+]
≈0
+x
x
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.12 M solution of
carbonic acid, H2CO3?
Ka1 for H2CO3 = 4.3 x 10−7
check if the
approximation is
valid by seeing if
x < 5% of
[H2CO3]init
[HA]
initial
change
equilibrium
0.12
−x
0.12
[A−]
0
+x
x
[H3O+]
≈0
+x
x
x = 2.27 x 10−4
the approximation is valid
Tro, Chemistry: A Molecular Approach, 2/e
184
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.12 M solution of
carbonic acid, H2CO3?
substitute x into the
equilibrium
concentration
definitions and
solve
[HA]
initial
change
equilibrium
0.12
−x
0.12−x
[A−]
0
+x
x
[H3O+]
≈0
+x
x
x = 2.3 x 10−4
Tro, Chemistry: A Molecular Approach, 2/e
185
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.12 M solution of
carbonic acid, H2CO3?
substitute [H3O+]
into the formula for
pH and solve
[HA]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
186
0.12
−x
0.12
[A−]
0
+x
[H3O+]
≈0
+x
0.00023 0.00023
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a 0.12 M solution of
carbonic acid, H2CO3?
check by substituting
the equilibrium
concentrations back
into the equilibrium
constant expression
and comparing the
calculated Ka to the
given Ka
Ka1 for H2CO3 = 4.3 x 10−7
[HA]
initial
change
equilibrium
0.12
−x
0.12
[A−]
0
+x
[H3O+]
≈0
+x
0.00023 0.00023
the values match
within sig figs
Tro, Chemistry: A Molecular Approach, 2/e
187
Copyright  2011 Pearson Education, Inc.
Practice – What is the [CO32−] in a 0.12 M
solution of carbonic acid, H2CO3?
(Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11)
Tro, Chemistry: A Molecular Approach, 2/e
188
Copyright  2011 Pearson Education, Inc.
Practice – What is the [CO32−] in a 0.12 M
solution of carbonic acid, H2CO3?
write the reactions
for the acid with
water one H at a
time
construct an ICE
table for the
reaction
enter the initial
concentrations for
the second
ionization using the
equilibrium
concentrations
from first ionization
H2CO3 + H2O  HCO3 + H3O+
HCO3− + H2O  CO32− + H3O+
[HCO3−] [CO32−]
0.00023
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
189
0
[H3O+]
0.00023
Copyright  2011 Pearson Education, Inc.
Practice – What is the [CO32−] in a 0.12 M
solution of carbonic acid, H2CO3?
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
HCO3− + H2O  CO32− + H3O+
[HCO3−] [CO32−]
0.00023
initial
change
x
equilibrium 2.3E−4 x
0
+x
x
[H3O+]
0.00023
+x
2.3E−4 +x
substitute into the
equilibrium
constant
expression
Tro, Chemistry: A Molecular Approach, 2/e
190
Copyright  2011 Pearson Education, Inc.
Practice – What is the [CO32−] in a 0.12 M solution of
carbonic acid, H2CO3?
determine the
value of Ka2
because Ka2 is very
small, approximate
the [HA]eq = [HA]init,
[H3O+]eq = [H3O+]init,
and solve for x
using this
approximation, it is
seen that x = Ka2.
Therefore [CO32−]
= Ka2
Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11
[HCO3−] [CO32−]
0.00023
initial
−x
change
equilibrium 2.3E−4
Tro, Chemistry: A Molecular Approach, 2/e
−x
191
[H3O+]
0
0.00023
+x
+x
xx
2.3E−4
+x
Copyright  2011 Pearson Education, Inc.
Ionization in H2SO4
• The ionization constants for H2SO4 are
H2SO4 + H2O  HSO4 + H3O+ strong
HSO4 + H2O  SO42 + H3O+ Ka2 = 1.2 x 10−2
• For most sulfuric acid solutions, the second
ionization is significant and must be accounted
for
• Because the first ionization is complete, use
the given [H2SO4] = [HSO4−]initial = [H3O+]initial
Tro, Chemistry: A Molecular Approach, 2/e
192
Copyright  2011 Pearson Education, Inc.
Example 15.18: Find the pH of 0.0100 M
H2SO4(aq) solution @ 25 °C
write the reactions
for the acid with
water
construct an ICE
table for the
second ionization
reaction
enter the initial
concentrations –
assuming the
[HSO4−] and [H3O+]
is ≈ [H2SO4]
H2SO4 + H2O  HSO4 + H3O+
HSO4 + H2O  SO42 + H3O+
[HSO4 ] [SO42 ] [H3O+]
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
193
0.0100
0
0.0100
Copyright  2011 Pearson Education, Inc.
Example 15.18: Find the pH of 0.0100 M
H2SO4(aq) solution @ 25 °C
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
[HSO4 ] [SO42 ] [H3O+]
initial
change
equilibrium
0.0100
0
0.0100
−x
+x
+x
0.0100
−x
x
0.0100
−x
substitute into the
equilibrium
constant
expression
Tro, Chemistry: A Molecular Approach, 2/e
194
Copyright  2011 Pearson Education, Inc.
Example 15.18: Find the pH of 0.0100 M
H2SO4(aq) solution @ 25 °C
Ka for HSO4− = 0.012
expand and solve
for x using the
quadratic formula
Tro, Chemistry: A Molecular Approach, 2/e
195
Copyright  2011 Pearson Education, Inc.
Example 15.18: Find the pH of 0.0100 M
H2SO4(aq) solution @ 25 °C
Ka for HSO4− = 0.012
substitute x into
the equilibrium
concentration
definitions and
solve
[HSO4 ]
[SO42 ]
[H3O+]
0.0100
0
0.0100
−x
+x
+x
0.0100
0.0055−x
0.0045
x
0.0100
0.0145−x
initial
change
equilibrium
x = 0.0045
Tro, Chemistry: A Molecular Approach, 2/e
196
Copyright  2011 Pearson Education, Inc.
Example 15.18: Find the pH of 0.0100 M
H2SO4(aq) solution @ 25 °C
Ka for HSO4− = 0.012
substitute [H3O+]
into the formula
for pH and solve
initial
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e
197
[HSO4 ]
[SO42 ]
[H3O+]
0.0100
0
0.0100
−x
+x
+x
0.0055
0.0045
0.0145
Copyright  2011 Pearson Education, Inc.
Example 15.18: Find the pH of 0.0100 M
H2SO4(aq) solution @ 25 °C
Ka for HSO4− = 0.012
check by
substituting the
initial
equilibrium
change
concentrations back
into the equilibrium equilibrium
constant expression
and comparing the
calculated Ka to the
given Ka
[HSO4 ]
[SO42 ]
[H3O+]
0.0100
0
0.0100
−x
+x
+x
0.0055
0.0045
0.0145
the answer
matches
Tro, Chemistry: A Molecular Approach, 2/e
198
Copyright  2011 Pearson Education, Inc.
Strengths of Binary Acids
• The more d+ H─X d− polarized
•
•
the bond, the more acidic the
bond
The stronger the H─X bond, the
weaker the acid
Binary acid strength increases to
the right across a period
 acidity: H─C < H─N < H─O < H─F
• Binary acid strength increases
down the column
 acidity: H─F < H─Cl < H─Br < H─I
Tro, Chemistry: A Molecular Approach, 2/e
199
Copyright  2011 Pearson Education, Inc.
Relationship between Bond Strength
and Acidity
Acid
Bond
Energy
kJ/mol
Type of
Acid
HF
565
weak
HCl
431
strong
HBr
364
strong
Tro, Chemistry: A Molecular Approach, 2/e
200
Copyright  2011 Pearson Education, Inc.
Strengths of Oxyacids, H–O–Y
• The more electronegative the Y atom, the stronger the
oxyacid
 HClO > HIO
 acidity of oxyacids decreases down a group
 same trend as binary acids
 helps weakens the H–O bond
• The larger the oxidation number of the central atom, the
stronger the oxyacid
 H2CO3 > H3BO3
 acidity of oxyacids increases to the right across a period
 opposite trend of binary acids
• The more oxygens attached to Y, the stronger the
oxyacid
 further weakens and polarizes the H–O bond
 HClO3 > HClO2
Tro, Chemistry: A Molecular Approach, 2/e
201
Copyright  2011 Pearson Education, Inc.
Relationship Between
Electronegativity and Acidity
Acid
H─O─Y
Electronegativity
of Y
Ka
H─O─Cl
3.0
2.9 x 10−8
H─O─Br
2.8
2.0 x 10−9
H─O─I
2.5
2.3 x 10−11
Tro, Chemistry: A Molecular Approach, 2/e
202
Copyright  2011 Pearson Education, Inc.
Relationship Between Number of Oxygens
on the Central Atom and Acidity
Tro, Chemistry: A Molecular Approach, 2/e
203
Copyright  2011 Pearson Education, Inc.
Practice – Order the Following
• By Acidity (Least to Most)
H3PO4
HNO3
H3PO3
H3AsO3
• By Acidity (Least to Most)
HCl
HBr
H2S
HS−
• By Basicity (Least to Most)
CO32−
NO3−
Tro, Chemistry: A Molecular Approach, 2/e
HCO3−
204
BO33−
Copyright  2011 Pearson Education, Inc.
Practice – Order the Following
• By Acidity (Least to Most)
H3PO4
HNO3
H3PO3
H3AsO3
H3AsO3 < H3PO3 < H3PO4 < HNO3
• By Acidity (Least to Most)
HCl
HBr
H2S
HS−
HS− < H2S < HCl < HBr
• By Basicity (Least to Most)
CO32−
NO3−
HCO3−
BO33−
NO3− < HCO3− < CO32− < BO33−
Tro, Chemistry: A Molecular Approach, 2/e
205
Copyright  2011 Pearson Education, Inc.
Lewis Acid–Base Theory
• Lewis Acid–Base theory focuses on transferring
an electron pair
lone pair  bond
bond  lone pair
• Does NOT require H atoms
• The electron donor is called the Lewis Base
electron rich, therefore nucleophile
• The electron acceptor is called the Lewis Acid
electron deficient, therefore electrophile
Tro, Chemistry: A Molecular Approach, 2/e
206
Copyright  2011 Pearson Education, Inc.
Lewis Bases
• Lewis Base has electrons it is willing to give
•
•
away to or share with another atom
Lewis Base must have lone pair of electrons on
it that it can donate
Anions are better Lewis Bases than neutral
atoms or molecules
N: < N:−
• Generally, the more electronegative an atom,
the less willing it is to be a Lewis Base
O: < S:
Tro, Chemistry: A Molecular Approach, 2/e
207
Copyright  2011 Pearson Education, Inc.
Lewis Acids
• Electron deficient, either from
 being attached to electronegative atom(s)
 not having an octet
• Must have empty orbital willing to accept the
•
•
•
•
electron pair
H+ has empty 1s orbital
B in BF3 has empty 2p orbital and an incomplete
octet
Many small, highly charged metal cations have
empty orbitals they can use to accept electrons
Atoms that are attached to highly electronegative
atoms and have multiple bonds can be Lewis Acids
Tro, Chemistry: A Molecular Approach, 2/e
208
Copyright  2011 Pearson Education, Inc.
Lewis Acid–Base Reactions
• The base donates a pair of electrons to the
•
•
•
acid
Generally results in a covalent bond forming
H3N: + BF3  H3N─BF3
The product that forms is called an adduct
Arrhenius and Brønsted-Lowry acid–base
reactions are also Lewis
Tro, Chemistry: A Molecular Approach, 2/e
209
Copyright  2011 Pearson Education, Inc.
Examples of Lewis Acid–Base Reactions
Tro, Chemistry: A Molecular Approach, 2/e
210
Copyright  2011 Pearson Education, Inc.
Examples of Lewis Acid–Base Reactions
Ag+(aq) + 2 :NH3(aq)  Ag(NH3)2+(aq)
Lewis
Acid
Tro, Chemistry: A Molecular Approach, 2/e
Lewis
Base
211
Adduct
Copyright  2011 Pearson Education, Inc.
Practice – Identify the Lewis Acid and
Lewis Base in Each Reaction
Lewis
Base
Lewis
Base
Lewis
Acid
Lewis
Acid
Tro, Chemistry: A Molecular Approach, 2/e
212
Copyright  2011 Pearson Education, Inc.
U.S. Fuel Consumption
• Over 85% of the energy use in the United
States comes from the combustion of fossil
fuels
 oil, natural gas, coal
• Combustion of fossil fuels produces CO2
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
• Natural fossil fuels also contain small amounts
of S that burn to produce SO2(g)
S(s) + O2(g) → SO2(g)
• The high temperatures of combustion allow
N2(g) in the air to combine with O2(g) to form
oxides of nitrogen
N2(g) + 2 O2(g) → 2 NO2(g)
Tro, Chemistry: A Molecular Approach, 2/e
213
Copyright  2011 Pearson Education, Inc.
What Is Acid Rain?
• Natural rain water has a pH of 5.6
naturally slightly acidic due mainly to CO2
• Rain water with a pH lower than 5.6 is called
acid rain
• Acid rain is linked to damage in ecosystems
and structures
Tro, Chemistry: A Molecular Approach, 2/e
214
Copyright  2011 Pearson Education, Inc.
What Causes Acid Rain?
• Many natural and pollutant gases dissolved in the
air are nonmetal oxides
 CO2, SO2, NO2
• Nonmetal oxides are acidic
•
CO2(g) + H2O(l)  H2CO3(aq)
2 SO2(g) + O2(g) + 2 H2O(l)  2 H2SO4(aq)
4 NO2(g) + O2(g) + 2 H2O(l)  4 HNO3(aq)
Processes that produce nonmetal oxide gases as
waste increase the acidity of the rain
 natural – volcanoes and some bacterial action
 man-made – combustion of fuel
Tro, Chemistry: A Molecular Approach, 2/e
215
Copyright  2011 Pearson Education, Inc.
pH of Rain in Different Regions
Tro, Chemistry: A Molecular Approach, 2/e
216
Copyright  2011 Pearson Education, Inc.
Weather Patterns
• The prevailing winds in the United States travel
west to east
• Weather patterns may cause rain to be
acidic in regions other than where the
nonmetal oxide is produced
• Much of the northeast United States has rain of
very low pH, even though it has very low sulfur
emissions, due in part to the general weather
patterns
Tro, Chemistry: A Molecular Approach, 2/e
217
Copyright  2011 Pearson Education, Inc.
Sources of SO2 from Utilities
Tro, Chemistry: A Molecular Approach, 2/e
218
Copyright  2011 Pearson Education, Inc.
Damage from Acid Rain
• Acids react with metals, and materials that contain
•
•
•
•
carbonates
Acid rain damages bridges, cars, and other
metallic structures
Acid rain damages buildings and other structures
made of limestone or cement
Acidifying lakes affects aquatic life
Soil acidity causes more dissolving of minerals
and leaching more minerals from soil
 making it difficult for trees
Tro, Chemistry: A Molecular Approach, 2/e
219
Copyright  2011 Pearson Education, Inc.
Damage from Acid Rain
Tro, Chemistry: A Molecular Approach, 2/e
220
Copyright  2011 Pearson Education, Inc.
Acid Rain Legislation
• 1990 Clean Air Act attacks acid rain
forces utilities to reduce SO2
• Result is acid rain in the Northeast stabilized
and beginning to be reduced
Tro, Chemistry: A Molecular Approach, 2/e
221
Copyright  2011 Pearson Education, Inc.