Chapter 16
Acids and Bases
Dr. Peter Warburton
peterw@mun.ca
http://www.chem.mun.ca/zcourses/1051.php
Arrhenius theory of acids and bases
The Arrhenius concept: acids dissociate in
water (aqueous solution) to produce
hydrogen ions H+, and bases dissociate in
water to give hydroxide ions, OH-.
Arrhenius acid: HA (aq)  H+ (aq) + A- (aq)
Arrhenius base: MOH (aq)  M+ (aq) + OH- (aq)
2
Neutralization reactions
Arrhenius acids and bases react with each
other to form water and aqueous salts in
neutralization reactions.
H+ (aq) + A- (aq) + M+ (aq) + OH- (aq) 
H2O (l) + M+ (aq) + A- (aq)
The net ionic equation is
H+ (aq) + OH- (aq)  H2O (l)
3
Aqueous salts
The aqueous salt in the reaction comes
from the spectator ions in the reaction.
These ions are present to balance the
proton positive charge or the hydroxide ion
negative charge in the acid or base.
If we evaporate all the water, we are left
with an ionic solid called a salt e.g. NaCl
4
Arrhenius theory of acids and bases
Many substances that do not contain OHact like bases!
The key to the Arrhenius description is that
we need water to act as a solvent to
promote the dissociation of the acid or
base.
5
Brønsted-Lowry theory of acids
and bases
Many substances (like NH3) that do not
contain OH- act like bases in water!
The Brønsted-Lowry Theory: an acid is
any substance that donates protons (H+)
while a base is any substance that can
accept protons.
This means that Brønsted-Lowry acidbase reactions are proton transfer
reactions.
6
Proton transfer reactions
Pairs of compounds are related to each other
through Brønsted-Lowry acid-base reactions. These
are conjugate acid-base pairs.
Generally, an acid HA has a conjugate base A- (a
proton has transferred away from the acid).
Conversely, a base B has a conjugate acid BH+ (a
proton has transferred toward the base).
7
Water as an acid in BL reactions
When a Brønsted-Lowry base is placed in water, it reacts
with the water (which acts as a Brønsted-Lowry acid)
and establishes an acid-base equilibrium.
8
BL base strength
The strength of a Brønsted-Lowry base can be
quantified by the equilibrium constant as it relates to
completeness of reaction with water. For the reaction of
ammonia with water

NH  OH 


Kb
-
4
NH3 
 1.8 x 10-5 at 25 C
9
BL base strength
We give the equilibrium constant a special name for
reactions like these – the base dissociation (or
ionization) constant Kb. Since the value of the constant
is less than one, the ammonia does not dissociate to a
great extent – it is a weak base!

NH  OH 


Kb
-
4
NH3 
 1.8 x 10-5 at 25 C
10
Water as a base in BL reactions
When a Brønsted-Lowry acid is placed in water, it reacts
with the water (which acts as a Brønsted-Lowry base)
and establishes an acid-base equilibrium.
11
BL acid strength
The strength of a Brønsted-Lowry acid can be
quantified by the equilibrium constant as it relates to
completeness of reaction with water. For the reaction of
acetic acid with water

CH COO  H O 


Ka
3

3
CH3COOH
 1.8 x 10 -5 at 25 C
12
BL acid strength
We give the equilibrium constant a special name for
reactions like these – the acid dissociation (or
ionization) constant Ka. Since the value of the constant
is less than one, the acetic acid does not dissociate to a
great extent – it is a weak acid!

CH COO  H O 


Ka
3

3
CH3COOH
 1.8 x 10-5 at 25 C
13
Strong BL acids and bases
Strong BL acids and bases have the exact same reaction
with water as do weak acids and bases, they just are
much more complete. This is reflected in the acid or base
dissociation constant – it is much larger than one!
Hydrochloric acid is a strong acid.

Cl  H O 


Ka

3
HCl
 1 x 106 at 25 C
14
Hydrated protons and hydronium ions
What is the strongest Brønsted-Lowry
acid there is? The strongest BrønstedLowry acid is H+.
The ultimate proton-donor is a proton itself!
In water there is no such thing as H+.

H  H 2O 
 H 3O

K  very large!
15
Hydrated protons and hydronium ions
Often more than one water molecule will crowd
around the hydronium ion (H3O+) to give hydrates
with the formula [H3O(H2O)n]+ where n is 1 to 4.
16
Requirements of Brønsted-Lowry bases
For a molecule or ion to accept a proton (to act
as a base) requires it to have an unshared pair of
electrons which can then be used to create a
bond to the H+.
All Brønsted-Lowry bases have at least one
lone pair of electrons.
In the previous reactions we’ve seen NH3 has a
lone pair and can act as a base. Also, water has
two lone pairs, and can act as a base.
17
Amphiprotic substances
Some substances, like water, have
protons that can be donated (BL acid),
and lone pairs of electrons that can
accept protons (BL base). This is why it
can act like an acid AND a base.
Such substances are said to be
amphiprotic.
18
Problem
Write a balanced equation for the
dissociation of each of the following
Brønsted-Lowry acids in water:
a) H2SO4
b) HSO4c) H3O+
d) NH4+
19
Problem
What is the conjugate acid of each of
the following Brønsted- Lowry bases?
a) HCO3b) CO32c) OHd) H2PO420
Problem
Of the following species, one is acidic, one
is basic, and one is amphiprotic in their
reactions with water: HNO2, PO43-, HCO3-.
Write the four equations needed to
represent these facts.
HNO 2  H 2 O
HCO 3  H 2 O
-






H 3 O   NO 2
H 3 O   CO 3

PO 4  H 2 O
2
HCO 3  H 2 O
3-
-



HPO 4  OH 



2-
H 2 CO 3  OH 
21
Problem
For each of the following reactions, identify
the acids and bases in both the forward
and reverse directions:
HF  H 2 O
Acid
Base



HSO 4  NH
-
Acid
F  H 3O
Base

3 

Base
C 2 H 3 O 2  HCl
-
Base

Acid
Acid
SO 4
2

 NH 4
Base




Acid
HC 2 H 3 O 2  Cl
Acid

Base
22
A 2nd look at acid and base strength
Acid-base equilibria are
competitions!
The equilibrium is a tug-of war between the two
bases in the system as they fight for protons given
away by the two acids.
The acid that is “better at giving away protons”
(or the base that is “better at taking protons”)
will be found in lesser amounts at equilibrium
than the other acid (or base).
23
Strong acids in water
A strong acid (HA) is one that almost completely
dissociates in water (which acts as a weak
base). The conjugate base of the strong acid
(A-) will be a very weak base.



HA
O


A

H
O
  H
2

3




strong acid
very weak base
weak base
weaker acid
At equilibrium, there will be very little to no HA
present in the system, and the concentration of
A- will essentially be the same as the initial
concentration of HA.
24
Strong bases in water
A strong base (A-) is one that almost completely
dissociates in water (which acts as an acid).
The conjugate acid of the strong base (HA) will
be a very weak acid.


A
O


HA

OH
  H
2



strong base
very weak acid
weaker base
-
weak acid
At equilibrium, there will be very little to no Apresent in the system, and the concentration of
HA will essentially be the same as the initial
concentration of A-.
25
Relationship of acid/base strengths
The stronger the acid, the weaker
its conjugate base.
The stronger the base, the weaker
its conjugate acid.
We’ll look at this a bit more in depth
later.
26
27
Differentiating strong acids
Very strong acids like HClO4 and HCl
dissociate in water so completely it is almost
impossible to find accurate Ka values, and
therefore determine which is the stronger acid.
We must place the acids in a solvent that is a
much weaker base than water (poorer at taking
protons).
Whichever reaction in the new solvent has a
higher K value tells us which is really the
stronger acid.
28
Differentiating strong acids
HClO4 must be a stronger acid than HCl
because it forces the very weak base diethyl
ether to accept protons much more readily
(reaction is essentially complete – large K)
than does HCl, which establishes an
equilibrium – smaller K.
29
Self-ionization of water
Water can act as an acid or a base because the molecule
has both protons and lone pairs available – its amphiprotic!
It is possible for one water molecule to act as an acid
while another water molecule acts as a base at the
same time. This leads to the auto-dissociation (or selfionization) of water equilibrium reaction:
H2O (l) + H2O (l)  H3O+ (aq) + OH- (aq)
The equilibrium constant for this reaction is called the
ion-product constant for water, Kw.
Kw = [H3O+] [OH-]
30
Autoionization of water
31
At 25 °C, Kw = 1.0 x 10-14
so [H3O+] = [OH-] = 1.0 x 10-7 mol/L
Relatively few water molecules are dissociated
at equilibrium at room temperature!
We will always assume that
[H3O+] [OH-] = 1.0 x 10-14 at 25 °C.
[H3O+] > 1.0 x 10-7 is acidic ([OH-] < 1.0 x 10-7)
[H3O+] < 1.0 x 10-7 is basic ([OH-] > 1.0 x 10-7)
[H3O+] = 1.0 x 10-7 is neutral ([OH-] = 1.0 x 10-7)
32
We also find, since
[H3O+] [OH-] = 1.0 x 10-14 = Kw
then
[H3O+] = 1.0 x 10-14 / [OH-]
and
[OH-] = 1.0 x 10-14 / [H3O+]
at 25 °C
33
34
Problem
The concentration of OH- in a
sample of seawater is 5.0 x 10-6
molL-1. Calculate the
concentration of H3O+ ions, and
classify the solution as acidic,
neutral, or basic.
Answer: [H3O+] = 2.0 x 10-9. Solution is basic.
35
Problem
At 50 °C the value of Kw is 5.5 x 10-14.
What are the [H3O+] and [OH-]
in a neutral solution at 50 °C?
Answer: Both are 2.3 x 10-7 molL-1
36
The pH scale
Because [H3O+] in water solutions can range from
very small (strongly basic) to very large
(strongly acidic) it is sometimes easier to use a
logarithmic (power of 10) scale to express [H3O+].
Additionally, since using a negative number is
sometime awkward, we actually use a negative
logarithmic scale to express the hydronium ion
concentration with a term we call the pH of a
solution.
pH = - log [H3
+
O]
37
pH and acidity
For [H3O+] = 1.0 x 10-7 molL-1 (neutral), pH = 7.00
For [H3O+] = 1.0 x 10-4 molL-1 (acidic), pH = 4.00
For [H3O+] = 1.0 x 10-11 molL-1 (basic), pH = 11.00
In general,
pH > 7 is basic
pH < 7 is acidic
pH = 7 is neutral
38
pOH and acidity
We occasionally see the pOH mentioned, which
is
pOH = - log [OH-]
or
[OH-] = 10-pOH
pOH < 7 is basic
pOH > 7 is acidic
pOH = 7 is neutral
39
Kw = 1.0 x 10-14 = [H3O+] [OH-]
14.00 = pH + pOH at 25 C
40
Problem
Calculate the pH of each of the following
solutions:
a) A sample of seawater that has an
OH- concentration of 1.58 x 10-6 molL-1
b) A sample of acid rain that has an
H3O+ concentration of 6.0 x 10-5 molL-1
Answers: a) pH = 8.19 b) pH = 4.22
41
Problem
Calculate [H3O+] and [OH-] in each of
the following solutions:
a) Human blood (pH 7.40)
b) A cola beverage (pH 2.8)
Answers:
a) [H3O+] = 4.0 x 10-8 M and [OH-] = 2.5 x 10-7 M
b) [H3O+] = 1.6 x 10-3 M and [OH-] = 6.3 x 10-12 M
42
Problem
The pH of a solution of HCl in water is
found to be 2.50. What volume of water
would you add to 1.00 L of this solution to
raise the pH to 3.10?
Answer: You would add 3.0 L of water
43
Strong acids and strong bases
Most of the strong acids are monoprotic acids,
that are capable of donating only one proton.
Sulphuric acid (H2SO4), which is a diprotic acid
capable of donating two protons, is also a strong
acid (for the first proton!).
Strong monoprotic acids (HA) essentially
dissociate 100% in water to give H3O+ and A-,
leaving virtually no HA in solution at equilibrium.
If we know the initial concentration of HA, then
the equilibrium concentration of H3O+ will be the
same, and we can calculate the pH.
44
pH and strong bases
Strong bases, such as the alkali metal (Li, Na,
K, etc.) hydroxides MOH completely dissociate in
water to give metal ions and hydroxide ions.
H2O
MOH (s)  M+ (aq) + OH- (aq)
Again, calculating the pH is straight-forward,
as the final concentration of OH- will be the
same as the initial concentration of MOH. Of
course, if we know [OH-], we can calculate
[H3O+], and the pH.
45
pH and strong bases
Alkaline earth (Ca, Mg, etc) metal
hydroxides M(OH)2 are strong
bases and will completely
dissociate in water up to the point
of their low solubility.
amount
M(OH) 2 (s) dissolve
 small


 M(OH) 2 (aq)
on
M(OH) 2 (aq) full
dissociati
 
 M 2 (aq)  2 OH  (aq)
46
pH and strong bases
Alkaline earth (Ca, Mg, etc) metal
oxides MO are stronger bases than
the equivalent hydroxides because
O2- is a very strong base. In fact,
much like bare H+ can’t exist in
water, neither can bare O2-.
O2- (aq) + H2O (l)  2 OH- (aq), so
H2O
MO (s)  M2+ (aq) + 2 OH- (aq)
47
Be careful!
We are assuming the pH of the solution
will be determined solely by the initial
concentration of the strong acid or
strong base. This is true if the initial
concentration is large enough that the
autoionization of water contributes
insignificant amounts of H3O+ and OH-!
What is the pH of 1.0 x 10-8 M HCl?
48
Common strong acids and bases
49
Problem
Calculate the pH of:
a) 0.050 M HClO4
b) 6.0 M HCl
c) 4.0 M KOH
d) 0.010 M Ba(OH)2
Answers:
a) pH = 1.30
c) pH = 14.60
b) pH = -0.78
d) pH = 12.30
50
Problem
If 535 mL of gaseous HCl at 26.5 C and
747 mmHg is dissolved in enough water to
prepare 625 mL of solution, what is the pH
of this solution?
The gas constant R = 0.08206 LatmK-1mol-1
and 760 mmHg = 1 atm exactly!
Answer: pH = 1.467
51
Problem
Milk of magnesia is a saturated solution of
Mg(OH)2. Its solubility is 9.63 mg Mg(OH)2
per 100.0 mL of solution at 20 C. What is
the pH of saturated Mg(OH)2 at 20 C?
The molar mass of Mg(OH)2
is 58.3197 gmol-1
Answer: pH = 11.52
52
Problem
Calculate the pH of an aqueous solution
that is 3.00% KOH by mass and has a
density of 1.0242 gmL-1.
The molar mass of KOH is 56.1056 gmol-1.
Answer: pH = 13.74
53
pKa and pKb
We can define
pKa = - log Ka and pKb = - log Kb
exactly like pH = - log [H3O+]
A very small Ka or Kb value is the same as a
large positive pKa or pKb which means the
acid or base is weak (partially dissociated in
water).
As Ka  (or pKa ) or as Kb  (or pKb )
acid strength or base strength increases.
54
pKa and pKb
55
Problem
The pH of 0.10 molL-1 HOCl is 4.27.
Calculate Ka for hypochlorous acid
HOCl (aq) + H2O (l)  H3O+ (aq) + ClO- (aq)
Answer: 2.9 x 10-8
56
Identifying weak acids
Generally there are three categories of
weak acids:
carboxylic acids
oxoacids
miscellaneous acids
57
Carboxylic acids
Carboxylic acids contain
the -COOH (carboxyl)
group. The hydrogen of this
group is the proton that is
donated.
58
Carboxylic acids
59
Oxoacids
Oxoacids are generally weak acids with
the formula
HmXOn where m = 1 to 3 and n = 1 to 4
This formula is often quite misleading
because the structure is actually
usually
(HO)mXOn where (m + n) = 1 to 4
60
Strong oxoacids
Nitric acid –
HNO3
Sulfuric acid –
H2SO4
Perchloric acid –
HClO4
61
Some weak oxoacids
Nitrous acid –
HNO2
Phosphoricic acid –
H3PO4
Chlorous acid –
HClO2
62
Miscellaneous weak acids
There are other weak acids that are not
carboxylic acids or oxoacids. Some of the
more common ones are
Hydrofluoric acid – HF
Hydrocyanic acid – HCN
Hydrazoic acid – HN3
63
Identifying weak bases
Many weak bases are amines, which contain
nitrogen. The lone pair on the nitrogen allow the
amine to be a proton acceptor (Brønsted-Lowry
base).
64
Identifying weak bases
65
Amino acids
Amino acids have
both a carboxyl group
and an amine group,
meaning different
parts of the molecule
can act as an acid and
a base.
66
Equilibrium in solutions of weak acids and bases
We can calculate equilibrium concentrations of
reactants and products in acid-base reactions
with known values for Ka or Kb.
We need to figure out what is an acid and what is
a base in our system. Water will be an acid or
base depending on whether we added a base or
an acid to it. For example, if we start with 0.10
molL-1 HCN, then HCN is an acid, and water is a
base.
HCN (aq) + H2O (l)  H3O+ (aq) + CN- (aq)
Ka = 4.9 x 10-10
67
Equilibrium in solutions of weak acids and bases
However, since the acid-base reaction we are
looking at takes place in water, we must include
the autoionization of water reaction as a
source of H3O+ and OH-!
For our example reaction (HCN is a weak acid)
HCN (aq) + H2O (l)  H3O+ (aq) + CN- (aq)
Ka = 4.9 x 10-10
H2O (l) + H2O (l)  H3O+ (aq) + OH- (aq)
Kw = 1.0 x 10-14
68
Equilibrium in solutions of weak acids and bases
The strongest acid or base (larger K
value) will dominate a system. We call this
equilibrium reaction with the larger K value
the principal reaction. Any other reactions
are subsidiary reactions.
69
Equilibrium in solutions of weak acids and bases
Since all reactions take place in one container,
then
[H3O+] = [H3O+] (principal) + [H3O+] (subsidiary)
OR
[OH-] = [OH-] (principal) + [OH-] (subsidiary)
If the K of the principal reaction is much greater
than K for the subsidiary reactions, then we
assume
[H3O+]  [H3O+] (principal reaction)
OR
[OH-]  [OH-] (principal reaction)
70
Equilibrium in solutions of weak acids and bases
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
HCN (aq)
0.10
-x
0.10 – x
+
H2O (l)
N/A
N/A
N/A

H3O+ (aq) +
0.0
+x
+x
CN- (aq)
0.0
+x
+x


[H
O
][CN
]
(x)(x)
10
10
3
K a  4.9 x 10 
so 4.9 x 10 
[HCN]
(0.10  x)
We can solve this equation using the quadratic formula
and get the right answer, but it might be possible to do
it more simply.
Let’s assume the initial concentration of the acid
and the equilibrium concentration of HCN are
essentially the same (that is x << 0.10 in this case).
71
Equilibrium in solutions of weak acids and bases
4.9 x 10-10 = x2 / 0.10
x2 = (4.9 x 10-10)(0.10)
x2 = 4.9 x 10-11
x = 4.9 x 10-11
x = 7.0 x 10-6 molL-1
Based on the assumption we’ve made, at
equilibrium [H3O+] = [CN-] = 7.0 x 10-6 molL-1 and
[HCN] = 0.10 mol/L.
However, any time we make an assumption, we
must check it! (5% rule)
72
Equilibrium in solutions of weak acids and bases
We also assumed that [H3O+]  [H3O+]
(principal reaction) so we must check this.
For the subsidiary reaction
H2O (l) + H2O (l)  H3O+ (aq) + OH- (aq)
Kw = 1.0 x 10-14
If the assumption we’ve made is good, then
[H3O+] = 7.0 x 10-6 molL-1 and so
[OH-]subsidiary = Kw / [H3O+]principal
[OH-]subsidiary = 1.0 x 10-14 / 7.0 x 10-6
[OH-]subsidiary = 1.4 x 10-9 molL-1
73
Equilibrium in solutions of weak acids and bases
From the subsidiary balanced equation we see if
[OH-]subsidiary = 1.4 x 10-9 molL-1, then
[H3O+]subsidiary = 1.4 x 10-9 molL-1.
Lets check our assumption
The ratio [H3O+]subidiary / [H3O+]principal,
is 1.4 x 10-9 molL-1 / 7.0 x 10-6 molL-1
= 0.02%.
Our assumption is valid (5% rule)!
74
Equilibrium in solutions of weak acids and bases
[H3O+]  [H3O+] (principal reaction)
= 7.0 x 10-6 molL-1
pH = - log [H3O+]
pH = - log 7.0 x 10-6
pH = 5.15
75
Generally it turns out that if our principal
acid-base calculation gives a [H3O+] or
[OH-] (depending on reaction type!) less
than 10-6 or so then the auto-dissociation of
water reaction will actually contribute a
significant amount of [H3O+] or [OH-] to our
system and we must include the subsidiary
reaction contribution to pH
See page 673 (9th ed.) or 682 (8th ed.) of
textbook
76
Problem
Acetic acid CH3COOH is the solute that
gives vinegar its characteristic odour and
sour taste. Calculate the pH and the
concentration of all species present in:
a) 1.00 molL-1 CH3COOH
b) 0.0100 molL-1 CH3COOH
77
Problem a)
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH3COOH (aq) +
1.00
-x
1.00 – x
H2O (l)
N/A
N/A
N/A

H3O+ (aq) + CH3COO- (aq)
0.0
0.0
+x
+x
+x
+x


[H
O
][CH
COO
]
(x)(x)
5
5
3
3
K a  1.8 x 10 
so 1.8 x 10 
[CH 3COOH]
(1.00  x)
Let’s assume that x << 1.00 and so
1.8 x 10-5 = x2 / 1.00
x2 = (1.8 x 10-5)(1.00)
x = 1.8 x 10-5
x = 4.2 x 10-3 molL-1 (must be +ve value since x = [H3O+])
CHECK ASSUMPTION
78
Problem a)
We also assumed that [H3O+]  [H3O+] (principal
reaction) so we must check this.
For the subsidiary reaction
H2O (l) + H2O (l)  H3O+ (aq) + OH- (aq)
Kw = 1.0 x 10-14
If the assumption we’ve made is good, then [H3O+]
= 4.2 x 10-3 molL-1 and so
[OH-]subsidiary = Kw / [H3O+]principal
[OH-]subsidiary = 1.0 x 10-14 / 4.2 x 10-3
[OH-]subsidiary = 2.4 x 10-12 molL-1
79
Problem a)
From the subsidiary balanced equation we see if
[OH-]subsidiary = 2.4 x 10-12 molL-1, then
[H3O+]subsidiary = 2.4 x 10-12 molL-1.
Lets check our assumption
The ratio [H3O+]subidiary / [H3O+]principal,
is 2.4 x 10-12 molL-1 / 4.2 x 10-3 molL-1
= 0.00000006%.
Our assumption is valid (5% rule)!
80
Problem a)
pH = - log [H3O+]
pH = - log 4.2 x 10-3
pH = 2.38
81
Problem b)


[H
O
][CH
COO
]
(x)(x)
3
K a  1.8 x 10 5  3
so 1.8 x 10 5 
[CH 3COOH]
(0.00100  x)
Let’s assume that x << 0.00100 and so
1.8 x 10-5 = x2 / 0.00100
x2 = (1.8 x 10-5)(0.00100)
x = 1.8 x 10-8
x = 1.3 x 10-4 molL-1
(must be +ve value since x = [H3O+])
CHECK ASSUMPTION
82
Problem b)


[H
O
][CH
COO
]
(x)(x)
3
K a  1.8 x 10 5  3
so 1.8 x 10 5 
[CH 3COOH]
(0.00100  x)
We can not assume that x << 0.00100 and so
5
[1.8 x 10 ](0.00100  x)  x  0
8
2
5
[1.8 x 10 ]  [1.8 x 10 ]x  x  0
2
83
Problem b)
 b  b 2  4ac
x
2a
 (1.8 x 10 5 )  (1.8 x 10 5 ) 2  4(1)(1.8 x 10 8 )
so x 
2(1)
1.8 x 10 5  3.2 4 x 10 10  7.2 x 10 8
1.8 x 10 5  3.2 4 x 10 10  7.2 x 10 8
x
or x 
2
2
4
1.8 x 10 5  2.68 x 10
1.8 x 10 5  2.68 x 10  4
x
or x 
2
2
2.87 x 10  4
 1.7 9 x 10 5
x
or x 
2
2
so x  1.43 x 10  4 mol/L
or x  1.25 x 10  4 mol/L
pH = - log [H3O+]
pH = - log 1.25 x 10-4
pH = 3.90
84
Problem
Piperidine (C5H11N; molar mass =
85.149 gmol-1) is a base found in small
amounts in black pepper. What is the
pH of 315 mL of an aqueous solution
containing 114 mg of piperidine if Kb is
1.6 x 10-3?
Answer: pH = 11.29
85
Percent ionization of weak acids
The pH of a solution of a weak acid like
acetic acid will depend on the initial
concentration of the weak acid.
Therefore, we can define a second
measure of the strength of a weak acid
by looking of the percent ionization (or
dissociation) of the acid.
%ionized = [H3O+]eqm / [HA]initial x 100%
86
Percent ionization
We saw on slide 78 that an acetic acid
solution with initial concentration of 1.00
molL-1 at equilibrium had
[H3O+] = 4.2 x 10-3 molL-1
%ionized = [H3O+]eqm / [HA]initial x 100%
%ionized = 4.2 x 10-3 molL-1 / 1.00 molL-1 x 100%
%ionized = 0.42%
87
Percent ionization
We saw on slide 84 that an acetic acid
solution with initial concentration of
0.0010 molL-1 at equilibrium had
[H3O+] = 1.25 x 10-4 molL-1
%ionized = [H3O+]eqm / [HA]initial x 100%
%ionized = 1.25 x 10-4 molL-1 / 0.0010 molL-1 x 100%
%ionized = 12.5%
88
89
Polyprotic acids
Acids that contain more than one dissociable protons are
polyprotic acids.
Each dissociable proton has its own Ka value.
Carbonic acid (H2CO3) regulates blood pH.
H2CO3 (aq) + H2O (l)  H3O+ (aq) + HCO3- (aq)
H O HCO   4.4 x 10



Ka
3
H 2CO3 
3
7
HCO3- (aq) + H2O (l)  H3O+ (aq) + CO32- (aq)

H O CO 

 4.7 x 10
2

Ka
HCO 
3

3
11
3
90
Polyprotic acids
Ka1 > Ka2 > Ka3 is always true!
91
Problem
Calculate the pH and the concentration
of all species, including OH- present in
0.10 molL-1 H2SO3. Values of Ka are
given in the table on the previous slide.
Ka1 = 1.3 x 10-2 and Ka2 = 6.2 x 10-8
92
Problem
Potential reactions that can occur in our system are
H2SO3 (aq) + H2O (l)  H3O+ (aq) + HSO3- (aq)
HSO3- (aq) + H2O (l)  H3O+ (aq) + SO32- (aq)
H2O (l) + H2O (l)  H3O+ (aq) + OH- (aq)
We’ll first have to assume
[H3O+]  [H3O+] (H2SO3 dissociation)
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
H2SO3 (aq) +
0.10
-x
0.10 – x
K a1  1.3 x 10  2
H2O (l)
N/A
N/A
N/A

H3O+ (aq)
0.0
+x
+x
+
HSO3- (aq)
0.0
+x
+x

[H 3O  ][HSO 3 ]
(x)(x)
2

so 1.3 x 10 
[H 2SO 3 ]
(0.10  x)
93
Problem
We can’t assume x << 0.10 so
1.3 x 102 (0.10  x)  x 2  0
(1.3 x 103 )  (1.3 x 102 ) x  x 2  0
 b  b 2  4ac
x
2a
 (1.3 x 10 2 )  (1.3x10 2 ) 2  4(1)(1.3x10 3 )
so x 
2(1)
1.3 x 10  2  1.69 x 10  4  (5.2 x 10 3 )
1.3 x 10  2  1.69 x 10  4  (5.2 x 10 3 )
x
or x 
2
2
1.3 x 10  2  7.33 x10  2
1.3 x 10  2  7.33 x 10  2
x
or x 
2
2
8.63 x 10  2
 6.03 x 10  2
x
or x 
2
2
so x  4.3 x 10  2 mol/L
or x  3.0 x 10  2 mol/L
94
Problem
x = [H3O+]principal = [HSO3-] = 3.0 x 10-2 molL-1
[H2SO3] = (0.10 – 3.0 x 10-2) molL-1 = 0.07 molL-1
The second dissociation reaction is
HSO3- (aq) + H2O (l)  H3O+ (aq) + SO32- (aq)
With our assumption that almost all H3O+ comes
from the first dissociation, if we substitute our
equilibrium concentrations from our first reaction
into the equilibrium constant expression for this
reaction we should see little change if the
assumption is true…
95
Problem
2
[H 3O ][SO 3 ]
8
K a2  6.3 x 10 

[HSO 3 ]
(3.2x10 2 )(x)
so 6.3 x 10 
(3.2x10 2 )
8
x = [SO32-] = [H3O+]sub = 6.3 x 10-8 mol/L
Our assumption is valid!
Since the second proton dissociation
contributes a insignificant amount of H3O+
then the autoionization of water won’t either,
because it has an even smaller K value.
96
Problem
pH = - log [H3O+]
pH = - log 3.0 x 10-2
pH = 1.52
[OH-] = Kw / [H3O+]
[OH-] = 1.0 x 10-14 / 3.0 x 10-2
[OH-] = 3.3 x 10-13 molL-1
97
Relation between Ka and Kb
The strength of an acid in water is expressed
through Ka. while the strength of a base can be
expressed through Kb
But a Brønsted-Lowry acid-base reaction involves
conjugate acid-base pairs so there should be a
connection between the Ka value and the Kb value.
HA (aq) + H2O (l)  H3O+ (aq) + A- (aq) K a
A- (aq) + H2O (l)  OH- (aq) + HA (aq) K b

H O A 



3
HA 

OH HA 

A 

98
Let’s add the reactions together
HA (aq) + H2O (l) + A- (aq) + H2O (l)  H3O+ (aq) + A- (aq) + OH- (aq) + HA (aq)
2 H2O (l)  H3O+ (aq) + OH- (aq)
The sum of the reactions is the dissociation of
water reaction, which has the ion-product
constant for water
Kw = [H3O+] [OH-] = 1.0 x 10-14 at 25 °C
Closer inspection shows us that

H3O A   OH HA 


14



Ka x Kb 

H
O
OH

K

1.0x10
at
3
w

A 
HA 
25 C
99
As the strength of an acid increases (larger Ka)
the strength of the conjugate base must
decrease (smaller Kb) because their product
must always be the dissociation constant for
water Kw.
Strong acids always have very weak conjugate
bases. Strong bases always have very weak
conjugate acids.
Since Ka x Kb = Kw
then Ka = Kw / Kb
and Kb = Kw / Ka
100
Problem
a) Piperidine (C5H11N) is an amine found
in black pepper. If Kb = 1.6 x 10-3 for
piperidine then calculate Ka for the
C5H11NH+ cation.
b) If Ka = 2.9 x 10-8 for HOCl then
calculate Kb for OCl-.
Answers: a) Ka = 6.3 x 10-12 b) Kb = 3.4 x 10-7
101
Ions as acids and bases
Consider the strong acid anions such as
Cl-, Br-, I-, NO3-, and ClO4-. They must
be very weak bases because their
conjugate acids are so strong. They
will not react with water to form the
strong acid…
Cl- (aq) + H2O (l)  no reaction!
102
Ions as acids and bases
Consider the strong base cations such
as Na+, K+, Ca2+, Mg2+, etc. They can
not be acids because they have no
protons. They will not react with water
to form the strong base…
H2O (l) + Na+ (aq)  no reaction!
103
Ions as acids and bases
Consider some weak acid anions such
as OCl-, CH3COO-, HCO3-, or NO2-. They
will be weak bases because their
conjugate acids are weak. They can
react with water to form the weak acid
and OH-…
OCl- (aq) + H2O (l)  OH- (aq) + HOCl (aq)
104
Ions as acids and bases
Consider some weak base cations such
as NH4+, C6H5NH+, CH3NH3+, etc. They
are weak acids because their conjugate
bases are weak. They can react with
water to form the weak base and H3O+ …
H2O (l) + NH4+ (aq)  H3O+ (aq) + NH3 (aq)
105
Salts that yield neutral solutions
Any ionic salt that contains neither
an acidic cation OR a basic anion
will give a neutral solution because
neither ion can react with water.
NaCl, KNO3, Ca(ClO4)2, MgI2, etc.
106
Salts that yield acidic or basic solutions
Any ionic salt that contains either
an acidic cation OR a basic anion
while the other ion can not react with water
will give an acidic or basic solution because
of the ability of the ion to react with water.
NH4Cl, CH3COONa, C6H5NHNO3, KOCl
etc.
107
Salts that yield acidic or basic solutions
If a salt is composed of an
acidic cation
and a
basic anion,
the acidity or basicity
of the salt solution
depends on the relative
strengths of the acid and base.
108
Salts that yield acidic or basic solutions
If the acid cation is “stronger”
than the base anion
the solution will be acidic.
If the base anion is “stronger”
than the acid cation
the solution will be basic.
109
Salts that yield acidic or basic solutions
Ka > Kb
the acid is better at
reacting with water than the base
and the solution is acidic.
Ka < Kb
the base is better at
reacting with water than the acid
and the solution is basic.
Ka  Kb
the solution is close to neutral.
110
Problem
Predict whether the following salt
solution is basic, neutral or acidic, and
calculate the pH.
0.25 molL-1 NH4Br
Kb of NH3 is 1.8 x 10-5
Answer: The solution is acidic and pH
is 4.93
111
Problem
Classify each of the following salts
as acidic, basic, or neutral:
a) KBr
b) NaNO2
c) NH4Br
d) NH4F
neutral
basic
acidic
Ka for HF = 6.6 x 10-4
Kb for NH3 = 1.8 x 10-5
acidic
112
Factors that affect acid strength
There are several factors that can affect
acid strength, and the importance of the
factors can be variable. However, two
trends are notable.
1) Bond strength – The strength of the
bond between the acidic proton and the
rest of the molecule will have an effect on
acidity. The weaker the bond, the stronger
the acid will generally be.
113
Factors that affect acid strength
2) Bond polarity – The polarity of a bond
is the distribution of the electrons
between the two bonded atoms.
A highly polar bond between an acidic
hydrogen and another atom tends to make
it more easy for the proton to leave the
molecule than would happen for a nonpolar bond.
114
Polarity versus strength
In the binary acids the bond strength is the more
important factor. Bond strength tends to decrease
down a column in the periodic table. HF is the weakest
binary acid even though it has the most polar bond
because it has the strongest bond.
115
Polarity versus strength
For acids of elements in the same row, the bond
strengths tend to be similar, and so the polarity of
the bond plays the greater role in determining acid
strength.
116
Polarity versus strength
Combining the decrease of bond strength down a
column and increase of bond polarity across a row
we find the strongest acids
tend to be those of the
elements in the bottom right
of the periodic table.
117
Oxoacids and acid strength
For oxoacids, acid strength tends to
increase with the electronegativity of
the central atom, and with an
increase in the central atom
oxidation number (which generally
increases with the number of other
atoms bonded to the central atom).
118
Oxoacids and acid strength
Here we see three oxoacids with different central
atoms.
Oxoacid strength increases with
electronegativity of X
119
Oxoacids and acid strength
Oxoacids with the same central atom X will be
strongest when many other atoms are bonded
to X (the oxidation number of X increases.)
120
Organic compounds and acid strength
The acid strength of organic compounds can be
rationalized by the bond strength between the
proton and the atoms it’s bonded to (weaker bond
- stronger acid) but more correctly it is
determined by the stability of the conjugate
base. Any factor that tends to stabilize the
conjugate base increases organic acid strength.
121
Organic compounds and acid strength
122
Organic compounds and acid strength
chlororoacetic acid Ka = 1.4 x 10-3
fluoroacetic acid Ka = 2.7 x 10-3
123
Amines and base strength
We require a lone pair of electrons to have a BL base, so
any factor that tends to reduce the availability of
the lone pair will weaken the base
while any factor that tends to increase the
availability of the lone pair will strengthen the
base.
124
Amines and base strength
Alkyl groups are slightly electron donating,
so secondary and tertiary amines tend to
be slightly stronger bases than ammonia
and primary amines.
125
Amines and base strength
Factors that stablize the structure of
amines will decrease the base strength
due to reduced electron availability.
126
Amines and base strength
Factors that stablize the structure of
amines will decrease the base strength
due to reduced electron availability.
127
Lewis acids and bases
A Lewis acid is an electron pair acceptor,
while a Lewis base is an electron pair
donor.
These definitions are more general than the
BL definitions because protons aren’t
involved which means there exist
Lewis acids that are not BrønstedLowry acids.
128
Lewis acids and bases
We’ve seen that all Brønsted-Lowry
bases must all have at least one lone
pair of electrons, so
any Brønsted-Lowry base
must also be a Lewis base,
and any Lewis base
must also be a Brønsted-Lowry base!
129
In general LA + :LB  LA-LB
130
In general LA + :LB  LA-LB
131
Coordination compounds
We’ve already seen
that coordination
compounds have
complex ions that are
formed from a central
metal ion (a Lewis
acid) surrounded by
ligands (Lewis
bases)
132
Acidic solutions of metal ions
Small and/or highly charged metal ions,
like Al3+, Be2+, and Li+
form acidic solutions because they form
complex ions with water. The protons of
the water ligands see less electron density
than in free water, and so the O-H bond is
weaker, leading to increased acidity!
133
Acidic solutions of metal ions
There is a second benefit as well because
the complex ion has a smaller charge in a
larger volume….
Ka = 1.74 x 10-5 – about the same as acetic acid!
134