Arrhenius theory of acids and bases
Chapter 16
The Arrhenius concept: acids dissociate in
water (aqueous solution) to produce
hydrogen ions H+, and bases dissociate in
water to give hydroxide ions, OH-.
Acids and Bases
Arrhenius acid: HA (aq) → H+ (aq) + A- (aq)
Arrhenius base: MOH (aq) → M+ (aq) + OH- (aq)
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1051.php
2
Neutralization reactions
Aqueous salts
Arrhenius acids and bases react with each
other to form water and aqueous salts in
neutralization reactions.
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.
H+ (aq) + A- (aq) + M+ (aq) + OH- (aq) →
H2O (l) + M+ (aq) + A- (aq)
The net ionic equation is
If we evaporate all the water, we are left
with an ionic solid called a salt e.g. NaCl
H+ (aq) + OH- (aq) → H2O (l)
3
4
1
Brønsted-Lowry theory of acids
and bases
Arrhenius theory of acids and bases
Many substances that do not contain OHact like bases!
Many substances (like NH3) that do not
contain OH- act like bases in water!
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.
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.
5
Proton transfer reactions
6
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.
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
8
2
BL base strength
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 =
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!
-
4
[NH ] [OH ]
+
[NH 3 ]
Kb =
= 1.8 x 10 -5 at 25o C
-
4
[NH 3 ]
= 1.8 x 10 -5 at 25o C
9
Water as a base in BL reactions
10
BL acid strength
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.
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
Ka =
[CH COO ] [H O ]
−
3
+
3
[CH 3COOH]
= 1.8 x 10-5 at 25o C
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12
3
BL acid strength
Strong BL acids and bases
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!
Ka =
[CH COO ] [H O ]
−
3
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.
+
3
[CH 3COOH]
= 1.8 x 10-5 at 25o C
Ka =
[Cl ] [H O ]
−
+
3
[HCl]
≈ 1 x 106 at 25o C
13
Hydrated protons and hydronium ions
14
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.
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!
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16
4
Amphiprotic substances
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.
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.
17
Problem
18
Problem
Write a balanced equation for the
dissociation of each of the following
Brønsted-Lowry acids in water:
What is the conjugate acid of each of
the following Brønsted- Lowry bases?
a) H2SO4
b) HSO4c) H3O+
d) NH4+
a) HCO3b) CO32c) OHd) H2PO419
20
5
Problem
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.
For each of the following reactions, identify
the acids and bases in both the forward
and reverse directions:
HF + H 2 O
Acid
⎯⎯→
←
⎯⎯
Base
Base
HSO 4 + NH
-
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-
−
Acid
H 2 CO 3 + OH −
F − + H 3O +
⎯⎯→
3 ←
⎯⎯
Base
C 2 H 3 O 2 + HCl
-
Base
Acid
SO 4
2−
+ NH 4
Base
⎯⎯→
←
⎯⎯
+
Acid
HC 2 H 3 O 2 + Cl −
Acid
Acid
Base
21
A 2nd look at acid and base strength
22
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.
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.
←
−
+
HA
⎯→
A
{ + H
{ + H
2O ⎯
3O
{
1
2
3
very weak base
strong acid
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.
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
24
6
Strong bases in water
Relationship of acid/base strengths
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.
The stronger the acid, the weaker
its conjugate base.
←
−
A
O ⎯⎯→
HA
+ OH
2
{ + H
{
{
{
strong base
very weak acid
weaker base
The stronger the base, the weaker
its conjugate 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-.
We’ll look at this a bit more in depth
later.
weak acid
25
26
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.
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28
7
Differentiating strong acids
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:
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.
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-]
29
30
Autoionization of water
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)
31
32
8
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
Problem
34
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?
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: Both are 2.3 x 10-7 mol⋅L-1
Answer: [H3O+] = 2.0 x 10-9. Solution is basic.
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36
9
The pH scale
pH and acidity
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+].
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,
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 > 7 is basic
pH < 7 is acidic
pH = 7 is neutral
pH = - log [H3O+]
37
38
Kw = 1.0 x 10-14 = [H3O+] [OH-]
pOH and acidity
14.00 = pH + pOH at 25 °C
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
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40
10
Problem
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
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
Answers: a) pH = 8.19 b) pH = 4.22
b) [H3O+] = 1.6 x 10-3 M and [OH-] = 6.3 x 10-12 M
41
Problem
42
Strong acids and strong bases
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?
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!).
Answer: You would add 3.0 L of water
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.
43
44
11
pH and strong bases
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) Æ 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.
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)
45
pH and strong bases
46
Be careful!
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-.
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?
O2- (aq) + H2O (l) Æ 2 OH- (aq), so
H2O
MO (s) Æ M2+ (aq) + 2 OH- (aq)
47
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12
Common strong acids and bases
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
49
Problem
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?
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 gas constant R = 0.08206 L⋅atm⋅K-1⋅mol-1
and 760 mmHg = 1 atm exactly!
The molar mass of Mg(OH)2
is 58.3197 g⋅mol-1
Answer: pH = 1.467
Answer: pH = 11.52
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13
Problem
pKa and pKb
We can define
pKa = - log Ka and pKb = - log Kb
exactly like pH = - log [H3O+]
Calculate the pH of an aqueous solution
that is 3.00% KOH by mass and has a
density of 1.0242 g⋅mL-1.
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).
The molar mass of KOH is 56.1056 g⋅mol-1.
As Ka ↑ (or pKa ↓) or as Kb ↑ (or pKb ↓)
acid strength or base strength increases.
Answer: pH = 13.74
53
pKa and pKb
54
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
55
56
14
Identifying weak acids
Carboxylic acids
Generally there are three categories of
weak acids:
Carboxylic acids contain
the -COOH (carboxyl)
group. The hydrogen of this
group is the proton that is
donated.
carboxylic acids
oxoacids
miscellaneous acids
57
Carboxylic acids
58
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
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60
15
Strong oxoacids
Some weak oxoacids
Nitric acid –
HNO3
Sulfuric acid –
H2SO4
Perchloric acid –
HClO4
Nitrous acid –
HNO2
Phosphoricic acid –
H3PO4
Chlorous acid –
HClO2
61
Miscellaneous weak acids
62
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).
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
64
16
Identifying weak bases
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.
65
Equilibrium in solutions of weak acids and bases
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
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
17
Equilibrium in solutions of weak acids and bases
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.
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)
69
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
K a = 4.9 x 10−10 =
+
H2O (l)
N/A
N/A
N/A
'
H3O+ (aq) +
0.0
+x
+x
70
Equilibrium in solutions of weak acids and bases
CN- (aq)
0.0
+x
+x
(x)(x)
[H 3O + ][CN − ]
so 4.9 x 10−10 =
(0.10 − x)
[HCN]
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
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
18
Equilibrium in solutions of weak acids and bases
Equilibrium in solutions of weak acids and bases
We also assumed that [H3O+] ≈ [H3O+]
(principal reaction) so we must check this.
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.
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
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)!
73
74
Equilibrium in solutions of weak acids and bases
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
[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
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19
Problem
Problem a)
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
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:
CH3COOH (aq) +
1.00
-x
1.00 – x
K a = 1.8 x 10 −5 =
H2O (l)
N/A
N/A
N/A
'
H3O+ (aq) + CH3COO- (aq)
0.0
0.0
+x
+x
+x
+x
[H 3O + ][CH 3COO − ]
(x)(x)
so 1.8 x 10 −5 =
[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
-3
-1
x = ±4.2 x 10 mol⋅L (must be +ve value since x = [H3O+])
CHECK ASSUMPTION
mol⋅L-1
a) 1.00
CH3COOH
b) 0.0100 mol⋅L-1 CH3COOH
77
Problem a)
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
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)!
79
80
20
Problem a)
Problem b)
pH = - log [H3O+]
pH = - log 4.2 x 10-3
pH = 2.38
K a = 1.8 x 10 −5 =
[H 3O + ][CH 3COO − ]
(x)(x)
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
81
Problem b)
K a = 1.8 x 10 −5 =
82
Problem b)
[H 3O + ][CH 3COO − ]
(x)(x)
so 1.8 x 10 −5 =
[CH 3COOH]
(0.00100 − x)
− (−1.8 x 10 −5 ) ± ( −1.8 x 10 −5 ) 2 − 4(−1)(1.8 x 10 −8 )
2(−1)
x=
− b ± b 2 − 4ac
2a
x=
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
or x =
−2
−2
x=
1.8 x 10 −5 − 2.68 x 10 − 4
1.8 x 10 −5 + 2.68 x 10
or x =
−2
−2
x=
2.87 x 10 − 4
− 1.7 9 x 10 −5
or x =
−2
−2
so x =
−4
We can not assume that x << 0.00100 and so
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
[1.8 x 10 −5 ](0.00100 − x) − x 2 = 0
[1.8 x 10 −8 ] − [1.8 x 10 −5 ]x − x 2 = 0
83
84
21
Problem
Percent ionization of weak acids
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
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%
85
Percent ionization
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
We saw on slide 84 that an acetic acid
solution with initial concentration of
0.0010 mol⋅L-1 at equilibrium had
[H3O+] = 4.2 x 10-3 mol⋅L-1
[H3O+] = 1.25 x 10-4 mol⋅L-1
%ionized = [H3O+]eqm / [HA]initial x 100%
%ionized = [H3O+]eqm / [HA]initial x 100%
%ionized = 4.2 x 10-3 mol⋅L-1 / 1.00 mol⋅L-1 x 100%
%ionized = 1.25 x 10-4 mol⋅L-1 / 0.0010 mol⋅L-1 x 100%
%ionized = 0.42%
%ionized = 12.5%
87
88
22
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)
Ka =
[H O ][HCO ] = 4.4 x 10
−
+
3
[H 2CO 3 ]
3
−7
HCO3- (aq) + H2O (l) ' H3O+ (aq) + CO32- (aq)
Ka =
[H O ][CO ] = 4.7 x 10
2−
+
[HCO ]
3
−
3
−11
3
89
Polyprotic acids
90
Problem
Ka1 > Ka2 > Ka3 is always true!
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
91
92
23
Problem
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
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
+
HSO3- (aq)
0.0
+x
+x
− b ± b 2 − 4ac
2a
x=
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 )
or x =
−2
−2
x=
1.3 x 10 − 2 + 7.33 x10 − 2
1.3 x 10 − 2 − 7.33 x 10 −2
or x =
−2
−2
x=
8.63 x 10 −2
− 6.03 x 10 − 2
or x =
−2
−2
−
[H 3O + ][HSO 3 ]
(x)(x)
so 1.3 x 10 − 2 =
[H 2SO 3 ]
(0.10 − x)
− (−1.3 x 10 −2 ) ± (−1.3x10 −2 ) 2 − 4( −1)(1.3x10 −3 )
2(−1)
x=
so x =
so x = −4.3 x 10 − 2 mol/L or x = 3.0 x 10 − 2 mol/L
93
94
Problem
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
K a2 = 6.3 x 10 −8 =
2−
[H 3O + ][SO3 ]
−
[HSO3 ]
so 6.3 x 10−8 =
(3.2x10−2 )(x)
(3.2x10−2 )
x = [SO32-] = [H3O+]sub = 6.3 x 10-8 mol/L
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
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
24
Problem
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
pH = - log [H3O+]
pH = - log 3.0 x 10-2
pH = 1.52
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.
[OH-] = Kw / [H3O+]
[OH-] = 1.0 x 10-14 / 3.0 x 10-2
[OH-] = 3.3 x 10-13 mol⋅L-1
HA (aq) + H2O (l) ' H3O+ (aq) + A- (aq) K a =
+
−
3
[HA]
[OH ][HA]
=
[A ]
−
A-
(aq) + H2O (l) '
OH-
(aq) + HA (aq) K b
97
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
[H O+ ][A − ] [OH− ][HA] = [H O+ ][OH− ] = K = 1.0x10−14
Ka x Kb = 3
3
w
[A − ]
[HA]
[H O ][A ]
−
98
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.
at 25 o C
99
Since Ka x Kb = Kw
then Ka = Kw / Kb
and Kb = Kw / Ka
100
25
Ions as acids and bases
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-.
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!
Answers: a) Ka = 6.3 x 10-12 b) Kb = 3.4 x 10-7
101
Ions as acids and bases
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…
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-…
H2O (l) + Na+ (aq) ' no reaction!
OCl- (aq) + H2O (l) ' OH- (aq) + HOCl (aq)
103
104
26
Ions as acids and bases
Salts that yield neutral solutions
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+ …
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.
H2O (l) + NH4+ (aq) ' H3O+ (aq) + NH3 (aq)
NaCl, KNO3, Ca(ClO4)2, MgI2, etc.
105
Salts that yield acidic or basic solutions
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
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
27
Salts that yield acidic or basic solutions
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.
If the acid cation is “stronger”
than the base anion
the solution will be acidic.
Ka < Kb
If the base anion is “stronger”
than the acid cation
the solution will be basic.
the base is better at
reacting with water than the acid
and the solution is basic.
Ka ≈ Kb
the solution is close to neutral.
109
Problem
110
Problem
Classify each of the following salts
as acidic, basic, or neutral:
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
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
28
Factors that affect acid strength
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.
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.
113
114
Polarity versus strength
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.
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.
115
116
29
Oxoacids and acid strength
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.
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).
117
Oxoacids and acid strength
118
Oxoacids and acid strength
Here we see three oxoacids with different central
atoms.
Oxoacid strength increases with
electronegativity of X
119
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
30
Organic compounds and acid strength
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
122
Amines and base strength
Organic compounds and acid 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.
chlororoacetic acid Ka = 1.4 x 10-3
fluoroacetic acid Ka = 2.7 x 10-3
123
124
31
Amines and base strength
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.
Factors that stablize the structure of
amines will decrease the base strength
due to reduced electron availability.
125
Amines and base strength
126
Lewis acids and bases
A Lewis acid is an electron pair acceptor,
while a Lewis base is an electron pair
donor.
Factors that stablize the structure of
amines will decrease the base strength
due to reduced electron availability.
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.
127
128
32
Lewis acids and bases
In general LA + :LB Æ LA-LB
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
130
Coordination compounds
In general LA + :LB Æ LA-LB
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)
131
132
33
Acidic solutions of metal ions
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!
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!
133
134
34