Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 17: Acid-Base Equilibria I
ACID-BASE EQUILIBRIA I
Chapt. 17:p.1
酸鹽基平衡 I
I. CONCEPT OF ACID/BASE
(A)
Acids typically have the following properties:
1. They are non-e1ectrolyte when pure but become electrolytes when dissolves in water.
2. Aqueous solutions of acids change the colour of indicators.
3. Aqueous solutions of acids undergo the following
characteristic reactions:
(a) they react with metal oxides to produce salts and water;
(b) they react with some reactive metals to produce hydrogen;
(c) they react with carbonates to produce salts, carbon dioxide and water.
Compounds which neutralize acids are called bases. A base which is soluble in water is called
alkali.
(B)
Ostwald-Arrhenius 阿列紐斯酸基鹽學說: Theory of Electrolytic Dissociation
An acid is a substance which dissociates in water to give hydrogen ions, H+.
A base is a substance which dissociates in water to give hydroixde ions, OH-..
Although this theory was still adequate for some purposes, it had its limitations. For example, it
did not account for the basic properties of some organic compounds such as triethylamine.
N(C2H5)3 which did not contain the group OH-.
(C) Bronsted-Lowry theory 布侖斯特-路里酸鹽基學說
An acid is a substance which donates protons.
A base is a substance which accepts protons.
The relationship between an acid and a base can be expressed as follow:
Acid  Base + H+
This equation does not represent an actual reaction in solution
since the proton, H+ , cannot exist in solution. The acid and base which are related in this way,
by the exchange of a proton, are called conjugate acid-base pair 共軛酸鹽基偶. Since
Acid 1

Base 2 + H
+
Conjugate base 1 + H+
and
 Conjugate acid 2
A reaction between an acid and a base is presented as follow:
Acid 1 + Base 2 
Conjugated base 1 + Conjugate acid 2
Note
<1> Acid 1 is transformed into its conjugate base and Base 2 is transformed into its conjugate
acid.
<2> No substance can act as an acid in solution unless a base is present to accept a proton. That
is
All reactions of acids or bases in solution are acid-base reactions.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 17: Acid-Base Equilibria I
Chapt. 17:p.2
Examples : <a> Bronsted-Lowry acids
HCl(g)
H2SO4(aq)
CH3COOH(aq)
____________
+
+
+
H2O(l)  H3O+(aq) + Cl-(aq)
H2O(l)  H3O+(aq) + SO42-(aq)
H2O(l)  H3O+(aq) + CH3COO-(aq)
_______
____________ ______________
<b> Bronsted-Lowry bases
NH3(g)
+ H2O(l)
RNH2 (l)
+ H2O (l)
HSO4 (aq)
+ H3O+(aq)
CH3COO- (aq) + H2O (1)
____________
_______
 NH4+(aq) + OH-(aq)
 RNH3+ (aq)
+ OH- (aq)
 H2SO4(aq)
+ H2O(l)
 CH3COOH ( aq) +
OH- (aq)
____________
______________
Note
<1> In <a>. water acts as a proton accepter. a base.
In <b>. water acts as a proton donor, an acid.
Water is described as an amphoteric solvent (substance).
The HSO4- ion is an amphoteric species.
<2> Acid base equilibria (reaction) are very often linked to redox equilibria. For example. the
acid base system
HCl(g)
+ H2O(l) 
H3O+(aq) + Cl-(aq)
can be considered as making up of two competing equilibria for protons
HCl(g)
H3O+(aq)
 Cl-(aq)
 H2O(l)
+ H+(aq)
+ H+(aq)
Since HCl(g) loses protons more readily than H3O+(aq). and
H2O(l) accepts protons more readily than Cl-(aq) ,the final position of the equilibrium,
as determine by this relative tendencies of the acid/base to lose/gain protons will be
HCl(g) + H2O(l) 
shift to product side.
H3O+(aq) +
Cl-(aq)
This is similar to the competition for electrons in the redox system.
Zn(s)

Zn2+(aq) + 2e
Cu(s)

Cu2+(aq) + 2e
reductant
oxidant
The final position of the equilibrium
Zn(s)
+
reductant
Cu (aq) 
oxidant
Zn (aq) +
oxidant
Cu(s)
reductant
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 17: Acid-Base Equilibria I
Chapt. 17:p.3
is determined by the relative tendencies of the two redox systems to from ions in solution.
(i.e. Since the Zn system has a higher reducing strength, or greater tendency to give up
electrons, the final position of the equilibrium will tend to shift in the forward direction.
II.
DISSOCIATION OF WATER
Ionic product of water , Kw
The electrical conductivity of even pure water never falls to exactly zero. This is due to the
self-ionization of water.
2H2O(l)
H2O(l)

 H3O+(aq)
+ OH-(aq) or simply.
H+(aq) + OH-(aq)
applying the equilibrium law to this equilibrium, the following expression can be obtained;
Kw
[ H  ][OH  ]

[ H 2 O]
Since the value of [H2O] is always approximately 55 mol dm3 The following expression can be
obtained;
Kw = K[H2O] = [H+][OH-]
Kw = [H+][OH-]
Kw is called the ionic product of water. Its unit is mol2 dm-6 . It is a constant which changes only
with temperature. At 250C its value is 1.0 x 10-14 mol2 dm-6.
III.
pH AND ITS MEASUREMENTS
Since the concentration of hydrogen ion in a solution is usually very small, it is quite
inconvenient to express it directly. a convenient method is to express it in terms of the index or
logarithm f the concentration.
The pH of a solution is the logarithm, to the base 10. of the : / reciprocal of hydrogen ion
concentration of the solution.
1
pH = log(
) = - log [H+]

[H ]
The pH of a neutral solution can be calculated directly from the ionic product of water.
Kw = [H+][OH-] = 1.0 x 10-14 mol2 dm-6
Since [H+] = [OH-] ,
[H+] = 1.0 x 10-7 mol dm-3
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 17: Acid-Base Equilibria I
and pH =
7
for neutral solutions
pH <
pH >
7
7
for acid solutions [H+] > 10-7 mol dm-3
for alkaline solutions
[H+] < 10-7 mol dm-3
Chapt. 17:p.4
The relationship between pH and [H+] , [H3O+] and [OH-] is illustrated in the following table:
Exercise 1:
What is the pH of (a) 0.1 M HCl
(b) 0.01 M NaOH
(A)
Limitations of pH value
Consider the pH of (a) 1M HCl and (b) 2M NaOH
(a) HCl(aq)  H+(aq) Cl-(aq)
[H+] = 1M
pH = - log 1
=0
(b) NaOH(aq)  Na+(aq) + OH-(aq)
[OH-] = 2M
[H+] =
pH =
10 14
2
14.3
Note
<1> It is obvious that the pH values can be out of the range 1-14.
<2> The pH values theoretically has no limitation. However, if the acids and bases are too
concentrated, they may not be completely ionized in spite of the fact that they are strong.
10M HC1 solution may not have higher hydrogen ion concentration than a 5 M HC1
solution because of the incomplete ionization of the 10M HC1.
For strong acids and alkalis,
concentration  6M
concentration  6M
complete ionization
incomplete ionization
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 17: Acid-Base Equilibria I
Chapt. 17:p.5
Exercise 2
Discuss whether the following statement is correct or not
“ The pH value of a 10-8 mol dm3 aqueous solution of HNO3 is 8”
(B)
Measurement of pH
The pH of a solution is usually measured by using a pH meter. Such pH meter consists of an
electrochemical cell with
1. a reference electrode whose electrode potential is constant.
2. an electrode response to the hydrogen ion concentrations. e.g. glass electrode.
The e.m.f. of these cell is proportional to the pH value of the solution in which it is measured.
Ecell  pH
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 17: Acid-Base Equilibria I
Chapt. 17:p.6
STRONG AND WEAK ACIDS/BASES
The strength of acids and bases should be considered in terms of their dilution instead of their
concentration.
Strong acids:
The dilution of the acid solution would cause a proportionate increase in pH. i.e. The acid is
completely dissociated at a certain range of concentration. e.g.
HCl(aq)  H+(aq) + Cl-(aq)
1M
0.1M
pH
=0
pH =1
0.01M
0.001M
pH
pH
=2
=3
Weak acids:
The increase in pH is not proportionate to dilute The acid solution is not completely dissociated and
the degree of dissociation increases as the concentration decreases.
Consider a weak acid HA, an equilibrium is set up between undissociated molecules HA and the ions
H+ and AHA(aq)  H+(aq) + A-(aq)
The equilibrium constant Ka (acid dissociation constant) is given by
[ H  ][ A  ]
Ka =
[ HA]
3
If one mole of acid is dissolved in V dm of solution, the amounts and concentration of each species at
equilibrium are as follows:
HA(aq)
 H+(aq) +
A-(aq)
amounts/mol
concentration/moldm3
1-
(1-)/V

/V

/V
(= degree of dissociation)
Putting these concentration into the expression for Ka gives
( / V ) 2
2
Ka 

(1   ) / V (1   ) / V
since V=volume containing 1 mol of solute, 1/V=c and
Ka 
 2c
1
For many weak acids, a is so small that the error involved in putting (1 - ) = 1 is negligible.
Then
Ka  c 2
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 17: Acid-Base Equilibria I
Chapt. 17:p.7
<1> From (*), the degree of dissociation increases as the concentration decreases.
<2> The degree of dissociation can be found by measuring the molar conductivities of the
solution at a known concentration and also at infinite dilution.
<3> The acid dissociation constant, Ka , is a measure of the strength of an acid. For an acid
solution such as HCl which is virtually fully dissociated in aqueous solution, its value is
extremely large. On the other hand, for weak acids, values for Ka can be extremely small.
It is often more convenient to compare the strengths of acids using pKa values.
pKa = -log Ka
Most acids gives a range of values from 1 to 14.
The higher the value of Ka, the lower the value of pKa, and the stronger is the acid.
Exercise 3
Conductivity experiments on aqueous ethanoic acid, CH3COOH, with concentration of
0.005 mol dm3 showed that the degree of dissociation at this concentration and a temperature of
25°C was 0.057.Calculate the values for Ka and pKa.
ANSWER
Exercise 4
Calculate the pH of a 1.0 x10-2 mol dm-3 solution of butanoic acid, CH3CH2CH2COOH for which
Ka = 1.51x10-5 mol dm3.
ANSWER
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 17: Acid-Base Equilibria I
Chapt. 17:p.8
Exercise 5
Calculate the pH of a 1.00x10-2 mol dm3 solution of ammonia, for which Kb = l.9x10-5 mol dm-3.
(Kw = l.0 x 10-14 mol2 dm-6 )
ANSWER
Exercise 6
Calculate the Ka of a 0.01 M methanoic acid solution at pH =2.90.
ANSWER
Exercise 7
A solution of dimethylamine, (CH3)2NH of concentration 1.0 x 10-2 mol dm-3 has a pH of 7.64 at 25 0C.
Calculate
(a) the dissociation constant of the base, Kb and
(b) the degree of dissociation, .
ANSWER
Exercise 8
Explain the following phenomena:
“Refrigerated lemon juices tastes less sour than un-refrigerated juice of the concentration.’
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 17: Acid-Base Equilibria I
V.
Chapt. 17:p.9
BUFFERS
(A) Principle
A buffer solution is one which will resist changes in pH due to the addition of small
amounts of acids and alkali.
A buffer solution can be made up by
1. a weak acid and its salts e.g. _________________________________________
(constant pH values of between 4 and 7)
2.
a weak base and its salt e.g. ________________________________________
(constant pH values of between 7 and 10)
Principle
The pH of a buffer solution does not change much on addition of acid or alkali because the ions
present in the solution will convert the acid or alkali into weak acid or alkali which is only
slightly ionized.
Example
<1> Buffer solution of CH3COOH and CH3COO-Na+
(i)
On addition of acid (H+), the CH3COO- ion will absorb the H+ to form CH3COOH
(ii)
On addition of alkali (OH- ), the CH3COOH will absorb OH- to form CH3COO-.
Thus, the pH of the buffers is roughly restored.
<2> Buffer solution of NH3 and NH4Cl
(i)
On addition of H+ (removed by NH3)
(ii)
On addition of OH- (removed by NH4+)
(B) Calculation of the pH of a buffer solution
The pH of a buffer solution consisting of a weak acid HA and its salt with a strong base is
calculated from the equation:
HA(aq)  H+(aq) + A-(aq)
Ka =
[ H  ][ A  ]
[ HA]
Rearranging,
[H+] = Ka
[ HA ]
[ A ]
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 17: Acid-Base Equilibria I
pH = pKa + log
Chapt. 17:p.10
[ A ]
[ HA]
It can be seen that
<1> Since the salt is completely ionized and the acid is only slightly ionized, it can be assumed that
all the anions come from the salt, and put
[A] = [salt]
[HA] = [acid]
[ salt ]
pH = pKa + log
[ acid ]
An effective buffering action is obtained at pH values fairly close to pKa.
<2>
The pH of a buffer is dependent on the ratio of the concentration of acid and base (salt), but not
on the actual values of these concentrations.
<3>
When [A-] = [HA] at equilibrium, the above expression would be reduced to pH = pKa. It
means that the Ka of an acid can be found by determining the pH value of a solution of the acid
which has been halt neutralized by a strong base.
Moreover. to prepare a buffer of a particular pH, it is only necessary to select a weak acid or
base with Ka=[H+] , and mixing equal concentrations of the acid and its salt together.
Exercise 9
Three solutions contain ethanoic acid (Ka = 1.8x10-5 mol dm3) at a concentration 0.10 mol dm-3 and
sodium ethanoate at a concentration
(a) 0.10 mol dm-3
(b) 0.20 mol dm-3 (c) 0.50 mol dm-3
ANSWER
Exercise 10
(a) How much sodium ethanoate must be dissolved in 1 dm3 of ethanoic acid with a concentration of
0.01 mol dm3 to produce a buffer solution of pH 5.(Ka of ethanoic acid 1.74x10-5 mol dm-3)
(b) How would be the pH of this buffer solution change if 1 cm3 of NaOH solution with a
concentration of 1M were added to 1 dm3 of the buffer.
ANSWER
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 17: Acid-Base Equilibria I
Chapt. 17:p.11
Exercise 11
Calculate the mass of amrnonium chloride which has to be added to 100 cm3 of 0.1 mol dm3 ammonia
solution to make its pH equal to 9.
(Given Kb, for ammonia solution is 1.9x10-5 mol dm-3)
ANSWER
(C)
(1)
Uses of buffers
In industrial processes
• electroplating and manufacture of dyes
• photographic materials and leather
(2)
In chemical analysis
•
(3)
calibration of pH meters
In biological system and other system
Such systems depend on buffer action to preserve a constant pH.
e.g.
The pH of human blood is maintained between 7.35 and 7.45 even though the
concentration of carbon dioxide and thus carbonic acid in the blood varies greatly. The
buffer in blood consists of a mixture of phosphate. hydrogencarbonate and proteins.
e.g.
The pH of tears is also maintained at 7.4 by use of buffers consisting of proteins.
e.g.
The use of buffer in bacteriological research to maintain the pH of culture media used
for the growth of bacteria.