15 Equilibria
Topics
15.1
.
Le Chatelier's principle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
Equilibrium constants. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
Acid±base equilibria. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
pH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
Buffers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
Acid±base titrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
Acid±base indicators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Introduction
Equations for chemical reactions are often written with a single arrow showing reactants going to products, for example, equation 15.1. The assumption
is that the reaction goes to completion, and at the end of the reaction, no
reactants remain.
Mg…s† ‡ 2HCl…aq†
"
MgCl2 …aq† ‡ H2 …g†
…15:1†
However, many reactions do not reach completion. For example, in the reaction of acetic acid with ethanol (equation 15.2), the reaction proceeds to a
certain point and gives a mixture of reactants and products. Such a system
has both forward and back reactions occurring concurrently and attains an
equilibrium position.
"
Carboxylic acids and esters:
see Chapter 32
CH3 CO2 H ‡ CH3 CH2 OH „ CH3 CO2 CH2 CH3 ‡ H2 O
Acetic acid
Ethanol
Ethyl acetate
…15:2†
Water
The equilibrium is denoted by the equilibrium sign „ instead of a full arrow
and the position of the equilibrium, i.e. to what extent products or reactants
predominate, is quanti®ed by the equilibrium constant, K. All reactions can
be considered as equilibria, but in some cases, the forward reaction is so
dominant that the back reaction can be ignored.
In this chapter, we consider di€erent types of equilibria, and show how to
calculate and use values of K. Equilibria are dynamic systems with both the
forward and back reaction in operation; the rate of the forward reaction
equals the rate of the back reaction. In the next section, we discuss how
changes in external conditions a€ect the position of an equilibrium.
A point that often causes confusion is the dual use of square brackets to
mean `concentration of ' and to signify a polynuclear ion such as ‰NH4 Š‡ .
In this chapter, therefore, we have, mainly, abandoned the latter usage of
square brackets for ions such as OH , H3 O‡ , NH4 ‡ and CH3 CO2 .
441
Le Chatelier's principle
15.2
Le Chatelier's principle
states that when an external
change is made to a system
in equilibrium, the system
will respond so as to oppose
the change.
Le Chatelier's principle
Consider the reaction of dichromate ion with hydroxide ion (equation 15.3).
‰Cr2 O7 Š2 …aq† ‡ 2OH …aq† „ 2‰CrO4 Š2 …aq† ‡ H2 O…l†
orange
…15:3†
yellow
On reaction with OH , orange ‰Cr2 O7 Š2 (dichromate) is converted into yellow
‰CrO4 Š2 (chromate), but in the presence of acid, ‰CrO4 Š2 is converted to
‰Cr2 O7 Š2 . By altering the concentration of acid or alkali, the equilibrium can
be shifted towards the left- or right-hand side. This is an example of Le
Chatelier's principle which states that when an external change is made to a
system in equilibrium, the system responds so as to oppose the change. In reaction
15.3, if OH is added to the equilibrium mixture, the equilibrium moves towards
the right-hand side to consume the excess alkali. Conversely, if acid is added, it
neutralizes some OH (equation 15.4), and so equilibrium 15.3 moves towards
the left-hand side to produce more OH , thereby restoring equilibrium.
OH …aq† ‡ H3 O‡ …aq†
"
2H2 O…l†
…15:4†
Gaseous equilibria: changes in pressure
In an ideal gaseous system at constant volume and temperature, the pressure
is proportional to the number of moles of gas present (equation 15.5).
Pˆ
nRT
V
…15:5†
By Le Chatelier's principle, an increase in external pressure is opposed by a
reduction in the number of moles of gas; conversely, the equilibrium
responds to a decrease in pressure by shifting in the direction that increases
the number of moles of gas. Consider reaction 15.6.
2CO…g† ‡ O2 …g† „ 2CO2 …g†
…15:6†
The forward reaction involves three moles of gaseous reactants going to two
moles of gaseous products; in the back reaction, the number of moles of gas
increases. In order to encourage the forward reaction, the external pressure
could be increased. Equilibrium 15.6 responds to this change by moving
towards the right-hand side, decreasing the number of moles of gas and
therefore decreasing the pressure. The back reaction is favoured by a
decrease in the external pressure.
Exothermic and endothermic reactions: changes in temperature
Changes to the external temperature of an equilibrium cause it to shift in a
direction that opposes the temperature change. For example, an increase
in the external temperature encourages an endothermic reaction to occur
because this takes in heat and so lowers the external temperature. Consider
equilibrium 15.7: the formation of NH3 from its constituent elements in
their standard states. The standard enthalpy of formation of NH3 (g) at
298 K is 45.9 kJ mol 1 and this refers to the forward reaction and to the
formation of one mole of NH3 .
1
2 N2 …g†
‡ 32 H2 …g† „ NH3 …g†
f H o …NH3 ;g;298 K† ˆ
45:9 kJ mol
1
…15:7†
442
CHAPTER 15 . Equilibria
The data reveal that the forward reaction is exothermic, and therefore the
back reaction is endothermic: if 45.9 kJ are liberated when one mole of
NH3 forms from N2 and H2 , then 45.9 kJ are required to decompose one
mole of NH3 into N2 and H2 . If the external temperature is lowered, the
equilibrium will move towards the right-hand side because the liberation
of heat opposes the external change. If the external temperature is increased,
the back reaction is favoured and yields of NH3 are reduced. The e€ects of
changes of temperature and pressure on equilibrium 15.7 have important
industrial consequences on the manufacture of NH3 by the Haber process
and this is discussed in Box 20.2.
15.3
Equilibrium constants
The discussion so far in this chapter has been qualitative. Quantitative
information about the position of an equilibrium can be obtained from the
equilibrium constant, K, which tells us to what extent the products dominate
over the reactants, or vice versa. The thermodynamic equilibrium constant is
de®ned in terms of the activities of the products and reactants, and before
proceeding with the discussion of equilibria, we must de®ne what is meant
by activity.
Activities
The activity, ai , of any pure
substance, i, in its standard
state is de®ned as 1;
activities are dimensionless
(no units).
In dealing with concentrations of solution species, it is common to work in
molarities: a one molar aqueous solution (1 M or 1 mol dm 3 ) contains one
mole of solute dissolved in 1 dm3 of solution. Another unit may also be
used for concentration: a one molal aqueous solution contains one mole of
solute dissolved in 1 kg of water (1 mol kg 1 ).
In the laboratory, we usually deal with dilute solutions and we can make
some approximations that let us work with concentrations when strictly
we should be using the activities of solutes. When the concentration of a
solute is greater than about 0.1 mol dm 3 , there are signi®cant interactions
between the solute molecules or ions. As a result, the e€ective and real
concentrations are not the same. It is necessary, therefore, to de®ne a new
quantity called the activity. This is a measure of concentration, but it takes
into account interactions between solution species. The relative activity, ai ,
of a solute i is dimensionless, and is related to its molality by equation 15.8
where i is the activity coecient of the solute, and mi and mi o are the
molality and standard state molality, respectively.
ai ˆ
i mi
mi o
…15:8†
The standard state of a solute in solution refers to in®nite dilution at standard
molality (mo ), 1 bar pressure, where interactions between solute molecules
are insigni®cant. In equation 15.8, mi o is de®ned as 1, and so the equation
becomes equation 15.9.
ai ˆ i mi
…15:9†
For dilute solutions, we can take i 1, and from equation 15.9, it follows
that the activity is approximately equal to the molality. For most purposes,
Equilibrium constants
concentrations are measured in mol dm
make the approximation that:
3
443
rather than molalities, and we can
For a dilute solution, the activity of the solute is approximately equal to the
numerical value of its concentration (measured in mol dm 3 ).
For an ideal gas, the activity
is equal to the numerical
value of the partial pressure,
the pressure being measured
in bar.
The activity of an ideal gas is equal to the numerical value of the partial
pressure, the pressure being measured in bars. For a real gas, i, equation
15.10 holds where i is the activity coecient and Pi is the partial pressure
of component i.
ai ˆ i Pi
…15:10†
Only for an ideal gas is i ˆ 1, and so it is only valid to take the activity of a
gas as being equal to its partial pressure if the gas is ideal.
The important points to remember from this section are:
.
.
.
the activity of any pure substance in its standard state is de®ned as 1 (no
units);
the activity of a solute in a dilute solution approximately equals the numerical value of its concentration (measured in mol dm 3 );
the activity of an ideal gas approximately equals the numerical value of its
partial pressure (measured in bar).
We put these approximations into practice in the following discussion of
equilibrium constants.
The thermodynamic
equilibrium constant, K, is
de®ned in terms of the
activities of the products
and reactants, and is
dimensionless.
Thermodynamic equilibrium constants
Consider the general equilibrium 15.11 in which x moles of X and y moles of
Y are in equilibrium with z moles of Z.
xX ‡ yY „ zZ
…15:11†
The thermodynamic equilibrium constant, K (referred to after this point
simply as the equilibrium constant), is de®ned in terms of the activities of
the components of the equilibrium as in equation 15.12.
Kˆ
…aZ †z
…aX †x …aY †y
…15:12†
Thus, for reaction 15.13, K is given by the equation shown.
2CO…g† ‡ O2 …g† „ 2CO2 …g†
Kˆ
…aCO2 †2
…aCO †2 …aO2 †
…15:13†
We now apply the approximation discussed in the previous section and
rewrite K (equation 15.14) in terms of the numerical values of the partial
pressures of the gases in equilibrium 15.13; we assume that the gases are ideal.
Kˆ
…PCO2 †2
…PCO †2 …PO2 †
…dimensionless†
…15:14†
It is understood that P, although indicating a pressure, is included in the
expression for K only as a number. Similarly, we can write an expression for
equilibrium 15.15 in terms of the activities (equation 15.16) or concentrations
444
CHAPTER 15 . Equilibria
(equation 15.17) of the solution species provided that the solution is dilute. The
expression is simpli®ed by taking the activity of solid Cu as 1.
2Cu‡ …aq† „ Cu2‡ …aq† ‡ Cu…s†
…15:15†
Kˆ
…aCu2‡ †…aCu † …aCu2‡ †
ˆ
…aCu‡ †2
…aCu‡ †2
…15:16†
Kˆ
‰Cu2‡ Š
‰Cu‡ Š2
…15:17†
…dimensionless†
It is understood that the concentrations are included in the expression for K
only as numbers.
The next set of worked examples shows how to determine values of
thermodynamic equilibrium constants when the composition of an equilibrium mixture is known, and how to use values of K to quantify the position
of an equilibrium in terms of its composition.
Worked example 15.1
Determination of an equilibrium constant
N2 O4 dissociates according to the following equation:
N2 O4 …g† „ 2NO2 …g†
At 350 K, the equilibrium mixture contains 0.13 moles N2 O4 and 0.34 moles
NO2 . The total pressure is 2 bar. Find the value of K at 350 K.
First ®nd the partial pressure of NO2 and N2 O4 .
Moles of X
Total pressure
Partial pressure of component X ˆ
Total moles
0:34
2 ˆ 1:45 bar
Partial pressure of NO2 ˆ
0:34 ‡ 0:13
0:13
2 ˆ 0:55 bar
Partial pressure of N2 O4 ˆ
0:34 ‡ 0:13
Kˆ
…PNO2 †2
…PN2 O4 †
Kˆ
…1:45†2
ˆ 3:8
0:55
This value is dimensionless because only numerical values of P are substituted into the equation.
Worked example 15.2
Determination of an equilibrium constant
In an aqueous solution of CH3 NH2 , the following equilibrium is established:
CH3 NH2 …aq† ‡ H2 O…l† „ CH3 NH3 ‡ …aq† ‡ OH …aq†
Determine K for this equilibrium at 298 K if the initial concentration of
CH3 NH2 is 0.10 mol dm 3 , and at equilibrium, the concentration of OH is
0.0066 mol dm 3 .
Let the total volume be 1 dm 3 . Write out the equilibrium composition in
terms of moles:
445
Equilibrium constants
CH3 NH2 …aq† ‡ H2 O…l† „ CH3 NH3 ‡ …aq† ‡ OH …aq†
Moles initially:
0.10
Moles at equilm: (0.10±0.0066)
excess
0
0
excess
0.0066
0.0066
Write out an expression for K, taking into account that the activity is
approximately equal to concentration in dilute solution, and that the
activity of the solvent water is 1.
Kˆ
‰CH3 NH3 ‡ Š‰OH Š ‰CH3 NH3 ‡ Š‰OH Š
ˆ
‰CH3 NH2 Š‰H2 OŠ
‰CH3 NH2 Š
Concentration ˆ
Number of moles
mol dm
Volume (in dm3 †
If the volume is 1 dm3 :
Worked example 15.3
3
Concentration ˆ Number of moles
Kˆ
‰CH3 NH3 ‡ Š‰OH Š
‰CH3 NH2 Š
Kˆ
…0:0066†…0:0066†
ˆ 4:7 10
…0:10 0:0066†
4
…dimensionless†
Finding the composition of an equilibrium mixture
At 400 K, the equilibrium constant for the following equilibrium is 40:
H2 …g† ‡ I2 …g† „ 2HI…g†
If 2.0 moles of H2 and 2.0 moles of I2 vapour are mixed at 400 K, and the total
pressure is 1 bar, what is the composition of the equilibrium mixture?
First write down a scheme to show the composition of the equilibrium mixture, using the reaction stoichiometry to work out the ratios of moles of
products : reactants.
H2 (g)
Moles initially:
2.0
Moles at equilm:
2:0
‡ I2 (g)
„ 2HI(g)
2.0
x
2:0
0
x
2x
At equilibrium, the total moles of gas ˆ …2:0 x† ‡ …2:0 x† ‡ 2x ˆ 4:0
Now ®nd the partial pressures of each gas at equilibrium:
Moles of X
Total pressure
component X ˆ
Total moles
2:0 x
2:0 x
H2 ˆ
1ˆ
bar
4:0
4:0
2:0 x
2:0 x
1ˆ
bar
I2 ˆ
4:0
4:0
2x
2x
1ˆ
bar
HI ˆ
4:0
4:0
Partial pressure of
Partial pressure of
Partial pressure of
Partial pressure of
Write down an expression for K in terms of the partial pressures of the
equilibrium components:
Kˆ
…PHI †2
ˆ 40
…PH2 †…PI2 †
446
CHAPTER 15 . Equilibria
Substitute in the numerical values of the partial pressures:
2x 2
4:0
ˆ 40
Kˆ
2:0 x
2:0 x
4:0
4:0
…2x†2
ˆ 40
…2:0 x†2
Now solve for x:
4x2 ˆ 40…2:0
x†2
4x2 ˆ 40…4:0
4:0x ‡ x2 †
4x2 ˆ 160
160x ‡ 40x2
0 ˆ 160
160x ‡ 36x2
The general solution for a quadratic equation of form ax2 ‡ bx ‡ c ˆ 0 is:
xˆ
b
p
b2 4ac
2a
For the quadratic: 36x2
160x ‡ 160 ˆ 0
p
‡160 1602 4…36†…160†
xˆ
2…36†
There are two solutions: x ˆ 2:9 or 1.5, but only x ˆ 1:5 is meaningful.
(Why?) The equilibrium mixture therefore has the composition:
…2:0
1:5† ˆ 0:5 moles H2
…2:0
1:5† ˆ 0:5 moles I2
…2 1:5† ˆ 3:0 moles HI
Check: Total number of moles ˆ 4:0
Values of K and extent of reaction
Values of K for di€erent reactions vary from being extremely small (reactants
predominate) to extremely large (products predominate) and examples are
listed in Table 15.1. In the ®rst two equilibria in the table, the large values
of K show that products predominate over reactants; in the case of the formation of HCl, the reaction lies so far to the right-hand side that negligible
amounts of reactants remain. An increase in temperature from 298 K to
Table 15.1 Values of equilibrium constants, K, for selected gaseous equilibria at 298 K and 400 K. Values of f H o refer to the
forward (formation) reactions and are per mole of compound formed.
f H o (298 K) / kJ mol
1
f H o (400 K) / kJ mol
Equilibrium
K (298 K)
K (400 K)
H2 …g† ‡ Cl2 …g† „ 2HCl…g†
2:5 1033
1:4 1025
92.3
92.6
N2 …g† ‡ 3H2 …g† „ 2NH3 …g†
5:6 10
36
45.9
48.1
N2 …g† ‡ 2O2 …g† „ 2NO2 …g†
4:7 10
19
5:0 10
16
‡34.2
‡33.6
N2 …g† ‡ O2 …g† „ 2NO…g†
2:0 10
31
2:9 10
23
‡91.3
‡91.3
5
1
Equilibrium constants
447
Table 15.2 The temperature dependence of K for the equilibrium:
2H2 …g† ‡ O2 …g† „ 2H2 O…g†
Temperature / K
K
400
500
600
700
3:00 1058
5:86 1045
1:83 1037
1:46 1031
400 K causes a decrease in K, i.e. the back reaction is favoured. This is
consistent with each of the formations of HCl and NH3 at 298 and 400 K
being exothermic (apply Le Chatelier's principle). The last two equilibria
listed in Table 15.1 have very small values of K, showing that the equilibrium
mixtures contain far more reactants than products. In both cases, increasing
the temperature from 298 K to 400 K favours the forward reaction, although
reactants still predominate at equilibrium. These data are consistent with the
formations of NO and NO2 being endothermic.
The dependence of K on temperature is quanti®ed in equation 15.18.
d…ln K† H o
ˆ
dT
RT 2
…15:18†
The integrated form of equation 15.18 (equation 15.19, the derivation of
which is given in Section 16.8) shows a linear relationship between ln K
and 1=T.
ln K ˆ
H o
‡c
RT
…c ˆ integration constant†
…15:19†
Table 15.2 shows the temperature dependence of K for equilibrium 15.20
over the temperature range 400±700 K.
2H2 …g† ‡ O2 …g† „ 2H2 O…g†
…15:20†
Figure 15.1 gives a plot of ln K against 1=T using the data from Table 15.2.
From equation 15.19, the positive gradient in Figure 15.1 indicates that the
forward reaction 15.20 is exothermic (negative H o ). An approximate value
Fig. 15.1 A plot of ln K against
1=T for the equilibrium:
2H2 …g† ‡ O2 …g† „ 2H2 O…g†.
448
CHAPTER 15 . Equilibria
of f H o (H2 O,g) can be obtained from the gradient of the line (equation
15.21).
H o
R
Gradient ˆ 59 000 K ˆ
H o ˆ
ˆ
59 000 …K† 8:314 10
490 kJ mol
1
…15:21†
3
…kJ mol
1
K 1†
(per two moles of H2 O…g††
The value is only approximate because H o is actually temperature
dependent, but the variation over the temperature range 400±700 K is not
enormous (about 8 kJ per two moles of H2 O) and this method of determining
enthalpy changes remains a useful experimental approach. The di€erence in
values of f H o (298 K) andf H o (400 K) for selected reactions can be seen
from Table 15.1 and these data give an indication of how approximate it is
to assume a constant value for f H o .
15.4
Acid±base equilibria
Brùnsted acids and bases
A Brùnsted acid is a proton donor (e.g. hydrochloric acid) and a Brùnsted
base is a proton acceptor (e.g. hydroxide ion). Equation 15.22 shows the
reaction between a general acid HA and H2 O in which a proton is transferred
from the acid to the base: HA acts as a Brùnsted acid and H2 O functions as a
Brùnsted base.
A Brùnsted acid is a proton
donor, and a Brùnsted base
is a proton acceptor.
HA
‡
H2 O
Bro= nsted base
Bro= nsted acid
„ H3 O ‡ ‡ A
…15:22†
Water can also act as a Brùnsted acid and equation 15.23 shows its general
reaction with a Brùnsted base B.
B
Bro= nsted base
‡
H2 O
„ BH‡ ‡ OH
…15:23†
Bro= nsted acid
If reaction 15.22 goes essentially to completion, then HA is a strong acid; an
example is HCl which is fully ionized in aqueous solution.
HCl…aq† ‡ H2 O…l†
"
H3 O‡ …aq† ‡ Cl …aq†
…15:24†
In aqueous solution, NaOH acts as a strong base because it is fully ionized;
the basic species in solution is OH (equation 15.25).
NaOH…aq†
"
Na‡ …aq† ‡ OH …aq†
…15:25†
The equilibrium constant, Ka
"
Carboxylic acids:
see Chapter 32
In many cases, dissociation of the acid in aqueous solution is not complete
and an equilibrium is established as shown for aqueous acetic acid in
equation 15.26. Acetic acid is an example of a carboxylic acid of general
formula RCO2 H.
CH3 CO2 H…aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ CH3 CO2 …aq†
Acetic acid
…15:26†
Acetate ion
Aqueous acetic acid is a weak Brùnsted acid and the equilibrium lies to
the left-hand side. The equilibrium constant is called the acid dissociation
Acid±base equilibria
449
constant, Ka , and for CH3 CO2 H, Ka ˆ 1:7 10 5 . This is expressed by equation 15.27 where the concentrations are those at equilibrium. Remember that
the equilibrium constant is strictly expressed in terms of activities, and for the
solvent H2 O, the activity is 1.
Ka ˆ
‰H3 O‡ Š‰CH3 CO2 Š ‰H3 O‡ Š‰CH3 CO2 Š
ˆ
ˆ 1:75 10
‰CH3 CO2 HŠ‰H2 OŠ
‰CH3 CO2 HŠ
5
…15:27†
Assuming that the contribution to ‰H3 O‡ Š from H2 O is negligible, then the
stoichiometry of equation 15.26 shows that the concentration of H3 O‡ and
CH3 CO2 ions must be equal and so equation 15.27 can be written in the
forms shown in equation 15.28.
Ka ˆ
‰H3 O‡ Š2
‰CH3 CO2 HŠ
or
Ka ˆ
‰CH3 CO2 Š2
‰CH3 CO2 HŠ
…15:28†
These equations can be used to calculate the concentration of H3 O‡ or
CH3 CO2 ions as shown in worked example 15.4.
Worked example 15.4
The concentration of H3 O‡ ions in an aqueous solution of CH3 CO2 H (acetic
acid)
Determine ‰H3 O‡ Š in a 1:0 10
…Ka ˆ 1:7 10 5 †.
2
mol dm
3
solution of CH3 CO2 H
The equilibrium in aqueous solution is:
CH3 CO2 H…aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ CH3 CO2 …aq†
and the acid dissociation constant is given by:
Ka ˆ
‰H3 O‡ Š‰CH3 CO2 Š
‰CH3 CO2 HŠ
The stoichiometry of the equilibrium shows that ‰CH3 CO2 Š ˆ ‰H3 O‡ Š,
and therefore:
Ka ˆ
‰H3 O‡ Š2
‰CH3 CO2 HŠ
This equation can be rearranged to ®nd ‰H3 O‡ Š:
‰H3 O‡ Š2 ˆ K a ‰CH3 CO2 HŠ
p
‰H3 O‡ Š ˆ K a ‰CH3 CO2 HŠ
The initial concentration of CH3 CO2 H is 1:0 10 2 mol dm 3 , but we
need the equilibrium concentration. Since CH3 CO2 H is a weak acid, only
a very small amount of the acid has dissociated, and therefore we can
make the approximation that the concentration of CH3 CO2 H at equilibrium
is roughly the same as the original concentration:
‰CH3 CO2 HŠequilm ‰CH3 CO2 HŠinitial ˆ 1:0 10
2
mol dm
3
Substituting this value into the equation for ‰H3 O‡ Š gives:
p
K a ‰CH3 CO2 HŠ
p
ˆ 1:7 10 5 1:0 10 2
‰H3 O‡ Š ˆ
ˆ 4:1 10
4
mol dm
3
450
CHAPTER 15 . Equilibria
Check that the assumption that ‰CH3 CO2 HŠequilm ‰CH3 CO2 HŠinitial is
valid: the initial concentration of acid was 1:0 10 2 mol dm 3 , and the
equilibrium concentration of H3 O‡ is found to be 4:1 10 4 mol dm 3 ,
i.e. the degree of dissociation of the acid is very small:
Degree of dissociation ˆ
4:1 10
1:0 10
4
2
100 ˆ 4:1%
pKa
Although we often use values of Ka , a useful and common way of reporting
this number is as the pKa value. This is de®ned as the negative logarithm (to
the base 10) of the equilibrium constant (equation 15.29).
For a weak acid:
pKa ˆ log Ka
"
log = log10
pKa ˆ
log Ka
…15:29†
If pKa is known, equation 15.29 can be rearranged to ®nd Ka (equation
15.30).
Ka ˆ 10
pKa
…15:30†
From equations 15.29 and 15.30, it follows that as Ka becomes smaller, pKa
becomes larger. For example, CH3 CO2 H is a weaker acid than HCO2 H and
the respective values of pKa and Ka are:
Worked example 15.5
CH3 CO2 H
pKa ˆ 4:77
Ka ˆ 1:7 10
5
HCO2 H
pKa ˆ 3:75
Ka ˆ 1:8 10
4
The relationship between Ka and pKa
pKa for hydrocyanic acid (HCN) is 9.31. Find the concentration of H3 O‡ ions
in an aqueous solution of concentration 2:0 10 2 mol dm 3 .
The appropriate equilibrium is:
HCN…aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ CN …aq†
and the equilibrium constant is given by:
Ka ˆ
‰H3 O‡ Š‰CN Š
‰HCNŠ
From the stoichiometry of the equilibrium, ‰H3 O‡ Š ˆ ‰CN Š, and so we
can write:
Ka ˆ
‰H3 O‡ Š2
‰HCNŠ
and therefore:
‰H3 O‡ Š2 ˆ K a ‰HCNŠ
p
‰H3 O‡ Š ˆ K a ‰HCNŠ
Ka can be found from pKa :
Ka ˆ 10
pKa
ˆ 10
9:31
ˆ 4:9 10
10
Acid±base equilibria
451
Using the approximation that ‰HCNŠequilm ‰HCNŠinitial ˆ 2:0 10 2 mol dm 3 , we can calculate ‰H3 O‡ Š:
p
K a ‰HCNŠ
p
ˆ 4:9 10 10 2:0 10 2
‰H3 O‡ Š ˆ
ˆ 3:1 10
6
3
mol dm
Check that the assumption that ‰HCNŠequilm ‰HCNŠinitial is valid: the
initial concentration of acid was 2:0 10 2 mol dm 3 , and the equilibrium
concentration of H3 O‡ is found to be 3:1 10 6 mol dm 3 , i.e. the degree
of dissociation of the acid is very small.
Neutralization reactions
A neutralization reaction is
the reaction of an acid with
a base to give a salt and
water.
Equations 15.31 and 15.32 summarize general neutralization reactions, and
examples are given in reactions 15.33±15.35.
H3 O‡ …aq† ‡ OH …aq†
Acid ‡ Base
"
2H2 O…l†
"
Base
…15:32†
KCl…aq† ‡ H2 O…l†
H2 SO4 …aq† ‡ 2NaOH…aq†
Acid
"
Base
CH3 CO2 H…aq† ‡ NaOH…aq†
Acid
…15:31†
Salt ‡ Water
HCl…aq† ‡ KOH…aq†
Acid
"
Na2 SO4 …aq† ‡ 2H2 O…l†
"
Base
Na‰CH3 CO2 Š…aq† ‡ H2 O…l†
…15:33†
…15:34†
…15:35†
Even if an acid is weak, the neutralization reaction goes to completion if
stoichiometric quantities of reagents are available.
Conjugate acids and bases
"
Nitrous acid is unstable
with respect to
disproportionation:
see equation 21.73
When a Brùnsted acid donates a proton, it forms a species which can, in
theory, accept the proton back again. Equation 15.36 shows the dissociation
of nitrous acid in water.
HNO2 …aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ NO2 …aq†
…15:36†
In the forward reaction, HNO2 acts as a Brùnsted acid and H2 O as a
Brùnsted base. In the back reaction, the Brùnsted acid is H3 O‡ and the
Brùnsted base is NO2 . The NO2 ion is the conjugate base of HNO2 , and
conversely, HNO2 is the conjugate acid of NO2 . Similarly, H3 O‡ is the
conjugate acid of H2 O, and H2 O is the conjugate base of H3 O‡ . The conjugate acid±base pairs are shown in scheme 15.37.
HNO2(aq) + H2O(l)
NO2–(aq) + H3O+(aq)
Conjugate
acid 1
Conjugate
base 1
Conjugate
base 2
Conjugate
acid 2
Conjugate acid–base pair
Conjugate acid–base pair
(15.37)
452
CHAPTER 15 . Equilibria
Formula
pKa
Conjugate base
Formula
Perchloric acid
Sulfuric acid
Hydrochloric acid
Nitric acid
Sulfurous acid
Hydrogensulfate(1 ) ion
Phosphoric acid
Nitrous acid
Acetic acid
Carbonic acid
Hydrogensul®te(1 ) ion
Dihydrogenphosphate(1 ) ion
Hydrocyanic acid
Hydrogencarbonate(1 ) ion
Hydrogenphosphate(2 ) ion
HClO4
H2 SO4
HCl
HNO3
H2 SO3
‰HSO4 Š
H3 PO4
HNO2
CH3 CO2 H
H2 CO3
‰HSO3 Š
‰H2 PO4 Š
HCN
‰HCO3 Š
‰HPO4 Š2
±
±
±
±
1.81
1.92
2.12
3.34
4.77
6.37
6.91
7.21
9.31
10.25
12.67
Perchlorate ion
Hydrogensulfate(1 ) ion
Chloride ion
Nitrate ion
Hydrogensul®te(1 ) ion
Sulfate ion
Dihydrogenphosphate(1 ) ion
Nitrite ion
Acetate ion
Hydrogencarbonate(1 ) ion
Sul®te(2 ) ion
Hydrogenphosphate(2 ) ion
Cyanide ion
Carbonate ion
Phosphate(3 ) ion
‰ClO4 Š
‰HSO4 Š
Cl
‰NO3 Š
‰HSO3 Š
‰SO4 Š2
‰H2 PO4 Š
‰NO2 Š
‰CH3 CO2 Š
‰HCO3 Š
‰SO3 Š2
‰HPO4 Š2
‰CNŠ
‰CO3 Š2
‰PO4 Š3
"
Acid
Increasing strength of conjugate base
Increasing strength of acid
"
Table 15.3 Conjugate acid and base pairs, and values of Ka for the dilute aqueous solutions of the acids; strong acids are fully
dissociated.
Table 15.3 lists selected conjugate acid±base pairs and pKa values for the
acids. Note the inverse relationship between the relative strengths of the
conjugates acids and bases. We return to this point later in the section
(equation 15.54).
The self-ionization of water
The self-ionization constant,
Kw , for water is 1:0 10 14 ;
pKw ˆ 14:00 (at 298 K).
Water itself is ionized (equation 15.38) to a very small extent; the form of the
equilibrium sign in equation 15.38 indicates that the reaction lies far to
the left-hand side. The equilibrium constant for the dissociation is called
the self-ionization constant, Kw , (equation 15.39) and the value reveals how
very few ions are present in pure water.
2H2 O…l†) * H3 O‡ …aq† ‡ OH …aq†
‡
K w ˆ ‰H3 O Š‰OH Š ˆ 1:0 10
14
…15:38†
…at 298 K†
…15:39†
It is convenient to de®ne the term pKw (equation 15.40) which has the same
relationship to Kw that pKa has to Ka .
pKw ˆ
log Kw ˆ 14:00
…15:40†
We make use of Kw and pKw later in the section.
Polybasic acids
Acids such as HCl, CH3 CO2 H, HNO2 and HCN are monobasic because they
lose only one proton per molecule of acid (e.g. equation 15.36). Some Brùnsted
acids may lose two, three or more protons. Sulfuric acid is an example of a
dibasic acid, and equations 15.41 and 15.42 show the two dissociation steps.
H2 SO4 …aq† ‡ H2 O…l†
"
H3 O‡ …aq† ‡ HSO4 …aq† fully dissociated
HSO4 …aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ SO4 2 …aq†
pK a ˆ 1:92
…15:41†
…15:42†
Acid±base equilibria
453
The ®rst dissociation step goes to completion, H2 SO4 being a strong acid, but
the second step is an equilibrium since HSO4 is a relatively weak acid. Reactions of sulfuric acid with a base may lead to the formation of a sulfate(2 )
or hydrogensulfate(1 ) salt as in equations 15.43 and 15.44.
H2 SO4 …aq† ‡ KOH…aq†
H2 SO4 …aq† ‡ 2KOH…aq†
In general, for a dibasic acid:
Ka …1† > Ka …2†
"
KHSO4 …aq† ‡ H2 O…l†
"
…15:43†
K2 SO4 …aq† ‡ 2H2 O…l†
…15:44†
It is a general trend that the ®rst dissociation constant, Ka …1†, of a dibasic acid
is larger than the second, Ka …2†, and this is seen in Table 15.3 for H2 SO3 and
H2 CO3 . Remember that if Ka …1† > Ka …2†, then pKa …1† < pKa …2†. The trend
in Ka values extends to tribasic and higher polybasic acids: it is harder to
remove a proton from a negatively charged species than from a neutral one.
Phosphoric acid is tribasic, and the three dissociation steps are shown in equations 15.45 to 15.47. The acid strength of H3 PO4 > H2 PO4 > HPO4 2 .
H3 PO4 …aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ H2 PO4 …aq†
‡
H2 PO4 …aq† ‡ H2 O…l† „ H3 O …aq† ‡ HPO4
HPO4
2
‡
…aq† ‡ H2 O…l† „ H3 O …aq† ‡ PO4
3
2
…aq†
…aq†
pK a ˆ 2:12
…15:45†
pK a ˆ 7:21
…15:46†
pK a ˆ 12:67 …15:47†
The dissociation constant, Kb
In an aqueous solution of a weak base B, proton transfer is not complete and
equation 15.48 shows the general equilibrium and expression for the associated equilibrium constant, Kb .
B…aq† ‡ H2 O…l† „ ‰BHŠ‡ …aq† ‡ OH …aq†
Kb ˆ
For a weak base:
pKb ˆ log Kb
Worked example 15.6
‰BH‡ Š‰OH Š ‰BH‡ Š‰OH Š
ˆ
‰BŠ‰H2 OŠ
‰BŠ
…15:48†
In the same way that we de®ned pKa and pKw , we can also de®ne pKb (equation 15.49). The weaker the base, the smaller the value of Kb and the larger
the value of pKb .
pKb ˆ
log Kb
…15:49†
Determining the concentration of OH ions in an aqueous solution of NH3
Calculate the concentration of OH in an aqueous solution of NH3 of concentration 5:0 10 2 mol dm 3 if pKb for NH3 ˆ 4:75.
The appropriate equilibrium is:
NH3 …aq† ‡ H2 O…l† „ NH4 ‡ …aq† ‡ OH …aq†
and the equilibrium constant is given by:
Kb ˆ
‰NH4 ‡ Š‰OH Š
‰NH3 Š
where the concentrations are those at equilibrium. From the stoichiometry of
the equilibrium, it follows that ‰NH4 ‡ Š ˆ ‰OH Š (assuming that the contribution to ‰OH Š from H2 O dissociation is negligible) and so we can write:
Kb ˆ
‰OH Š2
‰NH3 Š
454
CHAPTER 15 . Equilibria
Rearrangement of the equation gives:
‰OH Š2 ˆ K b ‰NH3 Š
p
‰OH Š ˆ K b ‰NH3 Š
We make the assumption that, because NH3 is a weak base and is largely
undissociated, ‰NH3 Šequilm ‰NH3 Šinitial ˆ 5:0 10 2 mol dm 3 .
We are given the value of pKb ˆ 4:75, and therefore:
Kb ˆ 10
Therefore:
‰OH Š ˆ
4:75
ˆ 1:8 10
5
p
1:8 10 5 5:0 10 2
‰OH Š ˆ 9:5 10
4
mol dm
3
Check that the assumption ‰NH3 Šequilm ‰NH3 Šinitial was valid: the initial
concentration of NH3 was 5:0 10 2 mol dm 3 , and the equilibrium
concentration of OH is much smaller than this, 9:5 10 4 mol dm 3 .
The relationship between Ka and Kb for a conjugate acid±base pair
We mentioned earlier that there is an inverse relationship between the relative strengths of a weak acid and its conjugate base (Table 15.3) and we
can quantify this by considering the relationship between Ka and Kb for a
conjugate acid±base pair. Consider the dissociation of acetic acid in aqueous
solution (equation 15.50) and the equilibrium established by the acetate ion
in aqueous solution (equation 15.51).
CH3 CO2 H…aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ CH3 CO2 …aq†
Ka ˆ
‰H3 O‡ Š‰CH3 CO2 Š
‰CH3 CO2 HŠ
…15:50†
CH3 CO2 …aq† ‡ H2 O…l† „ CH3 CO2 H…aq† ‡ OH …aq†
Kb ˆ
‰CH3 CO2 HŠ‰OH Š
‰CH3 CO2 Š
…15:51†
By comparing these equations for Ka and Kb with equation 15.39 for Kw , we
®nd that the three equilibrium constants are related by equation 15.52:
Ka Kb ˆ
‰H3 O‡ Š‰CH3 CO2 Š ‰CH3 CO2 HŠ‰OH Š
‰CH3 CO2 HŠ
‰CH3 CO2 Š
ˆ ‰H3 O‡ Š‰OH Š
Ka Kb ˆ Kw
…15:52†
If we now take logs of both sides of equation 15.52, we obtain equation 15.53.
log …Ka Kb † ˆ log Kw
Therefore:
log Ka ‡ log Kb ˆ log Kw
…15:53†
This equation can be rewritten in terms of pKa , pKb and pKw (equation 15.54).
pKa ‡ pKb ˆ pKw ˆ 14:00
log Ka
log Kb ˆ
log Kw
pKa ‡ pKb ˆ pKw ˆ 14:00
…15:54†
Acids and bases in aqueous solution: pH
455
Thus, for acetic acid for which pKa ˆ 4:77 (Table 15.3), we can use equation
15.54 to determine that pKb for the acetate ion is …14:00 4:77† ˆ 9:23.
Equation 15.54 allows us to quantify the relationships between the relative
strengths of the conjugate acids and bases in Table 15.3. For example, pKa
for HPO4 2 is 12.67, and from equation 15.54, pKb for the conjugate base
PO4 3 is …14:00 12:67† ˆ 1:33. Thus, while HPO4 2 is the weakest acid
listed in Table 15.3, PO4 3 is the strongest base.
15.5
Acids and bases in aqueous solution: pH
The concentration of H3 O‡ ions in solution is usually denoted by a pH value
as de®ned in equation 15.55.§ If the pH is known, then the hydrogen ion
concentration can be found from equation 15.56.
pH ˆ
pH ˆ
log ‰H3 O‡ Š
log ‰H3 O‡ Š
‡
‰H3 O Š ˆ 10
"
pH meter: see end of
Section 17.8
pH
…15:55†
…15:56†
For strong acids, determining the pH of the solution from the hydrogen ion
concentration, or ®nding the ‰H3 O‡ Š from a reading on a pH meter is
straightforward. The method is shown in worked example 15.7. However,
for a weak acid, we must take into account that not all the acid is dissociated
and this is explained in worked example 15.8.
ENVIRONMENT AND BIOLOGY
Box 15.1 pH and the gardener
Commercially available kits used to test the pH of soil.
E. C. Constable.
§
Walk around any garden centre, and you will ®nd kits
for testing the pH of garden soil. The bedrock and
sources of water run-off affect the pH of soil and in
the UK, soil pH varies in the approximate range 4.0
to 8.5. Many plants are quite tolerant of the pH conditions in which they grow, and may thrive within pH
limits of 1.0 of an optimum value. Other plants are
not at all tolerant, and many agricultural crops grow
poorly in acid soils. Rhododendrons and azaleas
require acidic soil and grow best where the soil is
peat-based. Bilberries and cranberries also need acidic
conditions: pH 4.5 is ideal. Anemones (e.g. Anemone
pulsatilla), clematis and daphne grow best in slightly
alkaline conditions (pH 7.5) and respond well to
the addition of lime. A well-documented example of
the effects of pH is the colour of hydrangea ¯owers:
alkaline soil tends to make the ¯owers pink, while
acid soil results in blue ¯owers, making the plant an
indicator of soil pH.
You will also ®nd equation 15.55 written in the form of pH ˆ log‰H‡ Š; in water, protons
combine with H2 O molecules and the predominant species are H3 O‡ ions.
456
CHAPTER 15 . Equilibria
Worked example 15.7
pH of an aqueous solution of HCl (hydrochloric acid)
What is the pH of an aqueous hydrochloric acid solution of concentration
5:0 10 2 mol dm 3 ?
Hydrochloric acid is a strong acid (Table 15.3) and is fully dissociated in
water:
HCl…aq† ‡ H2 O…l†
"
H3 O‡ …aq† ‡ Cl …aq†
Therefore, the concentration of H3 O‡ ions is the same as the initial concentration of HCl:
‰H3 O‡ Š ˆ ‰HClŠ ˆ 5:0 10
pH ˆ
ˆ
2
mol dm
3
log ‰H3 O‡ Š
log …5:0 10 2 †
ˆ 1:30
Worked example 15.8
pH of an aqueous solution of HCO2 H (formic acid)
Formic acid has a pKa of 3.75. What is the pH of an aqueous solution of concentration 5:0 10 3 mol dm 3 ?
Formic acid is a weak acid and is not fully dissociated in aqueous
solution:
HCO2 H…aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ HCO2 …aq†
Therefore, the equilibrium concentration of H3 O‡ ions is not the same as
the initial concentration of HCO2 H. To ®nd ‰H3 O‡ Š, we need the equilibrium constant:
Ka ˆ
‰H3 O‡ Š‰HCO2 Š
ˆ 10
‰HCO2 HŠ
‰H3 O‡ Š‰HCO2 Š
ˆ 10
‰HCO2 HŠ
3:75
pK a
ˆ 1:8 10
4
From the stoichiometry of the equation above, ‰H3 O‡ Š ˆ ‰HCO2 Š, and
therefore:
‰H3 O‡ Š2
ˆ 1:8 10
‰HCO2 HŠ
4
‰H3 O‡ Š2 ˆ 1:8 10 4 ‰HCO2 HŠ
q
‰H3 O‡ Š ˆ 1:8 10 4 ‰HCO2 HŠ
where ‰HCO2 HŠ is the equilibrium concentration of undissociated acid. If
only a very small amount of the initial acid were dissociated, we could
make the assumption that ‰HCO2 HŠequilm ‰HCO2 HŠinitial . But in this
case, the approximation is not valid as we can see below. Assume that
‰HCO2 HŠequilm ‰HCO2 HŠinitial , and so:
‰H3 OŠ‡ ˆ
p
1:8 10 4 5:0 10 3 ˆ 9:5 10
4
mol dm
3
457
Acids and bases in aqueous solution: pH
If you compare this concentration to the initial concentration of HCO2 H
of 5:0 10 3 mol dm 3 , it shows that 19% of the acid has dissociated, and
this is a signi®cant amount:
Degree of dissociation ˆ
4
9:5 10
5:0 10
3
100 ˆ 19%
Therefore, we need to solve this problem more rigorously:
+H2 O(l) „ H3 O‡ (aq) ‡ HCO2 (aq)
HCO2 H(aq)
mol dm
3
mol dm
3
initially: 5:0 10
3
excess
0
0
3
x excess
x
x
equilm: …5:0 10 †
K a ˆ 1:8 10
2
4
ˆ
‰H3 O‡ Š‰HCO2 Š
‰HCO2 HŠ
ˆ
x2
…5:0 10 3 †
4
3
x ˆ …1:8 10 †…5:0 10 †
x
1:8 10 4 x
x2 ‡ …1:8 10 4 †x
…9:0 10 7 † ˆ 0
q
…1:8 10 4 † …1:8 10 4 †2 ‡ 4…9:0 10 7 †
xˆ
2
ˆ 8:6 10
4
or
1:0 10
3
mol dm
3
Clearly, only the positive value is possible and therefore:
x ˆ ‰H3 OŠ‡ ˆ 8:6 10
4
mol dm
3
Now we can ®nd the pH of the solution:
pH ˆ
ˆ
log ‰H3 O‡ Š
log …8:6 10 4 †
ˆ 3:07
Figure 15.2 illustrates the pH scale with examples of strong and weak acids
and bases of di€erent concentrations. The pH of an aqueous solution of a
Fig. 15.2 The pH scale runs
below 0 to 14. The chart gives
some examples of aqueous acids
and bases covering a range of pH
values.
458
CHAPTER 15 . Equilibria
base can be determined by ®rst ®nding the concentration of hydroxide ions in
solution and then using relationship 15.39 to determine the concentration of
H3 O‡ ions. Worked examples 15.9 and 15.10 consider the cases of a strong
(fully dissociated) and weak (partially dissociated) base, respectively.
Worked example 15.9
pH of an aqueous solution of NaOH (sodium hydroxide)
What is the pH of an aqueous NaOH solution of concentration
5:0 10 2 mol dm 3 ?
NaOH is fully dissociated when dissolved in water:
NaOH…aq†
"
Na‡ …aq† ‡ OH …aq†
and the concentration of OH ions is equal to the initial concentration of
NaOH:
‰OH Š ˆ ‰NaOHŠ ˆ 5:0 10
2
mol dm
3
The pH of the solution is given by the equation:
pH ˆ
log ‰H3 O‡ Š
and so we need to relate ‰H3 O‡ Š to the known value of ‰OH Š:
K w ˆ ‰H3 O‡ Š‰OH Š ˆ 1:0 10
‰H3 O‡ Š ˆ
pH ˆ
ˆ
14
1:0 10 14 1:0 10
ˆ
‰OH Š
5:0 10
14
2
ˆ 2:0 10
13
mol dm
3
log ‰H3 O‡ Š
log …2:0 10
13
†
ˆ 12:70
Worked example 15.10
pH of an aqueous solution of CH3 NH2 (methylamine)
CH3 NH2 has a pKb value of 3.34. What is the pH of an aqueous solution of
CH3 NH2 of concentration 0.10 mol dm 3 ?
Methylamine is a weak base and is partially dissociated in aqueous
solution:
CH3 NH2 …aq† ‡ H2 O…l† „ CH3 NH3 ‡ …aq† ‡ OH …aq†
The equilibrium concentration of OH ions is not the same as the initial
concentration of CH3 NH2 . To ®nd the concentration of OH ions, we
require the equilibrium constant:
Kb ˆ
‰CH3 NH3 ‡ Š‰OH Š
‰CH3 NH2 Š
From the stoichiometry of the equilibrium, ‰CH3 NH3 ‡ Š ˆ ‰OH Š, and
therefore:
Kb ˆ
‰OH Š2
‰CH3 NH2 Š
where these are equilibrium concentrations. Make the assumption that
‰CH3 NH2 Šequilm ‰CH3 NH2 Šinitial ˆ 0:10 mol dm 3 , and therefore:
Acids and bases in aqueous solution: speciation
459
‰OH Š2 ˆ K b ‰CH3 NH2 Š ˆ K b 0:10
p
‰OH Š ˆ K b 0:10
We are given a value of pKb , and so can ®nd Kb :
Kb ˆ 10
ˆ 10
pK
3:34
ˆ 4:6 10
Therefore:
‰OH Š ˆ
4
p
4:6 10 4 0:10
ˆ 6:8 10
3
mol dm
3
Check that the assumption ‰CH3 NH2 Šequilm ‰CH3 NH2 Šinitial was valid: the
initial concentration of CH3 NH2 was 0.10 mol dm 3 and the equilibrium
concentration of OH is 6:8 10 3 mol dm 3 . The degree of dissociation
is therefore 6.8%, and this is at the limits of acceptability for the approximation. As an exercise, solve the problem rigorously (see worked example 15.8)
and show that the accurate value of ‰OH Š ˆ 6:6 10 3 mol dm 3 .
Taking the value of ‰OH Š ˆ 6:8 10 3 mol dm 3 , we now have to ®nd
‰H3 O‡ Š, and this is related to ‰OH Š as follows:
K w ˆ ‰H3 O‡ Š‰OH Š ˆ 1:0 10
‰H3 O‡ Š ˆ
pH ˆ
ˆ
1:0 10
‰OH Š
14
ˆ
14
1:0 10
6:8 10
…at 298 K†
14
3
ˆ 1:5 10
12
mol dm
3
log ‰H3 O‡ Š
log …1:5 10
12
†
ˆ 11:82
15.6
Acids and bases in aqueous solution: speciation
Dibasic acids
In the last section, we considered the pH of aqueous solutions of monobasic
acids. Many acids are polybasic (see, for example, equations 15.45±15.47 and
Table 15.3). To illustrate how to determine the pH of solutions of such acids,
we consider the dibasic acids listed in Table 15.4. We have already seen that
the ®rst dissociation step lies further to the right-hand side than the second
step, i.e. Ka …1† > Ka …2† or pKa …1† < pKa …2†. However, there is no general
Table 15.4 pKa values for selected dibasic acids.
Compound
Formula
pKa (1)
pKa (2)
Hydrogen sul®de
Carbonic acid
Oxalic acid (see 15.1)
Sulfuric acid
H2 S
H2 CO3
HO2 CCO2 H or H2 C2 O4
H2 SO4
7.04
6.37
1.23
±
19
10.25
4.19
1.92
460
CHAPTER 15 . Equilibria
way in which to treat the relative importance of the two dissociation steps in
terms of estimating the concentration of H3 O‡ ions in solution. First,
consider H2 S (equations 15.57 and 15.58).
H2 S…aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ HS …aq†
‡
…15:57†
2
HS …aq† ‡ H2 O…l† „ H3 O …aq† ‡ S …aq†
…15:58†
‡
From Table 15.4, pKa …2† pKa …1†, and most of the H3 O in an aqueous
solution of H2 S arise from equilibrium 15.57. In determining the pH of an
aqueous solution of H2 S, we can e€ectively ignore equilibrium 15.58 and
treat the system as though it were a monobasic acid. Thus, for an H2 S solution of concentration 0.010 mol dm 3 , the pH is found as follows:
K a …1† ˆ 10
7:04
ˆ 9:1 10
8
ˆ
‰H3 O‡ Š‰HS Š ‰H3 O‡ Š2
ˆ
‰H2 SŠ
‰H2 SŠ
Since Ka …1† is small, ‰H2 SŠequilm ‰H2 SŠinitial , and therefore:
‰H3 O‡ Š2 ˆ 9:1 10 8 0:010
p
‰H3 O‡ Š ˆ 9:1 10 8 0:010 ˆ 3:0 10
pH ˆ
5
mol dm
3
log ‰H3 O‡ Š ˆ 4:52
Now consider H2 SO4 for which the scenario is very di€erent from that of
H2 S. The ®rst dissociation step for dilute aqueous H2 SO4 goes to completion
(equation 15.41) and the concentration of H3 O‡ ions formed in this step
equals the initial concentration of acid. For the second dissociation step,
Ka …2† ˆ 1:2 10 2 and therefore the amount of H3 O‡ ions produced in
this step cannot be ignored (see problem 15.21). For carbonic and oxalic
acids (Table 15.4), the values of Ka …1† are 103 or 104 times greater than
the corresponding value of Ka …2† and in each case, the total concentration
of H3 O‡ can be approximated to that due to the ®rst dissociation step.
The examples in Table 15.4 illustrate that each polybasic acid must be treated as an individual case, although in many instances, the H3 O‡ ions produced in the ®rst dissociation step are the major contribution to the total
concentration of H3 O‡ .
Speciation in aqueous solutions of acids: the effect of pH
HO
O
C
C
O
OH
(15.1)
HO
HA…aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ A …aq†
O
C
C
O–
O
(15.2)
–O
O
C
C
O–
O
(15.3)
Consider equilibrium 15.59 where HA is a weak acid. How is the equilibrium
position a€ected by a change in pH? The addition of acid shifts the
equilibrium to the left-hand side, while the addition of alkali shifts it to the
right-hand side (Le Chatelier's principle).
…15:59†
The situation for polybasic acids is more complicated, with the positions of
two or more equilibria being altered by a change in pH. We can quantify what
happens using speciation curves of the type shown in Figure 15.3 for oxalic
acid. Oxalic acid, 15.1, is a dicarboxylic acid, and has two ionizable protons:
pKa …1† ˆ 1:23 and pKa …2† ˆ 4:19. The ®rst dissociation step gives 15.2, and
the second step produces 15.3. At very low pH, the dominant species is
HO2 CCO2 H, and as the pH is gradually increased, ®rst HO2 CCO2 becomes
dominant, and then O2 CCO2 2 . The concentrations of each species present at
a given pH can be determined as follows. Consider a general dibasic acid H2 A
(equilibria 15.60 and 15.61) for which the acid dissociation constants are Ka …1†
and Ka …2†. As you work through the calculations below, remember that, if the
Acids and bases in aqueous solution: speciation
461
Fig. 15.3 Speciation curves for
aqueous oxalic acid as a function
of pH. The red curve corresponds
to undissociated HO2 CCO2 H in
solution, the green curve to
HO2 CCO2 and the blue curve
to O2 CCO2 2 .
pH of the overall system is varied, equilibria 15.60 and 15.61 are interdependent.
H2 A…aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ HA …aq†
K a …1† ˆ
‰H3 O‡ Š‰HA Š
‰H2 AŠ
…15:60†
‰H3 O‡ Š‰A2 Š
‰HA Š
…15:61†
HA …aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ A2 …aq†
K a …2† ˆ
The total concentration of H2 A, HA
initial concentration of H2 A:
and A2
at equilibrium equals the
‰H2 AŠinitial ˆ ‰H2 AŠequilm ‡ ‰HA Šequilm ‡ ‰A2 Šequilm
For brevity, we shall write this in the form of equation 15.62, where it is
understood that the concentrations on the right-hand side refer to those at
equilibrium.
‰H2 AŠinitial ˆ ‰H2 AŠ ‡ ‰HA Š ‡ ‰A2 Š
…15:62†
By combining equations 15.60±15.62, we can derive equations for each
solution species as a function of Ka …1†, Ka …2† and ‰H3 O‡ Š. To ®nd ‰A2 Š:
‰H2 AŠinitial ˆ ‰H2 AŠ ‡ ‰HA Š ‡ ‰A2 Š
‰H3 O‡ Š‰HA Š
‡ ‰HA Š ‡ ‰A2 Š
ˆ
K a …1†
‰H3 O‡ Š
‡ 1 ‡ ‰A2 Š
ˆ ‰HA Š
K a …1†
‡
‰H3 O Š‰A2 Š
‰H3 O‡ Š
‡ 1 ‡ ‰A2 Š
ˆ
K a …2†
K a …1†
‡ ‰H3 O Š
‰H3 O‡ Š
‡1 ‡1
ˆ ‰A2 Š
K a …2†
K a …1†
‡ ‰H3 O Š
‰H3 O‡ Š ‡ K a …1†
2
‡1
ˆ ‰A Š
K a …2†
K a …1†
‰H3 O‡ Š2 ‡ K a …1†‰H3 O‡ Š
‡1
ˆ ‰A2 Š
K a …1†K a …2†
462
CHAPTER 15 . Equilibria
Therefore, as a fraction of the initial concentration of H2 A, the concentration of ‰A2 Š is:
‰A2 Š
K a …1†K a …2†
ˆ
‰H2 AŠinitial ‰H3 O‡ Š2 ‡ K a …1†‰H3 O‡ Š ‡ K a …1†K a …2†
We can similarly derive two further expressions (see problem 15.22) to ®nd
‰HA Š
‰H2 AŠ
and
:
‰H2 AŠinitial
‰H2 AŠinitial
‰HA Š
K a …1†‰H3 O‡ Š
ˆ
2
‡
‰H2 AŠinitial ‰H3 O Š ‡ K a …1†‰H3 O‡ Š ‡ K a …1†K a …2†
‰H2 AŠ
‰H3 O‡ Š2
ˆ
‰H2 AŠinitial ‰H3 O‡ Š2 ‡ K a …1†‰H3 O‡ Š ‡ K a …1†K a …2†
The curves in Figure 15.3 are constructed using the above expressions for
‰H2 AŠ
‰HA Š
‰A2 Š
,
and
for H2 A ˆ HO2 CCO2 H (15.1) and
‰H2 AŠinitial ‰H2 AŠinitial
‰H2 AŠinitial
pKa …1† ˆ 1:23 and pKa …2† ˆ 4:19. Points to note from the speciation curves
in Figure 15.3 are:
.
.
.
.
.
15.7
the dominance of HO2 CCO2 H under highly acidic conditions;
the dominance of ‰O2 CCO2 Š2 below pH 5.5;
the growth and decay of ‰HO2 CCO2 Š as the pH is gradually raised;
the crossing point of the HO2 CCO2 H and ‰HO2 CCO2 Š curves is at
pH ˆ pKa …1† ˆ 1:23;
the crossing point of the ‰HO2 CCO2 Š and ‰O2 CCO2 Š2 curves is at
pH ˆ pKa …2† ˆ 4:19.
Buffer solutions
What is a buffering effect?
A solution that consists of a
weak acid and its salt, or a
weak base and its salt
possesses a bu€ering e€ect
and can withstand the
addition of small amounts
of acid or base without a
signi®cant change in pH.
The maximum bu€ering
capacity is obtained when
the concentrations of the
weak acid and its salt (or the
weak base and its salt) are
equal.
A solution that possesses a bu€ering e€ect is one to which small amounts of
acid or base can be added without causing a signi®cant change in pH. A
bu€er solution usually consists of an aqueous solution of a weak acid and
a salt of that acid (e.g. acetic acid and sodium acetate) or a weak base and
its salt. To provide the maximum bu€ering capacity, the relative concentrations of the weak acid and its salt (or the weak base and its salt) must be
1 : 1; we return to the reasoning behind this later in the section. Bu€ers are
extremely important in living organisms where a constant pH is essential;
human blood has a pH of 7.4 and is naturally bu€ered. Consider a solution
that is made up aqueous acetic acid and sodium acetate. Equations 15.63 and
15.64 show that CH3 CO2 H is partially dissociated while its sodium salt is
fully dissociated.
CH3 CO2 H…aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ CH3 CO2 …aq†
Acetic acid
Na‰CH3 CO2 Š…s†
Sodium acetate
…15:63†
Acetate ion
Dissolve in water
"
Na‡ …aq† ‡ CH3 CO2 …aq†
…15:64†
Acetate ion
The combination of these two solutions produces a solution with a bu€ering
e€ect. Because CH3 CO2 H is a weak acid, most of it is in the undissociated
Buffer solutions
463
form. The amount of CH3 CO2 from the acid is negligible with respect to
that originating from sodium acetate. If a small amount of acid is added, it
will be consumed by CH3 CO2 and form more CH3 CO2 H in the back
reaction of equilibrium 15.63. Any OH added to the solution will be neutralized by H3 O‡ and equilibrium 15.63 will shift to the right-hand side. We now
consider why these shifts in the equilibrium do not cause signi®cant changes
to the solution pH, provided that appropriate relative concentrations of
acetic acid and sodium acetate are used.
Determining the pH of a buffer solution
Consider 250 cm3 of a solution of 0.10 mol dm 3 CH3 CO2 H. From equation
15.63 and the equilibrium constant …pKa ˆ 4:77†, we know that the concentrations of H3 O‡ and CH3 CO2 ions are equal and small. Now let us
add 250 cm3 0.10 mol dm 3 Na‰CH3 CO2 Š to the same solution. The sodium
salt is fully dissociated (equation 15.64) and so the concentration of
CH3 CO2 ions originating from the salt far exceeds that of CH3 CO2 arising
from the acid. Thus, we can make the approximation that the total concentration of CH3 CO2 in solution equals the initial concentration of
Na‰CH3 CO2 Š. Let this be designated [base] because CH3 CO2 acts as a
base in the bu€er solution. Equation 15.65 gives the expression for Ka for
CH3 CO2 H.
Ka ˆ
‰H3 O‡ Š‰CH3 CO2 Š
‰CH3 CO2 HŠ
…15:65†
We can now apply the following approximations:
.
.
‰CH3 CO2 Š ‰baseŠ;
‰CH3 CO2 HŠequilm ‰CH3 CO2 HŠinitial ˆ ‰acidŠ.
Substitution of these terms into equation 15.65 leads to equation 15.66 which
applies to a solution of a weak acid and its salt.
Ka ˆ
‰H3 O‡ Š‰baseŠ
‰acidŠ
…15:66†
If we now take negative logarithms of both sides of the equation, we derive a
relationship between pH, pKa and the initial concentrations of acid and base
(equation 15.67).
log K a ˆ
ˆ
log
‰H3 O‡ Š‰baseŠ
‰acidŠ
log‰H3 O‡ Š
pK a ˆ pH
log
log
‰baseŠ
‰acidŠ
‰baseŠ
‰acidŠ
pH ˆ pK a ‡ log
‰baseŠ
‰acidŠ
Henderson Hasselbalch equation
…15:67†
This is the Henderson±Hasselbalch equation and can be applied to solutions
consisting of a weak acid and its salt, or a weak base and its salt, as is shown
in worked examples 15.11 and 15.12.
464
CHAPTER 15 . Equilibria
For a solution consisting of a weak acid and a salt of the weak acid:
pH ˆ pK a ‡ log
‰baseŠ
‰acidŠ
For a solution consisting of a weak base and a salt of the weak base:
pH ˆ pK a ‡ log
‰baseŠ
‰acidŠ
where pKa refers to the conjugate acid of the weak base.
Worked example 15.11
pH of a buffer solution consisting of a weak acid and its salt
Determine the pH of a bu€er solution that is 0.050 mol dm
both acetic acid …pKa ˆ 4:77† and sodium acetate.
3
with respect to
Equation needed:
pH ˆ pK a ‡ log
‰baseŠ
‰acidŠ
The initial concentrations of both salt and acid are 0.050 mol dm 3 ; the salt
provides CH3 CO2 which is the base in the bu€er solution.
pH ˆ 4:77 ‡ log
0:050
0:050
ˆ 4:77
This answer illustrates that when the concentrations of acid and base are
equal, the pH of a bu€er solution is the same as the pKa of the acid.
Worked example 15.12
pH of a buffer solution consisting of a weak base and its salt
Determine the pH of a bu€er solution that is 0.027 mol dm 3 with respect to
aqueous NH3 …pKb ˆ 4:75† and 0.025 mol dm 3 with respect to NH4 Cl.
The equations that describe the system are:
NH3 …aq† ‡ H2 O…l† „ NH4 ‡ …aq† ‡ OH …aq†
NH4 Cl…s†
Dissolve in water
"
pK b ˆ 4:75
NH4 ‡ …aq† ‡ Cl …aq†
We can assume that ‰NH4 ‡ Štotal ‰NH4 ‡ Šsalt ˆ 0:025 mol dm 3 . In this
system, the salt provides NH4 ‡ which acts as an acid, and so in the
Henderson±Hasselbalch equation, ‰acidŠ ˆ 0:025 mol dm 3 .
The base in the bu€er solution is NH3 , and ‰baseŠ ˆ 0:027 mol dm 3 .
For NH3 , pKb ˆ 4:75, and for pKa of the conjugate acid NH4 ‡ :
pKa ˆ pKw
pKb ˆ 14:00
4:75 ˆ 9:25
The pH of the solution is therefore given by:
pH ˆ pK a ‡ log
ˆ 9:25 ‡ log
‰baseŠ
‰acidŠ
‰0:027Š
ˆ 9:28
‰0:025Š
465
Buffer solutions
What happens to the pH of a buffer solution on adding base
or acid?
When acid is added to a solution with a bu€ering e€ect, it is consumed by the
weak base present. When base is added to a bu€er solution, it reacts fully with
the weak acid present.
In order to understand why the pH of a solution with a bu€ering e€ect is
relatively insensitive to the addition of small amounts of acid or base, we consider the addition of 0.20 cm3 of HNO3 (0.50 mol dm 3 ) to a bu€er solution
consisting of 50 cm3 0.045 mol dm 3 aqueous CH3 CO2 H (pKa ˆ 4:77) and
50 cm3 0.045 mol dm 3 Na‰CH3 CO2 Š. The total volume before addition is
100 cm3 , and the addition of the HNO3 causes little increase in volume; let us
assume that the total volume remains 100 cm3 . The initial pH of the bu€er solution (before addition of extra acid) is found from the Henderson±Hasselbalch
equation, and because the concentrations of acid and base are equal:
pH ˆ pK a ‡ log
‰baseŠ
‰acidŠ
ˆ 4:77
Now consider the HNO3 . It is a strong acid, so fully dissociated, and H3 O‡
from 0.20 cm3 of a 0.50 mol dm 3 solution of HNO3 reacts completely with
CH3 CO2 in the bu€er solution:
HNO3 …aq† ‡ CH3 CO2 …aq†
"
CH3 CO2 H…aq† ‡ NO3 …aq†
Therefore, some CH3 CO2 is consumed from the bu€er solution, and an
equal amount of CH3 CO2 H is produced.
Amount of added H3 O‡ ˆ Amount of added HNO3
ˆ 0:20 0:50 10
3
4
ˆ 1:0 10
moles
Before HNO3 addition, amount of CH3 CO2 ˆ 50 0:045 10
3
ˆ 2:25 10
3
moles
After HNO3 addition, amount of CH3 CO2 ˆ …2:25 10 3 †
…1:0 10 4 † moles
3
ˆ 2:15 10
New concentration of CH3 CO2 ˆ
3
moles
3
2:15 10 10
100
ˆ 2:15 10
2
mol dm
3
Before HNO3 addition, amount of CH3 CO2 H ˆ 50 0:045 10
ˆ 2:25 10
3
3
moles
After HNO3 addition, amount of CH3 CO2 H ˆ …2:25 10 3 †
‡…1:0 10 4 † moles
ˆ 2:35 10
New concentration of CH3 CO2 H ˆ
2:35 10 3 103
100
ˆ 2:35 10
2
mol dm
3
3
moles
466
CHAPTER 15 . Equilibria
The new pH is therefore:
pH ˆ pK a ‡ log
ˆ 4:77 ‡ log
‰baseŠ
‰acidŠ
…2:15 10 2 †
…2:35 10 2 †
ˆ 4:73
Therefore, there is only a small change in pH (4.77 to 4.73) on adding the
HNO3 to the bu€er solution. If the nitric acid had simply been added to
100 cm3 of water, the pH would have been 3.00:
Concentration of H3 O‡ ˆ
pH ˆ
1:0 10 4 103
ˆ 1:0 10
100
3
mol dm
3
log…1:0 10 3 †
ˆ 3:00
The most e€ective bu€ering capacity is obtained when the bu€er solution has
‰baseŠ
ratio of 1 : 1. When acid or base is added to such solutions, minimal
a
‰acidŠ
‰baseŠ
changes to the
ratio occur and therefore the pH of the solution is little
‰acidŠ
perturbed. From the Henderson±Hasselbalch equation, it can be seen that
‰baseŠ
when
ˆ 1, the pH equals the pKa of the weak acid component of the
‰acidŠ
bu€er solution.
Making up a buffer solution of speci®ed pH
"
Enzymes: see Section 14.15
The uses of bu€er solutions are widespread, and biological and medical
uses are of special importance. For example, experiments with enzymes
(biological catalysts) require bu€ered media because enzyme action is pHspeci®c. Many bu€er solutions are available commercially. To prepare a
bu€er solution of a given pH, the procedure is as follows:
.
.
.
.
Worked example 15.13
Choose a weak acid with a pKa value close to the required pH of
the bu€er; the weak acid may be the salt of a polybasic acid, e.g.
NaH2 PO4 .
Choose an appropriate salt of the weak acid.
‰baseŠ
ratio
Use the Henderson±Hasselbalch equation to determine the
‰acidŠ
needed to attain the correct pH.
‰baseŠ
ˆ 1.
Remember that for maximum bu€ering capacity,
‰acidŠ
Making up a buffer solution
A bu€er solution of pH 7.23 is required. Choose a suitable weak acid from
‰baseŠ
required. Suggest what acid and
Table 15.3, and calculate the ratio of
‰acidŠ
base combination might be appropriate.
The acid of choice in Table 15.3 is ‰H2 PO4 Š …pKa ˆ 7:21). Phosphate
bu€ers are commonly used in the laboratory. From the Henderson±
Acid±base titrations
467
Hasselbalch equation:
pH ˆ pK a ‡ log
‰baseŠ
‰acidŠ
7:23 ˆ 7:21 ‡ log
log
‰baseŠ
‰acidŠ
‰baseŠ
ˆ 0:02
‰acidŠ
‰baseŠ
ˆ 100:02 ˆ 1:05
‰acidŠ
Possible components for the bu€er solution are NaH2 PO4 (acid) and
Na2 HPO4 (base).
15.8
Acid±base titrations
"
pH meter:
see end of Section 17.8
In this section, we look at the pH changes that accompany neutralization reactions. During an acid±base titration, the acid (or base) may be added from a
graduated burette to the base (or acid) contained in a ¯ask. The reaction can
be monitored by measuring the pH of the solution in the ¯ask using a pH meter.
Strong acid±strong base titration
Figure 15.4 shows the change in pH during the addition of 30.0 cm3 of a
0.10 mol dm 3 aqueous solution of NaOH (a strong base) to 25.0 cm3 of a
0.10 mol dm 3 solution of hydrochloric acid (a strong acid). Initially, the
¯ask contains a fully dissociated acid, and the pH is calculated as follows:
‰H3 O‡ Š ˆ ‰HClŠ ˆ 0:10 mol dm
pH ˆ
ˆ
3
log ‰H3 O‡ Š
log …0:10†
ˆ 1:00
This corresponds to the starting point of the titration curve in Figure 15.4. As
aqueous NaOH is added, OH neutralizes H3 O‡ (equation 15.68). As the
concentration of H3 O‡ decreases, the pH value rises.
H3 O‡ …aq† ‡ OH …aq†
3
"
2H2 O…l†
…15:68†
3
When 1.0 cm of 0.10 mol dm aqueous NaOH has been added, the value of
the pH can be determined as follows:
Initial amount of H3 O‡ ˆ 25:0 0:10 10
3
3
Amount of OH added ˆ 1:0 0:10 10
ˆ 1:0 10
Amount of H3 O‡ remaining ˆ …25:0 10 4 †
ˆ 24:0 10
4
ˆ 25:0 10
4
4
moles
moles
…1:0 10 4 †
moles
Total volume after the addition ˆ 26:0 cm3
Concentration of H3 O‡ after addition ˆ
24:0 10 4 103
26:0
ˆ 9:23 10
2
mol dm
3
468
CHAPTER 15 . Equilibria
Fig. 15.4 The variation in pH
during the addition of 30.0 cm3 of
an aqueous solution
(0.10 mol dm 3 ) of NaOH to
25.0 cm3 aqueous HCl
(0.10 mol dm 3 ).
pH ˆ
ˆ
log ‰H3 O‡ Š
log …9:23 10 2 †
ˆ 1:03
To ®nd the value of the pH after 5.0 cm3 of the 0.10 mol dm
NaOH has been added:
Initial amount of H3 O‡ ˆ 25:0 0:10 10
3
Amount of OH added ˆ 5:0 0:10 10
‡
Amount of H3 O remaining ˆ 20:0 10
4
Total volume after the addition ˆ 30:0 cm
Concentration of H3 O‡ after addition ˆ
3
ˆ 25:0 10
ˆ 5:0 10
ˆ
aqueous
moles
moles
moles
3
20:0 10 4 103
30:0
ˆ 6:67 10
pH ˆ
4
4
3
2
mol dm
3
log ‰H3 O‡ Š
log …6:67 10 2 †
ˆ 1:18
The equivalence point of a
strong acid±strong base
titration is at pH 7.00; the
salt formed is neutral.
The pH changes very gradually at ®rst, but then rises sharply (Figure 15.4).
The mid-point of the near-vertical section of the curve in Figure 15.4 is the
equivalence point, i.e. the point at which the alkali has exactly neutralized the
acid, and neither acid nor alkali is in excess. On either side of the equivalence
point, the pH is extremely sensitive to the composition of the solution, with
very small amounts of acid or alkali causing large changes in pH. In a strong
acid±strong base titration, the equivalence point is at pH 7.00. At this point,
the solution contains only a salt (NaCl in the example in Figure 15.4) and
water. The salt formed from the reaction of a strong base with a strong acid
is neutral. From Figure 15.4, the equivalence point occurs when 25.0 cm3 of a
0.01 mol dm 3 NaOH solution have reacted with 25.0 cm3 of a 0.01 mol dm 3
HCl solution, and this is consistent with the stoichiometry of reaction 15.69.
HCl…aq† ‡ NaOH…aq†
"
NaCl…aq† ‡ H2 O…l†
…15:69†
Weak acid±strong base
Figure 15.5 shows the change in pH during the addition of aqueous NaOH
(0.10 mol dm 3 ) to 25 cm3 aqueous acetic acid (0.10 mol dm 3 ). Initially the
Acid±base titrations
469
Fig. 15.5 The variation in pH
during the addition of 30.0 cm3 of
an aqueous solution
(0.10 mol dm 3 ) of NaOH to
25.0 cm3 aqueous CH3 CO2 H
(0.10 mol dm 3 ).
¯ask contains a weak acid (pKa ˆ 4:77) and the initial pH is determined as
follows:
CH3 CO2 H…aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ CH3 CO2 …aq†
Ka ˆ 10
4:77
1:7 10
5
ˆ
ˆ 1:7 10
5
‰H3 O‡ Š‰CH3 CO2 Š
‰H3 O‡ Š2
ˆ
‰CH3 CO2 HŠ
‰CH3 CO2 HŠ
Assume that ‰CH3 CO2 HŠequilm ‰CH3 CO2 HŠinitial .
‰H3 O‡ Š ˆ
p
1:7 10 5 0:10
ˆ 1:3 10
pH ˆ
3
mol dm
3
log ‰H3 O‡ Š ˆ 2:89
This value corresponds to the starting point of the titration curve in Figure
15.5. As alkali is added to the acetic acid, the pH rises gradually, and then
steeply as the equivalence point is reached. By comparing the shapes of the
titration curves in Figures 15.4 and 15.5, it is clear that the addition of a
strong base to a weak acid causes more signi®cant changes in pH early on
in the titration than does the addition of a strong base to a strong acid.
You can verify this by calculating the pH at several points along the
curves and checking the answers against Figures 15.4 and 15.5.
A signi®cant feature of Figure 15.5 is that the equivalence point is not at
pH 7.00. This is a characteristic result for a weak acid±strong base titration:
the equivalence point lies in the range 7:00 < pH < 14:00, the value depending on the pKa of the acid. For the CH3 CO2 H±NaOH titration, the pH at the
end point is found as follows.
At the equivalence point:
CH3 CO2 H…aq† ‡ NaOH(aq)
"
"
All consumed
All consumed
"
Na‰CH3 CO2 Š…aq† ‡ H2 O…l†
"
Salt which determines
the pH of the solution
Consider how the CH3 CO2 ion from the salt behaves in aqueous solution;
CH3 CO2 is the conjugate base of CH3 CO2 H:
CH3 CO2 …aq† ‡ H2 O…l† „ CH3 CO2 H…aq† ‡ OH …aq†
Kb ˆ
‰CH3 CO2 HŠ‰OH Š
‰OH Š2
ˆ
‰CH3 CO2 Š
‰CH3 CO2 Š
470
CHAPTER 15 . Equilibria
We need to ®nd the concentration of CH3 CO2 at the equivalence point:
Total volume of solution at the equivalence point ˆ 25:0 ‡ 25:0 ˆ 50:0 cm3
Amount of CH3 CO2 present ˆ Amount of CH3 CO2 H used in the titration
ˆ 25:0 0:10 10
ˆ 2:5 10
3
3
moles
moles
Concentration of CH3 CO2 at the equivalence point ˆ
2:5 10 3 103
50
ˆ 0:050 mol dm
3
Now, calculate ‰OH Š, and then ‰H3 O‡ Š. From above:
‰OH Š ˆ
p
K b 0:050
We know pKa for CH3 CO2 H and so pKb for the conjugate base is:
pKb ˆ pKw pKa ˆ 14:00 4:77 ˆ 9:23
p
‰OH Š ˆ 10 9:23 0:050 ˆ 5:4 10 6 mol dm
3
Therefore:
‰H3 O‡ Š ˆ
Kw
1:0 10
ˆ
‰OH Š
5:4 10
14
9
3
ˆ 1:9 10
pH ˆ
mol dm
6
log ‰H3 O‡ Š
ˆ 8:72
The equivalence point of a
weak acid±strong base
titration is in the range
7:00 < pH < 14:00; the salt
formed is basic.
This value corresponds to the pH of the midpoint of the near-vertical section
of the titration curve in Figure 15.5 and indicates that sodium acetate is a
basic salt. In solution, it dissociates (equation 15.70), giving the acetate ion
which establishes equilibrium 15.71.
Na‰CH3 CO2 Š
Dissolve in water
"
Na‡ …aq† ‡ CH3 CO2 …aq†
CH3 CO2 …aq† ‡ H2 O…l† „ CH3 CO2 H…aq† ‡ OH …aq†
…15:70†
…15:71†
Strong acid±weak base
Figure 15.6 illustrates the variation in pH during the addition of aqueous
HCl (0.10 mol dm 3 ) to 25 cm3 aqueous ammonia (0.10 mol dm 3 ). Initially
Fig. 15.6 The variation in pH
during the addition of 30.0 cm3 of
an aqueous solution
(0.10 mol dm 3 ) of HCl to
25.0 cm3 aqueous NH3
(0.10 mol dm 3 ).
Acid±base titrations
471
the ¯ask contains a weak base (pKb ˆ 4:75) which is partially dissociated
(equation 15.72).
NH3 …aq† ‡ H2 O…l† „ NH4 ‡ …aq† ‡ OH …aq†
…15:72†
The initial pH is determined as follows:
K b ˆ 10
4:75
1:8 10
5
ˆ
ˆ 1:8 10
5
‰NH4 ‡ Š‰OH Š ‰OH Š2
ˆ
‰NH3 Š
‰NH3 Š
Assume that ‰NH3 Šequilm ‰NH3 Šinitial :
‰OH Š ˆ
p
1:8 10 5 0:10
ˆ 1:3 10
‰H3 O‡ Š ˆ
3
Kw
1:0 10
ˆ
‰OH Š
1:3 10
ˆ 7:7 10
pH ˆ
3
mol dm
12
14
mol dm
3
3
log ‰H3 O‡ Š
ˆ 11:11
This pH value corresponds to the starting point of the titration curve in
Figure 15.6. As strong acid is added to the weak base, the pH falls gradually,
and then steeply as the equivalence point is reached. The shape of the curve
shown in Figure 15.6 is characteristic of a titration involving the addition of a
monobasic strong acid to a monobasic weak base. The end point lies in the
range 0 < pH < 7:00, indicating that the salt formed is acidic. In the reaction
of aqueous NH3 (i.e. NH4 OH) with HCl (equation 15.73), the salt formed
ionizes to give NH4 ‡ and Cl . The salt is acidic because equilibrium 15.74
is established; pKa for NH4 ‡ is 9.25.
The equivalence point of a
strong acid±weak base
titration is in the range
0 < pH < 7:00; the salt
formed is acidic.
"
See problem 15.27 for
con®rmation of the pH of
the equivalence point of this
titration
NH4 OH…aq† ‡ HCl…aq†
‡
"
NH4 Cl…aq† ‡ H2 O…l†
…15:73†
‡
…15:74†
NH4 …aq† ‡ H2 O…l† „ H3 O …aq† ‡ NH3 …aq†
Titrations involving polybasic acids
Each of Figures 15.4, 15.5 and 15.6 shows a single equivalence point because
each neutralization involves the reaction of a monobasic acid with a monobasic base. If NaOH is titrated against a dibasic acid, two equivalence
points are observed. Figure 15.7 shows the change in pH during the addition
Fig. 15.7 The variation in pH
during the addition of 30.0 cm3 of
an aqueous solution
(0.10 mol dm 3 ) of NaOH to
12.5 cm3 aqueous maleic acid
(15.4) (0.10 mol dm 3 ).
472
CHAPTER 15 . Equilibria
H
H
C
HO
C
C
C
O
OH
of a 0.10 mol dm 3 aqueous solution of NaOH to 12.5 cm3 of a
0.10 mol dm 3 aqueous solution of maleic acid, 15.4, which we abbreviate
to H2 A. The two equivalence points occur after the additions of 12.5 and
25.0 cm3 of alkali and correspond to reactions 15.75 and 15.76 respectively.
H2 A…aq† ‡ NaOH…aq†
O
NaHA…aq† ‡ NaOH…aq†
(15.4)
Overall:
"
NaHA…aq† ‡ H2 O…l†
"
H2 A…aq† ‡ 2NaOH…aq†
Na2 A…aq† ‡ H2 O…l†
"
…15:75†
…15:76†
Na2 A…aq† ‡ 2H2 O…l†
The pKa values for acid 15.4 are pKa …1† ˆ 1:92 and pKa …2† ˆ 5:79, i.e. two
weak acids are e€ectively present in the system. The pH of the ®rst equivalence point can be determined as follows.
At the ®rst equivalence point, HA is present in solution and this can
behave as a weak acid:
HA …aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ A2 …aq†
Acid
Base
HA …aq† ‡ HA …aq† „ H2 A…aq† ‡ A2 …aq†
Acid
Base
or as a weak base:
HA …aq† ‡ H2 O…l† „ H2 A…aq† ‡ OH …aq†
Base
Acid
HA …aq† ‡ HA …aq† „ H2 A…aq† ‡ A2 …aq†
Base
Acid
Two of these four reactions are identical because they show HA acting both
as an acid and a base, and it is this equilibrium that dominates in the solution.
(Think about the relative acid and base strengths of HA and H2 O.) Our aim
is to ®nd the pH of this solution, and so we must ®nd the concentration of
H3 O‡ ions. But, this term does not appear in the equilibrium:
HA …aq† ‡ HA …aq† „ H2 A…aq† ‡ A2 …aq†
However, since the position of the equilibrium depends upon the concentrations of H2 A, HA and A2 , it must depend upon Ka …1† and Ka …2†
(equations 15.77 and 5.78).
K a …1† ˆ
‰H3 O‡ Š‰HA Š
‰H2 AŠ
…15:77†
K a …2† ˆ
‰H3 O‡ Š‰A2 Š
‰HA Š
…15:78†
From these equations, we can write:
‰H3 O‡ Š ˆ
K a …1†‰H2 AŠ K a …2†‰HA Š
ˆ
‰HA Š
‰A2 Š
and:
K a …2† ‰H2 AŠ‰A2 Š
ˆ
K a …1†
‰HA Š2
From the stoichiometry of the equilibrium above, ‰H2 AŠ ˆ ‰A2 Š, and so we
can write:
K a …2† ‰H2 AŠ2
ˆ
K a …1† ‰HA Š2
…15:79†
473
Acid±base titrations
Rearranging equation 15.77 gives:
‰H2 AŠ ‰H3 O‡ Š
ˆ
‰HA Š
K a …1†
and substitution of this expression into equation 15.79 gives:
K a …2†
ˆ
K a …1†
‰H3 O‡ Š
K a …1†
2
Therefore:
‰H3 O‡ Š2 ˆ K a …1†K a …2†
p
‰H3 O‡ Š ˆ K a …1†K a …2†
…15:80†
For maleic acid, pKa …1† ˆ 1:92 and pKa …2† ˆ 5:79:
‰H3 O‡ Š ˆ
q
…10 1:92 †…10 5:79 † ˆ 1:40 10
4
mol dm
3
pH ˆ 3:86
This value corresponds to the ®rst equivalence point on Figure 15.7. The
pH can be found directly from the two pKa values. Taking the negative
logarithms of both sides of equation 15.80 gives equation 15.81.
log ‰H3 O‡ Š ˆ
1
log fK a …1†K a …2†g2
ˆ 12 f log K a …1†
log K a …2†g
pH ˆ 12 fpK a …1† ‡ pK a …2†g
The pH of the ®rst
equivalence point in the
titration of a strong base
(e.g. NaOH) against a
polybasic weak acid is
given by:
pH ˆ 12 fpKa …1† ‡ pKa …2†g
…15:81†
This is a general expression for ®nding the pH of the ®rst equivalence point in
the titration of a strong base (e.g. NaOH) against a polybasic weak acid.
Figure 15.7 shows that the second equivalence point in the titration of
NaOH against maleic acid, H2 A, is at pH 9. At this point, all HA has
been converted to A2 (equation 15.76) and a basic salt is present in solution.
To determine the pH of the solution, we must consider equilibrium 15.82.
A2 …aq† ‡ H2 O…l† „ HA …aq† ‡ OH …aq†
Kb ˆ
‰HA Š‰OH Š ‰OH Š
ˆ
‰A2 Š
‰A2 Š
…15:82†
2
To determine the concentration of A2 at the second equivalence point:
Total volume of solution ˆ 25:0 ‡ 12:5 ˆ 37:5 cm3
Amount of A2 present ˆ Amount of H2 A initially
ˆ 12:5 0:10 10
ˆ 1:25 10
3
3
moles
moles
Concentration of A2 at the second equivalence point ˆ
1:25 10 3 103
37:5
ˆ 0:033 mol dm
From above:
‰OH Š ˆ
q q
K b ‰A2 Š ˆ K b 0:033
pKa for HA is 5.79, and so pKb for the conjugate base is:
pK b ˆ pK w pK a ˆ 14:00 5:79 ˆ 8:21
p
‰OH Š ˆ 10 8:21 0:033 ˆ 1:4 10 5 mol dm
3
3
474
CHAPTER 15 . Equilibria
Therefore:
‰H3 O‡ Š ˆ
Kw
1:0 10
ˆ
‰OH Š
1:4 10
ˆ 7:1 10
pH ˆ
10
14
mol dm
5
3
log ‰H3 O‡ Š
ˆ 9:15
"
See problem 15.28
15.9
A titration of aqueous NaOH against a tribasic acid H3 A has three
equivalence points and the pH value of the ®rst equivalence point (where
the predominant solution species is H2 A ) can be found by using equation
15.81. At the second equivalence point, the predominant species is HA2 ,
and the pH can be found using equation 15.83 which can be derived in an
analogous manner to equation 15.81.
pH ˆ 12 fpK a …2† ‡ pK a …3†g
…15:83†
Acid±base indicators
Colour changes of acid±base indicators
OH
HO
When it is desirable to monitor the pH changes during an acid±base titration,
a pH meter is used as described in the previous section. In many titrations,
however, the aim is simply to ®nd the equivalence point and this is detected
by adding an acid±base indicator to the aqueous solution (usually the base) in
the ¯ask and observing a colour change at the end point. Clearly, it is important that the end point accurately coincides with the equivalence point, and
to ensure this is so, indicators must be chosen carefully as described below.
Some common acid±base indicators are listed in Table 15.5.
Acid±base indicators are generally weak acids of the general type HIn
(equation 15.84) in which the conjugate base In has a di€erent colour
from the undissociated acid HIn.
HIn…aq† ‡ H2 O…l† „ H3 O‡ …aq† ‡ In …aq†
Colour I
O
O
…15:84†
Colour II
An example is phenolphthalein, the acid form (15.5) of which is colourless
and the conjugate base of which is pink. By Le Chatelier's principle,
equilibrium 15.84 is sensitive to the pH of a solution. It shifts to the left-hand
side in acidic media and appears as colour I, and moves to the right-hand side
(15.5)
Table 15.5 Selected acid±base indicators in aqueous solution.
Indicator
pKa
Colour change from
basic solution
acidic
pH range in which
indicator changes colour
pink
Colourless
red
Yellow
purple
Yellow
blue
Yellow
yellow
Red
yellow
Red
blue
Yellow
8.00±10.00
6.8±8.2
5.2±6.8
3.8±5.4
3.2±4.4
1.2±2.8
8.0±9.8
"
Phenolphthalein
Phenol red
Bromocresol purple
Bromocresol green
Methyl orange
Thymol blue
9.50
8.00
6.40
4.90
3.46
n
1.65
9.20
"
"
"
"
"
"
"
Acid±base indicators
475
under alkaline conditions, thus appearing as colour II. The acid dissociation
constant for an indicator HIn is given by equation 15.85.
Ka ˆ
‰H3 O‡ Š‰In Š
‰HInŠ
…15:85†
The colour of the solution depends on the ratio of concentrations ‰In Š : ‰HInŠ
(equation 15.86). This expression is usually written in a logarithmic form
(equation 15.87).
For an acid±base indicator
for which the acid form is
HIn, the ‰In Š : ‰HInŠ ratio is
determined from:
log
‰In Š
ˆ
‰HInŠ
pK a ‡ pH
‰In Š
Ka
ˆ
‰HInŠ ‰H3 O‡ Š
log
‰In Š
ˆ log K a
‰HInŠ
log
‰In Š
ˆ
‰HInŠ
…15:86†
log ‰H3 O‡ Š
pK a ‡ pH
…15:87†
Application of this equation can be illustrated by determining the pH
range in which phenolphthalein (pKa ˆ 9:50) changes colour; the colourless
form is HIn, and the pink form is In . From equation 15.87 we can write:
log
‰In Š
ˆ
‰HInŠ
9:50 ‡ pH
We could randomly test values of pH in this equation to determine when In
or HIn dominates as the solution species. However, an important point
should be noted:
when pH ˆ pK a :
log
‰In Š
ˆ0
‰HInŠ
‰In Š
ˆ 1:00
‰HInŠ
The colour must therefore change around this pH value, so we should test
values of the pH on either side of pH ˆ 9:50.
At pH 8.00:
log
‰In Š
ˆ
‰HInŠ
At pH 10.00:
log
‰In Š
ˆ 0:50
‰HInŠ
1:50
‰In Š
ˆ 0:03
‰HInŠ
‰In Š
ˆ 3:16
‰HInŠ
The colourless to pink change occurs when the dominant solution species
changes from being HIn to In . At pH 8.00, ‰HInŠ 33 ‰In Š, and the
solution is colourless. At pH 10.00, ‰In Š ˆ 3:16 ‰HInŠ and the solution
is pink. The pH range from 8 to 10 may seem rather imprecise, but in a
strong acid±strong base titration such as that in Figure 15.4, the pH rises
sharply from 4 to 10 and, at this point, phenolphthalein changes colour
at the addition of a mere drop of base giving an accurate end point. Conversely, a pink to colourless change is observed if a strong acid is titrated
against a strong base.
Choosing an indicator for a titration
Table 15.5 lists the pH ranges over which selected indicators change colour.
If we compare these ranges with the near-vertical sections of the titration
curves in Figures 15.4±15.7, we can immediately see that not every indicator
is suitable for every titration. In order that the end point is accurate, the
indicator must undergo a sharp colour change, and so the pH range for the
change must coincide with the near-vertical part of the titration curve.
476
CHAPTER 15 . Equilibria
For a strong acid±strong base titration (Figure 15.4), the equivalence point
is at pH 7.00 and is in the middle of a large and rapid pH change. Of the indicators in Table 15.5, phenolphthalein, phenol red, bromocresol purple and
bromocresol green are suitable for detecting the end point of this experiment.
In a titration in which a strong base is added to a weak acid (Figure 15.5), the
choice of an appropriate indicator is fairly restricted because the change in
pH around the equivalence point is quite small. We showed in Section 15.8
that the equivalence point for the titration in Figure 15.5 is at pH 8.72. Of
the indicators in Table 15.5, phenolphthalein and thymol blue are suitable;
thymol blue is a dibasic acid and it is the second dissociation that is in
operation in this weak acid±strong base titration. In the strong acid±weak
base titration in Figure 15.6, the equivalence point is at pH 5.28, and the
near-vertical section of curve is relatively small. Since in Figure 15.6 we are
adding acid to a weak base, the pH changes from a higher to lower value
(contrast Figures 15.5 and 15.6 with 15.7) through the equivalence point
pH of 5.28. A suitable indicator for the titration is therefore bromocresol
purple. Many more indicators are available than the few selected in Table
15.5, making the choices less restricted than this discussion might imply.
SUMMARY
In this chapter we have discussed di€erent types of equilibria. Le Chatelier's principle can be used to assess qualitatively how an equilibrium responds to external changes. Quantitative treatments require the use of equilibrium
constants, K, which can be applied to gaseous or solution systems; equilibrium constants are temperaturedependent. For equilibria involving acids and bases, the constants Ka and Kb are de®ned; the self-ionization of
water is described by Kw . We have shown how to calculate the pH of solutions of acids and bases, illustrated
how the solution pH varies during acid±base titrations, and exempli®ed the use of selected acid±base indicators.
Constant pH can be maintained in a bu€er solution.
Useful equations
The meanings of all symbols are given in the chapter.
For:
xX ‡ yY „ zZ
Kˆ
For any pure substance, i:
…aZ †z
…aX †x …aY †y
ai ˆ 1 (dimensionless)
For a solute, i, in a dilute solution:
For an ideal gas, i:
ai ˆ Pi
ai ˆ ‰iŠ
(dimensionless)
(dimensionless)
o
d…ln K† H
ˆ
dT
RT 2
H o
‡c
RT
For an aqueous solution of a weak acid HA:
ln K ˆ
For an aqueous solution of a weak base B:
pKa ˆ
log Ka
Ka ˆ 10
pKa
pKw ˆ
log Kw ˆ 14:00
pKb ˆ
log Kb
Ka ˆ
Kb ˆ
‰H3 O‡ Š‰A Š
‰HAŠ
‰BH‡ Š‰OH Š
‰BŠ
Summary
Kb ˆ 10
477
pKb
Ka Kb ˆ Kw ˆ 1:0 10
14
pKa ‡ pKb ˆ pKw ˆ 14:00
pH ˆ
log ‰H3 O‡ Š
‰H3 O‡ Š ˆ 10
pH
pH ˆ pKa ‡ log
‰baseŠ
‰acidŠ
(Henderson±Hasselbalch equation for a bu€er solution)
Do you know what the following terms mean?
.
.
.
.
.
.
.
.
.
Le Chatelier's principle
activity
molarity and molality
equilibrium constant
(thermodynamic equilibrium
constant)
Brùnsted acid
Brùnsted base
strong acid or base
weak acid or base
acid dissociation constant, Ka
.
.
.
.
.
.
.
.
.
pKa
conjugate acid±base pair
neutralization reaction
base dissociation constant, Kb
pKb
self-ionization constant for
water, Kw
pKw
pH
speciation in an aqueous solution
of a weak, polybasic acid
.
.
.
.
.
.
buffer solution
Henderson±Hasselbalch
equation
acid±base titration
equivalence point (in an acid±
base titration)
acid±base indicator
end point (in an acid±base
titration)
You should be able:
.
.
.
.
.
.
.
.
.
.
.
.
.
to apply Le Chatelier's principle to an equilibrium
to appreciate the approximations made when
writing equilibrium constants in terms of
concentrations or partial pressures
to write down an expression for K given a
stoichiometric equation for an equilibrium
to calculate K given the composition of a system at
equilibrium
to ®nd the composition of a system at equilibrium
given K and amounts of initial reagents
to write down an equation that shows how K
depends on temperature, and to apply it to
determine an approximate value of H o for a
reaction
to discuss how the behaviour of a weak acid (or
base) differs from that of a strong acid (or base) in
solution
to write equilibria to show how weak acids and weak
bases behave in aqueous solutions
to write an expression for Ka or Kb given a
stoichiometric equation for an equilibrium
to calculate Ka or Kb given the composition of an
appropriate system at equilibrium
to ®nd the extent of dissociation of a weak acid or
base given the value of Ka or Kb
to determine Ka from a pKa value, and vice versa
to determine Kb from a pKb value, and vice versa
.
to interrelate pKw , pKa and pKb for a conjugate
acid±base pair
.
to write balanced equations for neutralization
reactions, and appreciate how different salts can
arise from a polybasic acid
.
to determine the pH of an acidic or basic solution
.
to discuss how the position of an equilibrium
involving a weak acid or base is affected by pH
.
to discuss what a buffer solution is, its importance
and how it functions
to write down the Henderson±Hasselbalch equation
and apply it to calculate the pH of a buffer solution
to illustrate why the pH of a buffer solution remains
almost constant when a small amount of acid or
base is added
to explain how to determine the composition of a
buffer solution of a speci®ed pH
.
.
.
.
to discuss the differences between strong acid±
strong base, strong acid±weak base and weak acid±
strong base titrations
.
to explain the meaning of the equivalence point in
an acid±base titration
to explain the variation in pH during the titration of
a dibasic or tribasic acid against a strong base
.
.
to discuss how acid±base indicators work and
illustrate their applications in titrations
478
CHAPTER 15 . Equilibria
PROBLEMS
15.1 Consider the equilibrium:
2SO2 …g† ‡ O2 …g† „ 2SO3 …g†
r H o …298 K† ˆ
96 kJ per mole of SO3
What are the e€ects of (a) increasing the external
pressure, and (b) lowering the external
temperature?
15.2 For NO2 (g), f H o …298 K† ˆ ‡34:2 kJ mol 1 .
(a) Write an equation for the reversible formation
of NO2 from its constituent elements. (b) What is
the e€ect on this equilibrium of raising the external
temperature? (c) If the external pressure is
increased, how is the yield of NO2 a€ected?
15.3 What is the e€ect on the following equilibrium of
(a) adding propanoic acid, and (b) removing
benzyl propanoate by distillation?
CH3 CH2 CO2 H ‡ C6 H5 CH2 OH „
Propanoic acid
Benzyl alcohol
CH3 CH2 CO2 CH2 C6 H5 ‡ H2 O
Benzyl propanoate
Water
15.4 Write down expressions for K in terms of the
activities of the components present in the
following gaseous equilibria:
(a) 2SO2 ‡ O2 „ 2SO3
(b) N2 ‡ 3H2 „ 2NH3
(c) Al2 Cl6 „ 2AlCl3
(d) Cl2 „ 2Cl
(e) H2 ‡ I2 „ 2HI
15.5 Write down expressions for K in terms of the
concentrations of the components present in the
following equilibria. What are the limitations of
using concentrations instead of activities?
(a) C6 H5 CO2 H…aq† ‡ H2 O…l† „
H3 O‡ …aq† ‡ C6 H5 CO2 …aq†
(b) ‰Fe…H2 O†6 Š3‡ …aq† ‡ 6CN …aq† „
‰Fe…CN†6 Š3 …aq† ‡ 6H2 O…l†
2
(c) Cr2 O7 …aq† ‡ 2OH …aq† „ 2CrO4 2 …aq† ‡
H2 O…l†
15.6 I2 is very sparingly soluble in water, and
laboratory solutions are usually made up in
aqueous KI in which the following equilibrium is
established:
I2 …aq† ‡ I …aq† „ I3 …aq†
2.54 g of I2 are added to 1 dm3 of a 0.50 mol dm 3
aqueous solution of KI, and the solution is allowed
to reach equilibrium. At this point, 9:8 10 3
moles of ‰I3 Š are present. Determine the
equilibrium constant, assuming no change in
solution volume on adding solid I2 .
15.7 Ammonia is manufactured in the Haber process:
N2 …g† ‡ 3H2 …g† „ 2NH3 …g†
If 0.50 moles of N2 and 2.0 moles of H2 are
combined at 400 K and 1 bar pressure and the
system is allowed to reach equilibrium, 0.80 moles
of NH3 are present in the system. Determine K
under these conditions.
15.8 The oxidation of SO2 is a stage in the manufacture
of sulfuric acid:
2SO2 …g† ‡ O2 …g† „ 2SO3 …g†
If 2.00 moles of SO2 react with 0.50 moles of O2 at
1100 K and 1 bar pressure and the system is left to
establish equilibrium, the ®nal mixture contains
0.24 moles of SO3 . Calculate K under these
conditions.
15.9 Consider the equilibrium:
H2 …g† ‡ CO2 …g† „ H2 O…g† ‡ CO…g†
At 800 K, K ˆ 0:29. If 0.80 moles of H2 and 0.60
moles of CO2 react under a pressure of 1 bar, how
many moles of CO2 will remain when the reaction
mixture has reached equilibrium?
15.10 The formation of HCl could be considered in
terms of the equilibria:
H2 …g† ‡ Cl2 …g† „ 2HCl…g†
or
1
1
2 H2 …g† ‡ 2 Cl2 …g† „ HCl…g†
What is the relationship between the values of the
equilibrium constants for these equilibria?
15.11 Write equations to show the dissociation in
aqueous solution of the following acids: (a)
CH3 CH2 CO2 H; (b) HNO3 ; (c) H2 SO3 ; (d) H2 SO4 .
15.12 Using data from Table 15.3, determine values of
Ka for HCN and HNO2 .
15.13 The pKa values for citric acid (15.6) are 3.14, 4.77
and 6.39. Write equations to show the dissociation
processes and assign a pKa value to each step.
What are the corresponding Ka values?
OH
H2C
HO2C
C
CO2H
CH2
CO2H
(15.6)
15.14 (a) The pKa values for acetic acid and chloroacetic
acid are 4.77 and 2.85 respectively. Which is the
weaker acid in aqueous solution? (b) The values of
Ka for HOBr and HOCl are 2:1 10 9 and
3:0 10 5 respectively. Which is the stronger acid
in aqueous solution?
15.15 Determine the concentration of CN ions in a
0.050 mol dm 3 aqueous solution of HCN
(pKa ˆ 9:31). How is the concentration of H3 O‡
related to that of CN ?
Problems
15.16 (a) Is KOH completely or partially dissociated in
aqueous solution? (b) What volume of a
0.20 mol dm 3 solution of HNO3 is needed to
completely neutralize 30 cm3 of a 0.40 mol dm 3
solution of KOH? (c) Amines of the type RNH2
behave in a similar manner to NH3 in aqueous
solution. Write an equation to show what happens
when ethylamine …CH3 CH2 NH2 † dissolves in
water. If pKb for ethylamine is 3.19, calculate Kb
and comment on the position of the equilibrium.
15.17 (a) To what equilibrium does a value of
pKa ˆ 9:25 for NH4 ‡ refer? (b) To what
equilibrium does a value of pKb ˆ 4:75 for NH3
refer? (c) Rationalize why, for NH4 ‡ and NH3 ,
…pKa ‡ pKb † ˆ 14:00:
15.18 (a) Find the pH of a 0.10 mol dm 3 solution of
aqueous HCl. (b) What is the change in pH upon
diluting the solution in part (a) by a factor of ten?
(c) What is the pH of a 1 dm3 aqueous solution
that contains 2.00 g of dissolved NaOH?
15.19 Calculate the pH of a 0.25 mol dm 3 aqueous
solution of CH3 CO2 H; pKa ˆ 4:77.
15.20 Find the concentration of OH ions in a
0.40 mol dm 3 aqueous solution of
NH3 …pKb ˆ 4:75†, and hence ®nd the pH of the
solution.
15.21 Determine the pH of an aqueous solution of
sulfuric acid of concentration 0.050 mol dm 3 .
Data: see Table 15.4.
15.22 By referring to equations 15.60±15.62, derive the
equations:
‰HA Š
K a …1†‰H3 O‡ Š
ˆ
2
‡
‰H2 AŠinitial ‰H3 O Š ‡ K a …1†‰H3 O‡ Š ‡ K a …1†K a …2†
and
479
15.26 Sketch a titration curve for the addition of 15 cm3
0.10 mol dm 3 aqueous KOH to 20 cm3
0.050 mol dm 3 aqueous HCl. Determine the pH
values at the start of the titration, at the
equivalence point, and at the end of the titration
when all the alkali has been added. What species
are present in solution at the end of the
experiment?
15.27 Determine the pH of the equivalence point of a
titration in which an aqueous solution
(0.10 mol dm 3 ) of HCl is added to 25.0 cm3
aqueous NH4 OH (0.10 mol dm 3 ); pKa for NH4 ‡
is 9.25.
15.28 (a) Determine the pH values of the ®rst and
second equivalence points during the titration of
20 cm3 aqueous NaOH (0.020 mol dm 3 )
against 20 cm3 H3 PO4 (0.020 mol dm 3 ). pKa
values for H3 PO4 are given in Table 15.3. (b)
Malonic acid has the formula CH2 …CO2 H†2 and
has pKa values of 2.85 and 5.67. Sketch a
titration curve to illustrate pH changes during the
addition of 45.0 cm3 of a 0.10 mol dm 3 solution of
NaOH to 20.0 cm3 of a 0.10 mol dm 3 solution of
malonic acid. Calculate the pH at the equivalence
points.
15.29 Copy Figure 15.5 and mark on it horizontal
bands to correspond to the pH ranges over which
the indicators in Table 15.5 change colour.
Con®rm the choices made in the discussion in the
last part of Section 15.9. Repeat the exercise for
Figures 15.6 and 15.7. What problem is
encountered when choosing an indicator for the
titration shown in Figure 15.7, and how can this be
overcome?
Additional problems
‡ 2
‰H2 AŠ
‰H3 O Š
ˆ
‰H2 AŠinitial ‰H3 O‡ Š2 ‡ K a …1†‰H3 O‡ Š ‡ K a …1†K a …2†
given in Section 15.6 for the speciation of a dibasic
acid H2 A.
15.23 (a) Rationalize the shapes of the curves in Figure
15.3. (b) Using data from Table 15.3, sketch
analogous speciation curves to those in Figure 15.3
to illustrate the behaviour of aqueous H3 PO4 as a
function of pH.
15.24 (a) Brie¯y explain what a bu€er solution is and
qualitatively explain how it functions. (b)
Calculate the pH of a bu€er solution that is
0.50 mol dm 3 with respect to both HCO2 H
…pKa ˆ 3:75† and Na‰HCO2 Š.
15.25 A bu€er solution is prepared by combining 50 cm3
0.025 mol dm 3 Na2 HPO4 and 50 cm3
0.018 mol dm 3 NaH2 PO4 . Find the pH of the
solution. Data: see Table 15.3.
15.30 Acetic acid and ethanol react to establish the
following equilibrium:
CH3 CO2 H ‡ C2 H5 OH „ CH3 CO2 C2 H5 ‡ H2 O
At 298 K, K ˆ 4:0. If 0.30 moles of CH3 CO2 H,
0.45 moles of C2 H5 OH and 0.20 moles of H2 O are
added together and the mixture is allowed to reach
equilibrium, what is the composition of the
equilibrium mixture?
15.31 Look at equation 21.17 in Chapter 21; it shows the
dissociation of ‰Al…H2 O†6 Š3‡ in aqueous solution.
Using data from the equation, calculate the pH of
a 1:20 10 3 mol dm 3 aqueous solution of
[Al(H2 O)6 ]3‡ .
15.32 Calculate the pH of an aqueous solution of HCl of
concentration 1:00 10 7 mol dm 3 .
15.33 What species are present in an aqueous solution of
Na2 CO3 ?