Unit 6 Conductance-II

UNIT 6
Conductance-II

Structure
6.1 Introduction Determination of Solubility and
Solubility Product of Sparingly
Expected Learning Outcomes
Soluble Salts
6.2 Ionic Mobilities and
Determination of Ionic Product of
Transference Number
Water
6.3 Determination of Transference
Determination of Hydrolysis
Numbers
Constant of a Salt
Hittorf’s Method
6.5 Conductometric Titrations
Moving Boundary Method
6.6 Summary
6.4 Applications of Conductivity
6.7 Terminal Questions
Measurements
6.8 Answers
Determination of Ionisation
Constant of a Weak acid

6.1 INTRODUCTION
In the previous unit you have learnt about some aspects of electrolytic
conduction. You have learnt about basic concepts like conductance,
conductivity, and molar conductivity in terms of interrelations between them,
their significance, determination, and factors affecting them. In this unit we
would continue our learning of electrolytic conduction and take up two main
aspects viz., the migration of ions and the application of conductance
measurements.

We would begin the unit by discussing migration of ions in an electrolytic

solution under the influence of applied potential. In this context we would
introduce the concepts of ionic mobility and transference numbers. We would
then take up the principle and experimental methods of determination of
transference numbers. This will be followed by different applications of
conductance measurements. These include determination of: ionisation
constant of weak acids, ionic product of water, solubility and solubility product
of sparingly soluble salts and degree of hydrolysis and hydrolysis constant for
a salt. In addition, we would also take up conductometric titrations as
application of conductance measurements. 143
Block 2 Conductance and Electrochemistry

In the next unit we would take up electrochemical cells in terms of their types,
representation and emf etc.

Expected Learning Outcomes

After studying this unit, you should be able to:

❖ define ionic mobility and outline its significance;

❖ list and explain factors affecting ionic mobility;

❖ define transference numbers and relate them to ionic mobilities;

❖ explain the principles of Hittorf’s and moving boundary methods for the
determination of transference numbers;

❖ describe the procedure of Hittorf’s method and moving boundary method

for the determination of transference numbers;

❖ define indicator electrolyte and list the requirements for an electrolyte to

be indicator electrolyte in the moving boundary method;

❖ solve problems based on Hittorf’s and moving boundary methods;

❖ list some important applications of conductivity measurements;

❖ explain the conductometric determination of ionisation constant of a

weak acid;

❖ determine solubility and solubility product of a sparingly soluble salt

using conductance measurement data;

❖ explain the method of determination of ionic product of water using

conductance measurements;

❖ state the importance of conductometric titrations; and

❖ explain the titration curves for the titration of weak (or strong) acids with
a strong base.

6.2 IONIC MOBILITIES AND TRANSFERENCE

NUMBERS
In any measurement of conductance of an electrolytic solution we get the
amount of current carried by all the ions present in the solution. However,
fraction of the total current carried by each type of ions is not necessarily the
same. It has been found that in dilute solutions of magnesium sulfate the
fraction of current carried by magnesium ion is about 0.38 (38%) of the total
current, whereas the sulfate ions carry the remaining 62%. The difference in
the amount of current carried by the cation and anion can also be inferred by
the limiting molar conductivities of ions about which you have learnt in the
previous unit. You would recall that in an aqueous solution of acetic acid at
infinite dilution the molar conductivities of hydrogen ions and acetate ions are
349.6 S cm2 mol−1 and 40.9 S cm2 mol−1 respectively. Since at infinite dilution,
144 the ions are free from interionic interactions, the difference in these values can
Unit 6 Conductance-II
be attributed to their different speeds or velocities. However, at ordinary
concentrations though the ions do not move independent of other ions, yet
they have different speeds and carry different amount of current. The speed of
different ions is identified in terms of their ionic mobilities. Let us understand
what is ionic mobility of an ion?
6.2.1 Ionic Mobility
In a solution of an electrolyte the ions move randomly. However, in an any act
of measurement of conductivity of the solution an electric field is applied
across the electrodes. The ions then move under the influence of the applied
electrical field. The applied field provides a direction to the ions and
accelerates them. This causes a net migration of ions in the solution towards
oppositely charged electrodes; cations towards cathode and anions towards
anode. In addition, as mentioned above, at moderate concentrations the ions
are not free from interionic interactions. Therefore, the ions experience a
frictional retardation from other ions. Thus, the ions experience two types of
forces viz., electrical field and frictional retardation and these two forces act in
opposite directions. When the acceleration due to the applied electric field
becomes equal to the frictional retardation force the ions drift towards the
opposite electrodes with a net speed called drift speed or drift velocity
denoted by𝑠. The drift speed is proportional to the electrical field, E and we
can write
s = uE …(6.1)
where, the proportionality constant, u is called ionic mobility. If the electrodes
are at a distance of 1m apart and a potential of 1V is applied across them i.e.,
the potential gradient is 1 Vm−1 then we can write
s =u …(6.2)
Thus, we can define ionic mobility to be the velocity with which an ion in a
solution would move under a potential gradient of 1 Vm−1. The SI unit for ionic
mobility can be obtained as follows
s m s −1
u= = = m2 V −1s −1 …(6.3)
E V m −1

The ionic mobilities of some common ions are given in Table 6.1.
Table 6.1: Ionic Mobilities of Some Ions in Water at 298 K

Cation u / (10–8m2s–1V–1) Anions u / (10–8m2s–1V–1)

H+ 36.23 OH– 20.64
Li+ 4.01 F– 5.74
Na+ 5.19 Cl– 7.92
K+ 7.62 Br– 8.09
Mg2+ 5.50 NO3– 7.41
Ca2+ 6.17 SO42– 8.29

The significance of ionic mobilities is that these provide a link between

measurable and theoretical quantities. The theoretical quantity, ionic mobility
(u) is related to measurable quantity, molar conductivity () of the ions as
follows: 145
Block 2 Conductance and Electrochemistry

 = zuF …(6.4)

Where, F is the Faraday’s constant, (F = NA e = 96485 C). For the solution of

an electrolyte at infinite dilution (when interionic interactions do not exist), we
know that

Λm0 = v + 0+ + v −0− …(6.5)

Substituting for molar conductivities of ions from Eq. (6.4) we get,

Λm0 = ( z+u+v + + z−u−v − ) F …(6.6)

Having learnt about ionic mobility and its significance, let us learn about the
factors affecting ionic mobilities

Factors affecting ionic mobilities

The ionic mobility of an ion depends on several factors, some of the important
factors are as under:

• Charge on the ion: an ion with a higher charge has higher ionic mobility
as it would experience a greater force due to electric field (F = zeE),
where z is the charge on the ion, e is the elementary charge and E is the
electric field strength.

• Radius of ion: smaller the ion higher the ionic mobility as a smaller ion
due to smaller surface area experiences lesser friction and can move
easily through the solvent.

• Viscosity of the medium: higher the viscosity, lower the ionic mobility
because the ion will experience a greater frictional force

We can make two important observations from the data given in Table 6.1.
First, that the ionic mobilities of hydrogen and hydroxide ions are exceptionally
high; more than that can be attributed to the small size. Secondly, the trend in
case of alkali metal ions and halide ions are contrary to that expected from
variation in their sizes. Let us learn about the explanation for the observations.
First, we take up high ionic mobilities of hydrogen and hydroxide ions.

The exceptionally high ionic mobilities of hydrogen and hydroxide ions was
explained initially by suggesting that in case of these ions in water there are
two mechanisms by which the conduction takes place. The first one is same
as discussed above for any other ion. The second mechanism called
Grotthus conduction mechanism and is specific to hydrogen and hydroxide
ions. According to this mechanism the proton binds to a neighbouring water
molecule and momentarily converts it to a hydronium ion. This in turn transfers
its proton to another molecule in neighborhood which does the same to
another molecule and this process continues. As a result, a proton in any
region of the solution is effectively transported very quickly to another part.
This mechanism is also known as proton hopping mechanism. A schematic
representation of such a mechanism for movement of proton and hydroxide
ions is given in Fig. 6.1. It is important to note here that the actual mechanism
146 is still under investigation.
Unit 6 Conductance-II

(a) (b)

Fig. 6.1: Schematic representation of Grotthus conduction mechanism for (a) H+

and (b) OH– ions.

Now, we take up the second observation. That is, the trend in the ionic
mobilities of alkali metal ions and halide ions. The observed trend is opposite
to the expected trend because these ions get hydrated and their effective size
increases. The extent of hydration depends on the charge density [charge /
(radius)2] of the ions. The charge on all alkali metal ions is same (1unit positive
charge) but since Li+ ion is the smallest it is heavily hydrated. This makes the
hydrated ion large, and hence it moves only slowly. In comparison though the
Na+ and K+ ions are larger in size; the extent of their hydration is lesser and
hence their effective size is smaller than that of hydrated lithium ion. The sizes
of some alkali metal ions in unhydrated and hydrated state and their ionic
mobilities are given below.

Ion Ionic radius of Approximate radius of Ionic mobility /

unhydrated ion / pm) hydrated ion / pm (10–8m2s–1V–1)

Li+ 76 340 4.01

Na+ 102 276 5.19

K+ 138 232 7.62

6.2.2 Transference Numbers

As discussed above, the current is transported through a solution by migration
of positive as well as negative ions and the ionic mobility or speed of ions are
different. The question comes up is that out of the total current carried in the
solution what is the contribution of different ions? Are these same or different?
How do we determine it? For this we define a new parameter called
transference number (also called transport number). It is defined as the
fraction of current carried by a given type of ion.
In order to derive an expression for transference number, we consider two
electrodes that are separated by a distance d and contain a solution of an
electrolyte between them. Let the solution have n+ positive ions having a
charge of z+ units each and have n− negative ions having a charge of z− units
each. When we apply a unit potential difference across the electrodes these
ions move towards the oppositely charged electrodes. Let us assume that the
cations have a mobility of u+ whereas the anions have a mobility equal to u−. A 147
Block 2 Conductance and Electrochemistry

schematic set up to establish a relationship between ionic mobility and

transference number is given in Fig. 6.2.

Fig. 6.2: A schematic set up to establish a relationship between ionic mobility

and transference number.

Now, let us try to calculate the current transferred by the cations and anions in
one second. We find that all the cations present in the volume of solution
enclosed between the negative electrode (cathode) and the plane A’B’C’D’ in
the figure (this plane is at a distance of u+ from the negative electrode) would
move to the electrode. The current carried by the cations can be calculated as
follows:
Total number of positive ions present in the solution: n+ …(6.7)
Volume of the solution = area ABCD  d …(6.8)
n+
Number of positive ions per unit volume = …(6.9)
area ABCD  d

Volume of the solution from which all the postive ions will reach the
negative electrode = area ABCD  u+ …(6.10)

The number of positive ions reaching electrode in one sec =

n+ nu
=  area ABCD u+ = + + …(6.11)
area ABCD  d d

Quantity of electricity associated with each positive ion = z+e …(6.12)

n+u+ z+e
➔ the current carried by positive ions in one sec = I + = …(6.13)
d
Similarly, in one second all the negative ions between the positive electrode
(anode) and the plane ABCD would reach the electrode. As above, we can
show that
n−u− z−e
➔ the current carried by negative ions in one sec = I − = …(6.14)
d
Total current conducted by the ions in one second =
n+u+ z+e + n−u− z−e
I = I+ + I− = …(6.15)
d
As the overall solution has to be neutral, n+ z+e = n−z−e …(6.16)

Substituting in Eq. (6.15), we get total current conducted in one sec =

n+ z+e(u+ + u− )
I= …(6.17)
148 d
Unit 6 Conductance-II
The fraction of current carried by cations =
I + n+ z+e(u+ ) d u+
t+ = =  = …(6.18)
I d n+ z+e (u+ + u− ) u+ + u−

Similarly, the fraction of current carried by anions =

I − n− z−e(u− ) d u−
t− = =  = …(6.19)
I d n− z−e (u+ + u− ) u+ + u−

u+ u−
Thus, t+ = and t− = …(6.20)
u+ + u− u+ + u−

There are two important corollaries of the expressions for transport numbers.
t+ u+ u + u− u+
=  + = …(6.21)
t − u+ + u− u− u−

i.e., the ratio of transport numbers of two ions in an electrolyte is same as the
ratio of their ionic mobilities
u+ u− u + u−
t+ + t− = + = + =1 …(6.22)
u+ + u− u+ + u− u+ + u−

i.e., the sum of transport number of the cation and the anion is equal to one.

You may note that in this derivation we have assumed that the potential
difference between the two electrodes is 1V therefore the expressions for the
transport numbers are in terms of ionic mobilities which are the ionic velocities
per unit potential. If we use the potential difference other than 1V, then the
expressions would have velocities in place of ionic mobilities; i.e.,
v+ v−
t+ = and t − = …(6.23)
v+ + v− v+ + v−

You would recall that in the beginning of Section 6.2 we mentioned that in a
measurement of electrical conductivity we get the total current carried by the
cations and the anions. To evaluate the contribution of individual ions we
defined the concept of ionic mobility and related it to the molar ionic
conductivity through Eq. (6.6). However, we cannot measure the ionic
mobilities directly. Therefore, we have defined a new parameter viz., transport
number that represents the fraction of total current carried by a given ion in
solution. The transference numbers are related to the ionic mobilities by
Eq. (6.20). The transference numbers can be measured experimentally and
provide a way of getting the individual ionic mobilities. Let us learn about their
experimental determination. However, before that solve the following simple
questions to gauge your understanding.

SAQ 1
Define ionic mobility and give its SI unit.

SAQ 2
What are transference numbers? How are these related to ionic mobilities?
149
Block 2 Conductance and Electrochemistry

6.3 DETERMINATION OF TRANSFERENCE

NUMBERS
As stated above, an experimental determination of transference numbers of
ions in an electrolyte are important to evaluate the contribution of individual
ions to the overall electrical conductivity of the solution of an electrolyte. Three
methods are generally employed for the experimental determination of
transference numbers. These are
• Hittorf’s method
• Moving boundary method
• Emf measurement method
We would take up the first two methods here. The Hittorf’s method is based on
the measurement of change in concentration in the vicinity of the electrodes
whereas in the moving boundary method we study the rate of movement of the
boundary between two solutions under the influence of current. Let us learn
about these methods one by one.
6.3.1 Hittorf’s Method
As mentioned above, in Hittorf’s method we measure the changes in the
concentration of electrolyte in the vicinity of electrode in an electrolytic cell. We
need to first understand why the concentration around the electrodes changes
and how is it related to the transport numbers. For this we consider an
electrolytic cell which has an equivalent number of positive and negative ions,
as shown schematically in Fig. 6.3 (A).
The arguments that follow will be valid for any kind of electrolyte but for
simplicity we are considering a simple uni-univalent electrolyte. This cell is
arbitrarily divided by the imaginary planes into three compartments, which can
be called cathodic, central, and anodic compartments. Before we apply
potential and the electrolysis begins, the condition of the cell is represented by
the upper part of Fig. 6.3 (A) in which each compartment has five equivalents
each of positive and negative ions.
In the first case we assume that only the cations can move under the influence
of an applied potential, and further, two equivalents of these ions move from
the anodic compartment to the cathode. The anions on the other hand are
practically immobile. The situation can be represented by lower part of
Fig. 6.3 (A), where we find that there are two equivalents of unpaired ions
each around the two electrodes. The anions get discharged on the anode by
giving their electrons and these electrons travel through the external circuit to
the cathode where they are taken up by the cations to get discharged. It is
interesting to note that though only the positive ions were able to move (i.e.,
carry current), equivalent amounts of the cation and anions are discharged at
the respective electrodes. This is in accordance with Faraday’s law which you
would have learnt in your school chemistry.
Let us now consider second case where we assume that the cations as well
as the anions can move and move with same velocity (mobility). The initial and
final situation in this case will be represented by Fig. 6.3 (B). Here we have
150 assumed that two equivalents of cations have moved from anodic
Unit 6 Conductance-II
compartment to the cathode and two equivalents of anions have moved from
the cathodic compartment to the anode. As a consequence, there are four
equivalents of unpaired ions each at the two electrodes and as in the first case
these would get discharged at the respective electrodes. Once again, we find
that the amounts of the cation and anions discharged at the electrodes are
equal.

Fig. 6.3: A schematic illustration of depletion around anode and cathode.

In the third case represented by Fig. 6.3 (C) we have assumed that the cations
and anions move at different velocities. The cation to anion velocity is in the
ratio of 2:3. In other words, when in certain time two equivalents of cations
have moved from anodic compartment to cathode, three equivalents of anions
have moved from cathodic compartment to anode. Thus, there are five
equivalents of unpaired ions each at the two electrodes and get discharged
there.

So, what do we find in these illustrative cases? First, we find that the current is This statement is
called as Hittorf’s
carried by those ions which move and secondly the amount of current carried
rule and forms the
by a given kind of ion is proportional to its speed. We can also say that the basis of determination
fraction of current carried by a given type of ion i.e., its transference number is of transference
proportional to its speed. In all these arguments we have made an important numbers by Hittorf’s
assumption that the discharged ions do not react with the material of the method.
electrodes. If we look at Fig. 6.3 again, we can make one more observation
that the loss of concentration of electrolyte around any electrode is
151
Block 2 Conductance and Electrochemistry

proportional to the speed of the ion moving away from it. We can, therefore,
write
Moles of electrolyte lost from the anodic compartmen t t
= + …(6.24)
Moles of electrolyte lost from the cathodic compartmen t t −

This is valid only for a 1:1 electrolyte. Now, the total moles of electrolyte lost
will be sum of the electrolyte lost from the anodic and cathodic compartments.
These can also be obtained from the total amount of current passed through
the cell. We can write,
Moles of electrolyte lost from the anodic compartmen t
t+ = …(6.25)
Total number of moles of electrolyte lost
Moles of electrolyte lost from the cathodic compartmen t
t− = …(6.26)
Total number of moles of electrolyte lost

Let’s learn about the experimental method of determination of transference

numbers by Hittorf’s method.
Experimental determination of transference numbers by Hittorf’s method
The Hittorf’s apparatus for the determination of transference numbers consists
of two vertical glass tubes joined through a U-tube with stopcocks at the tops
of the two limbs of the U-tube. The two vertical glass tubes represent the
anodic and cathodic compartments whereas the U-tube represents the middle
compartment. Each of these tubes is provided with stopcocks at the bottom.
The stopcocks in the two limbs of the U-tube are used to connect / disconnect
the solutions in the cathodic and anodic compartments with the middle
compartment.

Fig. 6.4: Hittorf’s experimental set up for the determination of transport numbers.

The three-tube apparatus (also called Hittorf cell or transport cell) is connected
in series with a silver coulometer through a variable resistance, a milli
ammeter and battery as shown in Fig. 6.4. The milliammeter allows to adjust
the current to any desired value. The anodic and cathodic tubes are provided
with an electrode each. (these could be platinum electrodes or metal
electrodes depending on the system). The cell is filled with the electrolyte for
which the transport numbers are to be determined. The current is turned on,
and the solution is allowed to undergo electrolysis for sufficient amount of time
152 such that there is appreciable change in the concentrations in the anodic and
Unit 6 Conductance-II
cathodic compartments. Thereafter, the solution is drained from each
electrode compartments and analysed. The total quantity of electricity passed
(or total amount of electrolyte discharged) is obtained from the increase in
weight of the cathode in the coulometer attached in series. The analysis of the
solution from the central compartment provides the initial concentration of the
electrolyte (as it is not expected to change). However, if the initial
concentration is known, the analysis of solution from the central compartment
may be skipped. The changes in the amount of electrolyte in the electrode
compartments are then used to determine the transference numbers as
discussed above. Let us understand it with the help of an example.

Example 6.1: A solution containing 19.23 g of silver nitrate in 50 cm3 of solution

was taken in a Hittorf cell using platinum electrodes for the
determination of transport numbers. After the experiment 50 cm3 of
anodic compartment solution contained 18.47 g of silver nitrate. If
0.982 g of metallic silver was deposited on the cathode, calculate
the transport number of the ions.
Solution: We know that the transport number of the cation is defined as
Moles of electrolyte lost from the anodic compartmen t
t+ =
Total number of moles of electrolyte lost

First, we would calculate the total number moles of electrolytes lost. You have
learnt above that it is equal to the total amount of electricity passed. This can
be calculated from the amount of silver deposited on the cathode.
0.982 g
The amount of silver deposited on cathode = = 0.00912 mol
107.88 g mol −1

Decrease in the amount of AgNO3 in the anode compartment =

(19.23 −18.47) = 0.76 g
0.76 g
In terms of number of moles = = 0.00447 mol
169.87 g mol −1
0.00447 mol
Substituting in the expression for t + : t Ag+ = = 0.49
0.00912 mol
We know that t + + t − = 1 So tNO− = 1 − 0.49 = 0.51
3

This was quite a simple and straightforward problem because platinum

electrodes are used, and it is not affected in the anode compartment.
However, if we use silver electrodes then the calculation becomes a little more
involved. Let us take one more example.

Example 6.2: In the determination of transport number a solution of silver nitrate

containing 0.00739 g of silver nitrate per gram of water was taken
in the Hittorf cell using silver electrodes. After a certain amount of
current was passed for enough time, 0.0624 g of silver was
deposited on the cathode. An analysis of the solution from anode
compartment showed that the solution contained18.5 g of water
and 189 mg of silver nitrate. Calculate the transport number of the
silver and nitrate ions.
Solution: We know that the transport number of the cation is defined as 153
Block 2 Conductance and Electrochemistry
Moles of electrolyte lost from the anodic compartmen t
t+ =
Total number of moles of electrolyte lost

Once again, first we would calculate the total number moles of electrolytes
lost. This can be calculated from the amount of silver deposited on cathode.

0.0624 g
The amount of silver deposited on cathode = = 0.000578 mol
107.88 gmol −1

Since in this case we are using silver electrodes, so an equivalent amount of

silver will dissolve from the silver anode i.e., 0.000578 moles. Thus, the
concentration of silver nitrate would increase by 0.000578 moles in the anode
compartment.

We are given the final amount of silver nitrate in the anode compartment as
189 mg of silver nitrate and 18.5 g of water i.e.,

0.189 g
Final number of moles of silver nitrate = = 0.00111mol
169.87 g mol −1

Initial number of moles of silver nitrate in the anode compartment can be

obtained by calculating silver nitrate in 18.5 g of water

18.5 g water  0.00739 g AgNO 3 per g water

= 0.0008 mol
169.87 g mol −1

The number of moles of silver nitrate transferred from anode compartment =

= Initial no. of moles + Moles formed by dissolution of electrode

– final no. of moles

= 0.0008 + 0.000578 – 0.00110 = 0.000278 mol

Substituting in the expression for transport number, we get

0.000278 mol
t A+ = = 0.48
g 0.000578mol

We know that

t+ + t− = 1

So tNO− = 1 − 0.48 = 0.52

3

Having learnt about the Hittorf’s method let us now take up moving boundary
method. However, before that answer the following questions to assess your
understanding.

SAQ 3
List different experimental methods for the determination of transference
numbers.

154
Unit 6 Conductance-II

SAQ 4
A 0.2000 molal solution of copper sulphate was taken in a Hittorf cell having
copper electrodes. After passing suitable current for sufficient time the
cathode solution from the cell was analysed. It was found to weigh 36.4340 g
and contained 0.44125 g of copper. On the other hand, the cathode in the
coulometer showed a deposition of 0.0405 g of silver. Calculate the
transference numbers of the copper and sulphate ions.

6.3.2 Moving Boundary Method

As the name suggests, in the moving boundary method the motion of ions
under the influence of an applied potential is observed in terms of a moving
boundary. It is relatively simple and accurate method and used extensively. In
one of the variants of this method we use two electrolytes, one whose
transference numbers are to be determined (called principal electrolyte) and
the other electrolyte that acts as indicator electrolyte.
The indicator electrolyte must meet the following criteria
• If we wish to measure the cation of the electrolyte, then the indicator
electrolyte must have an anion common with it. For example, if we wish to
determine the transference number of the cation of the electrolyte MA
then the indicator electrolyte would be MA. For anion, it would be MA.
• The speed of the M+ ion shall be less than that of M+ and that of A−
shall be less than that of the A − ion.
In a typical determination, the solutions of these electrolytes are taken in the
moving boundary apparatus. It is a glass tube having uniform diameter
mounted vertically with two electrodes at the ends. Similar to the Hittorf’s
method, these electrodes are connected in series with a silver coulometer
through a variable resistance, a milli ammeter and battery as shown in
Fig. 6.5 (a). The solutions are so arranged that the boundary between them is
quite sharp. The boundary between the solutions is generally located due to
the difference in their refractive indices or the colour. Further, to avoid mixing
of solutions and loss of the boundary, the solutions are placed in the order of
decreasing density.

Fig. 6.5: Schematic experimental set up for the determination of transport

numbers by moving boundary method. 155
Block 2 Conductance and Electrochemistry

Now, once the suitable indicator electrolyte is identified the two solutions are
loaded as shown schematically in the Fig. 6.5 (b), a distinct boundary is
formed at b. On applying potential, the M+ and M+ ions move towards cathode
and the boundary between them also moves.

After sufficient time the boundary reaches b'. As the mobility of M+ ions is
lesser than the M+ ions these are sometimes called following ions.

The distance bb’ would depend on the speeds of the ions being determined
(M+ ). As there is a uniform potential through the solution MA, distance bb’ will
be proportional to the ionic mobility u+ of the M+ ion and can be used to
determine the transference number. Let us understand it with the experimental
determination of the transference number of hydrogen ions.

Experimental determination of transference numbers by moving

boundary method

In the determination of transference number of hydrogen ions in HCl we use

cadmium chloride as the indicator electrolyte having a common anion. The cell
is filled first with cadmium chloride and HCl is added on top of it and a sharp
boundary is obtained. A platinum wire is inserted at the top of the cell to act
as the cathode and a stick of cadmium metal is used as the anode at the
bottom of the cell. A silver coulometer is connected with the cell in series
through a variable resistance, a milli ammeter and battery as shown in
Fig. 6.6.

Fig. 6.6: Schematic diagram showing experimental set up for the determination
of transference numbers of HCl by moving boundary method.

On applying potential, the hydrogen ions move towards the cathode at the top
of the cell and get discharged as hydrogen gas. As hydrogen ions move
toward the cathode their place is taken by cadmium ions, and the boundary
between two electrolytic solutions also moves upward. After suitable amount
of current is passed for sufficient amount of time the distance moved by the
boundary is used along with the cross-section area of the cell to obtain the
volume swept out by the moving boundary. This along with the total quantity of
electricity passed (as measured from coulometer) is then used to calculate the
transference number of hydrogen ion.

If the distance between b and b′, is 𝑙 cm and a cm2 is the cross-sectional area
156 of the tube, then the volume displaced by the moving boundary is 𝑙a cm3. If the
Unit 6 Conductance-II
–3
concentration of HCl is c mol dm , the number of moles of HCl contained in
this volume would be 10-3 lac. If Q Coulombs of charge is passed through the
cell, then the charge carried by hydrogen ions would be t+Q. Further, as the
charge carried by 10-3 lac moles of H+ ions will be10-3 lacF, we can write

t +Q = 10−3 l  a  c  F …(6.27)
Rearranging we get,
10−3 l  a  c  F
t+ = …(6.28)
Q
Let us take an example to see the application of this expression.

Example 6.3: In the determination of transference numbers of HCl by moving

boundary method a 0.10 M solution of HCl was taken in a cell
having uniform area of cross section of 1.25 cm2. At the end of
experiment the boundary moved by 7.5 cm and 0.12 g of silver
was deposited on the cathode of silver coulometer. Calculate the
transference numbers of ions of HCl.
Solution: We know that for determination of transference number by moving
boundary method,
10−3 l  a  c  F
t+ =
Q
We have been given the values as:
𝑙 = 7.5 cm a = 1.25 cm2 and c = 0.1 mol dm−3
Let us first find the quantity of charge (Q) passed from the coulometer data.
We know that 1 F of charge would have deposited 107.88 g of silver. In other
words
107.88 g Ag  96485 C

96485 C  0.12 g
 0.12 g Ag  = 107.32C
107.88 g

Substituting the data in the expression for t+, we get

10−3  7.5 cm  1.25 cm2  0.1 mol dm−3  96485 C mol −1
t+ = = 0.84
107.32 C

Once we know the transference number of hydrogen ions, we can calculate

the same for chloride ions as
 t − = 1 − 0.84 = 0.16

The data on measurements of the transference number over a range of

concentration of electrolyte in dilute solutions shows that the plot between the
transference number versus square root of concentration is linear. The plot
can be extrapolated to c = 0 to obtain the value of the transference number at
infinite dilution.
Having learnt about transference numbers and their experimental
determination let us take up different applications of conductivity
measurements. However, before that solve the following simple questions to
assess your understanding. 157
Block 2 Conductance and Electrochemistry

SAQ 5
What is an indicator electrolyte? What is its role in moving boundary method?

SAQ 6
In the determination of transference number of Li+ ion by moving boundary
method a 0.10 M solution of LiCl was taken in a cell having uniform area of
cross section of 1.17 cm2. At the end of experiment the boundary moved by
2.1 cm and 0.083 g of silver was deposited on the cathode of silver
coulometer. Calculate the transference numbers Li+ ion.

6.4 APPLICATIONS OF CONDUCTIVITY

MEASUREMENTS
Conductivity measurement finds application in several analytical
determinations. In Unit 5 you have learnt about Kohlrausch’s law of
independent migration of ions according to which at infinite dilution, where all
electrolytes (strong or weak) are completely ionised and the interionic effects
do not exist, each ion migrates independently of its co-ion. There are a
number of important applications of conductivity measurement based on
Kohlrausch’s law of independent migration of ions. Some of these are

• Determination of ionisation constant for a weak acid

• Determination of solubility and solubility product of sparingly soluble salts

• Determination of ionic product of water

• Determination of hydrolysis constant of a salt

Let’s learn about these applications of conductivity measurements one by one

6.4.1 Determination of Ionisation Constant

of a Weak Acid
As discussed in Unit 5, the molar conductivity of a weak electrolyte (e.g., a
weak acid) at a given concentration is low due to its partial ionisation. The
molar conductivity increases with dilution due to further ionisation and at
infinite dilution it becomes maximum where the ionisation is complete (100%).
We can therefore say that the ratio of molar conductivity of a weak electrolyte
at a given concentration to that at infinite dilution would be a measure of
degree of ionisation of the weak electrolyte. In other words, we can say that

Λmc
= …(6.29)
Λm0

The ionisation of a weak acid HA can be represented as

HA (aq) ⇌ H+ (aq) + A − (aq) …(6.30)

If the degree of ionisation of the acid having concentration as c mol dm−3 is 

158 then the equilibrium concentrations of different species would be
Unit 6 Conductance-II
HA (aq) ⇌ +
H (aq) + −
A (aq)
…(6.31)
c(1 −  ) c c

You would recall from Unit 8 of BCHCT-133 course that the ionisation constant
of a weak acid (a weak electrolyte) is given as
c 2
Ka = …(6.32)
(1 −  )
Where, c is the concentration and  is the degree of ionisation of the weak
acid. Substituting the expression for  in ionisation constant expression and
simplifying we get
2
 Λc 
c  m0 
Λ c( Λmc )2 c( Λmc )2
K a =  m c = = …(6.33)
 Λ   Λ0 − Λc  ( Λm0 ) ( Λm0 − Λmc )
1 − m0  ( Λm0 )2  m 0 m 
 Λm   Λ 
  m 
Let us take an example to see its application.

Example 6.4: The molar conductivity of a 0.1 M aqueous solution of acetic acid
is found to be 5.3 S cm2 mol−1. If the molar conductivity at infinite
dilution for acetic acid is 390.5 S cm2 mol−1, calculate the
ionisation constant of acetic acid.
Solution: We are given the following data:
c = 0.1 M, Λmc = 5.3 S cm2mol −1 and Λm0 = 390.5 S cm2mol −1

We know from Eq. (6.33) that ionisation constant is given as

c( Λmc )2
Ka =
( Λm0 ) ( Λm0 − Λmc )

Substituting the values in the above expression we get

0.1 (5.3 S cm2mol −1 )2
Ka =
(390.5 S cm2mol −1 ) (390.5 − 5.3) S cm2mol −1

Ka = 1.87  10−5

6.4.2 Determination of Solubility and Solubility

Product of Sparingly Soluble Salts
You would recall from Unit 9 of the BCHCT-133 course that when a sparingly
soluble salt MX is added to water an extremely small fraction of it dissolves
and generates ions by dissociation. These ions may combine back to give
solid salt. Over a period of time an equilibrium is established between the ions
obtained from the dissociation of dissolved salt and the solid salt i.e., we get a
saturated solution of the salt along with the undissolved solid salt. Such
equilibrium for a salt of MX type in aqueous solution can be represented as
MX(s) + H2O ⇌ M+ (aq) + A − (aq) …(6.34)

As the salt happens to be sparingly soluble, i.e., having extremely low

solubility and all the dissolved salt is ionised we can say that the solution can
159
Block 2 Conductance and Electrochemistry

be taken to be at infinite dilution. In other words, the measured molar

conductivity will be same as the molar conductivity at infinite dilution. We know
that
1000cm3dm−3 
Molar conductivity = S cm2 mol −1 …(6.35)
c
If, s is the molar solubility of the sparingly soluble salt i.e., c = s, we can write
1000 cm3dm−3 
Molar conductivity = Λm = S cm2mol −1 …(6.36)
s
Equating the measured molar conductivity with molar conductivity at infinite
dilution, we get
1000 cm3dm−3 
= Λm0 = S cm2mol −1 …(6.37)
s
Rearranging, the expression for solubility becomes
1000 cm 3 dm −3  ( S cm −1 ) 1000 
S= −1
= …(6.38)
0m 2
( S cm mol ) Λm0

The molar conductivity at infinite dilution can be obtained by using the molar
conductivities of the respective ions at infinite dilution by the following
expression.

Λm0 = v +0+ + v −0− …(6.39)

For the salt, MX we can write

Λm0 (MX) = 0m (M+ ) + 0m ( X− ) …(6.40)

The solubility product of the sparingly soluble salt (MX) can be written as
KSP = [ M+ ] [ X− ] = s 2 …(6.41)

Substituting the value of solubility in the solubility product expression, the

solubility product for the salt MX can be obtained as
2
 1000  
Solubility product = K SP =  0
 (mol 2 dm− 6 )
 …(6.42)
 Λm 

We have taken a 1:1 salt as an example; similar expressions can be obtained

for other types of salts.
Note: It is important to note here that since the solubility of sparingly soluble
salt is extremely low, its solution will have very low value of conductivity. In
such a case we cannot ignore the conductivity of water used for preparing the
solution. Therefore, the value of conductivity to be used in the expression for
solubility and solubility product must be corrected for water. That is,
 (salt ) =  (solution) −  ( water ) …(6.43)

Let us understand it with the help of an example.

Example 6.5: The conductivity of a saturated solution of AgCl in water (having

conductivity = 1.06  10−6 S cm−1) at a temperature of 291 K was
found to be = 2.47  10−6 S cm−1.The molar conductivities of
160 AgNO3, NaCl and NaNO3 at infinite dilution are given as 133.5,
Unit 6 Conductance-II
2 −1
126.3 and 121.6 S cm mol respectively. Calculate the solubility
and solubility product of AgCl.
Solution: According to Eq. (6.38) the solubility in mol dm−3, conductivity and
molar conductivity at infinite dilution are related as
1000 cm3 dm−3   (S cm −1 )
s=
Λm0 (S cm2mol −1)

Let us first calculate the conductivity of the salt. As per Eq. (6.43) the
conductivity is given as
 (salt) =  (solution) −  (water )
Substituting the values, we get
 ( AgCl) = (2.47  10−6 S cm−1) − (1.06  10−6 S cm−1) = 1.41 10−6 S cm−1

Now, the molar conductivity at infinite dilution can be obtained by using

Kohlrausch’s law of independent migration of ions as follows:
Λm0 ( AgCl) = Λm0 ( AgNO 3 ) + Λm0 (NaCl ) − Λm0 (NaNO 3 )

Substituting the values

Λm0 ( AgCl) = (133.5 + 126.3 − 121.6) S cm2 mol −1 = 138.2 S cm2 mol −1

Substituting the values of conductivity and molar conductivity at infinite dilution

in Eq. (6.38) we get
1000 cm3 dm−3   (S cm −1 ) 1000  1.41 10 −6 S cm−1
s= 0 2 −1
= 2 −1
= 1.02  10−5 mol dm−3
Λm (S cm mol ) 138.2 S cm mol

Now for AgCl the expression for solubility product can be written as
K sp = [ Ag + ] [Cl− ] = s 2

Substituting the value of solubility, we get

K sp = (1.02  10 −5 mol dm−3 )2 = 1.04  10 −10 mol 2 dm−6

SAQ 7
The molar conductivity of 0.01 M solution of acetic acid at 298 K was found to
be = 6.80 S cm2mol−1. If the molar conductivities of hydrogen and acetate ions
at infinite dilutions are = 349.8 and 40.9 S cm2mol−1, respectively. What
percentage of acetic acid is ionised at this temperature and concentration?

SAQ 8
Derive an expression for the solubility product of the salt MX2 from conductivity
measurement; assuming the solubility to be s mol dm−3.

6.4.3 Determination of Ionic Product of Water

You have learnt in the Unit-7 of BCHCT-133 course that water is amphoteric in
nature i.e., it can act as a weak acid as well as a weak base. In a sample of
water, a small number of water molecules donate a proton (thus acting as an 161
Block 2 Conductance and Electrochemistry

acid), which is accepted by equal number of other water molecules (acting as

base) as shown below

H2 O (𝑙) + H2O (𝑙) ⇌ H3O+ (aq) + OH− (aq) …(6.44)

As a consequence, a small concentration of H3O+ ions and OH− ions are

formed in a sample of water and this process is called autoprotolysis (or
autoionisation) of water. The product of the molar concentrations of H3O+ and
OH− ions in pure water sample at a given temperature is called the ionic-
product constant of water and is denoted as Kw

K w = [ H3O + ] [ OH− ] …(6.45)

We know that
The value of Kw can be experimentally determined from the measurement of
 electrical conductivity of highly purified samples of distilled water. From
Λ=
c Eq. (6.44) we can see that the concentrations of hydrogen and hydroxide ions
If we express  in are equal. So, we can write,
the units of S m−1
and c in mol m−3, [ H3O+ ] = [ OH− ] = (K w )1/ 2 …(6.46)
then,  = c
On the other hand, the conductivity of pure water is related to the
concentration of ions by the following equation.

( H2O) = (H+ ) + ( OH− )  ( H2O) = [H3O+ ] H + [OH − ] OH− …(6.47)

Substituting for the concentration of hydronium and hydroxide ions we can

write,

( H2O) = (K w )1/ 2 (H+ + OH− ) …(6.48)

Since the degree of ionisation of water is exceedingly small, the concentration

of the ions is also exceedingly small. Therefore, the conductivity of the ions
can be taken to be at infinite dilution. We can then write,

( H2O) = (K w )1/ 2 (0H+ + 0OH− ) …(6.49)

The value of  (H2O) has been found to be 5.5  10−6 (S m−1) by Kohlrausch
and Heydweiller. Substituting this value in the above equation along with the
molar ionic conductivities of hydrogen and hydroxide ions we get

5.5  10−6 (S m−1) = (K w )1/ 2 (349.81 10−4 + 198.3  10−4 ) S m2 mol −1 …(6.50)

1 mol Rearranging we get,

1 mol m − 3 =
3
m
5.5  10−6 (S m−1 )
1 mol (K w )1/ 2 = …(6.51)
= (548.1 10− 4 ) S m2 mol −1
3
(10 dm)
1 mol
= Taking squares, we get
1000 dm3
2
= 10 −3 mol dm−3  5.5  10 − 6 (S m−1 ) 
(K w ) =  −1 
= 1.007  10 − 8 mol 2m− 6 …(6.52)
(1 mol m−3 )2 ( 548. 1  10 −4
) S m 2
mol
 
= (10 −3 mol dm−3 )2
= 10 −6 mol2 dm−6 (K w ) = 1.007  10−8 mol 2m−6 = 1.007  10−14 mol 2 dm−6 …(6.53)

Thus, we find that the value of ionic product of water at 298 K is found to be
162 1.007  10−14 mol2 dm−6 on the basis of conductance measurements.
Unit 6 Conductance-II
−14 2 −6
However, generally 1.00  10 mol dm is used as the value of ionic
product of water at 298 K.

6.4.4 Determination of Hydrolysis Constant of a Salt

You have learnt in Unit 9 of BCHCT-133 course that the reaction of anions
or/and cations produced by the salts with water is called salt hydrolysis. On
hydrolysis, the anions of weak acids produce the conjugate acid and hydroxide
ion whereas the cations of weak bases give the conjugate base and
hydronium ion. For example, let us take the hydrolysis of the salt anilinium
hydrochloride-a salt of weak base and strong acid. The salt would ionise as

C6H5NH3Cl (aq) → C6H5NH3+ (aq) + Cl− (aq) …(6.54)

The anilinium ion so produced would react with water and get hydrolysed as

C6H5NH3+ (aq) + H2O(I) ⇌ C6H5NH3OH (aq) + H+ (aq) …(6.55)

The overall reaction of anilinium hydrochloride with water can be given as

C6H5NH 3Cl (aq) + H2O(I) ⇌ C6H5NH 3OH (aq) + HCl(aq) …(6.56)

If we start with 1 mol dm−3 of the salt and assume that the degree of
hydrolysis of the salt at this concentration is  , then at equilibrium the
concentration of different species would be
C6H5NH 3Cl (aq) + H2O(l ) ⇌ C6H5NH 3OH (aq) + HCl (aq)
…(6.57)
(1 −  )  

If we further assume that the base C6H5NH3OH obtained by hydrolysis is very

weak and does not ionise appreciably. Therefore, it will not contribute to the
conductivity of the solution. Thus, the conductivity () of the solution would be
due to (1−) moles of the salt that is ionised but not hydrolysed, and  moles
of the strong acid (HCl). We can write,
0
Λ = (1 − ) Λsalt +  ΛHCl …(6.58)
0
Where, ΛHCl is the molar conductivity of hydrochloric acid at infinite dilution. As
it is a strong acid, it is assumed to be completely ionised. The molar
conductivity (Λ) of the solution can be experimentally measured. On the other
hand, Λsalt, the molar conductivity of anilinium hydrochloride when it is ionised
but not hydrolysed can be obtained by measuring the conductivity of the salt in
presence of excess of aniline. This would completely suppress the hydrolysis
of the salt. Once these values are known, these can be substituted in the
expression and the value of degree of hydrolysis can be obtained as follows.
0
Λsolution = (1 − ) Λsalt + ΛHCl …(6.59)
By rearranging, we get
0
Λsolution − Λsalt =  ( ΛHCl − Λsalt ) …(6.60)
This gives,
Λsolution − Λsalt
= 0
…(6.61)
( ΛHCl − Λsalt )
Now, for the equilibrium reaction given in Eq. (6.57) we can write the
expression for hydrolysis constant as 163
Block 2 Conductance and Electrochemistry

2
Kh = …(6.62)
(1 −  )

Substituting the value of  we can calculate hydrolysis constant. Let’s

understand it with the help of an example.

Example 6.6: The molar conductivity of a 0.011 M solution of aniline

hydrochloride was observed to be 120 S cm2 mol−1. The solution
on saturating with excess of aniline reduced the molar
conductivity to 104 S cm2 mol−1. If the molar conductivity of HCl at
infinite dilution is 410 S cm2 mol−1, calculate the degree of
hydrolysis and the hydrolysis constant for aniline hydrochloride.
Solution: We are given
Λsolution = 120 S cm2 mol −1; Λsalt = 104 S cm2 mol −1; and ΛHCl
0
= 410 S cm2 mol −1
( Λsolution − Λsalt )
From Eq. (6.61) we know that  = 0
( ΛHCl − Λsalt )

Substituting the values, we get

(120 − 104) Scm 2mol −1
= = 0.052
( 410 − 104) Scm 2mol −1

2
From Eq. (6.62) we have, K h =
(1 −  )

Since the degree of hydrolysis is exceedingly small, we can ignore it in the

denominator and also the concentration in this problem is different from 1 so
we can write

Kh = c2

Substituting the value of , and c we get

K h = 0.011 (0.052)2 = 2.97  10−5

Thus, the degree of hydrolysis and hydrolysis constants would be 0.052 and
2.97  10 −5 respectively.

In addition, the conductance measurements can also be used for the

determination of equivalence point in a number of titrimetric determinations.
Let us learn about the principle of different titrimetric determinations using
conductance measurements.

6.5 CONDUCTOMETRIC TITRATIONS

As discussed in Unit 5, the electrical conductance of a solution is a measure of
its current carrying capacity of the ions present in the solution. It is determined
by the type and concentration of ions present in the solution. In the
conductometric titrations we measure the conductance of analyte solution as a
function of the volume of titrant added and use it for the determination of
equivalence point of the titration. This method of the determination of
164 equivalence point becomes relevant when either the solutions are of very low
Unit 6 Conductance-II
concentration or the analyte sample has a dark colour that does not permit the
use of common visual indicators.
In a typical conductometric titration the ionic species to be determined are
converted to non-ionic forms by a suitable reaction. For this purpose, a
measured volume of the solution to be titrated is taken in a beaker and a
conductivity cell is dipped into it. The titrant is added from a burette into the
solution and the conductance readings corresponding to the various additions
of the titrant are measured and plotted against the volume of the titrant. This is
then used to determine the equivalence point. Conductometric determination
of the equivalence point is commonly employed for acid-base and precipitation
titrations. Let us learn about these.
Acid-base titrations
In a typical determination of an acid, the hydrogen ions, mainly responsible for
the conductance, can be neutralised by reacting with a base in an acid-base
titration. Let us understand the changes in the conductance of the solution of a
strong acid say hydrochloric acid with a strong base like sodium hydroxide. To
begin with, in an aqueous solution of hydrochloric acid, the conductance will
be due to the hydrogen ions and chloride ions obtained by the ionisation of the
acid.
HCl (aq) → H+ (aq) + Cl− (aq) …(6.63)

As you know, the hydrogen ions have extremely high conductivity (due to
small size and Grotthus mechanism) whereas the chloride ions are bulkier and
relatively slow moving. Therefore, to begin with, the solution containing
hydrochloric acid would have high conductance. When we add a solution of
sodium hydroxide, it would provide slow moving sodium ions and hydroxide
ions with high mobility as follows.
NaOH (aq) → Na + (aq) + OH− (aq) …(6.64)

The hydroxide ions provided by NaOH would neutralise the hydrogen ions and
produce practically unionised water molecules as per the following reaction.
H+ (aq) + OH− (aq) ⇌ H2O(l) …(6.65)

The overall reaction can be given as

H+ (aq) + Cl− (aq) + Na + (aq) + OH− (aq) → Na + (aq) + Cl− (aq) + H2O …(6.66)

If we see the ions in the reaction carefully, we find that effectively the
hydrogen ions having high conductivity are replaced by relatively slow-moving
sodium ions. As a consequence, the conductance of the solution would
decrease. Adding more and more amount of NaOH solution would neutralise
more and more of hydrogen ions and the conductance would continue to
decrease. This trend would continue till all the hydrogen ions are neutralised
i.e., the equivalence point is reached.
Adding a drop of NaOH solution at this stage would provide additional sodium
ions and hydroxide ions. Therefore, the conductance of the solution would
start to increase; and the trend would continue with addition of more and more
of NaOH solution. If we plot a graph between the conductance of the solution
versus the volume of NaOH solution, we will get the titration curve as given in
Fig. 6.7 (a). 165
Block 2 Conductance and Electrochemistry

(a) (b)
Fig. 6.7: Schematic graphs showing the expected conductometric titration
curves for a) HCl versus sodium hydroxide and b) acetic acid versus
sodium hydroxide titration.

Let us consider another example, viz. conductometric titration between acetic

acid (a weak acid) and NaOH (a strong base). To begin with, in an aqueous
solution of acetic acid, the conductance will be due to the hydrogen ions and
acetate ions obtained from the ionisation of the acid as given below.

CH3COOH (aq) ⇌ CH3COO− (aq) + H+ (aq) …(6.67)

As the acid is weak there will be a small fraction of acetic acid in the ionised
form accordingly the solution will have a low concentration of hydrogen ions
and acetate ions. Therefore, the overall conductance will be low. On adding a
drop of sodium hydroxide, the few hydrogen ions present will combine with the
hydroxide ions provided by the base as shown below.
CH3COO− (aq) + H+ (aq) + Na + (aq) + OH− (aq) → Na + (aq) + CH3COO− (aq) + H2O …(6.68)

Effectively, what happens is that the fast-moving hydrogen ions are replaced
by slower sodium ions and the conductance decreases a bit. The acetate ions
will suppress the ionisation of acetic acid by common ion effect. Once all the
free hydrogen ions are neutralised, further addition of sodium hydroxide reacts
with the unionised acetic acid molecules. The reaction can be represented a
follow.

CH3COOH (aq) + Na + (aq) + OH− (aq) → CH3COO− (aq) + Na + (aq) + H2O …(6.69)

As a result of the reaction, the number of ions in the solution would increase
and the conductance would also increase.

With the addition of more and more of sodium hydroxide this process will
continue, and the conductance of the solution will continue to increase
gradually. This process will continue till the equivalence point where all the
undissociated acetic acid would have reacted with the base. The solution at
this stage will contain sodium and acetate ions in concentration equal to the
initial concentration of the acetic acid.

As in case of titration of HCl, any further addition of sodium hydroxide will

provide additional sodium and hydroxide ions. This would cause a further
increase in the conductance of the solution. However, as the hydroxide ions
166 have quite a high mobility, the increase in the conductance now will be much
Unit 6 Conductance-II
faster. This change in the rate of increase of the conductance marks the
equivalence point of the titration. A typical conductometric titration curve
expected for the titration of acetic acid and sodium hydroxide is given in
Fig. 6.7(b).

As you have learnt earlier that in the measurement of conductance, we

measure the conductance of all the ions present in the column of solution
contained within the electrodes of the conductivity cell. During the titration, the
measured conductance changes due to the change in the concentration of the
ions in the region due to the reaction taking place. In addition to the reaction,
the dilution of the solution during the titration also affects the conductance. It
is, therefore, advantageous to use a titrant solution in a relatively higher
concentration. Generally, in conductometric titrations the titrant is taken about
ten times stronger than the titrand to minimise the dilution effect.

Precipitation titrations

As a typical case of conductometric precipitation titrations, we take the

example of titration of BaCl2 with sulphuric acid. A schematic titration curve for
the titration of barium chloride with sulphuric acid is given in Fig.6.8.

Fig. 6.8: A schematic graph showing the expected conductometric titration

curve for barium chloride versus sulphuric acid titration.

Let us understand the changes in the conductance of the solution of barium

chloride during its titration with sulphuric acid. In a solution containing barium
chloride, initially the conductance of the solution is due to the barium and
chloride ions obtained from the dissociation of BaCl2. Barium chloride being a
strong electrolyte is completely dissociated; dissociation of one mole of barium
chloride generates one mole of barium ions and two moles of chloride ions as
per the following equation.

BaCl2 (aq) → Ba 2+ (aq) + 2 Cl− (aq) …(6.70)

Addition of sulphuric acid provides hydrogen ions and sulphate ions. The
sulphate ions precipitate out the barium ions as per the following equation.

Ba 2+ (aq) + 2Cl− (aq) + 2H+ (aq) + SO24− (aq) → BaSO 4  + 2Cl− (aq) + 2H+ (aq) …(6.71)

The net result of the reaction is that each barium ion present in the solution is
replaced by two hydrogen ions. As the conductivity of two hydrogen ions is
much more than that of a single barium ion; the conductance of the solution 167
Block 2 Conductance and Electrochemistry
increases. This trend continues with further addition of the titrant sulphuric
acid. Once all the barium ions are precipitated i.e., at the equivalence point,
further addition of the titrant provides additional ions in the form of sulphate
and hydrogen ions. This leads to a sharper increase in the solution
conductance. This switchover from a slower increase in conductance to a
sharper increase in the conductance marks the equivalence point. Let us sum
up what have we discussed in this unit. However, before that answer the
following simple question to assess your learning of conductometric titration.

SAQ 9
Draw a schematic conductometric titration curve for the neutralisation of oxalic
acid by NaOH.

6.6 SUMMARY
In this unit we continued our discussion on electrolytic conduction and took up
two main aspects viz., the migration of ions and the application of conductance
measurements. We started the unit by discussing about migration of ions in an
electrolytic solution under the influence of applied potential. In this context we
introduced the concepts of ionic mobility and transference numbers. The ionic
mobility has been defined as the velocity with which an ion moves in a solution
under a potential gradient of 1 V m−1 whereas transference numbers refer to
the fraction of total current carried by a given type of ion in solution. We
established relationship between the two. This was followed by a discussion
on the principle and experimental determination of transference numbers.
Herein, we have explained in detail the principles of Hittorf’s method and
moving boundary method and described the procedure for the determination
of transference numbers by these methods. We used examples to clarify these
methods.

In the context of applications of conductivity measurement, we took up a

number of applications, like determination of ionisation constant of a weak
acid, determination of solubility and solubility product of sparingly soluble salts,
determination of ionic product of water and determination of hydrolysis
constant of a salt. In addition to explaining the principle of these applications
we showed their importance with the help of examples. In the end we took up
another important application of conductivity measurement, viz.,
conductometric titrations and discussed their significance.

6.7 TERMINAL QUESTIONS

1. List and explain different factors affecting ionic mobility.
2. Explain exceptionally high ionic mobilities of hydrogen ions.
3. The molar ionic conductivities of Li+ ions and Na+ ions at 298 K are
3.87  10−3 S m2 mol−1 and 5.01  10−3 S m2 mol−1 respectively. Calculate
their ionic mobilities.
4. In the determination of transport number, a solution of silver nitrate having
1.19 mg of silver nitrate per gram of water was taken in the Hittorf cell
using silver electrodes. After a certain amount of current was passed for
168 enough time, 20.37 mg of silver was deposited on the cathode. An
Unit 6 Conductance-II
analysis of the solution from anode compartment showed that the solution
contained18 g of water and 38.8 mg of silver nitrate. Calculate the
transport number of the silver and nitrate ions.
5. The molar conductivity of 0.001028 M acetic acid solution at 298 K is
observed to be 48.15 S cm−2 mol−1. If the same at infinite dilution is found
to be 390.6 S cm−2 mol−1 calculate the degree of ionisation of acetic acid
at this concentration.

6. The conductivity of a saturated solution of AgCl at 298 K was found to be

3.41  10−6 S cm−1. The water used for dissolving AgCl had a conductivity
of 1.6  10−6 S cm−1. If the molar conductivity at infinite dilution for AgCl at
298 K is given as 138.2 S cm−2 mol−1 calculate the solubility and solubility
product of AgCl at 298 K.
7. Draw and explain the conductometric titration curve for the titration
between formic acid and NaOH.

6.8 ANSWERS
Self-Assessment Questions
1. The ionic mobility may be defined as the velocity with which an ion in a
solution would move under a potential gradient of 1 V m−1. It is denoted as
u and its SI unit is m2 V−1s−1.

2. The transference number is defined as the fraction of total current carried

by a given type of ion in solution. The transference numbers of cation and
anion are denoted as t+ and t− respectively. The transference numbers
and ionic mobilities are related as under
u− u−
t+ = and t− =
u+ + u− u+ + u−

where, u+ and u− respectively are the ionic mobilities of the cation and the
anion.
3. In this case copper sulphate is electrolysed using copper electrodes so
copper ions proportional to the total current passed in the solution (as
determined from the coulometer) would deposit on the cathode thereby
decreasing their concentration by same amount. However, copper ions
proportional to their transport number would move away from anode
compartment and move into the cathode compartment and increase their
concentration.
Let us first calculate the total amount of current passed:

The amount of silver deposited in the coulometer = 0.0405 g

We know that 107.88 g Ag ≡ 1 Faraday
1 0.0405
➔ 0.0405 g Ag = = 3.75  10− 4 F
107.88

As copper ion has a charge of 2 units, 3.75  10−4 F of current would

deposit
169
Block 2 Conductance and Electrochemistry

3.75  10−4
= = 1.877  10−4 moles of Cu2+ ion
2
i.e., 1.87  10−4 mol of copper sulphate will be lost from the cathodic
compartment.
It is given that the cathodic compartment after electrolysis has 0.44125 g
of Cu2+ ions; this is equivalent to
0.44125  159.6
= = 1.1090 g of CuSO 4
63.5

Mass of water in cathodic compartment = 36.4340 − 1.1090 = 35.3250 g

Initially, the solution was 0.2 molal i.e., 0.2  159.6 g of CuSO4 per kg of
water

The mass of CuSO4 initially in 35.3250 g of water was

0.2  159.6
=  35.3250 = 1.1276 g
1000

Loss in the mass of CuSO4 = 1.1276 −1.1090 g = 0.0186 g

0.0186 g
= = 0.0001165 mol
159.6 g mol −1

The expected loss of copper sulphate in the cathodic compartment due to

passage of current was 1.877  10−4 mol. However, the observed loss is
just 1.165  10−4 mol.

➔ (1.877−1.165)  10−4 mol = 7.12  10−5 mol of Cu2+ ions have migrated
into the cathodic compartment or have carried the current.
We know that transference number of cation is
Moles of electrolyt e lost from the anodic compartmen t
t+ =
Total number of moles of electrolyt e lost

Substituting the values,

0.712  10−4
t+ = = 0.379
1.877  10− 4
t − = 1 − t + = 1.0 − 0.379 = 0.621

Thus, the transport numbers of copper and sulphate ions have been found
to be 0.379 and 0.621 respectively

4. The methods generally employed for the experimental determination of

transference numbers are

• Hittorf’s method
• Moving boundary method
• Emf measurement method

5. In moving boundary method for the determination of transference number

of the ions of an electrolyte (called principal electrolyte) the additional
electrolyte that is used to create a boundary is called indicator electrolyte.
170 An indicator electrolyte must have anion or cation common with the
Unit 6 Conductance-II
electrolyte being determined and the other ion should not move faster
than the ion whose transport number is to be determined.

6. We know that for determination of transference number by moving

boundary method,

10−3  l  a  c  F
t+ =
Q

We have been given the values as:

l = 2.1 cm a = 1.17 cm2 and c = 0.10 mol dm−3

Let us first find the value of Q (amount of charge passed) from the
coulometer data. We know that 1 F of charge would have deposited
107.88 g of silver. In other words

96485
107.88 g Ag = 96485 C  1 g Ag = C
107.88

96485  0.083
0.083 g Ag = = 74.23 C
107.88

Substituting the data in the expression, we get

10−3  2.1 cm  1.17 cm2  0.10 mol dm−3  96485 C mol −1

t+ = = 0.319 = 0.32
74.23 C

t − = 1 − 0.32 = 0.68

7. We are given:

c = 0.01 M, Λmc = 16.8 S cm 2 mol −1

First, we calculate the molar conductivity of acetic acid at infinite dilution

by using Kohlrausch’s law. According to the law, the molar conductivity for
acetic acid can be calculated as given below

Λm0 (CH3COOH) = 0m (H+ ) + 0m (CH3COO− ) = (349.8 + 40.9) S cm2 mol −1

0
ΛCH3 COOH
= 390.7 S cm2 mol −1

Further, we know that the percent ionisation of a weak electrolyte is given

as

Λmc
=  100%
Λm0

Substituting the values, we get

16.8 S cm2 mol −1

=  100% = 4.30%
390.7 S cm2 mol −1

8. If we dissolve a sparingly soluble salt of MX2 type in water the solubility

equilibrium can be represented as

MX 2 (S) + H2O (𝑙) ⇌ M2+ (aq) + 2X− (aq) 171

Block 2 Conductance and Electrochemistry

The corresponding expression for Ksp would be : K sp = [ M2 + ] [X − ]2

As the salt has extremely low solubility, we can assume that all the
dissolved salt is ionised and the measured molar conductivity will be same
as the molar conductivity at infinite dilution. We can therefore write

1000 
Molar conductivity = Λm = = S cm2 mol −1 = Λm0
s
Where, s is the molar solubility of the sparingly soluble salt,
Rearranging and simplifying,
1000 cm3 dm−3   (S cm-1 )
s=
Λm0 (S cm2 mol −1 )

The molar conductivity at infinite dilution can be obtained by using the

molar conductivities of the respective ions at infinite dilution by the
following expression.
Λm0 = v + 0+ + v −0−

For the salt, MX2 we can write,

Λm0 (MX 2 ) = 0m (M+ ) + 20m ( X− )

Substituting the value of solubility in the solubility product expression, we

get
3
 1000 
K sp = s(2s ) = 4s = 4
2 3
0
 (mol 3 dm− 9 )

 Λ m 

9. A schematic conductometric titration curve for the titration between oxalic

acid and NaOH is given below

There will be two equivalence points corresponding to the stepwise

neutralisation of diprotic oxalic acid.

Terminal Questions
1. The ionic mobility of an ion depends on the following factors:

• Charge on the ion

• Size of the ion

172 • Viscosity of the medium

Unit 6 Conductance-II
An ion with a higher charge has higher ionic mobility because it would
experience a greater force due to electric field under which it is moving.
The force experienced is given as (F = zeE) where z is the charge on the
ion, e is the elementary charge and E is the electric field strength.
Secondly, a smaller ion has higher ionic mobility because it experiences
lesser friction with solvent molecules and other ions due to smaller
surface area. Thirdly, the ions slow down in a medium of higher viscosity
due to greater retardation force due to friction.

2. The exceptionally high ionic mobilities of hydrogen ions can be explained

in terms of Grotthus conduction mechanism. According to this mechanism
the proton in solution binds to a neighbouring water molecule and
momentarily converts it to a hydronium ion. This in turn transfers its proton
to another water molecule in neighbourhood which does the same to
another molecule and this process continues. As a result, a proton in any
region of the solution is effectively transported quickly to another part.
This mechanism is also known as proton hopping mechanism.

3. We know that the ionic conductivity 𝜆 is related to ionic mobility (𝑢) as

follows

 = zuF

Where, z is the charge on the ion and F is the Faraday’s constant.

The ionic mobilities can be calculated by rearranging the expression as,


u=
zF

Substituting the values.

3.87  10−3 S m2 mol −1

For Li+ ion: u = −1
= 4.011 10− 8 m2 V −1 s −1
1 96485 C mol

5.01 10 −3 S m2 mol −1
For Na+ ion: u = −1
= 5.193  10− 8 m2 V −1 s −1
1 96485 C mol

4. We know that the transport number of the cation is defined as

Moles of electrolyte lost from the anodic compartmen t

t+ =
Total number of moles of electrolyte lost

First, we calculate the total number of moles of electrolytes lost by using

the amount of silver deposited on cathode.

The amount of silver deposited on cathode

0.2037 g
= = 0.000189 mol
107.88 gmol −1

Since in this case we are using silver electrodes, so 0.000189 moles of

silver will dissolve from the silver anode.

We are given the final amount of silver nitrate in the anode compartment
as 38.8 mg of silver nitrate and 18 g of water i.e., 173
Block 2 Conductance and Electrochemistry
0.0388 g
Final number of moles of silver nitrate = = 0.000228mol
169.87 g mol −1

Initial number of moles of silver nitrate in the anode compartment can be

obtained by calculating silver nitrate in 18 g of water

18 g water  0.0119 g AgNO 3 per g water

= = 0.000126 mol
169.87 g mol −1

The number of moles of silver nitrate transferred from anode compartment

= Initial no. of moles + Moles formed by dissolution of electrode −

final no. of moles

= 0.000126 + 0.000189 − 0.000228 = 0.000087

Substituting in the expression for transport number, we get

0.000087mol
t Ag+ = = 0.46
0.000189mol

We know that t + + t − = 1 so tNO− = 1 − 0.46 = 0.54

3

5. We know that the degree of ionisation (𝛼) of the weak electrolyte is given
as

Λmc
=
Λm0

Where, Λmc and Λm0 are the molar conductivities of the weak electrolyte at
a given concentration and at infinite dilution, respectively.

Substituting the values, we get

48.15 S cm2 mol −1

= = 0.123
390.6 S cm2 mol −1

6. We know that the solubility, conductivity, and molar conductivity at infinite

dilution are related as

1000 cm3 dm−3   (S cm−1 )

s=
Λm0 (S cm2 mol −1 )

Let us first calculate the conductivity of the salt only by correcting given
conductivity of solution for water as follows

 (salt) =  (solution) −  (water)

Substituting the values, we get

( AgCl) = (3.41 10−6 ) − (1.60  10−6 ) = 1.81 10−6 S cm−1

The molar conductivity for AgCl at infinite dilution is given as

174 = 138.2 S cm2 mol −1

Unit 6 Conductance-II
Substituting the values of conductivity and molar conductivity at infinite
dilution in the expression for solubility (Eq. (6.38) we get

1000 cm3 dm−3   (S cm−1 )

s=
Λm0 (S cm2 mol −1 )

1000 cm3 dm−3  1.81 10 −6 S cm −1

s= 2 −1
= 1.31 10 − 5 mol dm− 3
138.2 S cm mol

Now for AgCl the expression for solubility product can be written as

K sp = [ Ag + ] [Cl− ) = s 2

Substituting the value of solubility, we get

K sp = (1.31 10 −5 )2 = 1.72  10 −10 mol 2 dm−6

7. Formic acid is a weak acid and its titration with a strong base like NaOH
will be similar to the one described for acetic acid in the text. A schematic
diagram for the conductometric titration between formic acid and NaOH is
given below.

Initially, the solution of formic acid would have a few hydrogen and
formate ions due to the poor ionisation of formic acid; the overall
conductivity is low. Initial addition of sodium hydroxide neutralises the fast
moving hydrogen ions and the conductivity decreases slightly. Further
addition of NaOH neutralises unionised formic acid and the conductivity
increases due to the formation of salt, sodium formate. Once all the formic
acid is neutralised i.e., at the equivalence point further addition of NaOH
increases conductivity sharply due to addition of sodium ions and
hydroxide ions (having high conductivity). The change in the slope
windicates equivalence point.

175