B.Sc II Sem III Chemistry College Name :- Late Ku.Durga K.Banmeru Science College Lonar University Name:- Sant Gadgebaba Amravati University Amravati

1. Mr.Shivshankar Purushottam More
Assistant Professor
Department of Chemisrty
Late Ku.Durga K.Banmeru Science
College,Lonar Dist.Buldana
Captor –VI
Liquid State and Electrochemistry

2. • Introduction
• Matter exists in three states namely solid, liquid and gas. The smallest structural unit of all
chemical substances in these three states may be atoms, ions or molecules. The solid exhibits
complete ordered arrangement of atoms, ions or molecules while the gaseous state exhibit complete
disorder or randomness and the liquid state exhibit only a short-range order. A liquid may be regarded
as a condensed gas or a molten solid.
In a liquid, the are not rigidly fixed as in solids. They have some freedom of motion which is much
more restricted than that in the gases. A liquid, therefore, has a definite volume finite shape.
• It is much less compressible and far denser than a gas. Since,the molecule in a liquid are not far
apart from one another, the intermolecular forces are
• rv strong. The characteristic properties of liquids arise from the nature and the magnitude of these
intermolecular forces
A.Liquid State

3. The surface tension (𝛾) is defined as, the downward force in Newton acting along the
surface of a liquid at right angle to any line I meter in length. The unit of surface
tension in CGS system is dynes per centimeter (dyne cm"). In SI system, the unit is
Newton per meter
(Nm-1). Both these units are related as: 1 dyne cm-1 = 10-3 N m-1
• Consider the molecule 'A' in the interior of liquid, which is surrounded from all
sides by other molecules. Hence, it is attracted equally in all directions. Therefore
molecule 'A’ will behave as if no force is acting on it. However, molecule 'B' which is
at the surface of liquid will experience inward pull because it is attracted sideways
and towards the interior.
A.Surface tension

4. • Thus, the surface of liquid tends to contract
to the smallest possible area and behaves as if
it under tension. This tension which acts
along the surface of liquid is called as surface
tension. It is the reason that drops of a liquid
has spherical shape because for a given
volume sphere has minimum surface area due
to surface tension
• Molecular attraction
A.Liquid State

5. • In drop number method, the number of drop formed for a fixed volume of liquid is
determined by using drop pitter or stalagmometer. The liquid under examination in
sucked up in capillary, Say up to upper mark A. then a definite number of drop say
20,are received in weight bottle and weight. form this weight of single drop is
calculated.
• Let ‘r’ is the radius of capillary tube and 𝛾 is surface tension of liquid. When drop
falls off at that time the weight of drop is equal to the force due to the surface
tension.
• W=2𝜋𝛾---------------(1)
Determination of surface tension by drop number method

6. For relative determinations, instead of determining the
weight of the single drop, number of drops falling between
two fixed marks (one above the bulb and other below the
bulb) are counted for water (reference). The stalagmometer
is dried and filled with another liquid of which surface
tension is to be determined.
Number of drops for the experimental liquid falling
between same two fixed marks is counted. If n, and n2 are the
number of drops falling between two fixed marks for water and
experimental liquid having densities d1 and d2 and surface
tensions liquid falling between same two fixed marks is counted.
Determination of surface tension by drop number method

7. 𝛾1& 𝛾2 then.
When drop falls, wt. of the drop = force due to surface tension
W = mg = vdg = 2𝜋r 𝛾-------------(2)
.. For water, we can write
2𝜋r 𝛾1= v1d1g--------------(3)
For experimental liquid
2𝜋r 𝛾2= v2d2g--------------(4)
Multiplying equation (3) and (4) by n, and n, respectively then,
2𝜋r 𝛾1n2=n1 v1d1g----------------(5)
2𝜋r 𝛾1n2 =v2d2g------------------(6)
But, n1v1=n2v2=V, the volume of liquid falling between two fixed marks
Equations (5) and (6) become
Determination of surface tension by drop number method

8. 2𝜋r 𝛾1n1=V d1 g---------------(7)
2𝜋r 𝛾2n2=V d2 g---------------(8)
Dividing equation (8) by (7), we get
𝛾2
𝛾1
=
𝑛1𝑑2
𝑛2𝑑1
Where ,
𝛾2
𝛾1
= 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛.
Hence by counting number of drop and determining densities, surface tension of any liquid relative to
reference (Water) can be calculated.
Determination of surface tension by drop number method

9. • For all liquid, surface tension (𝛾), decrease with increasing temperature due to
increased molecular agitation, these tend to decrease the effect of intermolecular
cohesive forces.
• At or near the critical temperature, a well defined surface cases to exits and the
surface tension become zero.
Table: Surface Tension values of some common liquids at various temperature.
Effect of temperature on surface tension
Liquid
Surface tension Nm-1 ×102
273K 293K 313K
Water 7.56 7.28 6.96
Banzene 3.16 2.89 2.63
Toulene 3.07 2.84 2.61
Acetone 2.62 2.37 2.12
Ethyl Alcohol 2.40 2.23 2.06

10. • i) It is important in the study of emulsion and colloid chemistry.
• ii). It is an essential factor in the concentration of ores by froth flotation process.
• iii) Surface tension measurements are of much importance in biological science, particularly in
bacteriology.
• iv) The movement of moisture of soil and passage of sap in plants involve the surface tension.
• v)It is used to determine parachor [P]=
𝑀
𝑑
𝛾1/4 a additive and constitutive property used to
investigate molecular structures of compounds.
• vi) in everyday life, soaps and detergents are used for cleaning purposes. Synthetic surfactants have
property of lowering the surface tension of water. Hence, they are used in preparations like tooth
paste, cream, toilet soaps, washing powders, medical emulsions, etc.
• vii) Surface tension measurements are useful in knowing the presence of air bubble in blood stream
and identifying the presence of bile salts in urine (Hay's test for bile salts)
Application of Surface Tension

11. The surface tension of toluene at 293K is 0.028Nm-1 and its density at this temperature is 0.866×103 if the
surface tension of water is 0.07275 Nm-1 and density 0.9982×103 kgm-3m calculate the ration of number of drops
of liquid to that of water.
Solution: - Given
Surface tension of toluene 𝛾 =0.028Nm-1
Surface tension of water 𝛾=0.07275Nm-1
Density of toluene dt=0.866×103kgm-3
Density of water dw=0.9982×103kg-3
Number of drops of toluene =nt
Number of drops of water=nw
We have
𝛾𝑡
𝛾𝑤
=
𝑛𝑤𝑑𝑡
𝑛𝑡𝑑𝑤
or
𝑛𝑡
𝑛𝑤
=
𝛾𝑤𝑑𝑡
𝛾𝑡𝑑𝑤
𝑛𝑡
𝑛𝑤
=
0.07275×0.866×103
0.028×0.9982×103 =2.25
Problem

12. • Every liquid exhibits some resistance to flow. This resistance to flow is called is
called viscosity. It is developed in liquid because of the shearing effect of moving
one layer of liquid past another.
1
2
3
4
5
Motion of liquid
Viscosity of different liquid
Viscosity

13. • A motion of liquid can be visualized as movement of one layer over another. A layer in contact with
stationary surface remains stationary. The second moves slowly, third faster than second and so on. This
type of flow is called as laminar flow or streamlines flow. In laminar flow, The force ‘f’ required to maintain
a steady difference of velocity dv between two parallel layers separated by distance dx proportional to the
area of contact (A) and velocity
𝑑𝑣
𝑑𝑥
• (-) f 𝛼 A.
𝑑𝑣
𝑑𝑥
----------------(10)
• f=ɳ𝐴.
𝑑𝑣
𝑑𝑥
-----------------(11)
• Where,f= Retarding force or viscous drag(acting in opposite direction of flow)
• ɳ=coefficient of viscosity of the liquid
• ɳ =
𝑓
𝐴
𝑑𝑣
𝑑𝑥
------------------(12)
• where, A=1sqm, dv=1msec-1, and dx=1m
• then, ɳ= 𝑓
• the coefficient of viscosity (ɳ) may be defined as the force that must be exerted between two parallel layer
1m2 in area and 1 meter apart in order maintain velocity difference of 1 m sec-1

14. • In CGS system the unit of ɳ is g cm-1.It is called Poise(P) in particle similar units
centipoise (10-2 ) and millipoise (10-3poise) are used.
• The SI unit of viscosity is kg m-1 s-1. then are related as
• 1 poise =1 g cm-1 =0.1 kg m-1s-1
Units of viscosity:

15. • The direct measurement of absolute viscosity of a liquid is very difficult, the
relative viscosity of liquid with respect to reference liquid (water) can be
conveniently determined with the help of apparatus called Ostwald's Viscometer.
• A definite quantity of the liquid under examination is put into the wider limb.
The quantity of liquid should be so taken that bigger bulb is filled more than half of
its volume (This generally requires 10 to 15 cm-3 of liquid). It is then sucked up into
the other limb through a capillary tube. The liquid is allowed to flow through the
capillary attached to the smaller bulb and the time of flow from mark A to mark B is
noted.
Measurement of viscosity by Ostwald’s viscometer method.

16. • When liquid flows through the capillary, the
time flown ‘ t’ is directly proportional to the
viscosity coefficient and inversely to the density ‘d’
of the liquid.
𝑡𝛼
ɳ
𝑑
or ɳ 𝛼 𝑡𝑑 or ɳ = ktd--------(13)
The whole process is then repeated for the same
volume of water, exactly under similar condition then
we have,
ɳ𝑤 = 𝑘𝑡𝑤𝑑𝑤-------------------(15)
When ɳ𝑤, 𝑘𝑡𝑤 𝑎𝑛𝑑 𝑑𝑤 are the coefficient of viscosity,
time of flow and density of water respectively.

17. Form the eq.no.(13) and (14) we get,
ɳ
ɳ𝑤
=
𝑡.𝑑
𝑡𝑤𝑑𝑤
-----------------(15)
ɳ
ɳ𝑤
=ɳ𝑟 ( i.e the relative viscosity of liquid)
Ostwald’s viscometer is a very convenient apparatus for the determination of viscosity at higher
temperature as it can be easily suspended in thermostat.
Effect of temperature on viscosity
It has been found that the viscosity of liquid decrease with rise in temperature. The variation of
viscosity with temperature is best expressed by the equation.
ɳ=𝐴. 𝑒+𝐸/𝑅𝑇
------------------(16)
Where, A and E (Activation energy for viscous flow) are constant for a given liquid taking logarithms.
In ɳ = 𝑖𝑛 𝐴 +
𝐸
𝑅𝑇
-----------------(17)
𝑙𝑜𝑔10ɳ=𝑙𝑜𝑔10A+
𝐸
2.303𝑅𝑇
-----------------(18)

18. Slop=
𝐸
2.303𝑅𝑇
log ɳ
1
𝑇
Hence a plot of log10 1/T should be straight line. the reason why viscosity decrease with temperature
is that as that, the temperature increase, the molecule agitations increase and hence the resistance to
flow may be expected to decrease.

19. 1.Viscosity measurements help in gradation of lubricant oils. In precision instruments such as
watches special kinds of lubricants are needed, which should not change their viscosities very much
with temperature. Therefore, all weather lubricants are manufactured by mixing long chain coiling
polymers with oil.
2.The viscosity measurements yield information regarding the movement of liquid Through Pipes
3. in determination of molecular weight of polymers by viscosity measurements,
4.The study of viscosity has been used by chemist for knowing the constitution of
molecules through Parachor R=
𝑑𝑣
𝑑𝑥
ɳ1/8
which is both additive as well as constative properties.
5. When carbon dioxide gets accumulated in blood, breathing becomes difficult. Due to absorption of
CO2, blood corpuscles swells which in turn, raise the viscosity of blood. This process quickly lead to
heart attack.
6.To the heart patients, doctors prescribe medicines to lower the viscosity of blood in
order to lower the pressure on heart.
Applications of viscosity measurements:

20. Example (2) : Water required 120.5 seconds to flow through a viscometer and the same he volume of acetone
required 49.5 seconds. If the densities of water and acetone at 293K are 9.982 x 102 kg m-3 and 7.92 x 102 kg m
respectively and the coefficient of viscosity of water at 293 K is 10.05 pascal second, calculate the coefficient of
viscosity of acetone at this temperature
Solution:
Given : Flow time of acetone ta = 49.5 sec.
Flow time of water tw = 120.5 sec.
Density of acetone da = 7.92 x 10² kg m
Density of water dw =9.982 x 10² kg m
Coefficient of viscosity of water ɳw= 10.05 Pascal sec.
Coefficient of viscosity of acetone ɳa=?
We have,
ɳ𝑎
ɳ𝑤
=
ɳ𝑎𝑑𝑎
𝑡𝑤𝑑𝑤
or ɳa=
ɳ𝑎𝑑𝑎
𝑡𝑤𝑑𝑤
× ɳ𝑤 ɳa=
49.5×7.92 ×10
.
05
×
10
2
120
.
5
×
9
.
982
×102
=3.257 Pascale Sec.
Problem

21. • Electrochemistry is the branch of physical chemistry which
deals with interdependence of chemical charges and electrical
energies. Electrochemistry holds a central position in chemistry
because it acts as a bridge between thermodynamic and rest of
chemistry and it enable to study ionic reaction in detail.
• A substance which decomposes on passing current through
it is known as electrolyte and phenomenon of decomposition is
called electrolysis.
• Spontaneous oxidation-reduction (redox) reaction can be used to
produce electric current under suitable condition. The device used
for this purpose is called electrochemical cell or Galvanic cell
Chemical reaction can be forced to proceed by passing electric
current. A device used for this purpose is called electrolytic cell.
B. Electrochemistry

22. The capacity of conductor to carry the electrical current (Energy) is
known as the conductance or conductivity. We generally come across
with two conductors. i.e metallic and electrolytic conductors. 𝐶
1
𝑅
It is measured in ohm-1 or mho or siemens (S)
Resistance of any uniform conductor varies directly as its length and
inversely to its area of cross section. 𝑅 𝛼
𝑙
𝑎
= 𝜌.
𝑙
𝑎
Where 𝜌 = constant called specific resistance or resistivity
l= length of conductor
a= cross section area of conductor
Conductance of electrolyte solution.

23. When, l=a=1, R=ρ
Hence, Specific resistance is the resistance of conductor of unit length and unit cross
sectional area or it is resistance of 1m3 material.
Specific conductance (k): “ The conductance of one centimeter cube (1cm3) or one
cubic meter(1m3) solution of an electrolyte is known as specific conductance.” It is
denoted by ‘k’ (Kappa).Specific conductance (k) is reciprocal of specific
conductance(𝜌).
∴ 𝑘 =
1
𝜌
𝑜𝑟 𝑘 = 𝐶.
𝑙
𝑎
𝑂𝑅 𝑘 =
1
𝑅
.
𝑙
𝑎
Where, C = conductance,R=Resistance ,l= length of conductance & a= cross section
area of conductor.
For electrolyte specific conductance is the conductance of one meter cube of solution.
Its CGS unit is ohm-1 or S cm-1 & SI unit is ohm-1 m-1 or S m-1

24. • Conductance of an electrolyte solution is the
reciprocal of resistance (C-1/R) Therefore,
measurement of conductance is done indirectly by
determining the resistance of the solution. The
resistance is measured by Wheatstone AC bridge
method. Direct current cannot be used in process
as this will give wrong result due to 1)Change in
concentration due to electrolyte
2)Change in resistance due to polarization at
electrode.
Determination of conductance of electrolyte solution

25. • These difficulties are overcome by using alternating current within audio
frequencies range and galvanometer in Wheatstone bridge is replaced by headphone.
The schematic diagram of the apparatus is shown in fig.
• The solution whose conductivity is to be determine is taken in a suitable
conductivity cell ‘C’. When current is flowing know resistance ‘R’ is introduced
through resistance box. The sliding contact ‘X’ is then moved along the wire ‘AB’ of
uniform thickness until a point of minimum sound is detected in headphone (G). At
this stage,
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑜𝑥
=
𝐿𝑒𝑛𝑔𝑡ℎ 𝐵𝑋
𝐿𝑒𝑛𝑔𝑡ℎ 𝐴𝑋

26. As all values from above equation are known, resistance of solution can be determined. Form this
resistance, Conductance of solution can be determined.
Determination of cell constant.
Cell constant (𝜃) is defined as the ratio of length( distance) between the electrode ‘l’ and area of cross
section ‘a’ of electrode.
∴ Cell constant (𝜃)=
𝑙
𝑎
In order to determine cell constant (𝜃) it is necessary to determine ‘l’ & ‘d’ but actually it is not
possible. So indirect method based on measurement of conductance of standard KCl solution is
employed as flows:
Specific conductance =
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒
to determine the cell constant, a standard solution of KCl of
known specific conductance at a given temperature is used. Its conductance is determined
experimentally at the same temperature. Substituting the two values is above equation, the cell
constant can be calculated.

27. A Conductivity cell was filled with 0.1M KCl which was known to have specific conductance of 0.1404 mhom-1 at
298K.Its measure resistance at 298K was 99.3 ohm. When the cell was filled with 0.02M AgNO3m,the resistance was
50.3 ohm. Calculate i) Cell Constant,(ii) specific conductance of AgNO3 solution.
Solution: -
For 0.01M KCl K= 0.1404mho-1
R= 99.3 ohm
Specific conductance (k) =cell constant × Observed conductance
∴ Cell Constant=
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒
=
0.1404
1/99.3
=0.1404 ×99.3
=13.94 m-1
ii)For 0.02 M AgNO3, R =50.3 ohm
Sp. Conductance =Cell Constant ×
1
𝑅
=13.94×
1
50.3
=0.2771 mho-1
Problem

28. Example (2): 0.5 N solution of salt occupying volume between two platinum electrodes 0.0172 m apart
and 0.04499 sq. m. area has resistance 25 ohm. Calculate equivalent conductance of solution.
Solution: Given data, Distance between electrodes, 1 = 0.0172 m
area of cross section of electrode a = 0.04499 sq. m, Resistance of salt solution R = 25 ohm
concentration of salt solution = 0.5N
Sp. conductance (k) = Cell constant x Observed conductance
=
0.0172
0.04499
×
1
25
=0.3823 × 0.04
= 0.01529 mho-1
Eq. Conductance 𝜆𝑣= Sp. Conductance × V
𝜆𝑣=k× 𝑉 =
𝑘×1000
𝑁
=
0.01529×1000
0.5
=38.58 S m2 equv-1

29. Example (3): The resistance of conductivity cell was 7.02 ohm when filled with 0.1 N
KCl solution (k=0.1480 ohm m') and 69.2 ohm, when filled with N/100 NaCl solution at same
temperature. Calculate the cell constant and equivalent conductance of NaCl
Solution:
(1) For 0.1 N KCI, R= 7.02 ohm, k = 0.1480 ohm’m
Cell constantCell Constant =
Specific conductance
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒
0.1480
1/7.02
=0.1480 × 7.02 = 1.038 m-1
(ii) For N/100 or 0.01 N NaCl, R = 69.2 ohm
Sp. Conductance k = Cell constant x Observed conductance
1.038 ×
1
69.2
= 0.015 𝑆 m-1
Equivalent conductance = Sp. conductance x V
𝜆𝑣=k× 𝑉 =
𝑘×1000
𝑁
=
0.015×1000
0.01
= 1500 S m² equiv-1

30. • Conductance of an electrolyte solution is the
reciprocal of resistance (C-1/R) Therefore,
measurement of conductance is done indirectly
by determining the resistance of the solution.
The resistance is measured by Wheatstone AC
bridge method. Direct current cannot be used in
process as this will give wrong result due to
1)Change in concentration due to electrolyte
• 2)Change in resistance due to polarization at
electrode.
Determination of conductance of electrolyte solution

31. These difficulties are overcome by using alternating current within audio frequencies range
and galvanometer in Wheatstone bridge is replaced by headphone. The schematic diagram of
the apparatus is shown in fig.
The solution whose conductivity is to be determine is taken in a suitable conductivity cell ‘C’.
When current is flowing know resistance ‘R’ is introduced through resistance box. The
sliding contact ‘X’ is then moved along the wire ‘AB’ of uniform thickness until a point of
minimum sound is detected in headphone (G). At this stage,
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑜𝑥
=
𝐿𝑒𝑛𝑔𝑡ℎ 𝐵𝑋
𝐿𝑒𝑛𝑔𝑡ℎ 𝐴𝑋
As all values from above equation are known, resistance of solution can be determined. Form
this resistance, Conductance of solution can be determined.

32. Cell constant (𝜃) is defined as the ratio of length( distance) between the electrode ‘l’ and area
of cross section ‘a’ of electrode.
∴ Cell constant (𝜃)=
𝑙
𝑎
In order to determine cell constant (𝜃) it is necessary to determine ‘l’ & ‘d’ but actually it is
not possible. So indirect method based on measurement of conductance of standard KCl
solution is employed as flows:
Specific conductance =
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒
to determine the cell constant, a standard solution
of KCl of known specific conductance at a given temperature is used. Its conductance is
determined experimentally at the same temperature. Substituting the two values is above
equation, the cell constant can be calculated.
Determination of cell constant.

33. A Conductivity cell was filled with 0.1M KCl which was known to have specific conductance of 0.1404 mhom-1 at
298K.Its measure resistance at 298K was 99.3 ohm. When the cell was filled with 0.02M AgNO3m,the resistance was
50.3 ohm. Calculate i) Cell Constant,(ii) specific conductance of AgNO3 solution.
Solution: -
For 0.01M KCl K= 0.1404mho-1
R= 99.3 ohm
i) Specific conductance (k) =cell constant × Observed conductance
∴ Cell Constant=
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒
=
0.1404
1/99.3
=0.1404 ×99.3
=13.94 m-1
ii)For 0.02 M AgNO3, R =50.3 ohm
Sp.Conductance =Cell Constant ×
1
𝑅
=13.94×
1
50.3
=0.2771 mho-1
Problem

34. Example (2): 0.5 N solution of salt occupying volume between two platinum electrodes 0.0172 m apart and
0.04499 sq. m. area has resistance 25 ohm. Calculate equivalent conductance of solution.
Solution: Given data
distance between electrodes, 1 = 0.0172 m
area of cross section of electrode a = 0.04499 sq. m.
resistance of salt solution R = 25 ohm
concentration of salt solution = 0.5N
Sp. conductance (k) = Cell constant x Observed conductance
=
0.0172
0.04499
×
1
25
=0.3823 × 0.04
= 0.01529 mho-1
Eq.Conductance 𝜆𝑣= Sp.Conductance × V
𝜆𝑣=k× 𝑉 =
𝑘×1000
𝑁
=
0.01529×1000
0.5
=38.58 S m2 equv-1

35. Example (3): The resistance of conductivity cell was 7.02 ohm when filled with 0.1 N KCl solution (k=0.1480 ohm m')
and 69.2 ohm, when filled with N/100 NaCl solution at same temperature. Calculate the cell constant and equivalent
conductance of NaCl
Solution:
(1) For 0.1 N KCI,
R= 7.02 ohm, k = 0.1480 ohm'm
Cell constant
Cell Constant =
Specific conductance
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒
0.1480
1/7.02
=0.1480 × 7.02 = 1.038 m-1
(ii) For N/100 or 0.01 N NaCl,
R = 69.2 ohm
Sp. Conductance k = Cell constant x Observed conductance
1.038 ×
1
69.2
= 0.015 𝑆 m-1
Equivalent conductance = Sp. conductance x V
𝜆𝑣=k× 𝑉 =
𝑘×1000
𝑁
=
0.015×1000
0.01
= 1500 S m² equiv-1

36. The solution of electrolyte conducts electricity due to the presence of ions. The conductivity
at constant temperature is approximately proportional to number of ions. According to ionic
theory the number of ions increases as the solution of the electrolyte is progressively diluted.
Therefore, equivalent conductivity increases as the dilution increases.
The specific conductivity decreases on dilution because on dilution the number of ions per
dm3 solution decreases in spite of increase in dissociation. The equivalent and molecular
conductivity increases because these are the product of specific conductivity and volume
containing on gram equivalent or one mole of an electrolyte.
𝜆𝑣= k. V On dilution, decrease in specific conductance is compensated by increase in volume
'V'.
Variation of specific and equivalent conductance with dilution

37. • Table: Specific and equivalent conductance of KCI solutions at different concentrations
(dilutions) at 18°C
• It is seen that equivalent conductance increases with dilution. The increase in case of
electrolytes like HCI, KCl and BaCl2, is not so large as in case of acetic acid or NH4OH.
The electrolytes of first category (HCI, KCl, BaCl2, etc.) are known as strong electrolytes
while in those of second category (CH3COOH, NH4OH, etc.) are known as weak
electrolytes.
This behavior can be illustrated from the table.
Concentration (Kg
equiv.per dm3)
Specific conductance
×10-4 (Sm-1)
Equivalent conductance
×10-4 (Sm2equiv-1
1.00 0.0982 98.2
0.10 0.01120 112.0
0.01 0.001223 122.3
0.001 0.0001273 127.3
0.0001 0.00001291 129.1

38. • In case of strong electrolyte the equivalent and molecular conductivity progressively
increase on dilution and rise to maximum limit. This limiting value of conductivity is
describe as equivalent conductivity at infinite dilution (𝜆∞)
• In case of weak electrolyte conductivity continues to increase with dilution for all practical
values of dilution.The maximum is reached when dilution is infinite. Therefore,for weak
electrolyte,There is no indication that limiting value can be attained even when
concentration approaches to zero.

39. • The titration in which end point is determine by measuring change in conductance
of solution upon addition of reagent is called conductometric titration. The
conductance of a solution depends largely on the number of ions and their
mobilities. Some examples of conductometric titrations are as given below.
• 1) Strong acid against strong base: (HCI against NaOH)
Consider the titration of strong acid (HCI) against strong base (NaOH). The acid is taken in
conductivity vessel and alkali in burette. The conductance of HCl is due to presence of H and
Cl ions. As alkali is added, gradually H ions are replaced by slow moving Nat ions as given
below:
H+ + CI-+ Na+ + OH- → Na+ +CI-+H2O
Until the complete neutralization, conductance decreases on addition of NaOH. Any
subsequent addition of alkali after end point will result in introduction of fast moving OH
ions. The conductance therefore increases on further addition of alkali.
Conductometric titrations

40. •
The variation of conductance is plotted
against volume of alkali added, we get two
straight lines meeting at point 'B' which
represents end point of titration.

41. • When acid is weak,
conductance is low, on addition
of strong base poorly conducting
acid is convert into highly
ionsed salt and hence
conductance increase slowly up
to the equivalence point. Beyond
the equivalence point addition of
alkali causes sharp increase in
conductance due to excess of
hydroxide ions. The graph is
represented as:
2)Weak acid against strong Base (CH3COOH against NaOH)

42. • In this case conductance initially
decrease due to the replacement of fast
moving H+ ions by slowly moving NH4+
ions. Beyond end point, further
addition of weakly ionized HN4OH will
not cause any appreciable change in
conductance. The point of intersection
of curve is the end point of titration.
3)Strong acid against weak base:(HCl against NH4OH)

43. • In this titration, conductance
initially increases because of
formation of salt
(CH3COONH4) which is strong
electrolyte. This increase
continues till end point. Beyond
end point, the conductance does
not change appreciable. The
graph is shown in fig.
4)Weak acid against weak base: (CH3COOH against NH4OH)

44. The iteration of silver nitrate against
potassium chloride involves precipitation
formation.
AgNO3 + KCl KNO3 + AgNO3
-
Since mobility of Ag+ and K+ ions is nearly
same, the conductance remains almost
constant till the equivalence point. After
equivalence point the added KCl, increases
The conductance rapidly as shown in graph
given below.
5)Precipitation titration: (AgNO3 against KCl)

45. The conductometric titration has many advantages over ordinary titration:
i) Small quantity of solutions is required for titrations.
ii) As end point is determined graphically, no special precautions are necessary
iii)Indicator is not required, so conductometric titrations are used in titration of colored and
turbid solutions.
iv)Conductometric titrations are used for analysis of dilute solutions as well as for
weak acids.
v) Conductometric titrations can be applied to mixture of acids, precipitation & other
types of titrations,
vi) Conductometric measurements give more accurate results.
Advantages of conductometric titration:

46. The conductometric titration has many advantages over ordinary titration:
i) Small quantity of solutions is required for titrations.
ii) As end point is determined graphically, no special precautions are necessary
iii)Indicator is not required, so conductometric titrations are used in titration of
colored and turbid solutions.
iv)Conductometric titrations are used for analysis of dilute solutions as well as for
weak acids.
v) Conductometric titrations can be applied to mixture of acids, precipitation & other
types of titrations,
vi) Conductometric measurements give more accurate results.
•
Advantages of conductometric titration:

47. Migration of ions under influence of electric field
• On passing electric current through
electrolyte solution, ions migrate and discharged
oppositely charged electrodes. The migration of
ions can be demonstrated by simple experiment.
• The lower portion of U-tube is filled with 5%
agar-agar solution in water with small quantity
of CuCr2O- (obtained by mixing equimolar
quantities of K2Cr2O, + CuSO4). It is allowed to
set by cooling as dark green jelly. Some charcoal
powder is sprinkled in both limbs. Then solution
of KNO3, and agar-agar is placed in each limb
and allowed to set as jelly.

48. • Finally, solution of KNO; in water is filled in each limb and platinum electrodes
are placed as shown in figure
• When electric current is passed, Cu+ ions migrate towards cathode (-ve
electrode).Due to this blue colour appears in cathode side and yellow color in anode
side by Cr207
-2. ions. From the movement of these colour bands, speed of ions can be
compared.

49. • Although most of ions differ in
their mobilities, the total number
of ion discharged at electrodes on
electrolysis is same. This can be
explained by Hittorf's theoretical
device as shown in figure. It
consists of an electrolytic cell
containing same number of
positive and negative ions with
same valency. The electrolytic cell
is divided into three compartments
by porous partitions B & C. Metal
electrodes A and D represent
cathode and anode respectively.
Hittorf's theoretical device

50. i) Represent initial state of electrolyte solution before electrolysis in which equal
number of positive and negative ions are present.
On passing electric current, three cases may be arise: -
i)suppose, only anions are migration and cations remains stationary.
If speed of anion (v=0) and speed on cation (u=0).then two anions migrate form
cathode to anode compartment and are discharge at anode.
ii)The unpaired cation in cathode compartment are discharge at
cathods.Therefore the number of ions discharge in both compartment is two.
iii)suppose,both ions are moving with same speed.
Let, u=v=2,then the number of ions discharge at respective electrode is four.
iv)suppose, both ions are migrating with different speeds (u=1,v=2).in this case
total number of ions discharge is three.

51. i) During electrolysis, ions are discharge in equivalent amount, Irrespective of their speeds of
migration.
ii) Concentration of electrolyte, around changes as a result of migration of ions.
iii)Fall in concentration around electrode is directly proportional to the speed of that ion which moves
away from that electrode.
∴
𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒
𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑐𝑎𝑡ℎ𝑜𝑑
=
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑐𝑎𝑡𝑖𝑜𝑛
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑛𝑖𝑜𝑛
𝑢
𝑣
= 𝑟(𝑠𝑝𝑒𝑒𝑑 𝑟𝑎𝑡𝑖𝑜)
iv) Total current carries by solution is measure of (u+v).This holds good when the electrodes or ions
are not attacked by solution.
In the above illustration concentration of central compartment remains constant. Whatever be
the speed of ions,the number of ions discharge on electrode is always equal.
Following conclusion can be drawn about the process of electrolysis:

52. During electrolysis current is carried by cation and anions. “The fraction of the total
current carried by an ionic species is called its transference number or transport
number.The transport number of cation(𝑡+) and anion (𝑡−) is given as.
𝑡+=
𝑢
𝑢+𝑣
𝑡−=
𝑣
𝑢+𝑣
u=Speed of cation v= speed of anion
“Sum of two transport number will be one”
𝑡+ + 𝑡− = 1 𝑜𝑟 1 − 𝑡+
If the speed ratio is r=
𝑢
𝑣
=
𝑡+
𝑡−
∴ r=
𝑡+
𝑡−
=
𝑡+
1−𝑡+
=
1−𝑡−
𝑡−
∴r𝑡−=1-𝑡− or 𝑡−+r𝑡−=1 or 𝑡−(1+r)=1 and 𝑡−=
1
1+𝑟
Transference number of transport number of Hittrof’s number of ions.

53. Problem(3)
The speed ratio of silver and nitrate ions in a solution of silver nitrate electrolysed
between silver electrode is 0.916.find the transport number of the two ion.
Solution. We have 𝑡−=
1
1+𝑟
Where 𝑡− is the transport number of anion and r is the speed ratio of cation and anion.
∴ 𝑡𝑁𝑂−
3
=
1
1+0.916
=0.521 and 𝑡𝐴𝑔+=1-𝑡𝑁𝑂−
3
=1-0.521=0.479

54. • This method is based on the fact that
change in concentration around the
electrodes is due to migration of ions. This
apparatus used in this method is shown in
figure .It consist of two vertical glass
tubes connected through U-tube. All the
three tubes are provided with stop cock at
bottom. Cathode is a portion of it is
exposed to solution in the form of spiral. .
The apparatus is connected to copper or
silver voltmeter in series. The apparatus
is filled with standard solution of silver
nitrate.
1.Hittrof’s Method

55. A steady current of 0.01 ampere is passed for nearly two to three hours. At the end of
this period suitable quantity of solution is drawn form lower portion of anode limb and
its weight is determined. It is then titrated with standard solution of potassium
thiocyanate solution to determine amount of silver present in it. The weight of silver
deposited in voltmeter is noted. If copper voltmeter us is used the weight of copper
deposited is multiplied by 108/31.05 to determined silver equivalent.
Precautions:
i) Steady current should be passed.
ii) There should be no change in concentration of middle compartment.
Observation and calculation:
Since, the change in concentration is accompanied by change in volume, loss of material
must be determined with reference to definite weight of solvent present after current
has passed. Two different cases may arise:

56. Case I. When electrodes are non-attackable (Pt electrodes are used): .
After passing electric current:
Let the weight of anodic solution taken out = a gm,Weight of AgNO3 present in it by titration=b gm
Weight of water = (a - b) gm
Before passing electric current:
Let the weight of AgNO3 in (a - b) gm of water before passing electric
Current = c gm
... Fall in concentration (c-b) gm of AgNO3
=
𝑐−𝑑
170
gm equvt.of AgNO3 = d (Say)
Let,the weight of silver deposited in silver coulometer= w1 gm
𝑤1
108
gm equvt of Ag
W= (say) gm equvt. of Ag
∴=Transport number of 𝐴𝑔+,𝑡+=
𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒
𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝐴𝑔 𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑 𝑖𝑛 𝑔𝑚 𝑒𝑞𝑢𝑖𝑣𝑡.
=
𝑑
𝑊
and transport number of NO3-,𝑡−= 1-
𝑑
𝑊
=
𝑊−𝑑
𝑊

57. Case II When electrode are attackable (Ag electrode are used)
In this case b>a because NO3
- ions react with Ag anode to produce AgNO3.Thus
concentration of AgNO3 or Ag is increased in anode compartment.
Increase in conc. of anodic solution=(b-c) gm of AgNO3
=
𝑏−𝐶
170
gm equvt of AgNO3
= e (say)
If no Ag+ ions had migration from the anode, the increase in concentration of Ag+ ions would
have been equal to W
∴ Fall in concentration due to migration of Ag+ ion =(W-e)
Hence, Transport number of Ag+.𝑡+ =
𝑊−𝑒
𝑊
And Transport number of NO3- 𝑡−=1-
𝑊−𝑒
𝑊

58. Hittorf's method does not give accurate results for dilute
solutions. Also small changes in concentration due to
passage of electric current may lead to experimental error.
Hence, direct observation method called "Moving
Boundary Method" was devised by Lodge and improved by
Whetham.
i)There are three conditions for determination of transport
number,
ii) Cation of indicator electrolyte should not move faster
than the cation whose transport number is to be
determined.
ii) Both have same anion
iii) Indicator electrolyte should have more density.
Moving Boundary Method:

59. • It consists of electrolytic cell with vertical tube of uniform cross section with two electrodes
at two ends. At lower side anode is made of cadmium rod and cathode is Pt foil.
• If we have to determine transport number of cation H+ in HCl, we have to choose another
electrolyte called "Indicator electrolyte' having common anion with the chemical species
under study (here Cl -ion). The cation of indicator electrode must be slow moving as
compared with the cation whose transport number is to be determined. Here serves as
indicator electrode because cd+ ion moves slowly than H' ion. This prevents blurring of
boundary line. The indicator solution CaCd2, is placed in lower half and over this the
solution of HCl is allowed to float, to produce sharp boundary at B, between two solution the
cadmium chloride produced during electrolysis will itself act as indicator electrolyte.
• A constant current is passed through apparatus for 5 to 6 hours. The H+ ion move
towards cathode, followed by Cd+2ions. The boundary gradually moves upwards up to
B2, i.e. through a distance T m.

60. Hence, quantity of current carried by H' ions = t. Q ; Hence, amount of H' ions migrated from B, to B2
=
𝑄.𝑡+
𝐹
kg equivalent--------------(1)
If 'x' sq. m. is cross sectional area of tube, then volume between B1, to B2, will be = x. l. dm3
If 'c' is the concentration of H+ ions in kg equivalent per liter (dm3) then amount of H+ ions in given
volume.
=
𝑥.𝑙.𝑐
1000
kg equivalent------------------(2)
From eq.(1) Ana (2)
𝑄.𝑡+
𝐹
=
𝑥.𝑙.𝑐
1000
Or 𝑡+=
𝑥.𝑙.𝑐.𝐹
1000.𝑄
If 𝑛
𝑄
𝐹
∴ 𝑡+=
𝑥.𝑙.𝑐.
1000.𝑛
And 𝑡−= 1-𝑡+
Where Q= Quantity of electricity passed= ampere x second
n= number Faradays of current passed
𝑡+= Transport number of H+ ion
(1F= 965000 coulomb)

61. Calculate the transport number of H+ and Cl-ions tor from electrode. Concentration of HCI solution c = 0.100 N. Mass
of the silver deposited in coulometer = 0.1209g Distance up to which boundary moves = 1.24 cmArea of cross section of
tube = E 7.5 cm.
Solution: Transport number of H+ ion 𝑡+
𝑡+=
𝑥.𝑙.𝑐.𝐹
1000.𝑄
= 𝑡+=
𝑥.𝑙.𝑐.
1000.𝑛
=
l=7.5m; x=1.24m2;c=0.1N
x= 1.24 m; c= 0.1 N
l = 7.5 m
Mass of silver in coulometer=0.1209g
Number of Faradays of current passed in circuit 'n' can be calculated as
108 g of silver = 1 Faraday (F)
0.1209 g of silver=
0.1209
108
F
i.e. n=
7.5×1.24×0.1×108
1000×0.1209
=0.8308
Transport number of Cl- ion 𝑡−=1-0.8308
=0.1692
Problem

62. • Problem A solution of LiCl of molarity 0.10 was placed in moving boundary cell cross-section area 1.17 m and
was electrolyzed for 131 minutes with a constant current of 9.42 x 10-3 A. The Li+ boundary was observed to move
a distance of 2.08 m What is the transference number of Li+ ions in this solution?
• Given data
Area of cross section of tube ‘x’=1.17m2
Distance up to which boundary moves l = 2.08 m.
Concentration of LiCl solution 'c' = 0.1 mol dm-3
Current passed = 9.42 x 10-3A
Time 't' = 131 min = 131 x 60 = 7860 sec.
Q = current x time=9.42 x 10-3 x 7860
Transport number of Li+
𝑡 +𝐿𝑖=
𝑥.𝑙.𝑐.𝐹
1000.𝑄
=
1.17×2.08×0.1×96500
1000×9.427×10−3×7860
𝑡 +𝐿𝑖=0.317
electrolysis between Platinum electrodes 100 dm cathode solution contained 0.06315

63. Problem The original strength of NaOH solution was 0.059 kg per 100 dm3. After Ag of NaOH and at same time
current deposited 0.05216 kg silver in voltmeter. Find out transport number of Na and OH ions. Solution: Given
data
Conc, of cathode solution before electrolysis = 0.059 kg NaOH/100 dm
Conc, of cathode solution after electrolysis = 0.0631 kg NaOH/100 dm'
Actual increase in conc, = (0,06315) -(0.059) = 0.00415 kg NaOH
If no hydroxide ion had migrated from cathode then,
Theoretical increase = Total deposition in voltmeter
= 0.05216 kg Ag = 0.05216 x 0.040/0.108 kg NaOH
= 0.01932 kg NaOH
Hence, decrease in conc. around cathode
= theoretical increase - actual increase
= 0.01932 -0.00415= 0.01517
∴ Transport number of OH-ion 𝑡 −𝑂𝐻 =
𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑐𝑜𝑛.𝑎𝑟𝑜𝑢𝑛𝑑 𝑐𝑎𝑡ℎ𝑜𝑑
𝑇𝑜𝑡𝑎𝑙 𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑖𝑛 𝑣𝑜𝑙𝑡𝑚𝑒𝑡𝑒𝑟
𝑡 −𝑂𝐻=
0.01517
0.01932
=0.785
𝑡 +𝑁𝑎=(𝑡 −𝑂𝐻) = (1-0.785)=0.215

64. Example (8) The speed ratio of silver and nitrate ions in a solution of AgNO3,
electrolyzed between two electrodes is 0.9. Find the transport number of Ag+
and NO3
-
Solution: Transport number of (NO3
-) 𝑡−=
1
1+𝑟
=
1
1+0.9
=0.5262
Transport number of (Ag+) 𝑡+= (1-𝑡−) = (1 -0.5262) = 0.4738

65. Kohlrausch (1875) observed that at infinite dilution, where ionization of electrolyte is
complete, each ion migrates independently and contributes to total equivalent conductance
of 1, & ne is the ionic conductance of anion and cation respectively. Kohlrausch electrolyte.
He has given a generalization called Kohlraush's law.
It states that, “the equivalent conductance of an electrolyte at infinite dilution is the
sum of equivalent conductance of anion and cation."
i.e. 𝜆∞= 𝜆𝑎 + 𝜆𝑐
• 𝜆𝑎& 𝜆𝑐 is the ionic conductance of ions and cation respectively. Kohlrausch studies studied
equivalent conductance at infinite dilution for various electrolyte having same cation or
anion which can be given in table.
Kohlrausch's law of independent migration of ions

66. Table: Equivalent conductance at infinite dilution at 18°C
It is evident from the values that difference in conductance of any two cations appears to be same
irrespective of nature of anion with which it is associated. Similarly,
difference in conductance of any two anions appears to be same irrespective of nature of cations with
which it is associated.
Pair of Electrolyte
𝜆∞
× 10−4
𝑆𝑚2
𝑒𝑞𝑢𝑖−1
Difference
× 10−4
𝑆𝑚2
𝑒𝑞𝑢𝑖−1
KCl
NaCl
130.0
108.9
21.1
KNO3
NaNO3
126.3
105.2
21.1
KCl
KNO3
130.0
126.3
3.7
NaCl
NaNO3
108.9
105.2
3.7

67. The ionic conductance's are proportional to velocities of ions.
𝜆𝑎 𝛼 𝑣 or 𝜆𝑎=k.v -----------(1)
& 𝜆𝑐 ∝ u or 𝜆𝑐=k.v--------------(2)
∴ 𝜆𝑎 + 𝜆𝑐 = k(v+u)------------(3)
Form Kohlrauch’s law
𝜆𝑎 + 𝜆𝑐 = 𝜆∞ 𝜆∞= k(u+v)-----------(4)
Dividing equation (1) by (3)
𝜆𝑎
𝜆∞
=
𝑣
𝑢+𝑣
=𝑡− ∴ 𝜆𝑎=𝑡−.𝜆∞
&
𝜆𝑎
𝜆∞
=
𝑣
𝑢+𝑣
=𝑡− ∴ 𝜆𝑎=𝑡−.𝜆∞ −−−−− −(5)
&
𝜆𝑐
𝜆∞
=
𝑢
𝑢+𝑣
=𝑡−
∴ 𝜆𝑐=(1 − 𝑡−).𝜆∞ OR ∴ 𝜆𝑐=𝑡+.𝜆∞--------(6)
Thus,the ionic conductance of any is the product of its own transport number and 𝜆∞value for any strong
electrolyte containing that ion. Its is express as mho m2, equiv-1
Relation between transport number & ionic conductance:

68. Weak electrolytes do not ionize to a sufficient extent in solution and are not beingcompletely ionized even at very
great dilution. The practical determination of equivalent conductance at infinite dilution '2' in such cases is
therefore, not possible. However, it can calculate with the help of Kohlrausch's law, e.g. the equivalent
conductance at infinite flation of CH3COOH (weak electrolyte) can be obtained from the equivalent conductance
at infinite dilution of HCI, CH3COONa and NaCl (all of which are strong electrolytes) as given below,
Now
i)𝜆∞(HCl) = 𝜆∞(H+) + 𝜆∞(Cl-) = 425.0 S m2 equvi-1
ii) 𝜆∞(CH3COONa)= 𝜆∞(Na+) + 𝜆∞ (CH3COO-) 91.6 S m2 equvi-1
iii) 𝜆∞(NaCl)= + 𝜆∞(Na+) + (Cl-)=128.1 S m2 equvi-1
From eq.(i),(ii) and (iii)
𝜆∞(H+)+ (CH3COO-)=𝜆∞(H+)+𝜆∞(Cl-)+𝜆∞(Na+)+(CH3COO-)-𝜆∞(Na+)+𝜆∞(Cl-)
𝜆∞(Ch3COOH)= 𝜆∞(HCl)+ 𝜆∞(CH3COONa)- 𝜆∞(NaCl)
𝜆∞(Ch3COOH)= 425.0+ 91.6-128.1= 388.5 S m2 equvi-1
In this manner, the equivalent conductance at infinite dilution of weak electrolyte can be calculated from the
equivalent conductance at infinite dilution of strong electrolytes containing similar ions.
Applications of conductivity measurement

69. 2. Determination of degree of dissociation (a) of weak electrolyte:
Degree of dissociation (𝛼) of a weak electrolyte at any dilution can be calculated by
the relationship
𝛼 =
𝜆𝑐
𝜆∞
Where, 𝜆𝑐 = equivalent conductance at given concentration 'c'
𝜆∞ = equivalent conductance at infinite dilution.
The dissociation constant (𝑘𝑎) is defined as, "equilibrium constant for dissociation of
m electrolyte obtained by applying law of mass action at given temperature."
eg. Consider dissociation of a weak acid HA
HA H+ + A-

70. For strong electrolyte ka has higher value i.e value ka indicates grater degree of dissociation (𝛼) can
be given as
HA H+ + A-
C 𝛼 𝛼
(1- 𝛼) 𝛼. 𝐶 𝛼. 𝐶
𝑘𝑎=
𝐻+ 𝐴−
𝐻𝐴
=
(𝛼.𝑐)(𝛼.𝑐)
1−𝛼 .𝑐
∴ 𝑘𝑎=
𝛼2.𝑐
1−𝛼
----------(1)
But, in case of weak electrolyte ,ka has lower value i.e degree of dissociation (𝛼) is negligible, hence
,the dissociation constant of weak electrolyte can be given can be given as ∴ 𝑘𝑎= 𝛼2
.c
From the value of equivalent conductance at given concentration (𝜆𝑐) and equivalent conductance at
infinite dilution (𝜆∞) degree of dissociation (𝛼) cab be determine.
𝛼=
𝜆𝑐
𝜆∞
Hence, form conductivity measurement dissociation constant Ka for weak electrolyte can be
determine.

71. Example (9): At 25°C the transport number of H+ ions In HCl and CH3COO- lon in CH3COONa are
0.81 and 0.47 respectively. The equivalent conductance at Infinite dilution of HCl and CH3COONa
are 426 ohm.-1cm2 equvt-1 and 91.0ohm-1 cm2 equvt-1 respectively. Calculate the equivalent
conductance of acetic acid at infinite dilution.
Solution: Conductance of H+ ion; Ac (H.T) = t. (H) x 200(HCI)
= 0.81 x 426 = 345.06
Conductance of CH3COO- ion;
1(CH3COO-) = t. (CH3COO-) x (CH3COONa)
= 0.47 x 91 = 42.77
Conductance of acetic acid at infinite dilution;
2 (CH3COOH) = 2, (CH3C00) + Rc (H)
= 42.77 + 345.06
= 387.83 mhos
Problem