Conductance

In case of metallic conductor, the electricity is conducted by the flow of electrons from negative
to positive potential. On the other hand in case of electrolytic conductor electric current is
carried by the charged ions moving towards the oppositely charged electrodes.
In case of metallic conductor no chemical change takes place. On the other hand conduction of
electricity by electrolytic conductor is accompanied by some definite chemical process.
From the Ohm’s law we can write resistance for metallic conductor that

𝛒𝐥
𝐑= where R is the resistance l is the length of metallic conductor, A is the area
𝐀
and ρ is resistivity. Similar equation is applied for electrolytic solution where
l is the length between two electrodes and A is the area of each electrodes
𝟏 𝟏𝐀 𝐀
𝐆= = =𝛋
𝐑 𝛒𝐥 𝐥
If A = 1 and l =1 then G = κ, it means when two electrode of unit area separated by unit
distance, the current conducted by the solution kept within two electrode is known as specific
conductivity
In another definition we can say that the conduction obtained by the electrolytic solution placed
within unit cube where electrodes are placed in opposite wall.
Here κ is known as specific conductivity having unit
 Unit of A cm2
Unit of G = Unit of κ × or, Ohm−1 = Unit of κ × or, Unit of κ = 𝐎𝐡𝐦−𝟏 𝐜𝐦−𝟏
Unit of l cm
In case of SI Unit of κ = 𝐎𝐡𝐦−𝟏 𝐦−𝟏 = 𝐌𝐡𝐨 𝐦−𝟏 = 𝐒𝐦−𝟏
Specific conductivity depends on
(i) Nature of the solvent
(ii) Temperature

(iii) Number of ions between two electrodes of conductivity cell

(iv) Speed of the ions between two electrodes.
Discussion
More electrolytic the nature of the solution, it gives higher values of conductivity. With increase the
temperature the velocity of the respective ions present with in electrolytic solution increases which
thereby increases the conductivities as the deposition tendency of them towards electrode increases.
More ions present between two electrodes higher will be its specific conductivity. Conductivity increases
with increasing the mobility of the ions.
Specific conductance (κ) variation with concentration
It is already to be notedthat specific conductance of an ionic
solution increases with increasing concentration. For strong
electrolytes, the increase in κ with increase in concentration
is sharp. However for weak electrolytes, the increase in κ is
more gradual. In both the cases the increase in κ with
concentration is due to an increase in the number of ions per
unit volume of the solution. For strong electrolytes, which
are completely ionized, the increase in κ is almost
proportional to the concentration. In case of weak
electrolytes the increae in κ is not large due to the low
ionization of electrolytes, and consequently κ does not go up
so rapidly as in case of strong electrolyte.

Additive nature
Conductance is an additive property. In case of an aqueous solution containing several
electrolytes, the total conductance is 𝐆𝐭𝐨𝐭𝐚𝐥 = ∑ 𝐆𝐢 + 𝐆𝐇𝟐 𝐎 , where ∑ is carried over all the
electrolytes I present in the solution and GH2 O is the conductance of H2 O (utilized for making the
solution) Now GH2O is often negligible in comparison to ∑ Gi
Again we can say that all the electrolytes as well as water present in same electrolytic cell then

A κi A A
Gtotal = ∑ Gi + GH2O or, κtotal =∑ + κH2 O or, κtotal = ∑ κi + κH2 O
l l l
Equivalent conductance
 Molecular wt wt taken
Equivalent wt = and gm equivalent =
valency Equivalent wt
wt taken
gm equivalent = × Valency = mole × Valency
Molecular wt
Equivalent conductance is defined as the conductance offered by the volume of solution
containing 1 gram equivalent of electrolyte. It is denoted by
𝚲 = 𝛋𝐕
where V is the volume of solution containing 1 gram equivalent of an electrolyte
If we consider the strength of the solution by c(N) so we can say that
𝐜 gm equivalent of electrolyte present in a solution of volume 1 lit
1 gm equivalent of electrolyte present in a solution of volume (𝟏⁄𝐜) lit
𝛋
𝚲 = 𝛋𝐕 = ( )
𝐜
If κ is given in ohm−1 cm−1 and c present in gm equivalent lit −1
κ ohm−1 cm−1 κ ohm−1 cm−1
Λ=( × 1lit) = ( × 1000 cm3 )
c gm equivalent c gm equivalent
1000κ
= ohm−1 cm2 (gm equivalent )−1
c

If κ is given in ohm−1 m−1 and c present in gm equivalent lit −1

κ ohm−1 m−1 κ ohm−1 cm−1 1
Λ=( × 1lit) = ( × m3 )
c gm equivalent c gm equivalent 1000
κ
= ohm−1 m2 (gm equivalent )−1
1000 c
In general
κ
Λ=( )
Normality

Molar conductance is defined as the conductance offered by the volume of solution containing 1
mole of electrolyte. It is denoted by
𝚲𝐌 = 𝛋𝐕𝐌
where VM is the volume of solution containing 1 mole of an electrolyte
If we consider the strength of the solution by D (M) so we can say that
D mole of electrolyte present in a solution of volume 1 lit
1 mole of electrolyte present in a solution of volume (1⁄D) lit
𝛋
𝚲𝐌 = 𝛋𝐕𝐌 = ( )
𝐃
If κ is given in ohm−1 cm−1 and c present in mole lit −1
κ ohm−1 cm−1 κ ohm−1 cm−1 1000κ
ΛM = ( × 1lit) = ( × 1000 cm3 ) = ohm−1 cm2 (mole )−1
D mole D mole D
In general
κ
ΛM = ( )
Molarity
gm equivalent of electrolyte mole of electrolyte × valency
Normality = =
Volume of the solution Volume of the solution
mole of electrolyte
=( ) × valency = Molarity × valency
Volume of the solution
κ κ
ΛM = ( )=( × valency) = 𝛬 × valency
Molarity Normality

Specific conductance (κ) and equivalent conductance (𝚲) with concentration

κ decreases with dilution. This is due to the fact that the number of current carrying particles
i.e, ions per ml of the solution becomes less. Ofcourse on dilution the degree of dissociaton is
also increases. But the number of ions per milliliter is decreased to such an extent that even
complete dissociation does not produce as many ions per milliliter as were present before
dilution.
The equivalent conductance on the otherhand increass with dilution and finally attains a
maximum value at infinite dilution. This is due to the fact that the value of Λ is a product of κ
and the volume of solution containing 1 gram equivalent of electrolyte. Decreasing the value of
κ is more than compensated by increasing the the value of V. Therefore Λ increases with
dilution.

Equivalent conductance of strong an weak electrolte with concentration

In cae of strong electrolytes such as KCl, HCl, NaCl etc the number of ions is same at all dilution
(since strong electrolytes are completely ionized) and the variation of equivalent conductance
with dilution is therefore due to the change in the speed of ions with dilution. In a concentrated
solution of such electrolyte, the interionic attraction may also form some ion pairs of type 𝐀+ 𝐁 −
which would not contribute to the conductance. This interionic forces considerably lowers the
speed of the ions and hence the conductivity of the solution. As the dilution is increased the
interionic attraction decrease with the result that the ion will move freely and independently of
their co-ions, thereby, increasing the equivalent conductance. With dilution the ions are quite
far apart, the interionic attractions are almost absent and each ion moves completely
independent of its co-ions. The equivalent conductance then approaches a limmiting value at
infinite dilution and represent the conducting power of 1 gm equivalent of electrolyte when it is
completely split up into ions. It is denoted by Λ0 or Λ∞
In case of weak electrolytes, the increase in equivalent
conductance with dilution is mainly due to (a) an increase in
the number of ions in solution (degree of ionization increases
with dilution) and (b) smaller interionic attractions at higher
concentrations. In these electrolytes, the extent of ionization
is small and the number of ions are relatively small. So the
interionic attraction between the ions is not of any
importance . Whatever ions are produced by ionization they
are free to move i.e, there speeds are not affected by interionic
attraction. In solution of ordinary concentration the degree of
ionization is quite small. The increase in Λ with dilution is due
to increase in degree of ionization.

An important relation can be obtained by extrapolating the curve for strong electrolytes to C →
0 where all interionic affects are absent. The limiting value obtained by this extrapolation is
called “ equivalent conductance at infinite dilution”.
The equation is
𝚲 = 𝚲𝟎 − 𝐛√𝐜

Kohlraussch law of independent migration of ions

This law states that at infinite dilution, where ionization of all electrolytes is complete and
where all interionic effects are absent, each ion migrate independently of its co-ion, and
contributes a definite share to the total equivalent conductance of the electrolyte. Such that
𝚲𝟎 = 𝛌𝟎(+) + 𝛌𝟎(−)
Where Λ0 is equivalent conductance of electrolyte at infinite dilution and λ0(+) be the equivalent
conductance of cation at infinite dilution and λ0(−) is that of anion. Kohlraussch again found that
when salts of potassium and sodium with a common anion are taken, the difference in their Λ0
values was found to be same irrespective of the nature of anion
𝚲𝟎𝐊𝐂𝐥 − 𝚲𝟎𝐍𝐚𝐂𝐥 = 𝚲𝟎𝐊𝐍𝐎𝟑 − 𝚲𝟎𝐍𝐚𝐍𝐎𝟑 = 𝚲𝟎𝟏 − 𝚲𝟎𝟏
𝐊 𝐒𝐎 𝐍𝐚 𝐒𝐎
𝟐 𝟐 𝟒 𝟐 𝟐 𝟒
(𝛌𝟎𝐊 + + 𝛌𝟎𝐂𝐥− ) − (𝛌𝟎𝐍𝐚+ + 𝛌𝟎𝐂𝐥− ) = (𝛌𝟎𝐊 + + 𝛌𝟎𝐍𝐎−𝟑 ) − (𝛌𝟎𝐍𝐚+ + 𝛌𝟎𝐍𝐎−𝟑 )
𝟏 𝟏 𝟏 𝟏
= ( × 𝟐𝛌𝟎𝐊 + + 𝛌𝟎𝐒𝐎𝟐− ) − ( × 𝟐𝛌𝟎𝐍𝐚+ + 𝛌𝟎𝐒𝐎𝟐− )
𝟐 𝟐 𝟒 𝟐 𝟐 𝟒

Ionic mobility
Ionic mobility is defined by
Velocity of the respective ion
mobility =
Field strength
So we can write that velocity of any species per unit field strength is known as mobility of the
respective ion. In CGS and SI unit the mobility is as follows
 cm s −1 m s −1
(mobility)CGS = = 𝐜𝐦𝟐 𝐕 −𝟏 𝐬 −𝟏 (mobility)SI = = 𝐦𝟐 𝐕 −𝟏 𝐬 −𝟏
V cm−1 V m−1
Speed of a single ion through the solvent

Consider a single ion placed within a solvent medium (considered to be infinitely dilute
medium) under field strength E (V/m). Now the driving force has been opposed slightly by
viscous force by stokes.
Fdriving = ZeE and Fdrag = 6πηrv
Fapp = (Fdriving − Fdrag ) = ZeE − 6πηrv
dv
ma = m= ZeE − 6πηrv
dt
dv ZeE 6πηr dv 6πηr ZeE
or, = − v or, + v =
dt m m dt m m
6πηrt
Multiply both sides by e m

dv 6πηrt 6πηr 6πηrt ZeE 6πηrt

e m + ve m = e m
dt m m
d 6πηrt ZeE 6πηrt
or, (ve m ) = e m
dt m

6πηrt ZeE 6πηrt 6πηrt ZeE m 6πηrt

or, ∫ d (ve m ) =∫ e m dt or, ve m = ( )e m +I
m m 6πηr
ZeE
At t = 0, v = 0, I = −
6πηr
Putting the value in the above formula
6πηrt ZeE m 6πηrt ZeE ZeE 6πηrt
ve m = ( )e m − or, v = [1 − e−( m ) ]
m 6πηr 6πηr 6πηr
6πηrt
or, v = vmax [1 − e−( )
m ]

Transport number with mobility

 Let us consider 𝐧+ number of cation (charge
𝐳+ ) and 𝐧− number of anion (charge 𝐳− )
present per unit cc of the electrolyte solution .
The velocity of cation and anion are u cms-1
and v cms-1.
If we consider a space of length u cm then we
can say those cation present in u cm3 will
deposited in cathode in 1 second.
1 cm3 of solution contain 𝐧+ number of cations
of u cm3 of solution contain (𝐧+ 𝐮) number of
cation. Hence in 1 second a total of 𝐧+ 𝐮
number of cation of charge 𝐳+ will deposited
towards cathode.

So total number of positive charge (n+ uz+ ) in coulomb (n+ uz+ e) will carried by cation in 1
second.
𝐈+ = 𝐧+ 𝐮𝐳+ 𝐞
Similarly if we consider a space of length v cm from anode then we can say those anion present
in v cm3 will deposited in anode in 1 second.
1 cm3 of solution contain 𝐧− number of anode so v cm3 of solution contain (𝐧− 𝐯)
number of anion. Hence in 1 second a total of 𝐧− 𝐯 number of anion of charge 𝐳− will deposited
towards anode.
So total number of positive charge (𝐧− 𝐯𝐳− ) in coulomb (𝐧− 𝐯𝐳− 𝐞) will carried by cation in 1
second.
𝐈− = 𝐧− 𝐯𝐳− 𝐞
Now the total current carried by cation and anion
𝐈 = 𝐈+ + 𝐈− = (𝐧+ 𝐮𝐳+ + 𝐧− 𝐯𝐳− )𝐞
For electronutrality of the solution,
𝐧 + 𝐳 + = 𝐧− 𝐳 −
So, I becomes,
𝐈 = 𝐈+ + 𝐈− = (𝐧+ 𝐮𝐳+ + 𝐧+ 𝐯𝐳+ )𝐞 = 𝐧+ 𝐳+ (𝐮 + 𝐯)𝐞
Current carried by cation to the total current
𝐈+ 𝐧+ 𝐮𝐳+ 𝐞 𝐮
transport number of cation = 𝐭 + = = =
𝐈 𝐧+ 𝐳+ (𝐮 + 𝐯)𝐞 (𝐮 + 𝐯)
𝐈− 𝐧− 𝐯𝐳− 𝐞 𝐧+ 𝐯𝐳+ 𝐞 𝐯
transport number of anion = 𝐭 − = = = =
𝐈 𝐧+ 𝐳+ (𝐮 + 𝐯)𝐞 𝐧+ 𝐳+ (𝐮 + 𝐯)𝐞 (𝐮 + 𝐯)

Mobility with Conductance

Let us consider c (N) of 1-1 strong electrolyte M+X- are taken in an electrolytic cell where two
electrodes of unit area are separated by unit distance apart. The applied field strength is 1Vcm -
1. Here current conduction is taken for 1 second. So in unit field strength whatever be the

velocity of the ions it will be their mobilities.

For 1 second total

current passed will be
E 1
I= = =G
R R
κA
=
l
=κ
Here E = 1V , A =
1cm2 , l = 1cm
Again we can write
that
cΛ
κ=
1000
Considering the
solution would be
infinitely dilute so we
can replace Λ by Λ0
cΛ0
I=κ=
1000
Now we can apply
Kohlraussch law of
independent migration
of ions
cΛ0 c(λ0+ + λ0− )
κ= = = I … … . . (1)
1000 1000
At infinite dilute condition M + X − will be totally dissociates
M+X− → M+ + X−
C (N) 0 0
0 C (N) C (N)
1000cm3 of an electrolyte contain c gm equivalent cation
c
1 cm3 of an electrolyte contain gm equivalent cation
1000
cu
u cm3 of an electrolyte contain gm equivalent cation
1000
Similarly
cv
v cm3 of an electrolyte contain gm equivalent anion
1000
 cu cv
So in 1 second of time a total of (1000 + 1000) gm equivalent of ions deposited in both cathod and
anode. Here 1 gm equivalent of ions carried F coulomb of charge so,
cu cv
( + ) gm equivalent of ions due to deposition in both electrode
1000 1000
c
will carry (u + v)F coulomb of charge in 1 second
1000
c
I= (u + v)F … … … (2)
1000
Equating equation (1) and (2)
0 0
c (λ+ + λ−) c 0 0
I= = (u + v)F or, (λ+ + λ− ) = (u + v)F
1000 1000
At infinitely dilute condition cation and anion both will move independently,
λ0+ = uF and λ0− = vF

Waldens rule
When a cation of radius r+ moves with a velocity u+ relative to solvent, then according to Stokes
law the frictional resistance offered by the solvent is Fd = 6πηru+
At infinitely dilute condition the retarding forces originating from interionic interaction
disappear. Only retarding force that acts at infinite dilution is the viscous force. The steady state
is attained only due to the balancing of electrical force and viscous force. Then
z+ eE = 6πηr+ u+ so ionic mobility of cation
u+ z+ e k k
u= = = for anion mobility, v =
E 6πηr+ ηr+ ηr−
At infinite dilution we can write
Λ0 = (u + v)F where u and v are the mobilities of cation and anion
kF 1 1
Λ0 = [ + ]
η r+ r−
For ions of big size the solvation can be neglected, hence r+ , r− can be regarded constant in any
solvent. However this consideration is not valid for ions of high charge and low radius.
Therefore for ions of large size, Λ0 η = constant, This is called Walden rule. This rule suggest
that the factors which lowers the value of the viscosity coefficient must increase the value of Λ0 .
This is supported by the fact that on increasing the temperature η decreases while Λ0 increases.

Hydration and ionic conductance

In solution ions remain hydrated and hence the size of the hydrated ion determines the
magnitude of ionic conductance. Except H + ion the ionic conduction at infinite dilution of
Li+ , Na+ , K + , Rb+ ions are in the order 𝜆0Rb+ > 𝜆0K+ > 𝜆0Na+ > 𝜆0Li+ . This is in agreement with the
size of the hydrated ion, which is in the order of Li+ > Na+ > K + > Rb+ . As the size of the
hydrated ion increases the ions suffers greater viscous resistance and hence its ionic
conductance decreases. Therefore from equation
𝐤
𝐮=
𝛈𝐫+

It can be stated that with the increase from the value of r, ionic mobility decreases. Thus ionic
conductance decreases with the increase of r since 𝜆0+ = u+ F and 𝜆0− = u− F where u+ and u− are
the ionic mobility of cation and anion respectively.

Abnormal mobility of hydrogen and hydroxyl ion

The abnormally high mobilities and high ionic conductances of H + and OH − ions in aqueous
solution have been ascribed to an unusual type of conductance mechanism proposed by T. van
Grotthus, H + ion due to its high ionic potential remains hydrated at least in the form H3 O+ . The
formation of H3 O+ ion favours abnormal rapid migration of H + ion through aqueous medium
under the influence of electric field, where a water molecule takes up the proton forming H3 O+
ion and then transfers the proton to the adjacent water molecule. The molecule of water first
forms H3 O+ ion and inturn transfers the H + ion to the adjacent molecule. In this way H + ion
being carried by H2 O molecules passes for one portion to another portion. The water molecule
after transferring the H + ion to the adjacent molecule rotates at an angle of 180 0 for the next
capture of H + ion and its successive transfer. The mechanism is called Grotthus mechanism.

Also the above mentioned fact can also be explained by chain mechanism of H + ion migration
where the linear chain molecule forms by hydrogen bond. By breaking and formation of
hydrogen bond consecutively H + ion easily migrates from one end to other end. This type of
mechanism observed at room temperature whereas at high temperature orientation of H2 O
molecules changes, which reduces proton jump mechanism as a result mobility decreases at
high temperature. Similar types of behavior is also observed for OH −

Determination of acid dissociation constant of an weak acid

HA ⇌ H+ + A−
At time t=0, concentration C 0 0
At time t=t, concentration C(1 − 𝛼 ) C𝛼 C𝛼

CH+ CA− Cα × Cα Cα2

Ka = = =
CHA C(1 − α) (1 − α)
Λ 2
CH+ CA− Cα × Cα C ( 0)
Ka = = = Λ
CHA (
C 1−α ) Λ
{1 − ( 0 )}
Λ
Λ0 Λ 2
=( 0 ) C ( 0)
Λ −Λ Λ
0 2
Λ −Λ Λ C Λ
( 0 ) = ( 0) or, (1 − 0 )
Λ Λ Ka Λ
2
Λ C 1 1 ΛC
= 2 or, ( − 0 ) = 2
Λ0 K a Λ Λ Λ0 K a
1 1 ΛC
or, = 0 + 2
Λ Λ Λ0 K a
 1
Now plot Λ vs ΛC
1 1
Here slope = and intercept = so, dissociation constant can be obtained by
Λ20 K a Λ0
1 (intercept)2
Ka = =
slope Λ20 slope
Conductance ratio

Assumption: The speed does not vary appreciably with dilution then the increase in
conductance with dilution is mainly due to the formation of more ions in solution. At infinite
dilution, the electrolyte is completely ionised and all the ions which can be derived from one
gram equivalent of the electrolyte takes part in conducting the current. At appreciable
concentrations, only a fraction of electrolyte is ionized and the degree of ionization is given by

Λ
α= where Λ = equivalent conductance at a given concentration and Λ0 = equivalent
Λ0
conductance at infinite dilution. This quantity (α) is termed as conductance ratio.

Conductance
Q. Calculate the ionic mobility of the cation in KCl at 250C, given that its transport number is 0.49
and the equivalent conductance of KCl at infinite dilution is 150

As the electrolyte is uni-univalent typre, so we can write equivalent conductance of electrolyte

𝚲𝟎 = 150 ohm−1 cm2 mole−1 = 150 ohm−1 cm2 (gm eq)−1
Now from the relation we can write that mobility of the cation related to equivalent conductance of
cation at infinite dilution 𝛌𝟎𝐊 +
𝛌𝟎𝐊+ 𝐭 + 𝚲𝟎
𝐮𝐊 + = = where t+ is transport number of cation
𝐅 𝐅
t+ Λ0 0.49 × 150 ohm−1 cm2 (gm eq)−1 0.49 × 150 cm2
uK+ = = = = 7.62 × 10−4 cm2 volt−1 s−1
F 96500 Amp s gm eq
( )−1 96500 Amp ohm s
( )

Q. The conductivities of aqueous NaCl, KCl, K2SO4 at 250C at infinite dilution are 126.45, 149.84
and 306.60 𝐨𝐡𝐦−𝟏 𝐜𝐦𝟐 𝐦𝐨𝐥𝐞−𝟏 respectively. If 𝛌𝟎𝐍𝐚+ at 250C is 50.11𝐨𝐡𝐦−𝟏 𝐜𝐦𝟐 𝐦𝐨𝐥𝐞−𝟏, calculate
the value of 𝐭 𝐒𝐎𝟐− in 𝐍𝐚𝟐 𝐒𝐎𝟒 at 250C
𝟒
 Λ0Na2 SO4 = 2Λ0NaCl + Λ0K2 SO4 − 2Λ0KCl
= (2 × 126.45 + 306.60 − 2 × 149.84)ohm−1 cm2 mole−1 = 259.82 ohm−1 cm2 mole−1
λ0SO2− = (Λ0Na2 SO4 − 2λ0Na+ ) = (259.82 − 2 × 50.11)ohm−1 cm2 mole−1 = 159.6 ohm−1 cm2 mole−1
4

λ0SO2− 159.6 ohm−1 cm2 mole−1

4
t SO2− = = = 0.61
4 Λ0Na2 SO4 259.82 ohm−1 cm2 mole −1

Q. A saturated solution of SrSO4 shows a specific conductance (𝛋) value 𝟏. 𝟓 × 𝟏𝟎−𝟒 𝐨𝐡𝐦−𝟏 𝐜𝐦−𝟏 at 250C.
The solubility is 𝟎. 𝟓 × 𝟏𝟎−𝟑 𝐦𝐨𝐥𝐞𝐥𝐢𝐭 −𝟏 and equivalent conductance 𝚲𝟎𝐒𝐫𝐒𝐎𝟒 =
𝟏𝟒𝟎 𝐨𝐡𝐦−𝟏 𝐜𝐦𝟐 (𝐠𝐦 𝐞𝐪)−𝟏 , calculate the approximate value of 𝛋 for water at the given
temperature

cm2 1 × 10−3 gm eq
κSrSO4 = Λ0SrSO4 × s = 140 × = 1.4 × 10−4 ohm−1 cm−1
ohm × (gm eq) 1000 cm3
As specific conductance κ is an additive property, so we can write that
κH2O = (κsolution − κSrSO4 ) = (1.5 − 1.4) × 10−4 = 10−5 ohm−1 cm−1

Q. Equivalent conductances at infinite dilution of HCl, NaCl and CH 3COONa are 426.2, 126.5 and 91
𝐨𝐡𝐦−𝟏 𝐜𝐦𝟐 𝐞𝐪−𝟏 at 250C. 𝐂alculate 𝚲𝟎 of CH3COOH
Using Kohlrauschs’ law of independent migration of ions

Λ0CH3 COOH = λ0CH3 COO− + λ0H+

Λ0CH3 COOH = λ0CH3 COO− + λ0Na+ − λ0Na+ − λ0Cl− + λ0Cl− + λ0H+ = (λ0CH3 COO− + λ0Na+ ) − (λ0Na+ + λ0Cl− ) +
(λ0Cl− + λ0H+ ) = Λ0CH3 COONa − Λ0NaCl + Λ0HCl = 91 − 126.5 + 426.2 = 390.7 ohm−1 cm2 eq−1

Q. At 250C, the equivalent conductance of 0.1(N) 𝐂𝐇𝟑 𝐂𝐎𝐎𝐇 is 5.20 𝐨𝐡𝐦−𝟏 𝐜𝐦𝟐 𝐞𝐪−𝟏 and 𝚲𝟎 of
CH3COOH is 391 𝐨𝐡𝐦−𝟏 𝐜𝐦𝟐 𝐞𝐪−𝟏. Find out the degree of ionization and dissociation constant of
𝐂𝐇𝟑 𝐂𝐎𝐎𝐇

Degree of ionization with respect to equivalent conductance can be represented by

Λ 5.20 ohm−1 cm2 eq−1

∝= = = 0.0133
Λ0 391 ohm−1 cm2 eq−1

CH3 COOH → CH3 COO− + H+

C 0 0
C(1 − α) Cα Cα

Dissociation constant

𝐶CH3 COO− 𝐶H+ Cα × Cα 𝐶𝛼 2 0.1 × 0.01332

𝐾𝑎 = = = = = 1.79 × 10−5
𝐶CH3 COOH C(1 − α) (1 − α) (1 − 0.0133)

Q. The mobility of 𝐍𝐇𝟒+ ion is 𝟕. 𝟔𝟐𝟑 × 𝟏𝟎−𝟖 𝐦𝟐 𝐕−𝟏 𝐬−𝟏, calculate

(i) the ion conductivity of 𝐍𝐇𝟒+ ion
(ii) the velocity of the ion if 15 volt is applied across electrodes 25 cm apart
(iii) The transport number of the ions in 𝐍𝐇𝟒 𝐂𝟐 𝐇𝟑 𝐎𝟐 solution if the mobility of 𝐂𝟐 𝐇𝟑 𝐎−
𝟐 ion is 𝟒. 𝟐𝟑𝟗 ×
−𝟖 𝟐 −𝟏 −𝟏
𝟏𝟎 𝐦 𝐕 𝐬

The ionic conductivity of NH4+ at infinite dilution

λ0NH+
uNH+4 = 4
or, λ0NH+ = uNH+4 F = (7.623 × 10−8 m2 V −1 s−1 )(96500 Amp s (gm eq)−1 ) =
F 4
Amp Amp
= 73.55 × 10−4 m2 (gm eq)−1 = 73.55 × 10−4 m2 (gm eq)−1
V V
= 73.55 × 10−4 m2 ohm−1 (gm eq)−1
15Volt
Now potential gradient = (0.025 m) = 60Vm−1
Now velocity of NH4+ ion = Ionic mobility × potential gradient = (7.623 × 10−8 m2 V −1 s−1 ) ×
(60Vm−1 ) = 4.574 × 10−6 ms−1
Now the transport number of NH4+
uNH+4 7.623 × 10−8 m2 V −1 s−1
t NH+4 = = −8 −8 2 −1 −1
= 0.643
(u + + u𝐶 𝐻 𝑂− ) (7.623 × 10 + 4.239 × 10 )m V s
NH4 2 3 2

Q. At 250C, 𝚲𝟎 of KCl is 149.9 and equivalent conductance of 0.1(N) KCl is 128.96 𝐨𝐡𝐦−𝟏 𝐜𝐦𝟐 𝐞𝐪−𝟏.
Find out the specific conductance of 0.01(N) KCl solution

For 1-1 electrolyte (strong) the equivalent conductance at any concentration is related to equivalent
conductance at infinite dilution
Λ0 − Λ
Λ = Λ0 − b√c or, b = ( )
√c
Now, b is fixed for KCl at two different concentration, at that two concentration equivalent conductances
are different
Λ0 − Λ1 Λ0 − Λ2 Λ0 − Λ1
b=( )=( ) or Λ2 = Λ0 − ( ) √c2
√c1 √c2 √c1
149.9 − 128.96
or Λ2 = 149.9 − ( ) √0.01 = 143.28 ohm−1 cm2 eq−1
√0.1
Now equivalent conductance related to specific conductivity as
eq
κ2 = Λ2 c2 = 143.28 ohm−1 cm2 eq−1 × 0.01 = 143.28 × 10−5 ohm−1 cm−1
1000 cm3

Q. The mobilities of 𝐇 + ion 𝐂𝐇𝟑 𝐂𝐎𝐎− ion at infinite dilution at 250C are 36.30× 𝟏𝟎−𝟒 and 4.23×
𝟏𝟎−𝟒 𝒄𝐦𝟐 𝐕 −𝟏 𝐬−𝟏. If the degree of ionization of 0.02M acetic acid solution at 250C is 0.03, calculate
the specific conductance of 0.02M acetic acid solution
Since acetic acid has uni-valent, so the concentration of the solution is 0.02(M) = 0.02(N)
Specific conductivity (κ) of a weak electrolyte
κ = αCF(u + v) where α is degree of ionization
0.03 × 0.02 gmeq × 96500 Amp s gmeq−1
κ= (36.30 + 4.23) × 10−4 𝑐m2 V −1 s−1
1000 cm3
 amp
κ = 2.3463 × 10−4 = 2.3463 × 10−4 ohm−1 cm−1
cm V

Q. In an infinitely dilute solution NaCl at 250C, the mobility of 𝐍𝐚+ ion is 5.19× 𝟏𝟎−𝟒 𝒄𝐦𝟐 𝐕 −𝟏 𝐬−𝟏
and viscosity coefficient of water is 0.01 poise. Find out the radius of 𝐍𝐚+ ion

Using Waldens rule

v ze ze 1.602 × 10−19 Amp s × Volt s
zeE = 6πηrv or, u+ = = or, r+ = =
E 6πηr+ 6πηu+ 6π × 0.001 Nsm−2 × 5.19 × 10−8 m2
1.602 × 10−8 J
r+ = = (1.636 × 10−2 m)
6π × 5.19 N

Q. Calculate the velocity of 𝐋𝐢+ in 0.1(M)LiCl when a current of 1 A is passed through a column of
solution with cross-sectional area 𝟒 × 𝟏𝟎−𝟒 𝐦𝟐. Given that equivalent conductance of 𝐋𝐢+ and
𝐂𝐥− are 𝟒 × 𝟏𝟎−𝟑 and 𝟕. 𝟓 × 𝟏𝟎−𝟑 respectively in mho 𝐦𝟐 𝐞𝐪−𝟏
uLi+ E λ0K+ E λ0K+ IR λ0K+ I λ0 + Il λ0 + I λ0 + I I λ0K+
velLi+ = = = = = K = K = K 0 =
l F l F l F lG F lκA F κA F Λ CA FCA (λ0K+ + λ0Cl− )
I λ0K+ 1 amp × 1000cm3 𝟒 × 𝟏𝟎−𝟑
velLi+ = =
FCA (λ0K+ + λ0Cl− ) 96500Amp s gmeq−1 × 0.1gmeq × 4cm2 (𝟒 + 𝟕. 𝟓) × 𝟏𝟎−𝟑

velLi+ = 9.013 × 10−3 cm s−1

Q. The resistance of a electrolyte solution A is 45 ohms in a given cell and that of a electrolyte
solution B is 100 ohms in the same cell. Equal volumes of solutions of both A and B are mixed.
Calculate the resistance of this mixture, again in the same cell.

From the conductance cell

1 A
G= = κ( )
R l
Here both A and B are present in the same cell, of same cell constant (l⁄A) = (1⁄x)cm−1
κ is specific conductivity in ohm−1 cm−1
In case of electrolyte solution A,
1 κA A l 1
= or κA = = cm−1
RA l RA A 45 ohm × x
In case of electrolyte solution B,
1 κB A l 1
= or κB = = cm−1
RB l RB A 100 ohm × x
In the mixture, equal volumes of A and B are mixed together, so specific conductance of two individual
electrolytes after mixing will be given by the following
1 1 1 1
κ1A = κA = cm−1 and κ1B = κB = cm−1
2 90 ohm × x 2 200 ohm × x
Since specific conductivity is an additive property so after mixture, specific conductivity is given by
1 1 1 1 1 1
κ = (κ1A + κ1A ) = ( + ) cm−1 = ( + ) ohm−1 cm−1
90 ohm 200 ohm x 90 200 x
Now using the specific conductance of the mixtures the conductance of the mixing solution,
 1 κA 1 1 1 1 1 1
= =( + ) ohm−1 cm−1 × x cm = ( + ) or, R = 62.07 ohm
R l 90 200 x 90 200 ohm

Q. The resistance of an electrolyte solution A is 10 ohms in a given cell and that of an electrolyte
solution B is 50 ohms in the same cell. 1 volume of A of the same concentration is mixed with the
3 volumes of solution B of the above concentration. Calculate the approximate resistance of the
mixture, again in the same cell.

From the conductance cell

1 A
G= = κ( )
R l
Here both A and B are present in the same cell, of same cell constant (l⁄A) = (1⁄x)cm−1
κ is specific conductivity in ohm−1 cm−1
In case of electrolyte solution A,
1 κA A l 1
= or κA = = cm−1
RA l RA A 10 ohm × x
In case of electrolyte solution B,
1 κB A l 1
= or κB = = cm−1
RB l RB A 50 ohm × x
In the mixture, one volumes of A and three volumes of B are mixed together, so specific conductance of A
(κ1A ) becomes one fourth of its initial (κA ) and specific conductance of B (κ1B ) becomes three fourth of its
initial (κB )
1 1 3 3
κ1A = κA = cm−1 and κ1B = κB = cm−1
4 40 ohm × x 4 200 ohm × x
Since specific conductivity is an additive property so after mixture, specific conductivity is given by
1 3 1 1 3 1
κ = (κ1A + κ1A ) = ( + ) cm−1 = ( + ) ohm−1 cm−1
40 ohm 200 ohm x 40 200 x
Now using the specific conductance of the mixtures the conductance of the mixing solution,
1 κA 1 3 1 1 3 1
= =( + ) ohm−1 cm−1 × x cm = ( + ) or, R = 25 ohm
R l 40 200 x 40 200 ohm

Application of conductometry

(A) Dissociation constant of weak acid

(B) Determination of hydrolysis constant
(C) Ionic product of water
(D) Conductometric titration

Determination of hydrolysis constant

Let us consider the hydrolysis of aniline hydrochloride

Let the degree of hydrolysis be h and the equivalent conductance of the salt solution be (considering
C=1(M))
Λsolu = (1 − h)Λsalt + hΛH3 O+ + hΛCl−
or, Λsolu = (1 − h)Λsalt + h(ΛH3 O+ + ΛCl− ) or, Λsolu = (1 − h)Λsalt + hΛHCl
where Λsalt is the equivalent conductance of the salt and ΛHCl is the equivalent conductance of acid
formed. Using the above expression we can write
Λsolu − Λsalt
h=
ΛHCl − Λsalt
(c) 𝐈𝐨𝐧𝐢𝐜 𝐩𝐫𝐨𝐝𝐮𝐜𝐭 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫
Water is a weak electrolyte. Very few amount of water molecules reacts togather to form H3 O+ and OH − .
With respect to all water very few amount of this ions acts as a solute in large water medium, so we can
considered the solution to be infinitely dilute.
H2 O + H2 O ⇌ H3 O+ + OH −
Here the two solutes H3 O+ and OH − will be present in infinitely dilute medium of same concentration
CH3O+ = COH− = C
Now from equivalent conductance relation, applying Kohlrausch law of independent migration
κ
Λ0 = (λ0H3 O+ + λ0OH− ) = κ⁄C or, C =
0
(λH3O+ + λ0OH− )
Now the ionic product is given by,

CH3 O+ COH− C C C 2
KW =( )( ) = ( )( ) = ( )
C0 C0 C0 C0 C0
2
κ
KW = [ ]
C0 (λ0H3O+ + λ0OH− )
At 250C, κ of pure water = 0.58 × 10−7 ohm−1 cm−1 and λ0H3 O+ = 349.8 and λ0OH− = 198 ohm−1 cm2 gmeq−1
2
0.58 × 10−7 ohm−1 cm−1 × 1000cm3 lit −1
K W,250 =[ ] = 1.12 × 10−14
1 gmeq lit −1 (349.8 + 198)ohm−1 cm2 gmeq−1

Now pH of water at 250C

CH O+ C κ
pH = − log ( 3 ) = − log ( ) = − log ( )
C0 C0 (λ0 H3 O+ + λ0OH− ) C0

0.58 × 10−7 ohm−1 cm−1 × 1000cm3 lit −1

pH = − log ( ) = − log (1.0587 × 10−7 ) = 6.97
1 gmeq lit −1 (349.8 + 198)ohm−1 cm2 gmeq−1
Q. 200 ml of 0.002(M) BaCl2 is added to 300 ml of 0.003(M) 𝐍𝐚𝟐 𝐒𝐎𝟒. Calculate approximate κ of
the resulting mixture. Determine transport number of 𝐂𝐥−
Given 𝛌𝟎 of 𝐂𝐥− = 𝟕𝟔, 𝐍𝐚+ = 𝟓𝟎, 𝟏⁄𝟐 𝐁𝐚𝟐+ = 𝟖𝟎, 𝟏⁄𝟐 𝐒𝐎𝟐−
𝟒 = 𝟖𝟎

BaCl2 will react with Na2 SO4 to give insoluble BaSO4 , So after mixing the concentration of BaCl2 and
Na2 SO4
200 300
CBaCl2 = × 0.002 (M) = 0.0008(M) and CNa2SO4 = × 0.003(M) = 0.0018(M)
500 500
CBa2+ = 0.0008 (M), CCl− = 0.0016(M), CNa+ = 0.0036(M), CSO2− 4
= 0.0018(M)

Here Ba2+ present in lesser amount, so all concentration of Ba2+ will react with SO2−
4 to form a complete ppt of
BaSO4

After forming ppt, the remaining strength of SO2−

4 will be,

CSO2−
4
= (0.0018 − 0.0008)(M) = 0.001(M) = 0.002(N)
𝜆0𝐂𝐥− = 76 ohm−1 cm2 gmeq−1 = 76 ohm−1 cm2 mole −1
𝜆0𝐍𝐚+ = 50 ohm−1 cm2 gmeq−1 = 50 ohm−1 cm2 mole−1

𝜆0𝟏 = 80 = 80 ohm−1 cm2 mole−1 or, 𝜆0𝐁𝐚𝟐+ = 160 ohm−1 cm2 mole−1 or, or, 𝜆0𝐁𝐚𝟐+ = 80 ohm−1 cm2 gmeq−1
𝐁𝐚𝟐+
𝟐
𝜆0𝟏 = 80 = 80 ohm−1 cm2 mole−1 or, 𝜆0SO2− = 160 ohm−1 cm2 mole −1 or, or, 𝜆0SO2− = 80 ohm−1 cm2 gmeq−1
SO2−
4 4 4
𝟐
0.0016 mole
κ𝐂𝐥− = 𝜆0𝐂𝐥− CCl− = 76 ohm−1 cm2 mole−1 × ( ) = 1.216 × 10−4 ohm−1 cm−1
1000 cm3
0.0036 mole
κNa+ = 𝜆0Na+ CNa+ = 50 ohm−1 cm2 mole−1 × ( ) = 1.8 × 10−4 ohm−1 cm−1
1000 cm3
0 −1 2 −1
0.002 gmeq
κSO2− = 𝜆 SO2− CSO2− = 80 ohm cm gmeq × ( ) = 1.6 × 10−4 ohm−1 cm−1
4 4 4 1000 cm3
Since specific conductivity is an additive property, so
κsolu = κ𝐂𝐥− + κNa+ + κSO2− 4
= (1.216 + 1.8 + 1.6) × 10−4 ohm−1 cm−1 = 4.616 × 10−4 ohm−1 cm−1
κ𝐂𝐥− 1.216
𝑡𝐂𝐥− = = = 0.26
κsolu 4.616

Q. Equivalent conductances at infinite dilution of HCl, NaCl and CH 3COONa are 426.2, 126.5 and
91 𝐨𝐡𝐦−𝟏 𝐜𝐦𝟐 𝐞𝐪−𝟏 respectively at 250C. A conductance cell filled with 0.01M KCl has a resistance
of 257.3 ohm at 250C. The same cell filled with 0.2N acetic acid has a resistance of 508.6 ohms.
Calculate the dissociation constant of the acid (specific conductance of 0.01M KCl at 25 0C =
𝟏. 𝟒𝟏 × 𝟏𝟎−𝟑 𝐨𝐡𝐦−𝟏 𝐜𝐦−𝟏
Using Kohlrausch law of independent migration of ions
Λ0CH3 COOH = λ0CH3 COO− + λ0H+
Λ0CH3 COOH = λ0CH3 COO− + λ0Na+ − λ0Na+ − λ0Cl− + λ0Cl− + λ0H+ = (λ0CH3 COO− + λ0Na+ ) − (λ0Na+ + λ0Cl− ) +
(λ0Cl− + λ0H+ ) = Λ0CH3 COONa − Λ0NaCl + Λ0HCl = 91 − 126.5 + 426.2 = 390.7 ohm−1 cm2 eq−1
Now cell constant = specific conductance × reistance
 (l⁄A) = κKCl R KCl = (1.41 × 10−3 ohm−1 cm−1 ) × 257.3 ohm = 362.79 × 10−3 cm−1
Specific conductance of 0.2N acetic acid present in the same electrolytic vessel
1 362.79 × 10−3 cm−1
(l⁄A) = κCH3COOH R CH3COOH or, κCH3 COOH = (l⁄A) =
R CH3COOH 508.6 ohm
κCH3 COOH = 7.133 × 10−4 ohm−1 cm−1
Equivalent conductance of 0.2N acetic acid solution
κCH3COOH (7.133 × 10−4 ohm−1 cm−1 ) × 1000cm3
ΛCH3 COOH = = = 3.567ohm−1 cm2 eq−1
CCH3COOH 0.2 gmeq
Now the degree of dissociation
ΛCH COOH 3.567ohm−1 cm2 eq−1
α= 0 3 = = 0.00913
ΛCH3 COOH 390.7 ohm−1 cm2 eq−1
Dissociation constant of CH3 COOH
Cα2 0.2 × 0.009132
Ka = ( )=( ) = 1.67 × 10−5
1−α 1 − 0.00913

Q. The ion conductances of 𝐋𝐢+ and 𝐊 + ions are 38.7 and 78.5 𝐨𝐡𝐦−𝟏 𝐜𝐦𝟐𝐞𝐪−𝟏 respectively. How
long would an ion of each type take to move from one electrode to another electrode at a
separation of 1.5 cm in a conductivity cell where a potential difference of 10 volt is applied.
Ionic conductance related to ionic mobility as
λ0K+ 73.5 ohm−1 cm2 eq−1
uK+ = = = 7.6 × 10−4 cm2 V−1 s−1
F 96500 amp s eq−1
λ0Li+ 38.7ohm−1 cm2 eq−1
uLi =
+ = = 4 × 10−4 cm2 V−1 s−1
F 96500 amp s eq−1
Now speed of respective ion = ionic mobility × Potential gradient
10
Speed of Li+ ion = 4 × 10−4 cm2 V−1 s−1 × ( ) Vcm−1 = 2.67 × 10−3 cms−1
1.5
10
Speed of K + ion = 7.6 × 10−4 cm2 V−1 s−1 × ( ) Vcm−1 = 5.1 × 10−3 cms−1
1.5
So time taken by Li+ ion to traverse from one electrode to another electrode
1.5 𝑐𝑚
= = 560.95 𝑠𝑒𝑐
2.67 × 10−3 cms−1

So time taken by K + ion to traverse from one electrode to another electrode

1.5 𝑐𝑚
= = 295.33 𝑠𝑒𝑐
5.1 × 10−3 cms−1
Q. The specific conductance of pure water at a certain temperature was found to be 𝟔 ×
𝟏𝟎−𝟕 𝐨𝐡𝐦−𝟏 𝐜𝐦−𝟏 . Find the pH of water at this temperature if 𝛌𝟎𝐇 + = 𝟑𝟓𝟎 𝐚𝐧𝐝 𝛌𝟎𝐎𝐇 − =
𝟐𝟎𝟎 𝐨𝐡𝐦−𝟏 𝐜𝐦𝟐 𝐠𝐦𝐞𝐪−𝟏
Here the two solutes H + and OH − will be present in infinitely dilute medium of same concentration
CH+ = COH− = C
Now from equivalent conductance relation, applying Kohlrausch law of independent migration
 κ
Λ0 = (λ0H+ + λ0OH− ) = κ⁄C or, C =
(λ0H+ + λ0OH− )
At the given specified temperature, κ of pure water = 6 × 10−7 ohm−1 cm−1 and λ0H+= 350 and λ0OH− =
200 ohm−1 cm2 gmeq−1
Now pH of water
CH+ C κ
pH = − log ( ) = − log ( ) = − log ( 0 )
C0 C0 (λH+ + λ0OH− )C0

6 × 10−7 ohm−1 cm−1 × 1000cm3 lit −1

pH = − log ( ) = 6.59
1 gmeq lit −1 (350 + 200)ohm−1 cm2 gmeq−1

Q. Find out the specific conductance of a solution formed by mixing equal volume of 0.01N 𝐍𝐇𝟒 𝐂𝐥
and 0.01N 𝐂𝐇𝟑 𝐂𝐎𝐎𝐍𝐇𝟒 solutions. Ionic mobilities of 𝐍𝐇𝟒+ , 𝐂𝐥− and 𝐂𝐇𝟑 𝐂𝐎𝐎− ions are respectively
𝟕. 𝟔 × 𝟏𝟎−𝟒, 𝟕. 𝟗𝟏 × 𝟏𝟎−𝟒 and 𝟒. 𝟐𝟑 × 𝟏𝟎−𝟒 𝐜𝐦𝟐 𝐕 −𝟏 𝐬−𝟏 at 250C

Specific conductance of the solution

κ = F ∑ Ci ui
i
After mixing equal volumes of 0.01N NH4 Cl and 0.01N CH3 COONH4 solutions, the concentration of the salts
become half of the initial concentration
0.01 0.01
𝐶NH4 Cl = (𝑁) = 0.005(𝑁 ) 𝐶CH3 COONH4 = (𝑁) = 0.005(𝑁)
2 2
Now concentrations of individual ions are
CNH+4 = (0.005 + 0.005) = 0.01(N), CCl− = 0.005(N) 𝑎𝑛𝑑 CCH3 COO− = 0.005(N)
Now the specific conductance
96500 𝐶𝑔𝑚 𝑒𝑞−1
𝜅= [0.01gmeq × 7.6 × 10−4 cm2 V −1 s−1 + 0.005gmeq × 7.91 × 10−4 cm2 V −1 s−1
1000 𝑐𝑚 3
+ 0.005gmeq × 4.23 × 10−4 cm2 V −1 s−1 ] = 1.319 × 10−3 𝑜ℎ𝑚−1 𝑐𝑚 −1

Q. The transport number of 𝐍𝐚+ ion in NaCl is 0.385 and the equivalent conductance at infinite
dilution of NaCl is 126.5 in the usual unit at 250C. Estimate the distance traversed in one hr by
𝐍𝐚+ ion in a very dilute solution kept in a cell with electrodes 5 cm apart, when a potential of 3
volt is applied between the electrode.
Now from the relation we can write that mobility of the cation related to equivalent conductance of
cation at infinite dilution λ0Na+
λ0Na+ t+ Λ0
uNa+ = = where t+ is transport number of cation
F F
t+ Λ0 0.385 × 126.5 ohm−1 cm2 (gm eq)−1
uNa+ = = = 5.01 × 10−4 cm2 V −1 s−1
F 96500 Amp s (gm eq)−1
3
Potential gradient = ( ) 𝑉𝑐𝑚 −1 = 0.6𝑉𝑐𝑚 −1
5
So the speed of the Na ions = 5.01 × 10−4 cm2 V −1 s−1 × 0.6𝑉𝑐𝑚 −1
+
Therefore the distance traversed by Na+ ions in one hr
5.01 × 10−4 cm2 V −1 s−1 × 0.6𝑉𝑐𝑚 −1 × 3600𝑠 = 1.09 𝑐𝑚

Q. 100 ml of 0.02(M) NaCl is added to 100 ml of 0.004(M) AgNO 3. Calculate an approximate value
for the specific conductance of the mixture
Given 𝛌𝟎 of 𝐂𝐥− = 𝟕𝟔. 𝟑, 𝐍𝐚+ = 𝟓𝟎. 𝟏, 𝐀𝐠 + = 𝟔𝟏. 𝟗, 𝐍𝐎−
𝟑 = 𝟕𝟏. 𝟒

NaCl will react with AgNO3 to give insoluble AgCl, So after mixing the concentration of NaCl and AgNO3
100 100
CNaCl = × 0.002(M) = 0.001(M) and CAgNO3 = × 0.004(M) = 0.002(M)
200 200
CNa+ = 0.001(M), CCl− = 0.001(M), CAg+ = 0.002(M), CNO3− = 0.002(M)

Here Cl− present in lesser amount, so all concentration of Cl−will react with Ag + to form a complete ppt of
AgCl
After forming ppt, the remaining strength of Ag + will be,
CAg+ = (0.002 − 0.001)(M) = 0.001(M)
λ0Cl− = 76.3 ohm−1 cm2 gmeq−1 = 76.3 ohm−1 cm2 mole−1
λ0Na+ = 50 ohm−1 cm2 gmeq−1 = 50 ohm−1 cm2 mole −1
λ0Ag+ = 61.9 ohm−1 cm2 gmeq−1 = 61.9 ohm−1 cm2 mole −1
λ0NO3− = 71.4 ohm−1 cm2 gmeq−1 = 71.4 ohm−1 cm2 mole−1
0.001 mole
κAg+ = λ0Ag+ CAg+ = 61.9 ohm−1 cm2 mole−1 × ( ) = 6.19 × 10−5 ohm−1 cm−1
1000 cm3
0.001 mole
κNa+ = λ0Na+ CNa+ = 50.1 ohm−1 cm2 mole −1 × ( ) = 5.01 × 10−5 ohm−1 cm−1
1000 cm3
0.002 mole
κNO−3 = λ0NO−3 CNO−3 = 71.4 ohm−1 cm2 mole−1 × ( ) = 14.28 × 10−5 ohm−1 cm−1
1000 cm3
Since specific conductivity is an additive property, so
κsolu = κ𝐴𝑔+ + κNa+ + κNO3− = (6.19 + 5.01 + 14.28) × 10−5 ohm−1 cm−1 = 25.48 × 10−5 ohm−1 cm−1

Conductometric titration of strong acid by a strong base

There are three stages of this conductometric titration
Stage 1. At the beginning: When no base has been added, the solution contain only H + and Cl− .
H + shows high mobility as well as high conductance,
Stage 2. When NaOH is added drop wise, then
equivalent amount of H + (highly mobile) is replaced
by weakly mobile Na+ , as a result conductance
gradually decreases with the addition of NaOH. If we
add v equivalent of NaOH then the conversion takes
place in the following way
(𝐇 + + 𝐂𝐥− ) + 𝐯(𝐍𝐚+ + 𝐎𝐇 − )
→ (𝟏 − 𝐯)𝐇 + + 𝐂𝐥− + 𝐯𝐍𝐚+ + 𝐯𝐇𝟐 𝐎
This decrement of conductance continues till the
equivalent point is achieved
Stage 3. After the equivalence point: With the
addition of further NaOH, new ion OH − will arise,
which is previously utilized to form water.
But after the equivalence point no H + is present within the medium, so requirement of OH − for
neutralization purpose is practically nil. As a result conductance increases very strongly. The
slope magnitude will be less than the slope magnitude before the equivalence point.
From the plot we get the drop of NaOH required at the equivalence point. From this we can get
the volume of NaOH (VNaOH ) used for neutralization. Let us consider for this titration
VHCl volume of acid used. Hence
SNaOH VNaOH = SHCl VHCl

If SNaOH is provided, then SHCl can be easily obtained.

Conductometric titration of weak acid by a strong base

There are three stages of this conductometric titration
Stage 1. At the beginning: When no base has been added, the solution contains only weak acid
HAc (say CH3COOH). The dissociation is very poor almost all the acid present in undissociated
form. Very few acetate and H+ is formed which shows very low value of conductance (Point A)
CH3 COOH ⇌ CH3 COO− + H +
Stage 2. Now few drops of NaOH is added which react with equivalent amount of CH3 COOH
CH3 COOH + (Na+ + OH − ) → CH3 COO− + Na+ + H2 O

The formed CH3 COO− acts as a common ion for the first dissociation step, so the reaction will be
suppressed, and it is assumed that all dissociated CH3 COO− and H + are accumulated to form
CH3 COOH. In this common ion effect H + is replaced by weakly mobile Na+ . As a result
conductance decreases slightly; this situation is represented by the point B.
Stage 3. At this point of B it is to be assumed
that due to previously explained common
ion effect all HAc present in non dissociated
form and this fall of point A to B will be
insignificant compared to the rise in
conductivity due to the added new Na+ . So
the increment of conductivity is due to the
formation of more and more
CH3 COO and Na until the equivalence
− +

point has been reached.

Stage 3. After the equivalence point: With
the addition of further NaOH, new ion OH −
will arises, which is previously utilized to
form water. The slopes of the two line
(before and after equivalence point) are
markedly different.
The point of intersection is obtained by extending the lines. In the vicinity of the end point, the
salt formed from weak acid will be somewhat hydrolyzed and such the equivalence point will
not be sharp. So, the lines are to be extended from the both directions to reach the equivalence
point C.

Conductometric titration of strong acid by a weak base

There are three stages of this conductometric titration
Stage 1. At the beginning: When no base has been added, the solution contain only H + and Cl− .
H + shows high mobility as well as high conductance,
Stage 2. When NH4 OH is added drop wise, then equivalent amount of H + (highly mobile) is
replaced by weakly mobile NH4+ , as a result conductance gradually decreases with the addition
of NH4 OH. If we add v equivalent of NH4 OH then the conversion takes place in the following
way

(𝐇 + + 𝐂𝐥− ) + 𝐯(NH4 OH)

→ (𝟏 − 𝐯)𝐇 + + 𝐂𝐥− + 𝐯NH4+ + 𝐯𝐇𝟐 𝐎

This decrement of conductance continues till the

equivalent point is achieved
Stage 3. At equivalence point sharp point is not
observed, generally a round in shape is formed due
to the salt hydrolysis of salt of weak base and
strong acid. After that theoretically conductance
remains to be fixed as NH4 OH is expected to be
unionized in absence of any acid. But some amount
of weak base is dissociates, which increase
conductivity to a slight portion.
Conductometric titration of weak acid by a weak base
There are three stages of this conductometric titration
Stage 1. At the beginning: When no base has been added, the solution contains only weak acid
HAc (say CH3COOH). The dissociation is very poor almost all the acid present in undissociated
form. Very few acetate and H+ is formed which shows very low value of conductance (Point A)
CH3 COOH ⇌ CH3 COO− + H +
Stage 2. Now few drops of NH4 OH is added which react with equivalent amount of CH3 COOH
CH3 COOH + (NH4 OH) → CH3 COO− + NH4+ + H2 O
The formed CH3 COO− acts as a common ion for the first dissociation step, so the reaction will be
suppressed, and it is assumed that all dissociated CH3 COO− and H + are accumulated to form
CH3 COOH. In this common ion effect H + is replaced by weakly mobile NH4+ . As a result
conductance decreases slightly, this situation is represented by the point B.
Stage 3. At this point of B it is to be assumed that
due to previously explained common ion effect
all HAc present in non dissociated form and this
fall of point A to B will be insignificant compared
to the rise in conductivity due to the furnished
new NH4+ . So the increment of conductivity is
due to the formation of more and more
CH3 COO− and NH4+ until the equivalence point
has been reached.
At equivalence point sharp point is not observed,
generally a round in shape is formed due to the
salt hydrolysis of salt of weak base and strong
acid. After that theoretically conductance
remains to be fixed as NH4 OH is expected to be
unionized in absence of any acid. But some
amount of weak base is dissociates, which
increase conductivity to a slight portion.

Conductometric titration of mixed acid (𝐇𝐂𝐥 + 𝐇𝐀𝐜) by strong base NaOH

At the beginning of the titration when no base has been added, solution contains H + , Cl− and
undissociated HAc. Here the titration progress in stepwise manner. Initially the solution shows
very high conductance due to presence of H + in the solution. When NaOH is added dropwise,
strong acid will be titrated first where strong mobile H + is replaced by weakly mobile Na+ and
conductance of the solution decreases gradually. This is continued till HCl is completely
exhausted. After that only HAc is remain so the plot looks like weak acid and strong base. First
common ion effect is observed then conductance increases then salt hydrolysis gives a round
shaped equivalence point after that conductance steadily increases with larger slope. Here the
volume of NaOH required for neutralization of HCl is V 1 and the total volume of NaOH required
for achieve the second equivalence point is V 2. Since HAc is weak its degree of dissociation will
be low, so larger amount (V2 − V1 ) of NaOH is needed to titrate HAc. Strength of strong and
weak acid can be obtained by the following way
SNaOH V1
SHCl V(mixed acid) = SNaOH V1 or, SHCl = (at first equivalence point)
taken V(mixed acid)
taken
SNaOH (V2 − V1 )
SHAc V(mixed acid) = SNaOH (V2 − V1 ) or, SHCl = (at first equivalence point)
taken V(mixed acid)
taken

Conductometric titration of mixed acid (𝐇𝐂𝐥 + 𝐇𝐀𝐜) by strong base N𝐇𝟒 OH

The titration curve looks like as above. The only difference observed after second round one
equivalence point (due to salt hydrolysis of salt comprised of weak acid and weak base) where
conductance very slowly increases due to very low dissociation of base.

Conductometric titration of dibasic acid by strong base NaOH

For weak dibasic acid such as succinic acid oxalic acid etc, first dissociation constant K a is very
high hence it behaves as a strong acid. But the second Ka is low and the monocarboxylate
behaves as a weak base. Let the weak dibasic acid is represented by H2 Ox2 . When no base has
been added the solution contains H + and Ox − and it shows high conductance. When NaOH is
added dropwise, H + is replaced by Na+ and conductance decreases gradually and the graph just
looks like strong acid vs strong base. This will continues till the first H + is exhausted (point A).
At that point conductance shows for the following

HOx − ⇌ Ox 2− + H +
Now few drops of NaOH is added which react with equivalent amount of HOx −
HOx − + (Na+ + OH − ) → Ox 2− + Na+ + H2 O
The formed Ox 2− acts as a common ion for the dissociation step, so the reaction will be
suppressed, and it is assumed that all dissociated Ox 2− and H + are accumulated to form HOx − .
In this common ion effect H + is replaced by weakly mobile Na+ . As a result conductance
decreases slightly, this situation is represented by the point B.
Stage 3. At this point of B it is to be assumed that due to previously explained common ion effect
all HOx − present in non dissociated form and this fall of point A to B will be insignificant
compared to the rise in conductivity due to the added new Na+ . So the increment of
conductivity is due to the formation of more and more Ox 2− and Na+ until the equivalence point
has been reached.
Stage 3. After the equivalence point: With the addition of further NaOH, new ion OH − will arises,
which is previously utilized to form water. The slopes of the two line (before and after
equivalence point) are markedly different. The point of intersection is obtained by extending
the lines. In the vicinity of the end point, the salt formed from weak acid will be somewhat
hydrolysed and such the equivalence point will not be sharp. So, the lines are to be extended
from the both directions to reach the equivalence point C.
The only difference with the mixed acid titration is that (V2 − V1 ) ≈ V1 , because at first due to C-
C bond rotation availability of both acid equal and those H + considered as weak, some % of that
will neutralize during the neutralization of strong H + . So the volume of NaOH required for those
H + neutralization which should be situated in second section are interestingly incorporated in
first section.

Conductometric titration of dibasic acid by weak base N𝐇𝟒 OH

It is same as previous the only difference is that after second equivalence point conductance
increases very slowly obtained from weak base dissociation.
Conductometric titration of sodium acetate by HCl

At first the solution contains only dissociated form of the salt of sodium acetate shows a
conductance which is not very high. When HCl is added dropwise the following reaction takes
place
H+ + Cl−
CH3 COO− + Na+ → CH3 COOH + Na+ + Cl−
Here high ionized sodium acetate is replaced by unionized CH3 COOH and Na+ , Cl− . Here Cl− has
higher ionic conductance than acetate so with the addition of HCl, conductance increases very
slowly upto first equivalence point is reached. At equivalence point the solution contains
CH3 COOH, Na+ andCl− . After that when further HCl is added conductance increases sharply
(since H+ shows abnormal mobility by Grotthus mechanism)

Conductometric titration of Silver nitrate by KCl

At first the solution contains only dissociated form of the salt of silver nitrate. It shows
sufficiently high conductance. When KCl is added dropwise AgCl is precipitated and Ag+ is
replaced by K+ . The ionic conductance of Ag+ ≈ K+ and hence conductance of the solution
remains more or less constant. This constancy continued till the equivalence point is reached.

+ K+ + Cl−
Ag + NO−
3 → AgCl ↓ +K+ + NO−
3
At equivalence point solution contains AgCl, K+ and NO−
3 . When further KCl is added, after the
equivalence point, no reaction takes place but conductance of the solution increases due to
addition of K+ and Cl−

Conductometric titration of 𝐁𝐚(𝐍𝐎𝟑 )𝟐 by H2SO4

At the initial stage, the solution contains Ba2+ and NO−

3 which show high conductance. When
gradual H2SO4 is added, BaSO4 is precipitated out and Ba2+ is being exchanged by 2H+ by the
following way
2H+ + SO2−
4
Ba2+ + 2NO− 3 → BaSO4 ↓ +2H + + 2NO− 3
Ionic conductances as well as mobility of H+ is much higher (shows Grotthus mechanism) than
Ba2+ and hence conductance will increase rapidly. This will continue till the equivalence point
is reached. At the equivalence point, the solution contains H+ and NO− 3 . After the equivalence
point, when H2SO4 is being added no reaction takes place but conductance increases due to
presence of H+ and SO2−4 . After the equivalence point the rate of increase of conductance is
even steeper than that obtained before the equivalence point due to new ion SO2− 4 arises along
with H .
+

Conductometric titration of MgSO4 and 𝐁𝐚(𝐎𝐇)𝟐

At the beginning, when no Ba(OH)2 is added the solution contains Mg 2+ and SO2− 4 . With the
gradual addition of Ba(OH)2 , sparingly soluble Mg(OH)2 and BaSO4 are formed due to which
conductance decreases. This decrement continues till equivalent point has been reached.
At the equivalence point the solution contains only Mg(OH)2 and BaSO4 both of which are
sparingly soluble and hence conductance attains a minimum value. After the equivalent point
further addition of Ba(OH)2 , conductance of the solution increases due to presence of free Ba2+
and OH − .

Temperature dependence of ionic conductance

Conductance of electricity by ions is a rate process like viscosity of liquid and needs energy of
activation. If E represents the energy of activation then the equivalent conductance at infinite
dilution (𝜆0 ) is related to temperature as follows
E
λ0 = Ae−(RT) where A is a constant
E
or, ln λ0 = ln A − ( )
RT
Differentiating with respect to temperature
d ln λ0 E 1 dλ0 E
= 2
or, =
dT RT λ0 dT RT 2
Except H + and OH − ions near 250C (E/RT2 ) varies from 0.018K −1 to 0.022K −1 . Here (E/RT2 )
known as temperature coefficient denoted by 𝛼
1 dλ0
=𝛼
λ0 dT
If we consider the finite change with respect to 298K
1 [(λ0 )TK − (λ0 )298K ]
= α 𝑜𝑟, (λ0 )TK = (λ0 )298K {1 + α(T − 298K)}
(λ0 )298K [T − 298K]
H + and OH − ions have a α of magnitude 0.014 and 0.016 𝐾 −1 respectively due to their different
mechanism of conduction of electricity compared to other ions in water.
Conductance
Moving Boundary method
This method involves the
principle of direct determination
of speed of the ions. For this the
electrolytic solution is taken
within a long tube of uniform
cross-section in between two
indicator solutions. For the
electrolyte MA, the two indicator
electrolytes are NA and MB. The
solutions are arranged in the
order of decreasing density from
the lower end so that electrolytes
NA remains at the lower end and
electrolyte MB remains at the
upper end.
The indicator electrolytes are so
chosen that the speed of 𝐌 + ion >
speed of 𝐍+ ion and the speed of
𝐀− ion > speed of 𝐁− ion. Due to
this under the influence of electric
field 𝐍+ and 𝐁− ions will not
overtake M + and A− ions
respectively. As a result
boundaries between MB and MA
and between MA and NA remain
sharp as the ions move under the
influence of electric field in the
respective directions.
Boundary bb moves to cc due to migration of 𝐌 + ion while the boundary aa moves to dd due to the
migration of anion A− . The displacement of the boundary is proportional to the velocity of the respective
ion. The boundary is detected either from the colors of the two electrolytic solutions in contact with the
boundary or from the refractive indices of the two solutions in contact.
If bc represents the displacement of the boundary between MA and NA and ad represents the
displacement of the boundary between MB and MA, then
 𝐮𝐌+ 𝐛𝐜 𝐮𝐀− 𝐚𝐝 𝐮𝐀− 𝐚𝐝 𝐮𝐀− + 𝐮𝐌+
= 𝐨𝐫, = 𝐨𝐫, +𝟏= +𝟏 𝐨𝐫,
𝐮𝐀− 𝐚𝐝 𝐮𝐌+ 𝐛𝐜 𝐮𝐌 + 𝐛𝐜 𝐮𝐌 +
𝐚𝐝 + 𝐛𝐜
=
𝐛𝐜
𝐮𝐌 + 𝐛𝐜
𝐭𝐌+ = =
𝐮𝐀− + 𝐮𝐌+ 𝐚𝐝 + 𝐛𝐜
Let the concentration of MA solution is C(N) or C eqlit −1 and the boundary bb moves by l cm
when Q faraday of electricity is passed through the solution. From 1-1 strong electrolyte
solution dissociation we can write that concentration of 𝐌 + is also C(N) or C eq𝐥𝐢𝐭 −𝟏 . It means
1000 cm3 contains C eq of 𝐌 + . Hence within la cm3 contains laC eq of 𝐌 + . So we can write laC
eq of M + ion migrated within that time range.
Moreover out of Q coulomb of electricity 𝐭 + 𝐐 coulomb is transported by M + ion. Hence 𝐭 + 𝐐/𝐅
equivalent of M + ion migrates towards the cathode
𝐭+ 𝐐 𝐥𝐚𝐂
= 𝐥𝐚𝐂 𝐨𝐫, 𝐭 + =
𝐅 𝐐⁄
𝐅
If the electrolyte is weak as well as 1-1, then
𝛂𝐥𝐚𝐂
𝐭+ = where α is degree of dissociation
𝐐⁄
𝐅
But if the electrolyte is not 1-1 type, in general we can represented by 𝐌𝛎+ 𝐀𝛎− then from the
total dissociation the concentration of M + ion will be 𝛎+ 𝐥𝐚𝐂
𝛎+ 𝐥𝐚𝐂
𝐭+ =
𝐐⁄
𝐅
But if the general electrolyte is weak one then
𝛂𝛎+ 𝐥𝐚𝐂
𝐭+ = where α is degree of dissociation and ν+ is stoichiometry of cation
𝐐⁄
𝐅
Debye-Huckel Onsager theory of interionic attraction

To discuss the effect of the forces on the speed of the ions let us consider the mobility of a cation
of charge z under the influence of an electric field of intensity E through a solvent of viscosity
coefficient η and dielectric constant D at a temperature T.
The electric force is zeE (driving force) under which the ion moves steadily. The retarding
forces are
(i) Viscous force:
Frictional force offered by the solvent (stokes law)
Fd = 6πηrvT
At infinite dilution when there is no interionic attraction between the ions, electrical driving force has
been balanced by the viscous force offered by the solvent
𝐳𝐞𝐄 = 𝟔𝛑𝛈𝐫𝐯𝐓 then after rigorous calculation considering both positive and negative ion
𝚲𝟎 𝛈 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 (Waldens rule) [see previous lecture]

(ii) Asymmetric effect:

Asymmetric effect In solution due to the presence of interionic attraction, each ion remains surrounded
by an ion atmosphere whose charge is equal in magnitude but opposite in sign to that of the central ion.
Ion atmosphere is rarefied with the lowering of ionic strength. It remains spherically symmetrical in
absence of external electric field and has no effect on ionic speed. Electric field distorts the ionic
atmosphere as central ion and the ion atmosphere move in opposite direction i.e., the ion atmosphere
becomes asymmetric under the influence of the applied field. Disappearance of the existing ion
atmosphere and the formation of a new one occur successively as the central ion moves under the
electric field. But in between the disappearance and formation of ion atmosphere a time lag exists for the
orientation of the ions as well as of the solvent. The asymmetric shape of the ion atmosphere reduces the
speed of the central ion. The effect of the asymmetric ion atmosphere on the speed of the central ion is
called asymmetric effect. The magnitude of asymmetric effect increases with the increase in the
magnitude of interionic attraction. Since the dielectric constant of the medium lowers the interionic
attraction and the ionic strength increases the interionic attraction, at a given temperature the
magnitude of asymmetric effect is proportional to √C and inversely proportional to the dielectric
constant of the medium.

The asymmetric effect can be

eliminated by using AC
current of very high
frequency. If the period of
oscillation is much less than
the time of relaxation of the
ion atmosphere, i.e, the time
lag between the
disappearance of the existing
ion atmosphere and the
appearance of a new
atmosphere, then the ion
atmosphere remains
spherically symmetrical and
under this condition the ions moves with the limiting speed that it would attain at infinite dilution. This
is called Debye-Falkenhagen effect.
The same effect can be caused by using an electric field of high intensity, e.g. 20,000 volt/cm. Under this
condition the speed of an ion is so fast that the oppositely charged ions find no time to form ion
atmosphere and hence the ion can be considered as moving freely. This effect is called Wien effect.
 (iii) Electrophoretic effect
Migration of ion through the solvent imparts momentum to the adjacent solvent molecules and hence
streaming of solvent is associated with the movement of ions. In the deformed ion atmosphere density of
negative charge is greater behind the cation than in front. So the streaming of the solvent is more in the
direction opposite to that of the central ion as greater number of anions remaining behind the cation
impart greater momentum to the solvent than do the central ion. The net result is that the overall
streaming of the solvent occurs in the direction opposite to the movement of the central ion.
Consequently the relative velocity of the central ion with respect to the solvent becomes greater than the
speed of the ions had there been no streaming of the solvent. The increase in relative speed increases
the viscous drag. The streaming of the solvent with the migration of ion is called electrophoresis and its
retarding effect on the speed of an ion is called electrophoretic. Like asymmetric effect electrophoretic
effect is also proportional to √C and inversely proportional to the dielectric constant of the medium.
Debye-Huckel-Onsagar conductance equation correlating variation of Λ with C is given by

𝟐𝟗. 𝟏𝟓(𝐳+ + 𝐳− ) 𝟗. 𝟗 × 𝟏𝟎𝟓

𝚲 = 𝚲𝟎 − [ 𝟏 + 𝟏 𝚲𝟎 𝛚] √𝐂(𝐳+ + 𝐳− )
(𝐃𝐓)𝟐 𝛈 (𝐃𝐓)𝟐

In a simple case of 1-1 electrolyte z+ and z− are uynity and the magnitude of ω is (2 − √2), the equation
then becomes
𝟖𝟐. 𝟒 𝟖. 𝟐 × 𝟏𝟎𝟓
𝚲 = 𝚲𝟎 − [ 𝟏 + 𝟏 𝚲𝟎 ] √𝐂
(𝐃𝐓)𝟐 𝛈 (𝐃𝐓)𝟐
The first term gives the effect due to the electrophoretic force and the second term represents the
influence of the asymmetric or relaxation force. The equation for 1-1 electrolyte can simply be expressed
as
𝚲 = 𝚲𝟎 − [𝐀 + 𝐁𝚲𝟎 ]√𝐂
Where A and B are constants and are dependent only on the nature of the solvent and temperature as
follows
𝟖𝟐. 𝟒 𝟖. 𝟐 × 𝟏𝟎𝟓
𝐀= 𝟏 and 𝐁 = 𝟏
(𝐃𝐓)𝟐 𝛈 (𝐃𝐓)𝟐

Specific conductivity with concentration based on Debye-Huckel Onsager theory of

interionic attraction
Specific conductance of a solution of strong electrolyte of concentration C eq/lit is given by the
expression

𝐂𝐅
𝛋= (𝐮 + 𝐯)
𝟏𝟎𝟎𝟎
With dilution C decreases proportionally. Speed of the ion is opposed by viscous drag of the solvent,
asymmetric effect of the ion atmosphere and electrophoratic effect. With dilution the magnitude of these
retarding forces except the viscous drag decreases but not at par with dilution and consequently the
ionic mobility increases. The increase in ionic mobility is also not at par dilution. Hence the overall effect
of dilution is the lowering of specific conductance value.

For weak electrolyte

𝛂𝐂𝐅
𝛋= (𝐮 + 𝐯)
𝟏𝟎𝟎𝟎
Where α is the degree of
ionization of the weak
electrolyte at concentration C.
For the electrolyte of type AB,
𝐊
𝛂 = √ 𝐚⁄𝐂
So with dilution the value of αC
will decrease. As α has a very
low value even at moderate
concentration, the influence of
interionic attraction on ionic
speed can be neglected even at
a
moderate concentration. So with dilution ionic mobility practically does not vary. Hence specific
conductance decreases with dilution.
𝐅√𝐊 𝐚
𝛋= (𝐮 + 𝐯)√𝐂
𝟏𝟎𝟎𝟎

Equivalent conductance with concentration based on Debye-Huckel Onsager theory of

interionic attraction

For a solution of strong electrolyte considering that the ionization is complete and no ionpair formation
occurs, equivalent conductance is given by
𝛋
𝚲 = = 𝐅(𝐮 + 𝐯)
𝐂
where u and v are the ionic mobilities. With dilution the
magnitude of interionic attraction decreases causing
lowering of the influence of relaxation of asymmetry
effect and electrophoretic effect. Consequently ionic
mobility increases which ultimately increases the
equivalent conductance
According to Debye-Huckel Onsagar equation for 1,1
electrolyte,
𝚲 = 𝚲𝟎 − [𝐀 + 𝐁𝚲𝟎 ]√𝐂
i.e., 𝚲 increases with dilution and the plot of 𝚲 against
√𝐶 is a straight line with a negative slope.The intercept
obtained on extrapolation gives the value of 𝚲𝟎 .

The equivalent conductance of a weak electrolyte at concentration C is given by the expression,

𝛋
𝚲 = = 𝛂 𝐅(𝐮 + 𝐯)
𝐂
where α is the degree of ionization of the electrolyte. As α has a very low value even at ordinary
concentration, interionic attraction is practically absent. So with further dilution ionic mobility does not
vary. However with dilution α increases but very slowly i.e., not at par with dilution.
𝐤 𝐤 𝟏
𝐅𝐨𝐫 𝟏, 𝟏 𝐰𝐞𝐚𝐤 𝐞𝐥𝐞𝐜𝐭𝐫𝐨𝐥𝐲𝐭𝐞, 𝛂 = √ 𝐚⁄𝐂 𝐬𝐨, 𝚲 = √ 𝐚⁄𝐂 × 𝐅(𝐮 + 𝐯) = 𝐜𝐨𝐧𝐬𝐭.
√𝐂

Therefore Λ vs √C plot, 1-1 weak electrolyte is a

rectangular hyperbola, i.e., at very low concentration Λ
decreases steeply with the increase of √C and at
moderate concentration Λ decreases insignificantly as
√C value increses. Therefore Λ0 value cannot be
obtained by extrapolation of the Λ vs √C plot. The steep
rise of equivalent conductance with lowering of
concentration at low value of C is due to the fact that in
this region of concentration, increase in the number of
ions through dissociation of the electrolyte becomes
less than the added H+ and OH− ions produced from
water, i.e, the conductance of the solution becomes
equal to conductance of the solvent.

Q. To determine transport number of 𝐊 + ion in 0.1(N) KCl solution by moving boundary method
CdCl2 solution is used as the indicator electrolyte. The solutions are taken in a capillary of
diameter 2.124 mm. The cationic boundary shifts by 100 mm when a steady current of 14 mA
was passed for 497 sec. find the transport number of 𝐊 + ion in 0.1(N) KCl

From moving boundary formula

 laC laCF (10 cm) × π(0.1062 cm)2 × (0.1 gmeq) × 96500 Amp s (gmeq)−1
t+ = = = = 0.491
Q⁄ A × time 1000 cm3 × (0.014 × 497 Amp s)
F

Q. In a moving boundary experiment on KCl the apparatus consisted of a tube of diameter 4.176
mm and it contained an aqueous solution of KCl at a concentration of 0.021 mole𝐥𝐢𝐭 −𝟏 . A steady
current of 1.82 mA was passed and the cationic boundary advanced 3.18 mm in 100 second.
Calculate the transport number of 𝐊 + and 𝐂𝐥−

From moving boundary formula

Here CKCl = CK+ = 0.021 (M) = 0.021(N)

2
0.4176
laC laCF (0.318 cm) × π ( cm) × (0.021 gmeq) × 96500 Amp s (gmeq)−1
t K+ = = = 2
Q⁄ A × time 1000 cm3 × (0.0182 × 100 Amp s)
F
= 0.485

t Cl− = 1 − t K+ = 1 − 0.485 = 0.515

Q. In a moving boundary experiment with 0.01N HCl solution enclosed between the two indicator
electrolytes LiCl and HAc in a capillary of radius 5mm, the cationic and anionic boundaries move
through 7.5 cm and 7.8 cm respectively when a steady current of 12 mA when passed for 11.5
minutes. Calculate the transport number of 𝐇 + and 𝐂𝐥− ion

Here both the ions H + and Cl− moves through the same capillary under same condition towards
opposite electrodes. Also from the total dissociation of 1-1 electrolyte the concentrations of ions
furnished H + and Cl− are same i.e., 0.01N
From moving boundary formula we can write

laC
t= ∝ l (other terms here const and fixed for both ions)
Q⁄
F
tCl− lCl− tCl− lCl− tCl− + tH+ lCl− + lH+ 1 lCl− + lH+
= or, +1= + 1 or, = or, =
tH+ lH+ tH+ lH+ tH+ lH+ tH+ lH+

lH+ 7.5
tH+ = =
lCl− + lH+ (7.5 + 7.8)

Q. In a moving boundary experiment on KCl indicator electrolytes are CdCl2 and HAc. In the
experiment 0.1(N) KCl solution is taken in a capillary of radius 1.01 mm and a steady current of
12 mA is passed for 400 sec, the cationic boundary shifts by 76.1 mm. calculate simultaneous
shifts of anionic boundary

Here both the ions K + and Cl− moves through the same capillary under same condition towards
opposite electrodes. Also from the total dissociation of 1-1 electrolyte the concentrations of ions
furnished K + and Cl− are same i.e., 0. 1N
From moving boundary formula we can write

laC
t= ∝ l (other terms here const and fixed for both ions)
Q⁄
F
tCl− lCl− tCl− lCl− tCl− + tK+ lCl− + lK+ 1 lCl− + lK+
= or, +1 = + 1 or, = or, =
tK + lK+ tK + lK+ tK + lK+ tK + lK+
lH+ lH+ aCF
tK + = =
lCl− + lK+ A × time
1 aCF 1
= or,
lCl− + lK+ A × time lCl− + 7.61 cm
π(0.5 cm)2 × (0.01 gmeq) × 96500 Amp s (gmeq)−1
=
1000 cm3 × (0.012 × 11.5 × 60 Amp s)
lCl− = 7.92 cm

Conductance

Determination of transport number by Hittorff method

Electrolysis of electrolytic solution takes place when a current is passed over a specified long,
concentration change occurs in the vicinity of the electrodes. Due to different velocities of cations and
anions the change in concentration in the vicinity of two electrode (anode and cathode) will be different.
Before the passage of current let us consider anode and cathode compartment contain equal number of
cation and anion. As current passes through the solution cation moves towards cathode and anode
moves towards anode. Now consider the following steps
I. Some cations move from anode compartment through central compartment to cathode
compartment. Let us assume at a given time three cations leave the anode chamber.
II. In step 2, Let at a given time two anions leave the cathode chamber and transported to anode
chamber.
 As a result of migration of ion
(cation and anion), anode
chamber contains five anions in
excess whereas cathode chamber
contains five cations in excess.
These excess charges will be
discharged and finally we get two
cation and two anion in anode
chamber and three cation and
three anion in cathode chamber.
This means there is a
concentration loss in both the
anode chamber and cathode
chamber by maintaining central
compartment intact.
At initial stage before electrolysis, all the three compartments contain an electrolytic solution of same
concentration. After passage of current for a specified time
Concentration loss in anode chamber ∝ ionic mobility of cation (u),
Concentration loss in cathode chamber∝ ionic mobility of anion (v)
Total loss of concentration ∝ (u+v).
u number of equivalent of electrolyte lost from the anode chamber
t+ = =
u+v Total number of equivalents deposited at each electrode
v number of equivalent of electrolyte lost from the cathode chamber
and t + = =
u+v Total number of equivalents deposited at each electrode
These are called Hittorff’s rule.
This rule can be used for the determination of transport number of cation and anion. The apparatus
consists of three compartments : anode comprtment, middle compartment and cathode compartment.
The tubes containing the solution of known concentration is connected in a series with a battery, a wire
of variable resistance, an Ammeter and a coulometer containing AgNO3 . A current of 20mA is passed
through the solution for a long time (several hours) so that concentration changes in anode and cathode
compartments are significant. After passing current during a particular time the solution is drained out
from one or both the chambers, weighted and then analysed. From the loss of concentration of cation
and anion chamber we can determine the transport number of cation anion respectively. Amount of
electricity passed can be measured from the coulometer.
 To determine the transport number by this method current density should be adjusted in such a
way that it can be neither very high nor very low. If current density is high then there will be local
heating. On the other hand for low current density there will be diffusion.
It is also assumed that solvent does not migrate with the ions. But in practice this cannot be true. As the
ions are solvated, then solvent will also migrate together with ions. Therefore a correction has to be
made. It can be shown that the true transport number is given by
𝐍𝐒
𝐭+ = 𝐭𝐇 + + 𝐧𝐖 ( ⁄𝐍 )
𝐖
Where t H + is the Hittorff’s transport number. n W is the number of moles of water migrated to cathode
from anode, NW is the equivalents of salt remain associated with NW moles of water.

Abnormal transference number

In dilute solution of CdI2
CdI2 ⇌ Cd2+ + 2I −
Amount lost from anode chamber is proportional to transport number of Cd2+ and amount lost from
cathode chamber is proportional to transport number of I − and t Cd2+ + t I− = 1
Where as in concentrated solution
2CdI2 ⇌ Cd2+ + CdI2−
4
As Cd2+ is transferred from anode to cathode chamber, there should be a concentration loss in the anode
chamber. But at the same time as CdI2−4 comes in anode chamber will be discharged to form CdI2 ,
concentration of CdI2 in anode chamber increases. Loss of CdI2 in cathode chamber will decrease and it
finally become zero or even negative when concentration is sufficiently high. On the same way transport
number of anion will increase and it approaches greater than 1. This is known as abnormal colligative
property.

Variation of transport number with concentration

Case 1: when 𝑢 and 𝑣 i.e, the mobility of cation and anion decrease to the same extent then 𝑡+ and 𝑡−
will remain unchanged. For KCl transport number of cation and anion remain remain practically
independent of concentration.
Case 2: when extent of decrease of ionic mobility of cation is greater than that of anion, then transport
number of cation decreases and that of anion increases with increase in concentration. For HCl, KNO 3,
AgNO3, etc. variation of transport number of cation and anion follows this type of variation.
Case 3: when extent of decrease of ionic mobility of cation is less than anion, then transport number of
cation increases and that of anion decreases with increases in concentration. For LiCl, NaCl, etc. variation
of transport number of cation and anion with concentration follows this type of variation.

Derivation of Hittorf’s rule (𝐆𝐞𝐧𝐞𝐫𝐚𝐥 𝐭𝐲𝐩𝐞𝐬 𝐨𝐟 𝐥𝐞𝐜𝐭𝐮𝐫𝐞)

Let in an electrolytic solution n+ z+ and n− z− be the respective charge densities of cation and anion of
speed u and v respectively. Let t + and t − be the transport number of the cation and anion respectively in
the given solution of the electrolyte. If z faradey electricity is passed through the solution then z
equivalent of ion is discharged at each electrode during electrolysis. Out of z faradey, t + z faraday of
electricity is transported by cation and t − z faradey of electricity is transported by anion. Due to
migration of t + z equivalent of cation from anode compartment t + z equivalent of anion becomes excess
and to maintain electrical neutrality t + z equivalent of anion discharged at anode. Hence the loss at
anode compartment = t + z equivalent of catopn + t + z equivalent of anion = t + z equivalent electrolyte and
this loss of electrolyte at anode compartment is solely due to the migration of cation.
t + z = Loss of electrolyte in equivalents at anode
Loss of electrolyte in equivalents at anode chamber
or, t + =
z
 Loss of electrolyte in equivalents at anode chamber
or, t + =
Number of equiv. of ion discharged at any electrode
Loss of electrolyte in equivalents at anode chamber
=
Number of faradeys of electricity passed
Number of eqv. of the cation migrated from anode
=
Number of faradeys of electricity passed
This is Hittorf’s rule

Variation of transport number with temperature

Transport number of an ion depends upon temperature as ionic speed depends on temperature.
It is observed that transport number greater than 0.5 decreases with the increase in temperature tends
to be 0.5, while transport number less than 0.5 increases with the rise in temperature. This may be due
to (i) the greater degree of hydration of the slower moving ion and (ii) the decrease in the degree of
hydration with the rise in temperature.
With decrease in the degree of hydration associated with the rise in temperature the speed of ion
increases, but the relative increase being greater for the slower moving ion. Consequently transport
number less than 0.5 increases with the rise in temperature. For example, let us consider that 𝑡+ < 0.5,
i.e., 𝑢+ < 𝑢− . With the rise in temperature relative increase in speed is greater in case of 𝑢+ .
Let us consider that u+ changes to 1.2u+ while u− changes to 1.1u− .
𝟏. 𝟐𝐮+
𝐭 ′+ =
𝟏. 𝟏𝐮− + 𝟏. 𝟐𝐮+
i. e., 𝒕′+ > 𝒕+
which means 𝑡+ increases with the increase of temperature. Obviously in this case 𝑡− decreases with the
increase of temperature.

Q. Onsagar equation is represented by

𝚲 = 𝚲𝟎 − (𝐀 + 𝐁𝚲𝟎 )√𝐂
How does A/B related to viscosity of the medium
In Onsagar equation the constant part is denoted by
𝟏
𝟖𝟐. 𝟒 𝟖. 𝟐 × 𝟏𝟎𝟓 𝐀 𝟖𝟐. 𝟒 (𝐃𝐓) ⁄𝟐 𝟏
𝐀= 𝟏⁄ 𝐁= 𝟏⁄ 𝐡𝐞𝐧𝐜𝐞 ( ) = 𝟏 × 𝟓
∝
(𝐃𝐓) 𝟐𝛈 (𝐃𝐓) 𝟐 𝐁 (𝐃𝐓) 𝟐 𝛈 𝟖. 𝟐 × 𝟏𝟎
⁄ 𝛈

So the ratio is inversely proportional to the viscosity coefficient of the medium. With dilution, viscosity
of the medium (solvent) increases which exerts a resisting force towards the mobility of ions.

Q. The equivalent conductance at infinite dilution of AgNO 3, KNO3 and KCl are 133.36, 144.86 and
149.86 𝐎𝐡𝐦−𝟏 𝐜𝐦𝟐 𝐞𝐪−𝟏. Calculate
(𝐚) 𝚲𝟎 (𝐀𝐠𝐂𝐥) (𝐛)𝚲𝟎 (𝐍𝐚𝐂𝐥) − 𝚲𝟎 (𝐍𝐚𝐍𝐎𝟑 )
State the law used
Λ0 (AgCl) = Λ0 (AgNO3 ) − Λ0 (KNO3 ) + Λ0 (KCl) = (133.36 − 144.86 + 149.86 )Ohm−1 cm2 eq−1
Λ0 (NaCl) − Λ0 (NaNO3 ) = λ0 (Na+ ) + λ0 (Cl− ) − {λ0 (Na+ ) + λ0 (Cl− )}
= λ0 (K + ) + λ0 (Cl− ) − {λ0 (K + ) + λ0 (Cl− )} = Λ0 (KCl) − Λ0 (KNO3 ) = 149.86 − 144.86
The above phenomenon is based on Kohlrausch law of independent migration of ions.

Q. Arrange the following electrolytes in the order of their 𝚲𝟎 values in water, with proper
justification LiCl, HCl, KCl
Here the anionic part is same in all electrolyte, so relative order of Λ0 values depends on cataionic part.
As we know for ions having the same charge, the ionic mobility and hence the ionic conductivity
increase with decrease in size of the moving ion due to lower resistance involved in moving through
medium. Smaller ions having higher polarizing power so hydration by the solvent molecule is more as a
result apparent size increases which experience greater resisting force. Based on this phenomenon
Cataionic size order K+ > Li+ > H+
Apperent size order due to hydration K+ < Li+ < H+
Mobility order K+ > Li+ > H+
But the exact order of mobility as well as ionic conductance H+ > K+ > Li+, where H+ shows greater
mobility as well as greater ionic conductance due to its Grotthus mechanism offered at room
temperature. Hence order of Λ0 values in water
HCl > LiCl > KCl

Q. Arrange the following electrolytes in the order of their 𝚲𝟎 values in water, with proper
justification KCl, HCl, NaOH
Whatever be the associated ion , we will consider those species first which give Grotthus mechanism.
Cationic part in HCl and anionic part in NaOH will shows this special type of chain mechanism. Between
H + and OH − , H + shows greater mobility as compared to hydroxyl ion. Hence order of Λ0 values in water
HCl > 𝑁𝑎𝑂𝐻. After that KCl will be count.

Q. A conductivity cell of cell constant 1𝐜𝐦−𝟏 shows a resistance of 6667 ohms when filled with
0.001M KCl solution at 250C. The same cell records a resistance of 2353 ohms when filled with
0.001M HCl solution at 250C
a. Calculate equivalent conductance values for KCl and HCl
b. Calculate ion conductance of 𝐇+ ion assuming that 𝐊 + and 𝐂𝐥− ions have same mobilities.
Consider the solutions to be very dilute so that the condition of infinite dilution may be applied.
c. How far will the 𝐇+ ion move in 10 sec when a potential difference of 2 volts is applied between
two electrodes placed 2 cm apart.
d. calculate 𝐭 𝐇 + and 𝐭 𝐂𝐥− in the HCl solution. What will be the effect of temperature on these values

1 κA (l⁄A) κ (l⁄A)
= or, κ = now equivalentconductance Λ = =
R l R C RC
1cm−1 × 103 cm3 1cm−1 × 103 cm3
ΛKCl = = 150 ohm−1 cm2 eq−1 ΛHCl =
0.001 eq × 6667 ohm 0.001 eq × 2353 ohm
= 425ohm−1 cm2 eq−1
Now the solution is assumed to be infinitely dilute
Λ0HCl = 425 ohm−1 cm2 eq−1 Λ0KCl = 150 = λ0K+ + λ0Cl− = 2λ0Cl− or, λ0Cl− = 75
 λ0H+ = 425 − 75 = 350 ohm−1 cm2 eq−1
Now mobility of H+
λ0H+ 350 ohm−1 cm2 eq−1
uH+ = = = 0.036 cm2 V −1 s−1
F 96500 amp seq−1
E cm2 2V
velocity = uH+ × = 0.036 × = 0.036 cms−1
l Vs 2 cm
Distance travelled by H + ion = velocity × time = 0.036 cms−1 × 10 s = 0.36 cm
λ0H+
t H+ = 0 = 0.829 t Cl− = 1 − 0.829 = 0.171
ΛHCl