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Prepared By: Dipen Shah B.Sc. / MATERIAL / SEM-VI / Chemistry - 603 / Unit-2 Page: 1 of 14

B.Sc. Semester – VI

Subject: - CHE - 603: Electrochemistry (conductometry)

Prepared By: - Dipen Shah

Contents:

Electric transport, conductance in metals and in electrolyte solution

Specific conductance, equivalent conductance

Importance of conductivity electrodes and platinization of electrodes

Variation of specific conductance with dilution as well as area of cross section of dip type

electrode and distance between two plates of electrodes etc.

Kohlrausch law and its importance, cell constant and its importance

Conductometric Titration:

1. Strong acid – Strong base

2. Strong acid – Weak base

3. Weak acid – Strong base

4. Weak acid – Weak base

5. Mixture of strong acid + Weak acid – Strong base

Precipitation Titration:

1. AgNO3 – NaCl

2. BaCl2 – K2SO4

3. Ba(OH)2 – MgSO4

Replacement Titration:

1. Salt of weak acid – strong acid

2. Salt of weak base – strong base

Degree or hydrolysis and Hydrolysis constant

Determination of solubility and solubility product of sparingly soluble salt, for the

measurement of conductivity

Importance of conductivity water and temperature for the measurement of conductivity

Introduction

Conductometric analysis is based on the measurement of the electrical conductivity of the

solution. The electrical conductivity of the solution is due to the movement of ions.

Therefore, when electrical current passed from the solution, the transport of positive ions

and negative ions can be obtained in opposite direction which is shown in following figure.

The movement of ions occurs in such a way that the solution remains electrically neutral.

The solution remains electrically neutral because the number of positive ions and negative

ions are constant and equivalent in solution.

Explain the following terms:

1. Ohm’s law: Ohm’s law is obeyed by metallic as well as by electrolytic conductors.

According to this law “the strength of current (I) flowing through a conductor is directly

proportional to the potential difference or EMF (E) applied to the conductor and inversely

proportional to the resistance (R) of conductor “thus mathematically, the law can be

represented as,


𝐈 =𝐄

𝐑 The current is measured in amperes, potential difference or EMF in volts and

electrical resistance in ohms.

2. Conductance: In the case of electrolytes, the term conductance (c) is generally used.

Conductance is the reciprocal of resistance therefore mathematically 𝐂 =𝟏

𝐑 mhos

(reciprocal of ohms).

3. Resistance and specific resistance: The resistance R of a conductor is directly

proportional to its length (l) and inversely proportional to its area (a) of cross section

therefore, 𝐑 ∝ 𝐥 and 𝐑 ∝𝐥

𝐚

∴ 𝐑 ∝𝐥

𝐚

∴ 𝐑 = 𝐏 ×𝐥

𝐚 Where P = specific resistance

If l=1 cm and a = 1 sq.cm then P = R ohms. Therefore specific resistance is defined as “the

resistance of the conductor having a length of 1 cm and area of cross section of 1 sq.cm”

We know that

∴ 𝐏 = 𝐑 ×𝐚

𝐥= 𝒐𝒉𝒎. 𝒄𝒎

Thus unit of the specific resistance is expressed in ohm.cm.

4. Specific conductance: “The specific conductance of any conductor is the reciprocal of

specific resistance” and it is denoted by k.

Now we know that ∴ 𝐑 = 𝐏 ×𝐥

𝐚

∴ 𝐑 =𝟏

𝐊.

𝐥

𝐚 ∴ 𝐊 =

𝟏

𝐑.

𝐥

𝐚

If l=1 cm and a=1 sq.cm, then the above equation can be written as

∴ 𝐊 =𝟏

𝐑=

𝐥

𝐏 𝒎𝒉𝒐/𝒄𝒎

Thus unit of the specific conductance is expressed in mho/cm.

5. Equivalent conductance: This may be defined as “the conductance of a solution

containing 1 gm equivalent of an electrolyte when placed between two sufficiently large

electrodes which are 1 cm apart ” it is represented by dc and is measured in mhos.

Mathematically it can be represented as

∴ 𝛌𝐜 = 𝐤 . 𝐕 ……. (1)

Where V = volume of solution in ml

Suppose C gm equivalent electrolyte is dissolved in 1000 ml solution then,

∴ 𝐕 =𝟏𝟎𝟎𝟎

𝐂 …….. (2)

Where C = normality of solution

∴ By equation (1) and (2)

∴ 𝛌𝐜 =𝐤 .𝟏𝟎𝟎𝟎

𝐂 𝒎𝒉𝒐 ……… (3)

6. Molecular conductance: It is denoted as λm and it’s measured in mho.

It is explain similar to the equivalent conductance.

Effects of Dilution:

On the conductance: The conductance of a solution is due to the presence of lions.

Moreover the degree of dissociation of weak electrolyte is increase on dilution which

means ions are produced in solution therefore it is clear that the conduction of a solution

should be increase on dilution.

On the specific conductance: The specific conductance depends on the number of ions

present per ml (c.c.) of the solution. The degree of dissociation of weak electrolyte is

increase on dilution but the number of ions per ml (c.c.) decreases. Therefore it is

expected that the specific conductance of solution should be decrease on the dilution.

On the equivalent conductance and molecular conductance: The equivalent

conductance increase on dilution this increase in because of that the equivalent

conductance is the product of the specific conductance and the volume V of the solution

containing 1 gm equivalent of the electrolyte the decreasing value of specific conductance

is less than the increasing value of V, therefore, the equivalent conductance will increase

on dilution.

Similarity molecular conductance will also increase on dilution.


Conductance Cell:

The electrodes (conductance cell) are generally made from two

parallel plates of platinized platinum plate, which do not bend

readily. The relative positions of these electrodes are fixed by

sealing with connected glass tubes into the side of the

conductivity cell.

High quality borosilicate glass is used in the high quality

borosilicate glass is used in the conductivity cell which is

inactive with electrolyte solution and temperature co-efficient of

the glass should be lower.

Importance of Cell Constant:

Distance between two platinum plate and cross section area of

the two platinum plate play important role in the measurement

of the conductance. For eg.

1. When the distance between two plates are 1 cm and the cross section area is 1 cm, the

conductivity cell is known as standard cell, thus the value of observed conductivity is

equal to specific conductance.

2. When the distance between two plates is 1 cm. and the cross section area of plates is 2

cm, the conductance comes double than it is expected. Thus the conductance cell does

not called standard cell and the value of observed conductance is not equal to specific

conductance but proporti5onal to it. Therefore, it is important to calculate a factor for the

conductance cell. Known as cell constant the observed conductance when multiplied by

the cell constant gives the value of specific conductance.

3. When the distance between the two plates is 2 cm and the cross section area of plates is 1

cm, the conductance comes half than it is expected. Thus the conductance cell does not

called standard cell and the value of observed conductance is not equal to specific

conductance but proportional to it. Therefore, it is important to calculate a factor for the

conductance cell, known as cell constant, the observed conductance when multiplied by

the cell constant gives the value of specific conductance.

Determination (equation) of the cell constant:

The platinum plates in the cell are not exactly one cm apart and may not have surface area of

one sq.cm. Thus the value of observed conductivity is not equal to specific conductivity but

proportional to it. therefore it is important to calculate a factor for the conductivity cell,

known as cell constant the cell constant when multiplied by the observed conductivity gives

the value of specific conductance we known that,

∴ 𝐑 = 𝐏 ×𝐥

𝐚 but

𝐥

𝐚= 𝒙 = 𝒄𝒆𝒍𝒍 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕

∴ 𝐑 = 𝐏 . 𝒙

∴ 𝐱 =𝐑

𝐏=

𝟏𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒂𝒏𝒄𝒆

𝟏𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒂𝒏𝒄𝒆

∴ 𝐱 =𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒂𝒏𝒄𝒆

𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒂𝒏𝒄𝒆=

𝐊

𝐂

∴ 𝐊 = 𝐂 . 𝒙

The cell constant is determined by substituting the value of specific conductance of KCl

solution at 25 °C. This value as determined by kohlrausch was round to be 0.002765 mhos.

The value of conductivity is then observed with the given cell using 𝐍

𝟓𝟎 KCl, solution. The cell

constant is then calculated by using the following relation.

∴ 𝐂𝐞𝐥𝐥 𝐂𝐨𝐧𝐬𝐭𝐚𝐧𝐭 =𝟎. 𝟎𝟎𝟐𝟕𝟔𝟓

𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒂𝒏𝒄𝒆


Kohlrausch Law:

Conductance of electrolyte solution depend on the number of ions and types of ions. Strong

electrolytes is completely ionized in the solution at any concentration but ionization ionized

of weak electrolytes is incomplete in the solution moreover the ionization of weak electrolytes

is increase on dilution. Thus conductance also increase.

Therefore according to kohlrausch rule, at infinite dilution all the positive and negative ions

are independent to each other in the solution. Thus all the ions gives contribution in

conductance of the solution. Therefore conductance at infinite dilution (λ∞) is equal to the

sum of conductance of positive ion (λ+) and negative ion (λ-)

𝛌∞ = 𝛌+ + 𝛌−

Conductance at infinite dilution of some ions at 25 °C temp are given below.

Ion λ∞ Ion λ∞

H+ 350 Cl- 76

Na+ 50 OH- 198

NH4+ 73.5 CH3COO- 40.9

Ag+ 61.9 NO3- 75.5

Ba+2 64 SO4-2 80

Platinlzation of electrode:

Electrodes are coated with finally divided platinum black it’s called platinized electrodes and

this process is called platinization of electrodes.

First of all platinum electrode of conductance cell is clean by the chromic acid and D.W. and

then the platinization is carried out by taking 2 to 3.1 solution of chloroplatinic acid (PtCl4)

and 0.02 to 0.03 gm at lead acetate in the cell. For this process conductance cell is

connected to a rheostat, source of current and two way switch.

Now pass the 4 volt current from the source about 15-20 minutes. Due to this electrolysis of

PtCl4 is occurs and platinum electrodes of conductance cell is platinized.

In the above process electrical current controlled in such a way that gas bubbles obtained

continue by rheostat. Every 30 sec. electrical current alternate by the two way switch, in this

way 15-20 minutes current is pass. Thus electrodes of the conductance cell is Platonized

reaction is given below.

PtCl4 → Pt+4 + 4Cl-

4Cl- → 2Cl2 + 4e-

Pt+4 +4e- → Pt

Reduction of polarization of electrode:

when electrical current pass from the platinum electrode of conductance cell, the gas

bubbles of H2 and O2 are deposited on the platinum electrode and they are behave as a gas

electrode it is called polarization. Gas electrode produced current in opposite direction to the

current applied it is called back emf, due to this, efficiency of the platinum electrode of the

conductance cell is decrease.

There are for the reduction of polarization platinum electrodes of conductance cell is

platinized electrodes of conductance cell is platinized because on the black surface very less

amount of gas bubbles are deposited.


There are for the reduction of polarization platinum electrodes of conductance cell is

platinized because on the black surface very less amount of gas bubbles are deposited.

Therefore for the reduction of polarization A.C. current is used in the conductometry.

Because polarization is reduced when A.C. current is used.

Care taken in conductometric titration:

Following care should be taken in conductometry titration.

i) Conductance of the water should be less than 5.0 x 10-8 mho/cm which used in

conductometry analysis.

ii) Pure conductivity water is used in the conductometry analysis because they have no

impurity of NH3, CO2 and other ions.

iii) During the conductometry titration, concentration difference between reactant and titrant

should be more than 10 times thus dilution effect can be reduced and accurate end point

can be taken.

iv) Cell constant should be taken with help of standard KCl solution.

v) Platinum plate of the conductivity cell should be platinized and A.C. current should be

used in the conductometry titration.

vi) It is necessary to keep the temp. of the solution Constant throughout the experiment.

This is achieved by placing the conductance cell in the thermostat.

Factors affecting conductance of solution:

Following factors are affect to the conductance

i) Type of solvent it means polar or non-polar solvent as well as viscosity.

ii) No of ions present in the solution.

iii) Electrical charge (oxidation state) on the ions and size of the ions.

iv) Concentration of the solution and extent of ionization of the electrolytes.

v) Temperature of the solution during experiment.

vi) Electro motive force (EMF) applied to the conductor.

Speciality of conductometric titration:

When the volumetric titration is difficult then the conductometric titration is very useful.

i) When proper indicator is not available than the conductometric titration is very useful.

ii) Titration of very weak acid-base are carried out by conduct metric method with more

accurately then other volumetric method.

iii) Titration of very dilute solution are carried out by conductometric method with more

accurate than other volumetric method.

iv) Redox titration can be carried out with more accurately than indicator method by the

conductometric method.

v) It is used in the ppt. titration and they are very fast.

vi) When both of the solutions, reactant and titrant are colored, no any indictor are

applicable. In this type of titration conductometric method is applicable.

Conductivity water:

It is observed that the conductivity of solution of any electrolyte is very sensitive to the

presence of impurities therefore it is essential that water used in the measurements of

conductance should be of high degree of purity it means has no conductance due to

dissolved impurities for this reason ordinary water is not suitable for the conductance


measurements because it may contain dissolved CO2, NH3 and others ion, thus conductivity

water is used in conductance measurements.

It can be prepared by distilling distilled water containing small amounts of sodium hydroxide

and potassium permanganate. The distillation is generally carried out in a pyrex glass

apparatus and conductivity water is collected in a receiver fitted with soda lime tube.

Temperature effect:

It is known that the conductance of the solution varies with temp, the temp. Co-efficient for

conductance measurements is about 2% per degree celsius. Therefore it is necessary to keep

the temperature of the solution constant throughout the experiment. This is achieved placing

the conductivity cell is the thermostat.

Neutralization titration:

1. Strong Acid with Strong Base:

Let us strong acid with strong base titration consider as the titration of HCl with NaOH

neutralization reaction is given as

HCl + NaOH → NaCl + H2O

Acid and base both are strong electrolyte, therefore they are very fast and completely ionized

the ionic reaction is given as

[H+ + Cl-] + [Na+ + OH-] → [Na+ + Cl-] + H2O

λ = 350 76 50 198 50 76

Take definite volume of acid solution in the conductivity cell and conductivity cell connect

with conductometer.

Before the start of the titration the acid solution has a high conductivity due to highly mobile

hydrogen ions and chloride ions. When NaOH is added to the HCl, the highly mobile

hydrogen ions are replaced by less mobile sodium ions. This will result in the decrease of

conductance rapidly, the solution at neutralization it means at end point containing only

sodium and chloride ions, will have a minimum conductivity.

Now if more NaOH is added, the conductivity will increase due to the presence of OH- and

Na+ ions.

Therefore if the titration is carried out at constant temperature and conductance is plotted

against the volume of NaOH added, following graph will be obtained, the point of intersection

of the two lines, give the end point,

Then by following equation we can calculated the

concentration of the acid solution.

Acid solution = Base solution

N1 V1 = N2 V2

∴ 𝐍𝟏 =𝐍𝟐 . 𝐕𝟐

𝑽𝟏 (From graph)

Note:

In actual practice the lines may be slightly curved because of

i) Variation in temperature due to heat of neutralization.

ii) Interionic effect.

iii) Increase in the volume of the solution because of the addition of the reagent.


In spite of these, the inflection is sharp enough to get the exact end point.

An important advantage of the conductometric titration is that the relative change in

conductance is almost independent of the concentration of strong acid being titrated. Thus,

dilute solution can be titrated with almost the same accuracy as concreted solution.

2. Strong Acid with Weak Base:

Let us strong acid with weak base titration consider as the titration of HCl with NH4OH

neutralization reaction is given as

HCl + NH4OH → NH4Cl + H2O

Acid HCl is strong electrolyte, thus they are completely ionized in the solution while base

NH40H is weak electrolyte, thus they are partly ionized the ionic reaction is given as

[H+ + Cl-] + [NH4+ + OH-] → [NH4

+ + Cl-] + H2O

λ = 350 76 73.5 198 73.5 76

Take definite volume of acid solution in the conductance cell and conductance cell connect

with conductometer.

Before the start of the titration the acid solution has high conductivity due to highly mobile

hydrogen ions and chloride ions, when NH4oH is added to HCl the highly mobile ammonium

ions are replaced by less mobile ammonium ions. This will result in the decrease of

conductivity rapidly. The solution at neutralization point means at end point, ail have a

minimum conductance.

Now if more NH4OH is added after the end point, the conductance will not change because

NH4OH does not ionized, because NH4OH is weak electrolyte and in presence of NH4Cl strong

electrolyte, common ion effect is obtained.

Then by following equation we can calculated the

concentration of the acid solution


N1 V1 = N2 V2

∴ 𝐍𝟏 =𝐍𝟐 . 𝐕𝟐


3. Weak Acid with Strong Base:

Let us weak acid with strong base titration consider as the titration of CH3COOH with NaOH.

Neutralization reaction is given as,

CH3COOH + NaOH → CH3COONa + H2O

Acid CH3COOH is weak electrolyte, thus they are slowly and partly ionized in the solution

while base NaOH are strong electrolyte, thus they are fast and completely ionized. The ionic

reaction is given as,

[CH3COO- + H+] + [Na+ + OH-] → [CH3COO- + Na+] + H2O

λ = 40.9 350 50 198 40.9 50

Take definite volume of acid solution in conductivity cell and conductivity cell connect with

conductometer.

Before the start of the titration, the acid solution has a low conductivity due to partly

ionization of weak acid. When a small amount of NaOH is added to CH3COOH solution, the

conductivity decrease initially, because highly mobile H+ ions replace by less mobile sodium

ions and then conductance of the solution slowly increase with further addition of NaOH till

end point or neutralization point does not come.


Now, if after the neutralization of CH3COOH, more NaOH is added, the conductance of the

solution is increase rapidly due to the more NaOH is added, the conductance of the solution

is increase rapidly due to the more Na+ and OH- ions produced.

Therefore it the titration is carried out of constant temperature and the conductance is

plotted against the volume of NaOH added following plotted will be obtained the point of inter

section of the two lines gives the end point.

Then by following equation we can calculate the



N1 V1 = N2 V2

∴ 𝐍𝟏 =𝐍𝟐 . 𝐕𝟐


4. Weak Acid with Weak Base:

Let us weak acid with weak base titration consider as the titration of CH3COOH with NH4OH.

Neutralization reaction is given as,

CH3COOH + NH4OH → CH3COONH4 + H2O

Acid and base both are weak electrolyte thus they are slowly and partly ionized. The ionic

reaction is given as,

[CH3COO- + H+] + [NH4+ + OH-] → [CH3COO- + NH4

+] + H2O

λ = 40.9 350 73.5 198 40.9 73.5

Take definite volume of acid solution in conductance cell and conductance cell connect with

conductometer.

Before the start of the titration, the acid solution has a low conductance due to partly

ionization of weak acid, when a small amount of NH4OH is added to CH3COOH solution, the

conductance decrease initially, because highly mobile H+ ions replace by less mobile

ammonium ions and then conductance of the solution slowly increase with further addition

of NH4OH till end point or neutralization point does not come.

Now if more NH4OH is added after the end point, the conductance will not change because

NH4OH does not ionized. Because NH4OH is weak electrolyte and in the presence of

CH3COONH4 strong electrolyte common ion effect is obtained.

Therefore if the titration is carried out at constant temperature and the conductance is

plotted against the volume of NH4OH added, following plotted will be obtained. The point of

intersection of the two lines gives the end point.




N1 V1 = N2 V2

∴ 𝐍𝟏 =𝐍𝟐 . 𝐕𝟐


5. Strong Acid and Weak Acid mixture with Strong Base:

Let us mixture of strong acid and weak acid with strong base titration consider as the

titration of mixture of HCl and CH3COOH with NaOH, neutralization reaction is given as,


HCl + CH3COOH + NaOH → NaCl + CH3COONa + H2O

The ionic reaction is given as,

[H+ + Cl-] + [CH3COO- + H+] + [Na+ + OH-] → [Na+ + Cl-] + [CH3COO- + Na+] + H2O

λ = 350 76 40.9 350 50 198 50 76 40.9 50

Take definite volume of mixture of strong acid and weak acid in the conductance all and

conductance all connect with conductorneter.

In the mixture strong acid HCl completely ionized but in the presence of strong acid HCl

weak acid CH3COOH cannot ionized. Therefore before the start of the titration, acid solution

has high conductivity due to highly mobile hydrogen ions and chloride ions. When NaOH is

added to the mixture, the highly mobile H+ ions are placed by less mobile Na+ ions. This will

be result in the decrease of conductance rapidly, which is shown by AB line in fig.

After completely neutralization of HCl, weak acid CH3COOH can be ionized, when a small

amount of NaOH is added to the mixture, the conductivity decrease initially, because highly

mobile H+ ions replaced by less mobile Na+ ions and then conductivity of the solution slowly

increase with further addition of NaOH till end or neutralization point does not come which is

shown by BC line in fig.

Now if after the neutralization of both acid of mixture more NaOH is added, the conductance

of the mixture solution is increase rapidly due to the more OH- and Na+ ions produced which

is shown by CD line in fig.

Therefore if the titration is carried out at constant temperature and the conductance is

plotted against the volume of NaOH added, following graph will be obtained. The point of

intersection of the two lines gives the end points. In the graph first section shows the end

point of strong acid while second section show the end point of weak.


concentration of the mixture of both acid.

∴ 𝐍𝟏 =𝐍𝟐 . 𝐕𝟐

𝑽𝟏 (From graph for HCl)

∴ 𝐍𝟏 =𝐍𝟐 . (𝐕𝟑− 𝐕𝟐)

𝑽𝟏 (From graph for CH3COOH)

6. Strong Acid and Weak Acid mixture with Weak Base:

Answer is similar to the previous question to the neutralization. after neutralization of both

acid, if more amount of NH4OH is added, the conductance is not change because weak base

NH4OH cannot ionized in the presence of salt due to common ion effect, which shown by CE

line in the fig. Fig. is given in the previous question.

Precipitation titration:

1. Titration of NaCl against AgNO3:

Let us consider the precipitation titration between KCl and AgNO3. The reaction which is

taking place can be represented as

KCl + AgNO3 → KNO3 + AgCl


[K+ + Cl-] + [Ag+ + NO3-] → [K+ + NO3

-] + AgCl

λ = 73.5 76 61.9 75.5 73.5 75.5

Take a specific volume of KCl solution in the conductivity cell and conductivity cell is connect

with conductometer.


In the initial stage of the titration, with the addition of the silver nitrate, the conductance

does not change much. This is because the Cl- ions are replaced by NO3- ions and both have

almost same ionic conductance. Therefore conductance is constant till end or neutralization

point does not come.

Now if after the end point, the excess of silver nitrate added, the conductance is increase

because number of Ag+ and NO3- ions are increase in the solution.

When the conductance is plotted against the volume of

AgNO3 added following graph will be obtained. The point of

intersection of two lines gives the end point.


concentration of the KCl solution.

KCl solution = AgNO3 Solution

N1 V1 = N2 V2

∴ 𝐍𝟏 =𝐍𝟐 . 𝐕𝟐


2. Titration of BaCl2 against Na2SO4:

During the titration between BaCl2 and Na2SO4 the precipitation titration can be represented

as

BaCl2 + Na2SO4 → BaSO4 + 2NaCl


[Ba+2 + 2Cl-] + [2Na+ + SO4-2] → [2Na+ + 2Cl-] + BaSO4

λ = 64 76 50 80 50 76

Take definite volume of BaCl2 solution in conductance cell and conductance cell connect with

conductometer.

In the initial stage of the titration with addition of the sodium sulfate, the conductance slowly

decrease. This is because the Ba+2 ions are replaced by the Na+ ions. The mobility of the Na+

ions are less than Ba+2 ions but difference between them is small. The solution at the end

point containing only sodium and chloride ions, will have minimum conductance.

Now if more Na2SO4 is added, the conductance is increase rapidly, because number of Na+

and SO4-2 ions are increase in the solution.

When conductivity of solution is plotted against the

volume of Na2SO4 added, following graph will be

obtained. The point of intersection of the two lines gives

the end point.


concentration of BaCl2 solution.

BaCl2 solution = Na2SO4 Solution

N1 V1 = N2 V2

∴ 𝐍𝟏 =𝐍𝟐 . 𝐕𝟐


3. Titration of Ba(OH)2 against MgSO4 (titration of co-precipitation of two compounds):

Let us both the products of the titration are sparingly soluble salt consider as the titration of

Ba(OH)2 with MgSO4. Precipitation reaction is given below.

Ba(OH)2 + MgSO4 → BaSO4 + Mg(OH)2


[Ba+2 + 2OH-] + [Mg+2 + SO4-2] → BaSO4 + Mg(OH)2

λ = 64 198 53 80


When the co-precipitation take place of two sparingly soluble salt, by the reaction between

two reactant, when which compound precipitated first is determined by the value of their Ksp.

Compound which have less Ksp is precipitated first here BaSO4 is precipitated first due to

having less Ksp value than Ksp of Mg(OH)2.

Before the start of the titration conductivity of solution is very high due to the Ba+2 and OH-

ions. In the initial stage of the titration with addition of the MgSO4, the conductance rapidly

decrease. This is because the number of Ba+2 and OH- ions are decrease due to precipitation

of both products Baso4 and Mg(OH)2. The solution at the end point containing less number

of ions (Ba+2, OH-, Mg+2 and SO4-2) due to the solubility of sparingly soluble salt thus will

have a minimum conductance.

Now if more MgSO4 is added, the conductance of the soluble is increase rapidly because No.

of Mg+2 and SO4-2 ions are increase in the solution.

When Conductance of the solution is plotted against the

volume of Mgso4 added the following graph will be

obtained the point of intersection of the two lines gives the

end point.


concentration of Ba(OH)2 solution.

Ba(OH)2 solution = MgSO4 solution

N1 V1 = N2 V2

∴ 𝐍𝟏 =𝐍𝟐 . 𝐕𝟐


Replacement Titration:

Replacement Titration are of two types

1. Titration of salt of weak acid with strong acid.

2. Titration of Salt of weak base with strong base.

1. Titration of salt of weak acid with strong acid.

Titration of salt of weak acid with strong acid consider as the titration of CH3COONa with

HCL replacement reaction is given below.

[CH3COO- + Na+] + [H+ + Cl-] → CH3COOH + [Na+ + Cl-]

Take specific volume of salt of weak acid solution in the conductance cell and conductance

cell connect with conductometer.

Before start of the titration solution will shows specific conductance due to the CH3COO- and

Na+ ions. In the initial stage of the titration with addition of HCl solution, the conductance

slowly increase up to end point because CH3COO- ions are replaced by CI- ions. The mobility

of CH3COO- ions are less than CI- ions but difference between them is small.

Now is more HCl is added the conductance is increase rapidly, because high mobility of H+

ions and CI- ions.

When conductance of the solution is plotted against the

volume of HCl added following graph will be obtained. The

point of intersection of the two lines gives the end point.


concentration of CH3COONa solution.

CH3COONa Solution = HCl Solution

N1 V1 = N2 V2

∴ 𝐍𝟏 =𝐍𝟐 . 𝐕𝟐


2. Titration of Salt of weak base with strong base:

Titration of salt of weak base with strong base consider as the titration of NH4Cl with NaOH

replacement reaction is given below.

[NH4+ + Cl-] + [Na+ + OH-] → NH4OH + [Na+ + Cl-]


Take specific volume of salt of weak base solution in the conductivity cell and conductivity

cell connect with conductometer.

Before the start of the titration solution will show specific conductance due to the NH4+ and

CI- ions. In the initial stage of the titration with addition of NaOH solution, the conductance

slowly decrease up to end point, because NH4+ ions are replaced by Na+ ions. The mobility of

NH4+ ions are greater than Na+ ions but difference between them is small Thus at the end

point, solution will have minimum conductance.

Now if more NaOH is added, the conductance is increase rapidly, because high mobility of

OH- ions and Na+ ions.

When conductance of the solution is plotted against the

volume of NaOH added, following graph will be obtained.

The point of intersection of two lines gives the end point.


concentration of NH4Cl solution.

NH4Cl solution = NaOH solution

N1 V1 = N2 V2

∴ 𝐍𝟏 =𝐍𝟐 . 𝐕𝟐


Application of Conductometry:

1. The solubility and solubility product of sparingly soluble salts:

AgCl, BaSO4, PbSO4, Mg(OH)2 etc. are sparingly soluble salts with help of conductance

measurement we can calculate the solubility product of the sparingly soluble salts with

greater accuracy than other chemical methods.

Suppose find out the solubility and solubility product of AgCl at 25 °C. The soluble impurity

in the AgCl are readily removed by washing with conductivity water then prepare saturated

solution of AgCl.

Now the conductance of this solution is measured by usual method and then specific

conductance will be calculated by following equation.

Specific conductance = measured conductance ⨉ Cell constant

K = C ⨉ X …..….. (1)

We know that the saturated solution of a sparingly soluble salt AgCl is so dilute, Therefore

electrolyte present in it may be regarded as completely ionized. Thus the value of equivalent

conductance λv can be considered to be equal to equivalent conductance at infinite dilution

λ∝ therefore,

λv = λ∝ (But λv = k ⨉ V)

k ⨉ V = λ∝

Where, V = Volume of the saturated solution having gm equivalent weight of salt.

V =λ∝

k …..….. (2)

V ml saturated solution having gm equivalent weight salt, therefore 1000 ml saturated

solution will have equal to their solubility therefore.

Solubility (S) = 1000×E

V ….…… (3)

Where, E = Equivalent weight of salt

Put the value of equation (2) in equation (3).

Solubility (s) = 1000×𝐸⨉𝑘

𝜕𝛼 ….…… (4)

Where, E = equivalent weight of salt

K = specific conductance

λ∝ = λ+ + λ- (According to Kohlrausch)

By above equation (4) we can calculated the solubility (S) of the sparingly soluble salt.

With the help of relation between solubility and solubility product we can determine the

solubility product of the sparingly soluble salts for example.

AgCl(S) ⇌ 𝐴𝑔(𝑎𝑞)+ + 𝐶𝑙(𝑎𝑞)

−


S𝑀

𝐿 S

𝑀

𝐿

Ksp of AgCl = [Ag+][Cl-]

= [S] [S]

Ksp of AgCl = S2 ...……. (5)

By above equation we can calculate the solubility product of sparingly soluble salt.

2. Degree of dissociation and dissociation constant of Weak acid:

Take specific volume of solution of weak acid with any concentration in conductance cell and

measure conductance (C). Then determine specific conductance by following equation.

Specific Conductance = measured conductance ⨉ Cell Constant

K = C ⨉ X ………. (1)

Now calculate the equivalent conductance by following equation.

𝛌𝑽 =𝟏𝟎𝟎𝟎 ⨉ 𝒌

𝒄 .……… (2)

Where, C = concentration in normality Now conductance at infinite dilution can be calculated

by kohlrausch rule.

∴ 𝛌∞ = 𝛌+ + 𝛌− ….…… (3)

Where, λ+ = conductance of positive ion, λ- = Conductance of negative ion.

With the help of equation (2) and (3) we can calculate degree of dissociation α by following

equation.

𝜶 =𝝀𝒗

𝝀𝜶 ..…….. (4)

When weak acid dissolved in water, it is partly ionized and following equilibrium obtained.

HA ⇌ H+ + A-

Suppose Initial Mole 1 0 0

Suppose Degree of dissociation is (1-α) α α

α/Mole ∴(1-α) at equilibrium

Concentration at equilibrium (1-α)c αc αc

∴ Dissociation constant (Ka) =[H+][A−]

[HA]=

α∙c×α∙c

(1−α)C

∴ (Ka) =𝛼2 ∙ 𝑐

1−𝛼

………. (5)

By the equation (4) we can calculate degree of dissociation and by equation (5) we can

calculate dissociation constant of weak acid.

3. Degree of hydrolysis and hydrolysis constant of salt:

When a salt of weak acid or weak base is dissolved in water, the conductance of solution will be

partly due to the ions of the salt and partly due to free H+ or OH- ions of the acid or base formed

by hydrolysis. An excess of weak acid or base, in the presence of its salt can be regarded as

completely unionized.

Let us consider NH4Cl, which is a salt of weak base and a strong acid. When this compound is

dissolved in water, it is hydrolyzed. If degree of hydrolysis is h, the h equivalents of free acid and

base are produced and (1-h) equivalents of salt is not hydrolyzed. Here h is degree of Hydrolysis.

This can be represented as,

NH4Cl + H2O ⇌ NH4OH + HCl

Now common ions are removed from both side

NH4+ + H2O ⇌ NH4OH + H+

If Starting mole is 1 - - -

Suppose degree of hydrolysis (1-h) - h h

Concentration at equilibrium (1-h)C - hc hc

∴ Hydrolysis constant (kh) =[H+][NH4OH]

[NH4+]

=ℎ∙𝑐 × ℎ∙𝑐

[1−ℎ]𝑐


(kh) =ℎ2𝑐

(1−ℎ) …..….. (1)

But extent of hydrolysis is very small, therefore we can take [1-h] ≅ 1

kh = h2 C ..…….. (2)

Now,

Total conductance of solution = conductance of salt + conductance of free acid

∴ λ = (1-h)λc + hλH+

λ = λc – hλc + hλH+

λ – λc = hλH+ - hλc

∴ h = 𝜆−𝜆𝑐

𝜆𝐻+−𝜆𝑐 ….…… (3)

Where, λ = total conductance of solution of salt

λc = Equivalent conductance of unhydrolyzed salt

λH+ = Equivalent conductance of free acid

By above equation degree of hydrolysis can be calculated.

Now total conductance of solution of salt can be calculated by experimentally, λc can be

determined by measuring the equivalent conductance of NH4Cl in presence of excess of NH4OH.

The value of λH+ are 350.

Thus, H can be calculated. Now hydrolysis constant Kh is calculated by equation (1) or (2).

Best of Luck

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