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PHYSICAL CHEMISTRY LAB MANUAL

EXPERIMENT: 1

CONDUCTOMETRIC TITRATION OF A STRONG ACID WITH A STONG BASE

AIM:

To determine the concentration of a strong acid by a

conductometric titration using a strong base

INTRODUCTION:

The conductivity of an electrolytic solution depends on 1. ionic

concentration and 2. Ionic mobility. The changes in the number and

nature of the ions during an acid base titration, which affect the

total conductance of the solution, can be utilized for the

determination of end point. Now let us consider an acid base

titration in which 50.0 ml of 0.1 N hydrochloric acid is titrated with

1.0N sodium hydroxide solution. In aqueous solution a strong acid

can be considered to be completely in ionized form.

Although both chloride and hydrogen ions contribute to the total

conductance of the solution, hydrogen ion has a major share due to

its high mobility (349.8 mhos.cm2.mol-1) when compared to chloride

ion (76.4 mhos.cm2.mol-1). When a small portion of a strong alkali

such as sodium hydroxide is added to the solution, the OH - ion of

the alkali reacts with the hydrogen ion forming water. Further, a

sodium ion is introduced in its place. It means addition of sodium

hydroxide results in a replacement of high mobile proton with a low

mobile sodium ion (50.1 mhos.cm2.mol-1). This results in a steep

decrease of conductance until the equivalence point is reached.

1


After the equivalence point the excess of sodium hydroxide causes

the conductance to rise a little bit slowly when compared to the

decrease before the equivalence point. Both sodium ions and

hydroxyl ions contribute to the conductance after the equivalence

point. The major contribution, however, is due to OH -, which has an

ionic mobility of 198.6. mhos.cm2.mol-1. Therefore the titration

curve contains two lines intersecting at the end point. As

conductance is sensitive to dilution, a volume correction needs to be

applied to the meter readings. The volume of the analyte solution

goes on increasing with the addition of sodium hydroxide solution

during the titration.. If ‘C’ is the measured conductance, ‘V’ is the

initial volume of the analyte solution and ‘v’ is the volume of sodium

hydroxide added, then the corrected conductance is given by,

The corrected conductance in milli ohms should be plotted against

volume of the sodium hydroxide for detecting the end point of the

titration.

SOLUTIONS:

1. Sodium hydroxide solution : Prepare 250 ml of

approximately 1.0 N Sodium hydroxide by dissolving around

10 gms of analytical grade sodium hydroxide (gm.equivalent

weight=40.0) in carbon dioxide free distilled/ deionized water

2


or conductivity water. Use a rough balance for weighing

sodium hydroxide.

2. Oxalic Acid (gm.equivalent weight = 63.03) solution:

Prepare 100 ml of 1.0 N oxalic acid solution by dissolving

exactly 6.303 gms of analytical grade oxalic acid in distilled

water and make it up to the mark. Calculate the strength of

the oxalic acid solution using the actual weight of the

substance transferred in to the flask if it is not exactly 6.303

gms.

3. Hydrochloric acid solution: Prepare 100.0 ml of

approximately 1.0 N hydrochloric acid stock solution by

diluting concentrated hydrochloric acid (11.4N).

PROCEDURE:

Standardize sodium hydroxide solution using standard oxalic

acid solution with phenolphthalein as indicator.

Also standardize the hydrochloric acid using the standard

sodium hydroxide solution with phenolphthalein as indicator.

Prepare 100 ml of 0.1N hydrochloric acid solution by the exact

dilution of the stock solution.

Switch on the conductivity meter for at least one hour before

taking any measurements.

Standardize the conductivity meter (using internal standard)

following the instructions given in the Instruction Manual

supplied by the manufacturer along with the Instrument.

Clean the conductivity cell thoroughly with distilled water and

then with conductivity water.

Prepare the experimental solution by taking 25.0 ml each of

0.1N hydrochloric acid and conductivity water with a burette

in to a 100 ml beaker. Mix the contents of the beakers

thoroughly with a glass rod.

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Insert the conductivity cell in to the experimental solution and

note down the meter reading after selecting an appropriate

range using the range switch of the meter.

Fill the burette with the standardized solution of sodium

hydroxide and titrate the experimental solution by adding 0.2-

0.5 ml portions.

Note down the meter readings each time after thoroughly

stirring the contents of the beaker.

Tabulate the results in the Table format shown below

Concentration of the titrant (Sodium hydroxide)= ________ N

Total initial volume of the Analyte solution, V= ________ ml

S.N

o.

Volume of Sodium

hydroxide added,

v ml

Observed

Conductance

in milli mhos,

Corrected

conductance,

C’= C X (V+v)/V

milli mhos

1 0.0

2 1.0

3 2.0

4 3.0

etc.

Draw a graph of corrected conductance versus volume of

sodium hydroxide added.

Join the points linearly before and after the equivalence point

and extend them to get the intersection point.

Note down the volume of the sodium hydroxide ‘x’

corresponding to the intersection point (end point).

Calculate the concentration of hydrochloric acid using the

formula V1.N1=V2.N2, where V1 is the volume of hydrochloric

acid taken = 25.0 ml, N1 is the concentration of H Cl to be

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determined, V2 is the volume of sodium hydroxide at the end

point = x ml and N2 is the strength of the standardized sodium

hydroxide

EXPERIMENT: 2

CONDUCTOMETRIC TITRATION OF A MIXTURE OF STRONG

ACID AND WEAK ACID WITH A STRONG BASE

AIM: To determine the concentrations of strong acid and weak acid

in a mixture by a conductometric titration using a strong base.

INTRODUCTION:

When a mixture of strong acid and weak acid is taken, the

strong acid exists almost completely in the ionic form as shown

below.

Whereas as the weak acid such as acetic acid mostly exists in the

un-dissociated molecular form. In the presence of the strong acid,

due to common ion effect, the dissociation is further suppressed and

the molecule almost exists in the un dissociated molecular form.

Therefore, the initial conductivity of a mixture of strong acid and

weak acid is only due to the ions of the strong acid. The initial

conductivity of the solution is therefore, high, due to the high

mobility and abundance of the hydrogen ions in solution. When

sodium hydroxide is added during the titration, these hydrogen ions

are replaced with less mobile sodium ions resulting in a rapid

decrease of conductance. This happens until all the strong acid is

neutralized. Further addition of sodium hydroxide results in the

formation of

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sodium acetate due to the neutralization of acetic acid. As sodium

acetate exists in ionic form, its formatioAn raises the conductance of

the solution slightly due to the low ionic mobilities of both Na+ (50.1

mhos.cm2.mol-1) and CH3COO- (40.9 mhos.cm2.mol-1) ions. This

continues until all the acetic acid is neutralized. The excess alkali

added over and above this again raises the conductance but now

with different and high rate due to the relatively high mobility of

hydroxyl ions. The titration curve therefore contains three linear

portions with two intersection points. The first intersection point ‘x’

corresponds to the neutralization of the strong acid and the second

at ‘y’ corresponds to the total of strong and weak acids. From the

values of ‘x’ and ‘y’ the concentrations of both strong and weak

acids in a mixture can be determined.

SOLUTIONS:

1. Sodium hydroxide solution: Prepare 250 ml of approximately

1.0 N Sodium hydroxide by dissolving around 10 gms of

analytical grade sodium hydroxide (gm.equivalent

weight=40.0) in carbon dioxide free distilled/ deionized water

or conductivity water. Use a rough balance for weighing

sodium hydroxide.

2. Oxalic Acid (gm.equivalent weight = 63.03) solution: Prepare

100 ml of 1.0 N oxalic acid solution by dissolving exactly

6.303 gms of analytical grade oxalic acid in distilled water and

make it up to the mark. Calculate the strength of the oxalic

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acid solution using the actual weight of the substance

transferred in to the flask if it is not exactly 6.303 gms.

3. Hydrochloric acid solution: Prepare 100.0 ml of approximately

1.0 N hydrochloric acid stock solution by diluting concentrated

hydrochloric acid (11.4N).

4. Acetic acid solution: Prepare 100 ml of stock solution of

approximately 1.0 N acetic acid using conductivity water. This

can be done by the dilution of approximately 6.0 ml (use

measuring jar) of glacial acetic acid (17.4N) to 100 ml using

conductivity water.

PROCEDURE:

1. Standardize sodium hydroxide solution using standard oxalic

acid solution with phenolphthalein as indicator.

2. Standardize the stock solutions of hydrochloric and acetic acid

using the standardized sodium hydroxide solution.

3. Prepare 0.1 N solutions of both hydrochloric and acetic acid by

exact dilution of the stock solutions.

4. Switch on the conductivity meter for at least one hour before

taking any measurements.

5. Standardize the conductivity meter (using internal standard)

following the instructions given in the Instruction Manual

supplied by the manufacturer along with the Instrument.

6. Clean the conductivity cell thoroughly with distilled water and

then with conductivity water.

7. Prepare the experimental solution by taking 25.0 ml each of

0.1N hydrochloric acid and 0.1N acetic acid with a burette in

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to a 100 ml beaker. Mix the contents of the beakers

thoroughly with a glass rod.

8. Insert the conductivity cell in to the experimental solution and

note down the meter reading after selecting an appropriate

range using the range switch of the meter.

9. Fill the burette with the standardized solution of sodium

hydroxide and titrate the experimental solution by adding 0.2-

0.5 ml portions. Note down the meter readings each time

after thoroughly stirring the contents of the beaker.

10. Tabulate the results in the Table format shown below

Concentration of the titrant (Sodium hydroxide) = ____________

N

Total initial volume of the Analyte solution, V= ________

S.N

o.

Volume of Sodium

hydroxide added,

v ml

Observed

Conductance

in milli mhos,

Corrected

conductance,

C’= C X (V+v)/V

milli mhos

1 0.0

2 1.0

3 2.0

4 3.0

etc.

Draw a graph of corrected conductance versus volume of

sodium hydroxide added.

Join the points in the three portions of the curve linearly and

extend them to get the intersection points.

Note down the volumes of the sodium hydroxide ‘x’ and ‘y’

corresponding to the intersection points.

8


Calculate the strengths of the strong acid and weak acid in the

mixture as follows

EXPERIMENT : 3

THE POTENTIOMETRIC TITRATION OF FERROUS AMMONIUM

SULPHATE WITH POTASSIUM DICHROMATE

AIM:

To determine the concentration of a ferrous ammonium sulphate

solution, will be determined quantitatively by a potentiometric

titration against a standard solution of potassium dichromate.

THEORY:

Potassium dichromate is a strong oxidizing agent. Its principle

advantage is its availability as a primary standard. The long-term

stability of its solutions is another advantage for its use in red-ox

titrations. It is not however, as strong an oxidizing agent as MnO4- or

Ce4+. Its reduction half reaction is,

On the other hand, the solutions of ferrous ammonium sulphate

[Fe(NH4)2(SO4)2.6H2O] are normally very susceptible to aerial

oxidation, but when prepared in 0.5 to 1.0 M sulphuric acid may

remain stable for as long as a month. Its reduction half reaction is,

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The potential of the cell, given by,

PtFe3+, Fe2+Cr2O72-,Cr3+,H+Pt

Ecell = Ecathode – Eanode = 1.33-1.771 =0.559 V

The spontaneity of a chemical reaction requires G= -nFE, to be

negative. If the emf of the cell is positive, G, is negative and the

reaction is spontaneous. This indicates that the reaction between Cr

(VI) and Fe (II) in which Fe (II) is oxidized to Fe (III) and Cr (VI) is

reduced to Cr (III) is possible. Potentiometric methods of analysis

can be employed to detect the end point of the titration. The

electrochemical cell before the equivalence point is

Hg(liq) Hg2Cl2(solid), KCl (aq, satd.) Fe(II), Fe(III) Pt

Note that the Pt cathode is an inert indicator electrode that carries

electrons to the reduction half-cell. The electrode itself does not

undergo either oxidation or reduction. The reference electrode is a

saturated calomel electrode. The cell potential before the

equivalence point depends on the relative concentrations of the

oxidized and reduced forms of the iron.

Where, is the standard reduction potential (=0.771V) of

Fe(III)/ Fe(II) system

After the equivalence point the indicator electrode assumes the

potential of chromium system. The potential of the cell after the

equivalence point can therefore be represented by the equation,

Where, is the standard reduction potential (= 1.33V) of the

chromium system. The end point of the titration is indicated by a

large change (jump) in the cell potential due to the difference in the

standard reduction potentials of iron and chromium systems.

10


SOLUTIONS:

1. Ferrous ammonium sulphate solution: Prepare 100.0 ml

of 0.1N ferrous ammonium sulphate solution in 1.0 N sulphuric

acid as follows. Weigh accurately by the method of difference

about 3.9 gms of ferrous ammonium sulphate in to a 100 ml

volumetric flask. Add 5 ml of 1:1 sulphuric acid to it and make

it up to the mark with distilled water after a proper dissolution

of the substance. The concentration of the solution can be

calculated using the equation, 0.1 X W /3.9214, where ‘W’ is

the weight of the substance.

2. Potassium Dichromate solution: Prepare 250 ml of 0.1 N

potassium dichromate solution as follows. Weigh accurately

about 1.2 gms of the substance in to a 250 ml volumetric flask

and make it up to the mark with distilled water after proper

dissolution of the sample. The concentration of the solution

can be calculated using the equation, 0.1 X W/ 1.226, where

‘W’ is the weight of the substance.

3. 1:1 Sulphuric acid solution: Prepare 50 ml of 1:1 sulphuric

acid solution as follows: Mix approximately 25.0 ml of

concentrated sulphuric acid and 25.0 ml of water taking the

following precautions. Always add acid to water (not the

reverse) slowly in small portions with constant stirring and

cooling

PROCEDURE:

Prepare the experimental solution as follows: Take 10.00 ml of

ferrous ammonium sulphate solution using a burette in to a

100 ml beaker and add 5 ml of 1:1 sulphuric acid and 35 ml of

water with measuring jar. Mix the solution well using a glass

rod or magnetic stirrer

11


Wash the external surface of the saturated calomel electrode

(reference electrode) using distilled water

Now introduce both the electrodes in to the experimental

solution and connect them to the potentiometer, platinum

electrode to the red colored terminal and reference electrode

to the black terminal.

Fill the Burette with 0.1N potassium dichromate solution

Now perform a Pilot or Rough titration to locate the

equivalence point, by adding 1.0 ml portions of dichromate to

the analyte solution and noting down the potentiometer

readings in milli volts after thorough stirring of the solution. A

large change in the cell potential indicates the position of the

equivalence point. Note down the corresponding burette

readings.

Now an accurate titration can be performed by repeating all

the steps above but adding only 0.1 ml portions of dichromate

in the volume range where jump in cell potential has occurred

in the previous pilot titration.

Tabulate the results of the titration in the format shown in the

observations

Draw a sigmoid graph of observed cell potential versus

volume of the dichromate added. Note down the position (x)

of the end point. Also draw a first derivative graph i.e. E/V

versus Vaverage and note down the end point (x), which is the

volume of the dichromate corresponding to the maximum of

the curve.

The concentration of the ferrous ammonium sulphate solution

can now be determined using the equation, V1.N1=V2.N2 i.e.

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[Fe(II)]= x X 0.1/10.0 , where ‘x’ is the volume of the

dichromate at the end point.

OBSERVATIONS:

For both Pilot and Accurate titrations use the following Table format

for recording your observations.

Volume of Ferrous ammonium sulphate: ________ml

VOLUME OF K2Cr2O7

ADDED, ml (V)

POTENTIAL OF THE

ELECTROCHEMICAL CELL

AGAINST SCE, milli volts, (E)

E/V Vaverage

0.0

1.0

2.0

etc.

SAMPLE DATA AND RESULTS:

Volume of Ferrous ammonium sulphate: 10.0 ml

VOLUME OF K2Cr2O7 ( 0.1

N) ADDED, ml (V)

POTENTIAL OF THE

ELECTROCHEMICAL CELL

AGAINST SCE, milli volts,

(E)

E/V Vaverage

0.00 370

1.00 391 21 0.50

2.00 408 17 1.50

3.00 425 17 2.50

4.00 437 12 3.50

5.00 449 12 4.50

6.00 462 13 5.50

7.00 476 14 6.50

8.00 503 27 7.50

8.10 507 40 8.05

8.20 512 50 8.15

8.30 518 60 8.25

8.40 526 80 8.35

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8.50 536 100 8.45

8.60 565 290 8.55

8.70 691 1260 8.65

8.80 717 260 8.75

8.90 728 110 8.85

9.00 737 90 8.95

10.00 766 29 9.50

SIGMOID PLOT

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FIRST DERIVATIVE PLOT

Concentration of ferrous ammonium sulphate = 0.1 X 8.65/10.0 =

0.08650 N

EXPERIMENT: 4

KINETICS OF ESTER HYDROLYSIS

AIM: Determination of relative strengths of acids by Ester hydrolysis

INTRODUCTION: The hydrolysis of an ester, for example ethyl

acetate can be given by, is catalyzed by acids.

The rate of hydrolysis is proportional to the concentration of the

acid. Thus a measure of the rate (rate constant) in the presence of

different concentrations of acids can be used to study the relative

15


strengths of acids. The acid catalyzed hydrolysis of ester follows

pseudo first order and thus the rate constant is given by,

where ‘a’ is the initial concentration of the ester and ‘(a-x)’ is the

concentration of un-hydrolyzed ester at time ‘t’. The reaction is

followed by titrating fixed volume of the reaction mixture at fixed

intervals of time with sodium hydroxide. Titre value is a measure of

acetic acid produced with time. Let ‘T0’ is the titre value at the start

of the reaction,’Tt ‘ is the titre value at different time intervals of the

reaction is started and ‘T’ is the titre value when the reaction is

assumed to be completed.. Then ‘T-T0’ corresponds to ‘a’ and ‘T-

Tt’ corresponds to ‘a-x’. Hence the rate constant ‘k’ is

The hydrolysis is conducted at two different acid concentrations.

The corresponding rate constants k1 and k2 are obtained. The ration

‘k1/k2 ‘gives relative strengths of acids.

SOLUTIONS/ CHEMICALS:

1. Sodium hydroxide solution: Prepare 500 ml of 1N sodium

hydroxide solution by dissolving 20 gms of sodium hydroxide

(use rough balance) in carbon dioxide free distilled water.

2. Hydrochloric acid solution: Prepare 250 ml each of 1N and 2N

hydrochloric acid solutions by exact dilution of a pre-

standardized (~4N ) hydrochloric acid solution.

3. Ethyl acetate

4. Ice cold water

5. Phenolphthalein Indicator

PROCEDURE:

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Measurement of T0 : Take 10.0 ml of 1.0N HCl in to a 250 ml

conical flask. Add 30 ml distilled water and a drop of

phenolphthalein indicator. Titrate with 1.0N sodium hydroxide

until a permanent pink color is obtained. Note down the

volume of sodium hydroxide consumed as the value of the T0.

This is to be repeated for 2.0 N HCl also.

Study of the kinetics: Take 100.0 ml of 1.0N HCl in a reaction

bottle. Keep the bottle in a constant temperature water bath

to attain thermal equilibrium. Add 5.0 ml of ethyl acetate to

the contents of the reaction bottle simultaneously starting a

stop-watch when the pipette is half empty. Swirl the contents

thoroughly to obtain a uniform solution. For every ten minutes

withdraw 10.0 ml of reaction mixture in to a 250 ml conical

flask containing 40 ml of ice-cold water and one drop of

phenolphthalein indicator and titrate with standard sodium

hydroxide. This is to be continued for 70 mts. The titre values

for different time intervals represent ‘Tt ‘ values. Preserve the

remaining reaction mixture for 24 hrs and then titrate again

10.0 ml of reaction mixture with sodium hydroxide. Note down

the volume of the alkali consumed as T value.

Using T0, Tt and T values thus obtained calculate he rate

constant. Repeat the same procedure for 2N HCl .

Tabulate the observations in the format shown below

Concentration of acid = ______________

Titre value at zero time (T0) = __________________

Titre value at infinite time (T) = _______________

TIME

(Minutes)

Tt (T- Tt) log(T- Tt)

17


Average, k = _____________ min -

1

Graphical method: Make a plot of log(T- Tt) versus Time ‘t’as

shown. Calculate the slope of the straight line obtained. Slope

multiplied with 2,303 gives the rate constant.

If ‘k1’ and ‘k2’ are the rate constants obtained for 1N and 2N

hydrochloric acid solutions respectively, the relative strengths

of acids is given by k1/ k2.

EXPERIMENT: 5

DETERMINATION OF EQUILIBRIUM CONSTANT OF

POTASSIUM TRIODIDE EQUILIBRIUM

AIM:

Determination of the equilibrium constant of the equilibrium,

.

THEORY: Potassium iodide combines with iodine to form potassium

tri iodide, the equilibrium between KI, I2 and KI3 is represented by,

18


and the corresponding equilibrium constant Ke is given by,

Where the subscript ‘e’ stands for the equilibrium molar

concentrations. The concentrations at equilibrium can be calculated

using the distribution law (or partition law). As an example take

iodine as a solute and the two immiscible liquids as water and

chloroform. Iodine distributes between these two and one can

observe that iodine distributes between water and chloroform is a

definite ratio. This constant ratio is described as partition coefficient

( ) of Iodine between chloroform and water and expressed as

Partition coefficient =

i.e Partition coefficient =

This ratio (Which is found to be 125) is independent of the

total amount of dissolved iodine. If I2 and also KI are added to the

binary liquid system chloroform and water, value as expressed

above can be assumed to remain the same in spite of KI3 formation

according to the reaction in aqueous phase. This facilitates in

obtaining the

equilibrium concentration of I2, KI and KI3 and thus equilibrium

constant of the above equilibrium can be obtained.

Solutions/Chemicals

1. 250ml of 0.1M KI : It is prepared by weighing of KI and

dissolving in water in a 250ml volumetric flask and making it

up to the mark with distilled water.

2. 250ml of 0.1 N sodium thiosulfate : (Hypo) : Dissolve of

sodium thiosulfate using distilled water in a 250ml volumetric

flask and make it up to the mark. It is standardized against

standard dichromate solution.

19


3. 1%. Iodine solution (100ml) : Take 1g. of iodine in a beaker

and add 15 – 20ml of chloroform and stir it with a glass rod.

Transfer the I2 solution formed in to a 100ml volumetric flask.

Then add another 15-20ml of chloroform to the solid iodine

left in the beaker and transfer the I2 solution formed leaving

the un dissolved solid I2 in the beaker. Repeat the procedure

until all the 100ml chloroform is added.

4. 250ml of 0.1N K 2 Cr2 O7 : Dissolve of K2 Cr2 O7 using distilled

water in a 250ml volumetric flask and the solution is made up

to the mark.

5. 100ml of 0.5M oxalic acid : It is prepared by weighing and

dissolving of oxalic acid in distilled water in a 100ml

volumetric flask.

6. 100ml of 2N Hydrochloric acid : About 20ml of concentrated

Hydrochloric acid is diluted to 100ml using distilled water to

give approximately 2N Hydrochloric acid.

7. 1% starch solution : 100ml of distilled water is taken in a

beaker and set to boil. When it is about to boil 1g.of starch is

made in to a paste and put in to the beaker.

8. 5% potassium iodide : 250ml of 5% KI is prepared by

dissolving 12.5g of KI in 250ml distilled water.

Procedure :

Take 50ml of 0.1N KI in a 200ml bottle and add 15ml of 1%

Iodine (prepared in chloroform) and 10ml of chloroform. Close the

bottle with a cap and seal it, with liquid wax. Shake the bottle for 45

minutes, remove the wax seal and take the contents in to a

separating funnel. The organic layer and the aqueous layer are

separated and they are separately titrated with 0.01N Hypo solution

to find out iodine present in them following the procedure described

below.

20


Titration of organic layer : 5ml of organic layer is taken in to a

250ml conical flask, add 10ml of 5% potassium iodine and add 10ml

distilled water. The contents are titrated against 0.01N hypo

solution from a burette. When the contents in the conical flask

appear light yellow add 0.5ml of freshly prepared starch solution.

Then it turns to blue in color and continue the titration to see it is

colorless. The color change from blue to colorless is the end point.

Volume of hypo consumed is noted. Repeat the titration until

concurrent values are obtained.

Titration aqueous layer : Take 10ml of aqueous layer in to a

250ml conical flask and add 20ml distilled water. Titrate the

contents against 0.01N hypo solution from a burette. When the

contents appear light yellow in color add 0.5ml of freshly prepared

starch solution and continue the titration and till it turns from blue

to colorless indicating end point. Volume of hypo consumed is

noted.

The entire experiment described above is repeated by taking

different amounts of iodine and chloroforms ;

Calculation :

Concentration of iodine in organic layer =

Where V1 is volume of hypo consumed for 5ml organic layer

Concentration of iodine in aqueouslayer =

V2 is volume of aqueous layer taken

C2 = [I2]e + [KI3]e …………….(3)

[I2]e = (Where 125 is the partition coefficient of iodine

between Chloroform and water)

From equation (3) [KI3]e = C2 – [I2]e = C2 -

Concentration of KI consumed in KI3 formation is equal to [KI3]e

21


[KI]e = [KI]initial – [KI3]e

[KI]initial = 0.05M

[KI]e = 0.05 -

Equilibrium constant for

is

Ke =

Substituting [I2]e, [KI3]e and [KI]e from equations (4) , (5) and (6)

respectively

Equation (7) becomes

C1 and C2, calculated using (1) and (2) are substituted and thus K

the equilibrium constant can be calculated.

The above experiment can be used to find out the unknown

concentrations of KI after knowing Ke value from equation (8). The

above described procedure is used, C1 and C2 are obtained. 0.05 in

equation (8) is replaced with x (to be determined). Now with Ke

known and C1 and C2 obtained from the experiment x can be

calculated.

EXPERIMENT: 6

DETERMINATION OF CRITICAL SOLUTION TEMPERATURE

OF PHENOL-WATER SYSTEM

AIM:

To determine the critical solution temperature of phenol-water

system.

22


INTRODUCTION:

Some liquids are completely miscible and some are completely

immiscible. In between these extremes, there is an important type

of system consisting of two liquids which are partially miscible. An

example of this is phenol and water. If a little phenol is added to

water at room temperature, it dissolves completely, but if the

addition is continued a point is reached when no further dissolution

possible and two liquid layers form. One of these layers is a

saturated solution of phenol in water and the other one is water in

phenol. The two layers in equilibrium are called conjugate solutions.

Rise in temperature can cause mutual solubilities of phenol and

water and there exists a temperature at which phenol and water are

miscible in all proportions. This is known as critical solution

temperature of phenol-water system. At this temperature and

beyond this temperature phenol and water are miscible in all

proportions.

Determination of critical solution Temperature :

CHEMICALS:

80% phenol : 80 gms of phenol is dissolved in distilled water and

100gms of a solution of phenol in water is obtained.

APPARATUS:

1. A boiling tube made of corning glass. It is of about 3cm in

diameter and of about 30ml capacity.

2. 600 ml beaker of corning glass

3. Two glass stirrers

4. A sensitive thermometer

5. A spirit lamp

PROCEDURE:

Take the 600ml beaker with sufficient volume of water and place it

over a wire gauge kept over a tripod stand. A glass stirrer is kept in

to the beaker and the boiling tube is suspended in to this beaker of

23


water. A small glass stirrer is kept in the boiling tube. The sensitive

thermometer is suspended in to the boiling tube. Phenol-water

mixture is taken in the boiling tube and the miscibility temperature

is found by heating the water bath with a spirit lamp. The detailed

procedure consists of taking 10ml of 80% phenol and 2ml of water

in the boiling tube. At room temperature this composition does not

give a clear solution and cloudiness is observed. Heat the water

bath with a spirit lamp. Cloudiness disappears. Note down the

clearing temperature. Put off the spirit lamp and allow it cool down.

Cloudiness reappears. Note down the clearing temperature. Average

of clearing and clouding temperatures is taken as mean miscibility

temperature. Now the composition is varied by adding another

1.0ml of water in to the same boiling tube which already contains

10ml phenol and 2ml water. Heat the water bath and find the

clearing and clouding temperatures and the mean miscibility

temperature. It is continued up to 10ml addition of water. Remove

the apparatus and wash the boiling tube, the stirrers and the

thermometer. Repeat the above procedure, now fixing volume of

water as 10ml and varying phenol each time by 1.0ml. Mean

miscibility temperature for each composition is recorded. Now the

density of phenol is determined using specific gravity bottle and the

percentage of phenol (wt/wt) in all compositions is calculated using

the formula

% phenol (wt/wt) =

V1 and V2 are volumes of phenol and water, d1 and d2 are respective

densities.

Now a graph is drawn using mean miscibility temperature versus %

phenol (wt/wt) as shown here.

A curve is obtained with a maximum. The maximum in the curve

represents the critical solution temperature. At this temperature and

beyond this temperature one can see phenol and water are miscible

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in all proportions. The critical solution temperature (CST) and %

phenol corresponding to CST are noted.

EXPERIMENT: 7

EFFECT OF AN ELECTROLYTE (NaCl ) ON THE MISCIBILITY

TEMPERATURE OF PHENOL – WATER SYSTEM

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Addition of a foreign substance to the phenol – water system has a

definite effect over the miscibility temperature of phenol and water.

This property can be used to determine unknown concentrations of

sodium chloride.

PROCEDURE:

About 250ml of N/10 sodium chloride is made and diluted to give

N/20, N/30, N/40, N/50 and N/60 solutions of sodium chloride. 5ml of

each of these is mixed with 5ml of 80%. Phenol in the boiling tube

and the miscibility temperature are found in each case. A graph is

made between miscibility temperature and concentration of NaCl. A

straight line with a positive slope is obtained as shown here.

The above study of effect of NaCl can be used to estimate the

concentration of NaCl in a given solution. The procedure consists of

taking 5ml of the sample and add 5ml of 80% phenol and find the

miscibility temperature. The above graph is used as a calibration

graph to read the unknown concentrations.

Caution : Phenol is highly corrosive and so take care not to come in

direct contact with even a phenol – water mixture. If any thing falls

on the skin wash with plenty of water and put some sodium

bicarbonate over that part.

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