practical chemistry (ch - 223) b. sc g. s. gugale a. v. nagawade r. a. pawar s. s. jadhav v. d....

A Book Of

PPRRAACCTTIICCAALL

CCHHEEMMIISSTTRRYY ((CCHH -- 222233))

For S.Y.B.Sc. As per New Revised Syllabus with Effect from June 2014

Dr. S. S. Jadhav Vice Principal,

New Arts, Commerce and Science College,

AHMEDNAGAR

Dr. G. S. Gugale Dr. A. V. Nagawade Head and Associate Professor Associate Professor

Department of Chemistry, Department of Chemistry,

H. V. Desai College Ahmednagar College,

PUNE – 411002. AHMEDNAGAR.

Dr. A. D. Natu Dr. V. D. Bobade Ex. Head, Associate Professor

Department of Chemistry, Department of Chemistry,

Abasaheb Garware College, H.P.T. Arts, R. Y. K Science College,

PUNE – 411020. NASHIK – 422210.

N2258

S.Y.B.Sc. Practical Chemistry ISBN 978-93-5164-008-0

Fourth Edition : July 2018

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Preface …

It gives us an immense pleasure to place this text book of S.Y.B.Sc. Practical

Chemistry (CH-223) in the hands of S.Y.B.Sc. students. This book has been presented to

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be highly appreciated.

AUTHORS

Contents …

Page

(A) Physical Chemistry Practicals (Any Five)

1. To determine the critical solution temperature of phenol-water system. 1-4

2. To determine molecular weight of given organic liquid by steam

distillation method.

5-9

3. Determination of solubility of benzoic acid at different temperature and

to determine ∆H of dissolution process.

9-14

4. To study neutralization of acid (HCl) by base (NaOH), CH3COOH by

NaOH and H2SO4 by NaOH.

14-22

5. To determine the rate constant (or to study kinetics) of acid catalysed

ester hydrolysis.

26-29

6. To determine the rate constant of base catalysed ester hydrolysis. 29-31

7. Partition coefficient of iodine between water and carbon tetrachloride. 32-38

(B) Inorganic Chemistry Practicals (Minimum Five Mixtures)

Inorganic Qualitative Analysis 39-58

1. One simple mixture (without phosphate or borate).

2. Two mixtures containing PO3−

4 (with PO

3−

4 removal).

3. Two mixtures containing BO3−

3 (with BO

3−

3 removal).

Inorganic Qualitative Analysis of Binary Mixtures (including phosphate and

borate removal).

Sodium carbonate extract is to be used wherever necessary for detecting

acidic radicals.

(C) Organic Chemistry Practicals

(a) Organic qualitative analysis of binary mixtures without ether

separation (Four only).

Two: solid-solid, one: solid-liquid, one; liquid-liquid.

59-76

(b) Organic Preparations :

(Any two including Crystallization, MP, TLC)

1. Phthalic anhydride to phthalimide 77-78

2. Glucose to glucosazone 78-80

3. Acetanilide to p-bromoacetanilide 81-82

4. Benzaldehyde to dibenzylidene acetone 82-84

(D) Analytical Chemistry Practicals (Any Five)

1. Estimation of sodium carbonate content of washing soda. 85-90

2. Determination of calcium in the presence of magnesium using EDTA. 90-94

3. (a) Preparation of standard 0.05 N oxalic acid solution and

standardisation of approx. 0.05 N KMnO4 solution.

(b) Determination of the strength of given H2O2 solution with

standardised 0.05 N KMnO4 solution.

95-101

4. Estimation of Aspirin from a given tablet and to find errors in

quantitative analysis.

102-107

5. Estimation of Al (III) from the given aluminium salt solution by using

Eriochrome Black-T indicator (Black titration method)

107-111

6. Iodometric estimation of copper. 112-117

7. Report on one day industrial educational visit. 118-120

Appendix - I 121-123

Appendix - II 124-128

− − −

(1)

SECTION - A

PHYSICAL CHEMISTRY PRACTICALS (Any Five)

Experiment No. 1

PHENOL-WATER SYSTEM

AIM :

To determine the critical solution temperature of phenol-water system.

THEORY :

Phenol and water are partially miscible liquids with each other at room temperature.

Thus, when water is mixed with phenol, following two separate layers are formed :

(i) Saturated solution of phenol in water, (ii) Saturated solution of water in phenol. We know

that the solubility increases with increase in temperature. Consequently at a particular

temperature, two liquids become completely soluble (miscible) with each other. The

miscibility temperature depends on the percent composition of two liquids present in the

solution. In case of phenol-water system, miscibility temperature is highest for a particular

composition. This temperature is called as critical solution temperature of phenol-water

system. Above the critical solution temperature, phenol and water are completely miscible

with each other for all the compositions. Thus, above critical solution temperature, only one

phase (homogeneous solution) is obtained when phenol and water are mixed with each

other in any proportions. Below this temperature the system may exist in miscible form or

immiscible form depending upon the composition and temperature both.

In this experiment, phenol and water proportions are varied in their solutions and for

each proportion miscibility temperature is experimentally measured by heating. Graph of

% phenol versus miscibility temperature is plotted from which, critical solution temperature

and critical composition (% of phenol and % of water) are obtained.

S.Y.B.Sc. Practical Chemistry 2 Physical Chemistry Practicals

Stirrer

Burner

Tripod stand

Wire gauze

Water bath

Rubber cork

Mixture of phenol and water andit should be completely dippedin the water of water bath

Iron stand

th

thermometer110

Fig. 1.1: Experimental set up of phenol-water system

APPARATUS :

250 ml beaker, hard glass test tube with rubber cork and stirrer, 1/10th thermometer, measuring cylinder, burette etc.

CHEMICALS :

Distilled water, pure phenol.

PROCEDURE :

1. Take 10 g (9.5 ml) of phenol in a hard glass test tube using measuring cylinder. Add to it 4 g (4 ml) distilled water using a burette.

2. Fit the cork with a thermometer and a stirrer to this tube. Mount the test tube in a water bath or a sand bath. Refer Fig. 1.1 (Sand heats up and cools faster).

3. Heat the mixture on the bath slowly with constant stirring. While stirring mixture appears turbid until it is immiscible indicating the presence of two distinct phases. At an elevated temperature, the solution becomes miscible and turbidity disappears. As soon as the turbidity disappears, note this temperature (t1). But as heating is a fast process and temperature of the system increases very rapidly one cannot note down the correct mixing temperature. Therefore as cooling is slow process and temperature decrease very slowly. Therefore to note exact temperature of mixing


remove the burner and allow the total system to cool down. But it will take a very long time. Therefore the hard glass tube is taken outside the water bath along with the clamp and fix it properly. Now the mixture comes in direct contact with the open atmosphere and the cooling is faster. Let the solution cool while stirring constantly. As soon as few white shine crystals appear, note down the temperature (t2) at which turbidity reappears, t2 is noted as the miscibility temperature.

4. Add 3 ml of water to the contents of the tube. This will form the second composition

given in the table. Repeat the same procedure for this mixture and measure its

miscibility heating temperature (t1) and cooling temperature (t2). After completion of

set I for 10 gms of phenol, throw this mixture carefully in the basin and go for the

set II.

5. Likewise prepare the various compositions of phenol and water as given in the

observation table for set II i.e. 5 gms and set III for 2.5 gms of phenol. Repeat the

same procedure as set I for set II and III and report the heating (t1) and cooling (t2)

temperature in the observation table.

6. Plot the graph of % phenol versus miscibility temperature. It is a bell-shaped curve

as shown in Fig. 1.2. The temperature tc on Y-axis corresponding to point H on the

graph is critical solution temperature and the composition corresponding to this

point on X-axis is the critical composition C of the phenol-water system.

The expected critical solution temperature is 68.10°C and expected critical composition

is 37% phenol and 63% water.

Graph :

10 20 30 40 50 60 70 80 90 100

10

20

30

40

50

% Phenol

Temperature

Htc

Fig. 1.2 : Graph of % phenol versus temperature

Note : Phenol is highly corrosive to skin, hence it should be handled with care.


OBSERVATIONS :

Miscibility TemperaturesSr. No.

Amount of

Phenol (g)

Amount of

Water (g) % Phenol % Water

Heating (t1) Cooling (t2)

1. 10 6 62.50 37.5

2. 10 9 52.60 57.4

3. 10 12 45.40 54.6

4. 10 15 40.00 60.0

5. 5 10 33.3 66.7

6. 5 15 25.0 75.0

7. 2.5 15 14.3 85.7

8. 2.5 20 11.1 88.9

Set I

Set II

Set III

9. 2.5 25 9.1 90.9

Note : 10 g phenol = 9.5 ml (density = 1.053 g/ml)

5 g phenol = 4.8 ml

2.5 g phenol = 2.4 ml

CALCULATIONS : % Phenol = Amount of phenol

Amount of phenol + Water × 100

% Water = 100 − % Phenol

RESULTS :

1.

2.

3.

Critical solution temperature

% Phenol at critical solution temperature

% Water at critical solution temperature

= ……………… °C

= ……………… %

= ……………… %

QUESTIONS

1. What do you mean by critical solution temperature ?

2. How to find critical solution temperature ?

3. Whether phenol is completely soluble in water or not at room temperature ?

4. List the different types of immiscible liquids.

5. What is the effect of temperature on solubility ?

6. What is the density of phenol ?

7. What is the percentage of phenol at critical solution temperature ?

8. Explain the graph of % phenol against miscibility temperature.

9. What are conjugate solutions ?

✍ ✍ ✍


Experiment No. 2

DETERMINATION OF MOLECULAR WEIGHT OF GIVEN IMMISCIBLE LIQUID BY STEAM DISTILLATION METHOD

AIM :

To determine molecular weight of given organic liquid by steam distillation method.

THEORY :

If two liquids are miscible with each other completely, then addition of one liquid to the

other does not affect the properties of other liquid. Equilibrium exists between vapours and

mixture of immiscible liquids at a constant temperature. For a system of two liquids, total

vapour pressure at a temperature is the sum of vapour pressures of pure liquids at the same

temperature. Thus,

P = Po

A + Po

B … (1)

Total vapour pressure is always greater than the vapour pressure of either of these two

liquids. If distillation of such immiscible liquids is carried out then distillate contains two

layers of each liquid in definite proportions. The proportion of more volatile liquid in

distillate (i.e. having greater vapour pressure or lower boiling point) is more and vice versa.

The ratio of amounts of two liquids in collected fractions of distillate is constant and is in the

proportion of vapour pressures of two liquids. This is given by the equation

P

o

A

Po

B

= nA

nB

… (2)

where Po

A = V.P. of liquid A

Po

B = V.P of liquid B

nA = moles of A

nB = moles of B

But nA

nB

= WA MB

WB MA

… (3)

where, WA = weight of liquid A in distillate

WB = weight of liquid B in distillate

MA = molecular weight of A

MB = molecular weight of B

P

o

A

Po

B

= WA MB

WB MA


MA

MB

= WA P

o

B

WB Po

A

∴ MA = MB × P

o

B

Po

A

× WA

WB … (4)

By using equation (4), molecular weight of given liquid can be calculated by measuring

the distilled fractions from a miscible liquid mixture.

Diagram :

Water

Condenser

Distillationflask

Mixture Water

Distillate

Thermometer

Fig. 2.1 : Steam distillation unit

APPARATUS :

Steam distillation unit with steam generator, measuring cylinder, specific gravity bottle,

thermometer, separating funnel etc.

CHEMICALS :

Distilled water, organic liquid (chlorobenzene, carbon tetrachloride, toluene).

PROCEDURE :

1. Take about 200 ml of given organic liquid and nearly about 50 ml distilled water in

distillation flask.

2. Take appropriate amount of water in steam generator and put 3-4 porcelain pieces

in it. Arrange the distillation assembly as shown in Fig. 2.1.

3. Heat the steam generator and allow the steam to pass into distillation flask at a

constant rate.

4. The temperature of mixture of water and given organic liquid increases initially and

becomes constant. At this temperature, boiling of liquids starts in the distillation

flask. The vapours of two liquids generated pass through the condenser, where

condensation of vapours takes place and mixture of two liquids (distillate) is


collected at the end. Initial portion of about 3 - 4 ml is rejected. Now collect an

aliquot of 25 ml distillate into the measuring cylinder.

5. Record the constant temperature as boiling temperature of the mixture by

thermometer. Collect two more fractions of distillate of 25 ml, while the mixture

boils.

6. Measure the total volume V1 of each fraction accurately.

7. By using separating funnel, separate water and organic layer from each other and

measure their volume separately, viz. Vw and Vo.

8. Determine the density of organic liquid with reference to the distilled water at room

temperature by weighing definite volumes of organic liquid and water using specific

gravity bottle. See the observation Table 2.1. Note down the barometric pressure

from the barometer.

OBSERVATIONS :

Table 2.1

Sr. No. Description Value

1. Barometric pressure (B.P.) = ……… cm of Hg

2. Boiling point of mixture (Tm) = ……… °C

3. Weight of empty specific gravity bottle (W) = ……… gm

4. Weight of specific gravity bottle + Organic liquid (W1) = ……… gm

5. Weight of specific gravity bottle + Water (W2) = ……… gm

6. Density of organic liquid (do) = ………

7. Density of water (dw) = ………

Density of organic liquid = W1 − W

Volume of specific gravity bottle

Density of water = W2 − W

Volume of specific gravity bottle

Table 2.2

Obs.

No.

Total volume of

distillate, V1 ml

Volume of organic

liquid, Vo ml

Volume of water,

Vw ml = Vt −−−− Vo

WA

WB =

Vo dw

Vw do

Mol. wt.

gm

1.

2.

3.


CALCULATIONS :

For calculation of molecular weight, vapour pressure of organic liquid and water is

necessary. To obtain vapour pressure of water at the boiling temperature of mixture,

calibration curve method is used. Construct the calibration curve by using the data given in

Table 2.3. Obtain the vapour pressure of water at boiling point of mixture (tm) by drawing

perpendicular from tm to point X on calibration curve and then from point X to PA on vapour

pressure axis. Vapour pressure of organic liquid is calculated by subtracting vapour pressure

of water from the barometric pressure i.e. PA = B.P. − PB.

Table 2.3

Vapour pressure

/ mm of Hg

Temperature / °°°°C

93

118

149

184

235

289

355

434

525

634

760

50

55

60

65

70

75

80

85

90

95

100

PBV

.P.

of

wa

ter

in m

m

Temperature in Co

tm

X

Fig. 2.2

Calculate molecular weight of given organic liquid by using the following equation :

MA = PB

PA

× WA

WB

× MB

where, MA = Molecular weight of organic liquid

MB = 18 gm

PB = Vapour pressure of water

PA = Vapour pressure of organic liquid

WA = Weight of organic liquid

WB = Weight of water

Calculate the molecular weight for three fractions of distillate and take the average.

RESULT :

1. Molecular weight of given organic liquid (MA) = …………… gm


QUESTIONS

1. Discuss the basic principle of steam distillation.

2. What is vapour pressure and partial vapour pressure ?

3. Define 'Boiling Point'.

4. What is barometric pressure ?

5. Explain Raoult's law.

6. Why the mixture is heated in water bath ?

7. How molecular weight of liquid is determined by steam distillation method ?

8. What are the applications of this method ?

9. What is the relationship between boiling point of liquid, its vapour pressure and

atmospheric pressure ?

✍ ✍ ✍

Experiment No. 3

SOLUBILITY OF BENZOIC ACID

AIM :

Determination of solubility of benzoic acid at different temperature and to determine ∆H of

the dissolution process.

THEORY :

When two substances are mixed to form a homogeneous mixture, then it is known as

true solution. The substance which is present in greater amount is called the solvent and the

other substance as the solute. The solubility of a solid in a liquid is governed by many

factors like nature of the solute, nature of the solvent, temperature, pressure, concentration

of the solute, presence of other solutes etc. The maximum amount of the solute that can

dissolve in a fixed quantity of the solvent at a given temperature and pressure is called the

solubility of the solute. Such a solution which contains the maximum amount possible of the

solute at that temperature and pressure is called a saturated solution.

The dissolution of a solute in a solvent is accompanied by a heat change. A chemical

reaction, during which heat is lost by the reacting system and gained by the surroundings is

called an exothermic reaction. On the other hand, if, during the reaction heat is absorbed by

the reacting system and lost by the surrounding, the reaction is called endothermic. When

the heat change is measured at constant pressure, it is called the enthalpy change ∆H of the

reaction. The enthalpy change is measured at standard conditions i.e. one atmospheric

pressure and temperature 298 K. The state of solution is important because the enthalpy

change on dissolution depends upon the composition of the solution. The integral enthalpy

of solution at a given concentration is defined as the change in enthalpy when one mole of


solute in its standard state is completely dissolved in enough of the solvent to produce a

solution having the desired composition. The heat of solution is defined as the heat change

when one mole of the solute is dissolved in a solution. If the process of dissolution is

endothermic, the solubility of the solute increases with temperature. But if the process of

dissolution is exothermic, the solubility of solute decreases with temperature. In general it is

observed that solubility of a substance rises with temperature.

The quantitative relationship between solubility and temperature is given by the Van't

Hoff isochore.

∂ln S

∂TP

= ∆H

RT2

where S is the solubility of the solute

∆H is the heat of solution

R is the universal gas constant

T is the absolute temperature

Integrating above equation

⌡⌠

S1

S2

d ln S = ⌡⌠

T1

T2

∆H

R ×

dT

T2

ln S2

S1 =

∆H

R

1

T2 −

1

T1 or ln

S1

S2 =

−∆H

R

1

T2 −

1

T1

log S1

S2 =

∆H

2.303 R

T1 − T2

T1 T2

∆H = 2.303 R

T1 T2

T2 − T1 log

S2

S1

where S1 and S2 are the solubilities at T1 and T2 temperatures. Thus heat of solution can be

determined by measuring the solubility at different temperatures.

AIM :

To determine the solubility of benzoic acid in water at different temperature and to

determine heat of solution of benzoic acid.

APPARATUS :

500 ml beaker, hard glass tube, 1/10th thermometer, burette, rubber cork, stirrer etc.

CHEMICALS :

Benzoic acid, distilled water etc.


Stirrer

Burner

Tripod stand

Wire gauze

Beaker

th

thermometer110

Rubber cork

Iron stand

Hard glass tube

Fig. 3.1 : Experimental set up

PROCEDURE :

1. Clean and dry the hard glass tube and take 2 gms of accurately weighed benzoic

acid in it.

2. Clean the burette and fill it with distilled water upto the mark. Add 10 ml of water

from the burette into the hard glass tube.

3. Cork the hard glass tube having stirrer and 1/10th thermometer. Warp the filter

paper around the hard glass tube and clamp it properly to the iron stand. The 500 ml

beaker is filled with sufficient amount of tap water so that the solution level from the

hard glass tube is below the water level of the beaker.

4. Arrange the apparatus as shown in Fig. 3.1.

5. Now heat the beaker and with the help of stirrer, stir the solution in the hard glass

tube. Continue heating and stirring till all the benzoic acid completely dissolves.

As soon as clear solution is formed, note down this temperature as t1°C.


6. Remove or make the burner off and allow the solution to cool down. While cooling

stir the solution continuously and observe the solution in the hard glass tube

carefully. As soon as white shining crystals start appearing note down this

temperature as crystallization temperature t2°C.

7. From the burette, add two ml of water in the hard glass tube. Now total quantity of

water will become 12 ml. Again repeat the same procedure and note down the

temperature at which benzoic acid forms homogeneous solution and when the

solution is allowed to cool down, then note down the crystallisation temperature of

benzoic acid.

8. Thus, go on recording the heating or mixing and crystallization temperature by

adding 2 ml of water every time, till the total addition of water becomes 20 ml.

Report the observations in the following table.

Table 3.1

Temperature Obs.

No.

Weight of solute

(Benzoic acid) in gm

Volume of water

added ‘ml’ Heating

t1°C

Cooling

t2°C

Mean

tm°C

Temperature

T (K)

1.

2.

3.

4.

5.

6.

2

2

2

2

2

2

10

12

14

16

18

20

Now calculate the solubility S in gm per 100 ml of water by using the following

equation.

Solubility, S (g/100 ml) = 100 × Weight of benzoic acid

Volume of water

Thus, S1 = 100 × 2

10 = 20

S2 = 100 × 2

12 = 16.66 and so on

S3, S4, S5 and S6 etc.

Put these calculations in the following table.


Table 3.2

Obs. No. Solubility

S g/100 ml water

log S Temperature T (K)

i.e. (t°C + 273)

1

T K−1

1.

2.

3.

4.

5.

6.

CALCULATIONS :

(A) Heat of solution of benzoic acid (Calculation Method). Select any two solubilities

from the observation table say S1 and S2, at two different temperatures, say T1 and T2

respectively. Thus, ∆H is obtained by using the following equation

∆H = 2.303 × R

T2 T1

T2 − T1 log

S2

S1

(B) Graphical Method : Plot the graph of log S versus 1

T , which is a straight line graph.

Slope = −∆H

2.303 R

Thus, ∆H = −2.303 × R × Slope

As slope is negative, ∆H is positive and R = 1.987 cal K−1 mole−1

log S

1T

Fig. 3.2 : Plot of log S versus 1

T


RESULT TABLE :

Sr. No. Solubility S

g/100 ml water

Temperature

(t°C + 273) = T°K

Heat of solution of benzoic acid

∆∆∆∆H kcal mole−1

Calculations Graphical

Method

1.

2.

3.

4.

5.

6.

QUESTIONS

1. What is true solution ?

2. Define the term 'Solubility'.

3. Give the factors affecting the solubility of a solute in a solvent.

4. Define the integral enthalpy of solution.

5. Give the relationship between enthalpy and solubility.

6. How is the behaviour of solubility with temperature ?

7. What are the applications of solubility ?

✍ ✍ ✍

Experiment No. 4

HEAT OF NEUTRALIZATION

AIM :

To determine the heat of neutralization by continuous variation method for the reactions

between

(a) HCl and NaOH.

(b) CH3COOH and NaOH.

(c) H2SO4 and NaOH.

THEORY :

Arrhenius proposed the theory of electrolytic dissociation in 1887 to account the

properties of aqueous solutions of electrolytes. According to him, 'the acid is defined as


hydrogen containing compound which in aqueous solution gives hydrogen ions and the base is

an hydroxy compound which in aqueous solution gives hydroxide ions. e.g. HCl, H2SO4,

CH3COOH are acids which give hydrogen ions in aqueous solutions. Thus,

HCl(aq) H+

(aq) + Cl−

(aq)

While KOH, NaOH, NH4OH, LiOH are bases which give hydroxide ions in aqueous

solutions. Thus,

KOH(aq) K+

(aq) + OH−

(aq)

In 1923, Bronsted-Lowry putforth the theory of acids and bases, as follows :

A Bronsted acid is a proton donor.

A Bronsted base is a proton acceptor.

Thus, hydrogen chloride is an acid because it can donate a proton to another molecule.

Methane is not a Bronsted acid because despite its hydrogen atoms, it is not a proton

donor. Ammonia is a base because it can accept a proton from another molecule and

becomes NH+

4 . The definition makes no mention of the solvent (apply even if no solvent is

present), however the most important medium is aqueous solution.

The acids and bases are classified as strong and weak on the basis of their extent of

ionization. The acids or bases having maximum or hundred percent ionization in aqueous

solution are called as strong acids and strong bases e.g. HCl, H2SO4, HNO3 are strong acids

while KOH, LiOH, NaOH are strong bases.

The acids and bases having very low ionization in aqueous solution and produce very

small number of H+ or OH− ions are called as weak acids and weak bases. e.g. CH3COOH,

HCN etc. are weak acids while NH4OH, Ca(OH)2 are weak bases.

All the chemical reactions undergo change in enthalpy. Thus, the chemical reaction

which takes place with an evolution of heat is known as exothermic reaction. It is possible

only when the enthalpy of the product (Hp) is less than that of reactants (Hr). Thus, if the

exothermic reactions are carried out in an isolated system, the temperature of the system

rises. All the acid-base reactions are neutralization reactions and exothermic in nature. While

the chemical reactions which take place with absorption of heat are known as endothermic

reactions. These type of reactions are possible only when sum of enthalpies of products (Hp)

is more than that of reactants (Hr). If endothermic reactions are carried out in an isolated

system, then the temperature of the system falls down.

The reactions between acids and bases give neutral water molecule, therefore these

reactions are called as neutralization reactions. The heat is evolved during neutralization

reaction, therefore it is exothermic in nature. Thus, heat of neutralization is defined as the


'amount of heat evolved when one gram equivalent of an acid is completely neutralized by one

gram equivalent of a base in its dilute solution at room temperature'. The one gram

equivalent of acid or base means the solution containing equivalent quantity of it. Suppose

if we want to take one gram equivalent of hydrochloric acid, then we have to dissolve

36.5 gm of gaseous hydrogen chloride gas (one gram equivalent, the molecular weight and

equivalent weight of HCl is same) in liquid water at room temperature to make the final

volume of the solution 1000 ml. Thus, this 1000 ml solution (1N) is called as one gram

equivalent of an acid. Similarly, we can also prepare solution of a base containing one gram

equivalent of it. Suppose for sodium hydroxide, one gram equivalent means 40 gram of it to

be dissolved in distilled water to make final volume 1000 ml. Thus, this 1000 ml solution

containing 40 gm of NaOH is called as one gram equivalent of a base. The 1000 ml HCl

solution containing 36.5 gm of HCl gas when mixed with 1000 ml solution containing 40 gm

of NaOH, the amount of heat evolved when the acid is completely neutralized by base at

room temperature is called as heat of neutralization. Following are some examples of

neutralization reactions.

HCl(aq) + LiOH(aq) = LiCl(aq) + H2O(l); ∆Ho

298K = − 13.7 kcal

H2SO4(aq) + NaOH(aq) = NaHSO4(aq) + H2O(l); ∆Ho

298K = − 14.75 kcal

HCl(aq) + NH4OH(aq) = NH4Cl(aq) + H2O(l); ∆Ho

298K = − 12.3 kcal

The heat of neutralization between strong acids and strong bases is observed to be

nearly constant and is equal to −13.7 kcal. This fact can be well explained by writing the

above equation in the ionic form and the net ionic equation

H+

(aq) + Cl−

(aq) + Li+

(aq) + OH−

(aq) = Li+

(aq) + Cl−

(aq) + H2O(l); ∆Ho

298K = − 13.7 kcal

Cancelling common ions on both sides and writing net ionic equation as follows.

H+

(aq) + OH−

(aq) → H2O(l); ∆Ho

298K = − 13.7 kcal

Thus, the net ionic equation for all strong acids and strong bases is found to be same

because of which the heat of neutralization must be same and constant i.e. −13.7 kcal.

Neutralization is the process of formation of water molecules by combination of equal

number of H+ ions from acid and OH− ions from base. Since strong acids and strong bases

dissociate completely in dilute solutions, therefore, heat of neutralization involving them is

nearly constant. This behaviour gives alternative definition as "when dilute solutions of strong

acids are neutralized with dilute solutions of strong bases at room temperature, the heat of

neutralization per mole of water formed is essentially constant and independent of the nature

of acid and base."


This constancy of heat of neutralization does not carry over to the neutralization of weak

acids by strong bases, weak bases by strong acids, or weak acids by weak bases. The heat of

neutralization in this case differs widely. For example,

HCN(aq) + NaOH(aq) → Na+

(aq) + CN−

(aq) + H2O(l) ; ∆H = − 2.46 kcal

weak strong

HCl(aq) + NH4OH(aq) → NH+

4(aq) + Cl−

(aq) + H2O(l) ; ∆H = − 12.3 kcal

strong weak

The heat of neutralization is less than expected value i.e. −13.7 kcal. This is because when

one of them either acid or base is weak or both are weak then the weak acid or base is

feebly ionised. Therefore, some of the heat energy evolved is utilized to do the further

ionization of the weak component because of which the net heat of neutralization is

observed to be less than the expected value. The amount of heat utilized to do the

ionization of weak acid or base is called as heat of ionization and can be evaluated as

follows :

1. HCN º H+ + CN− ; ∆H' = x kcal

weak

2. H+ + Na+ + OH− º H2O + Na+ ; ∆Ho

298K = −13.36 kcal

By adding equations (1) and (2), we get

HCN + Na+ + OH− º CN− + Na+ + H2O ; ∆H’ + ∆H = ∆H’ − 13.36 = x kcal

By the experimental observed value of ∆H’ for the reaction between HCN and NaOH is

− 2.46 kcal i.e. x kcal.

∆H − 13.36 = − 2.46

Heat of dissociation of HCN, ∆H = + 10.90 kcal

∆H is the heat of ionization of the hydrocyanic acid per mole. The heat of ionization or

dissociation is the amount of heat absorbed by a mole of weak acid or base for its

dissociation. It is different for different substances.

Heat of neutralization is obtained by mixing equal volumes of 0.5 M solutions of acid

and base in calorimeter and to measure the rise in temperature ∆t. But the continuous

variation study is much better approach to obtain heat of neutralization of a 1 M solution of

an unknown acid by 1 M solution of a strong base. Thus by using the volume ratio like

180/20, 160/40, 140/60, 120/80 … and so on the values of ∆t are readily obtained. The

values of ∆t are plotted against the volume ratio of base/acid. At certain volume, ratio of

base/acid maxima is obtained and the corresponding ∆t for that maxima is noted. From such

volume ratio and corresponding ∆t value, heat of neutralization is calculated.

Heat Capacity or Water Equivalent of Calorimeter :

In order to obtain the heat of neutralization accurately, it is necessary to evaluate the amount of heat gained by calorimeter, thermometer and stirrer. When we perform the experiment, some amount of heat is absorbed by calorimeter. Hence, determination of heat


capacity or water equivalent of calorimeter is must. Water equivalent of calorimeter is the amount of heat required to increase the temperature of calorimeter by 1°C. In general, the glass bottles, or polythene or polystyrene vessels are used to perform the experiment. In this case, water equivalent is obtained for those parts of the vessels which are in actual contact with the reacting system. In this case, the usual method of obtaining the water equivalent is not practicable. The water equivalent in such cases is found by carrying out an experiment similar to the experiment to be performed latter in the vessel.

APPARATUS :

Thermos flask, rubber-cork, 1/10th thermometer, measuring cylinder, beaker, tripod

stand, wire gauze, burner etc.

PROCEDURE :

The experiment is performed in two parts.

Part - (A) : To determine the water equivalent of calorimeter (Thermos flask).

1. Clean and dry the thermos flask.

2. With the help of measuring cylinder take exactly 100 ml distilled water in the

thermos flask. Wait for 10 minutes and record the accurate and constant

temperature with the help of 1/10th thermometer. Let it be t1 °C.

3. Take 100 ml distilled water in 250 ml beaker. Make necessary set up to heat it near

about 60°C. Stop heating and record the temperature of hot water accurately. Let it

be t2 °C.

4. Remove the cork of thermos flask and quickly add 100 ml hot water into it at

temperature t2 °C. Then immediately replace the cork, and stir the contents of the

flask continuously. Record the steady maximum temperature of the mixture. Let it be

t3 °C.

Fig. 4.1 : Typical thermos flask


Part - (B) : To determine the heat of neutralization for the reaction between acid and base

by continuous variation method.

1. Clean and dry the thermos flask, stirrer, 1/10th thermometer etc.

2. With the help of measuring cylinder, take 180 ml of 1 M NaOH in the same clean and

dry thermos flask. Record its constant temperature as tB oC.

3. In a clean and dry beaker, take 20 ml of 1 M HCl solution and record its constant

temperature as tA °C. (Throughout the experiment we are going to use same 1/10th

thermometer. Therefore one has to take precaution that whenever we are going to

dip the thermometer in an acid or base solution, it should be properly cleaned and

dried by using filter paper.)

4. Now remove the cork of thermos flask and transfer 20 ml of 1 M HCl solution to

180 ml of 1 M NaOH solution. Replace the cork immediately.

5. Then stir the mixture continuously and record the constant maximum temperature of

the mixture as tn oC.

6. Repeat the same procedure for different volumes of NaOH (x ml) and HCl (y ml)

solution as shown in the observation table below. In every case, find the exact

temperature of neutralization (tn oC).

7. Repeat the procedure as above for CH3COOH − NaOH and H2SO4 − NaOH.

OBSERVATIONS :

Table for reaction between …………… and NaOH.

Volume of 1 M solution Temperature of solution

°°°°C

Rise in

temperature

in °°°°C

Sr.

No.

Base NaOH

x ml

Acid ………… y ml NaOH

'tB'

Acid

'tA'

Mean

tm = tA + tB

2

*tn ∆∆∆∆t = tn −−−− tm

1.

2.

3.

4.

5.

6.

7.

8.

9.

180

160

140

120

100

080

060

040

020

020

040

060

080

100

120

140

160

180

* After neutralization


Graph : Plot the graph of ∆t against the base-acid volume ratio for each of the pair of

acid and base. The expected nature of plots are shown below.

Dt

200/0 180/20 160/40 140/60 120/80 100/100 80/120 60/140 40/160 20/180 0/200

13664

= 2:1100100

= 1:1

Base (x ml)Acid (y ml)

Dt

NaOH - H SO2 4

NaOH - HCl

NaOH - CH COOH3

Fig. 4.2

The information obtained from the graph is tabulated as follows.

Sr.

No.

Reaction between

Base and Acid Plot is maximized at

∆∆∆∆t in °°°°C ml of NaOH

'x ml'

ml of HCl

'y ml'

Volume

ratio 'x/y'

1.

2.

3.

NaOH − HCl

NaOH − CH3COOH

NaOH − H2SO4

CALCULATIONS :

Part - (A) : Calculation of water equivalent (e) of the thermos flask. The density of water is

1 gm cm−3, therefore,

100 ml of water + 100 g of water.

Heat gained by thermos flask

+ stirrer + thermometer +

Heat gained

by cold water =

Heat lost by

hot water

e (t3 − t1) + 100 (t3 − t1) = 100 (t2 − t3)

e = 100 (t2 − t3) − 100 (t3 − t1)

(t3 − t1) cal/g


Part - (B) : As the solutions are dilute, assume that

1. 100 ml acid = 100 ml base (NaOH) = 100 g.

2. Heat evolved =

Heat gained by

200 ml mixture +

Heat gained by flask,

stirrer and thermometer

Therefore, heat evolved = 200 × ∆t + e × ∆t = 'q' calories.

The '∆t' value is taken corresponding to the maximum on the plot.

3. Let 'x' ml of 1 M NaOH after neutralization liberate 'q' calories of heat, therefore,

1000 ml 1 M NaOH (1 g equivalent) after complete neutralization will liberate

1000 × q

x calories of heat. (The 'x' ml of 1 M NaOH is to be taken corresponding

to the maximum in the plot.)

∴ ∆H = Heat of neutralization =

1000 × q

x calories =

q

x kcal.

According to the sign convention, heat evolved is taken as negative, therefore,

∆H = − q

x kcal.

By using above calculations/steps, find out heat of neutralization for

1. NaOH − CH3COOH. 2. NaOH − H2SO4.

RESULTS :

Sr. No. Description Value with unit

Observed Expected

1. Water equivalent (e) of the

thermos flask.

………… cal g−1 ………… cal g−1

2. ∆t at the maximum in

(a) NaOH − HCl

(b) NaOH − CH3COOH

(c) NaOH − H2SO4

………… °C

………… °C

………… °C

____________

____________

____________

3. Volume ratio of base/acid at the

maximum

(a) NaOH − HCl


(c) NaOH − H2SO4

100/100 = 1 : 1

100/100 = 1 : 1

136/64 = 2 : 1

4. Heat of neutralization

(a) NaOH − HCl


(c) NaOH − H2SO4

∆H1 kcal

∆H2 kcal

∆H3 kcal

− 13.7 kcal

− 13.2 kcal

− 14.5 kcal

5. Heat of dissociation of

CH3COOH = ∆H2 − ∆H1

…………… kcal

0.5 kcal


QUESTIONS

1. Define heat of neutralization and heat of ionization.

2. Explain strong acid and strong base with the help of example.

3. Why the heat of neutralization is same for neutralization reaction between strong

acid and strong base ?

4. The heat of neutralization is found to be less for reaction between acid and base if

one of them is weak or both are weak than if both are strong. Explain.

5. Explain the exo and endo thermic reactions on the basis of enthalpy change.

6. Define : (i) Equivalent weight (ii) Molecular weight

(iii) Normality (iv) Molarity

(v) Molality (vi) Atomic weight

7. What is water equivalent of calorimeter ? Why is it necessary to use it for calculating

heat of neutralization ?

✍ ✍ ✍

CHEMICAL KINETICS

Introduction :

Thermodynamic parameters like free energy, entropy, heat of reaction etc. give an idea

about the direction of chemical reaction. Chemical kinetics deals with measurement of rate

and mechanism of chemical reaction and factors affecting it. All the reactions cannot be

studied at normal laboratory conditions because some of the reactions are so slow that

months or years are required for their completion. The reactions having moderate rates are

studied in ordinary laboratory conditions.

Rate of Reaction :

The rate of a chemical reaction mainly depends on nature, concentration and

temperature of the reactants. Kinetic studies are generally carried out at constant

temperature (except in the determination of energy of activation). A desired composition of

reactants is mixed to start the reaction and reaction is monitored for the decrease in

concentration of reacting substance or the increase in product concentration as a function

of time. The rate of the reaction is change in concentration per unit time of reactants or

products. During the course of reaction concentration of reactant decreases. The rate of a

chemical reaction is always a positive quantity. Change in the concentration of a reactant is

negative quantity (decrease). Thus when rate is expressed in terms of reactant concentration,

negative sign is used and when concentration of product is used for this purpose sign is

positive.


The rate of reaction represented by –dC/dt where dC is small change in concentration at

infinitesimally small time interval dt. Negative sign indicates that concentration decreases as

time passes.

Order of reaction : It is the sum of all the exponents to which the concentration terms

are raised in the rate equation. Thus when rate of reaction is given by

− dC

dt = k ⋅ C

n11

⋅ Cn22

⋅ Cn33

…

where, k = constant, and C1, C2, C3 etc. are concentrations of reactants 1, 2, 3, etc. then

order of reaction = n1 + n2 + n3 + …

Molecularity of reaction : Molecularity of a reaction is defined as the number of

molecules or atoms of reactant taking part in a reaction.

Pseudo-molecular reaction : Whenever in a reaction order is not equal to molecularity

the reactions are called as pseudo-molecular reactions. Generally we come across

molecularity greater than one but order equal to one, such reactions are called pseudo-

unimolecular reactions.

Zero order reaction : When the rate of reaction is independent of the concentration of

the reactant then it is said to be zero order reactions. The rate expression is

Rate = k

where, k is rate constant.

First order reaction : In first order reaction a single molecule reacts to give products.

These are generally decomposition reactions.

A → Product

The rate law for first order reaction is

− dC

dt = k ⋅ CA

Let ‘a’ be the initial concentration in moles/lit of reactant ‘A.’ If ‘x’ moles/lit of ‘A’ react in

time ‘t’ then concentration of unreacted ‘A’ at time ‘t’ will be (a – x) moles/lit.

∴ Rate = − dC

dt =

d(a − x)

dt

= − da

dt +

dx

dt

but − da

dt = 0 since ‘a’ is constant

∴ Rate = dx

dt


According to law of mass action the rate of a reaction is directly proportional to the

concentration of reactant ‘A’ at that instance.

∴ Rate ∝ (a − x)

Combining above two equations,

dx

dt = k (a − x)

where, k = rate constant

Separating the variables and integrating above equation gives the rate constant k for

first order reaction as

k = 2.303

t log

a

a − x

Thus the rate constant can be calculated by determining initial concentration and

change in concentration in the course of reaction. The rate constant can be found by using

graphical method as

k = 2.303

t log

a

a − x

∴ log a

(a − x) =

k

2.303 ⋅ t

This is a straight line equation of the form y = mx with variables

a

(a − x) and t.

Graph of log a

(a − x) against time is a straight line passing through origin having slope.

∴ Slope (m) = k

2.303

∴ Rate constant, k = 2.303 × slope

Similarly, log

a

a − x =

k

2.303 ⋅ t

∴ log a − log (a − x) =

k

2.303 ⋅ t

∴ log (a − x) =

− k

2.303 ⋅ t + log a

Graph of log (a – x) against time ‘t’ is a straight line with negative slope.

Thus, slope = − k

2.303

∴ Rate constant, (k) = − 2.303 × Slope

Units of rate constant k are time–1 (min–1 or sec–1).


Second Order Reaction :

In case of second order reaction there are different possibilities :

(1) Two molecules of same substance react to give products. Thus,

2A → B + C

(2) One molecule of one substance and one molecule of other substance reacts to give

products

A + B → C + D

Again there are two possibilities in this case :

(i) Both the reactants have same initial concentrations or

(ii) The two reactants have different initial concentration.

1. Expression for second order reaction with equal initial concentration

Let the general reaction be

A + B → C + D

Let ‘a’ be the initial concentration of both the reactants ‘A’ and ‘B’ in moles/lit.

If x moles/lit of A and B react in time t, then concentration of unreacted A and B in time t will

be (a – x) moles/lit.

∴ Rate = dx

dt = k (a − x)2

where, k = rate constant

∴ The rate constant k for second order reaction is given by,

k = 1

a ⋅ t

x

a − x

2. Expression for second order reaction with unequal initial concentration

Let the general reaction be

A + B → Product

Let ‘a’ moles/lit be the initial conc. of ‘A’ and ‘b’ moles/ lit be the initial conc. of ‘B’.

If the part x out of ‘a’ and the part ‘x’ out of ‘b’ are consumed upto time t, then

concentration of unreacted A and B at time t will be (a – x) and (b – x) moles/lit respectively.

∴ Rate = dx

dt = k (a − x) (b − x)

The rate constant k for second order reaction is given by equation

k = 2.303

t (a − b) log

b (a − x)

a (b − x)


This equation can be rearranged to

k (a − b)

2.303 t = log

b

a + log

(a − x)

(b − x)

∴ log (a − x)

(b − x) =

k (a − b)

2.303 t − log

b

a

Comparison of this with the equation y = mx + c, implies that this equation represents a

straight line.

∴ By plotting the graph of log (a − x)

(b − x) against t, the nature of graph is a straight line

with intercept log b

a .

Slope = k (a − b)

2.303

∴ k = 2.303 × slope

(a − b)

Units of k are conc−1 time−1.

Activation energy is an important quantity from chemical kinetics. It has been discussed

in experiment six.

✍ ✍ ✍

Experiment No. 5

RATE CONSTANT OF ACID CATALYSED ESTER HYDROLYSIS

AIM :

To determine the rate constant (or study kinetics) of acid catalysed ester hydrolysis.

APPARATUS :

Stoppered bottles, burette, 5 ml pipettes, measuring cylinder, beakers, stop watch, water

bath etc.

CHEMICALS :

Methyl acetate, 0.5 N HCl (or 0.5 NH2SO4), 0.1 N NaOH, phenolphthalein indicator, ice

etc.

THEORY :

The hydrolysis of methyl acetate in aqueous solution is very slow and is catalyzed by

strong acids like HCl or H2SO4. The hydrolysis takes place as

CH3COOCH3 [H+]

→ CH3COOH + CH3OH


In this reaction, concentration of water is very high and practically remains constant

relative to methyl acetate. Therefore, the rate of reaction is determined by concentration of

methyl acetate alone. This is an example of pseudo-unimolecular reaction. The rate of this

reaction is given by

dx

dt = k ⋅ [CH3COOCH3]

The rate constant for the first order reaction is

k = 2.303

t log

a

a − x

Since acetic acid is formed during the reaction, its concentration and hence the progress

of reaction can be studied by titrating known volume of the reaction mixture with standard

alkali at suitable interval of time from the start of reaction.

PROCEDURE :

Perform the experiment as follows.

1. Take 5 ml of methyl acetate using pipette and 100 ml 0.5 N HCl in two separate

clean and dry bottles and stopper them. Place the bottles in water bath to attain the

uniform temperature.

2. Rinse and fill a clean burette with 0.1 N NaOH solution upto the zero mark.

3. Take few pieces of crushed ice or around 40 ml ice cold water in a conical flask and

add 2 - 3 drops of phenolphthalein indicator to it.

4. After 10 to 15 minutes the liquids will have attained the temperature of bath, add

HCl solution to methyl acetate (1st bottle) and shake the reaction mixture well. Start

the stop watch and note the time of mixing as zero time.

5. Immediately pipette out 5 ml of reaction mixture into a conical flask containing ice

cold water. Stopper the bottle again.

6. Titrate the reaction mixture in the conical flask against 0.1 N NaOH and record the

titration reading (T0), when faint pink colour appears and persist for 30 sec. to the

solution.

7. Shake the reaction mixture from time to time and titrate the 5 ml of reaction mixture

with NaOH at the successive intervals of 10, 20, 30, 40 and 50 minutes, (Tt).

8. Infinite time readings : Take 25 ml of the reaction mixture in a flask. Cork it and

keep in water bath at around 50 to 60 °C temperature, for 15 to 20 minutes to

complete the hydrolysis. Finally, titrate 5 ml of it against 0.1 N NaOH and record the

readings as T∞.


OBSERVATIONS :

For set I : 5 ml methyl acetate + 100 ml 0.5 N HCl

Initial reading (T0) = ……..ml

Infinite reading (T∞) = ……..ml

∴ Initial concentration of methyl acetate

a = T∞ – T0 = ………..ml

Time (t)

min

Titration

reading

(Tt) ml

Tt – T0 = x T∞ – Tt = a – x log a – x log a

a −−−− x

k/min

0

10

20

30

40

50

CALCULATIONS :

(a) Rate constant (k) by calculations :

For first order reaction

k = 2.303

t log

a

a − x

where, a = Initial concentration of methyl acetate

a – x = Amount of methyl acetate remaining unreacted at time t

Calculate the values of k at 10, 20, 30, 40 and 50 minutes time interval. Calculate mean

value of k for each set.

(b) Rate constant (k) by graph :

(i) Plot the graph of log (a – x) against t.

Slope = −k

2.303

∴ k = − 2.303 × slope

log (a x)-

time/min

Fig. 5.1 : log (a −−−− x) versus t

(ii) Plot the graph of log a

(a − x) against t.

Slope = k

2.303

∴ k = 2.303 × slope

log a/(a x)-

time/min

Fig. 5.2 : log a

a −−−− x versus t


Straight line nature of both the graphs indicates that the reaction of hydrolysis of methyl

acetate is first order one.

RESULT TABLE :

k (min–1)

By calculations By graph

CONCLUSION :

k values are nearly constant. This shows that the rate constant is independent of initial

concentration of reactant.

✍ ✍ ✍

Experiment No. 6

RATE CONSTANT OF BASE CATALYSED ESTER HYDROLYSIS

AIM :

To determine the rate constant of base catalysed ester hydrolysis.

APPARATUS :

Stoppered bottles, burette, 10 ml pipette, conical flask, measuring cylinder, beakers, stop

watch, water bath etc.

CHEMICALS :

0.1 N ethyl acetate solution, 0.1 N NaOH, 0.1 N HCl, phenolphthalein indicator, ice etc.

THEORY :

Hydrolysis of ester by NaOH is also known as saponification. The velocity of this

hydrolysis reaction is approximately proportional to the concentration of OH ions.

The hydrolysis takes place as

CH3COOC2H5 + NaOH → CH3COONa + C2H5OH

The rate of this reaction depends on concentration of both the reactants, therefore this

is a second order reaction. Thus, the rate of this reaction is given by

dx

dt = k [CH3COOC2H5] [NaOH]


or dx

dt = k [CH3COOC2H5] [OH]

If initial concentration of both the reactants is same, then the rate constant is calculated

by the equation

k = 1

a ⋅ t

x

a − x

where ‘a’ moles/litre is the initial concentration of either of the substances and (a − x) is the

concentration left behind unreacted after time ‘t’.

The concentration of NaOH decreases with time, therefore titration reading with HCl also

decreases. Titration reading is taken as (a − x) because this is a back titration.

Initial concentration ‘a’ of NaOH can be determined by titrating 0.1 N NaOH directly with

0.1 N HCl.

PROCEDURE :

1. Take 50 ml 0.1 N ethyl acetate solution and 50 ml 0.1 N NaOH solution with the help

of measuring cylinder in two separate clean and dry stopper bottles labelled as

bottle No. 1 and 2 respectively. Keep these bottles in a water bath to attain uniform

constant temperature.

2. Rinse and fill the clean burette with 0.1 N HCl upto the zero mark and see that there

is no air gap.

3. In a clean conical flask take approximately 50 ml ice, cold water or few pieces of

crushed ice and add in it 2 - 3 drops of phenolphthalein indicator.

4. After 10-15 minutes, the solutions kept in the water bath attain the constant

temperature. Now add the solution from bottle No. 1 into 2 and mix the solution

vigorously for 5 - 6 times and note the time of mixing as zero time.

5. Immediately, pipette out 10 ml of this reaction mixture in the conical flask containing

ice cold water and phenolphthalein indicator. Titrate this reaction mixture in the

conical flask against 0.1 N HCl solution till faint pink colour disappears. Record the

titration reading as T0.

6. Shake the reaction mixture from time to time and it should be always kept in the

water bath and titrate 10 ml of the reaction mixture with 0.1 N HCl at successive

intervals of 10, 15, 20, 25 and 30 minutes (Tt).

7. To determine the initial concentration ‘a’ pipette out 10 ml of 0.1 N NaOH solution in

a conical flask and titrate with 0.1 N HCl by using phenolphthalein indicator.

Record the readings in the observation table as follows.


OBSERVATION TABLE :

Time Titration reading

Tt = (a − x)

x = a−(a−x) 1

a − x k =

1

a ⋅⋅⋅⋅ t

x

a − x

0

5

10

15

20

25

30

To calculate ‘x’ take difference between two readings with respect to time.

For graphical method rearrange the following equation for second order equal to

initial concentration.

k = 1

a ⋅ t

x

a − x

∴ 1

a − x −

1

a = kt

∴ 1

a − x = kt +

1

a

1a

1(a x)-

Slope = k

Intercept =1a

Time

Fig. 6.1

Thus, plotting 1

a − x versus time, we get a straight line having slope equal to rate

constant k and intercept equal to 1

a .

RESULT TABLE :

Calculation Method Graphical Method

‘k’ lit. mole−1 min−1

✍ ✍ ✍


Experiment No. 7

PARTITION COEFFICIENT

AIM :

Partition coefficient of iodine between water and carbon tetrachloride.

THEORY :

We can prepare the solution of iodine in water as well as in carbon tetrachloride because

the iodine is soluble in both these solvents. If we shake vigorously the solution of iodine in

water with that of carbon tetrachloride which is immiscible with water; it is found that the

iodine distributes itself between the water and carbon tetrachloride layer in such a way that

at equilibrium the ratio of concentration of iodine in the two layers is a constant at any given

temperature. Suppose when a solute is present in two immiscible liquids 1 and 2 and is in

equilibrium with them, ratio of its concentrations in the two layers of the liquids is constant

at a particular temperature. This constant is called the distribution coefficient or partition

coefficient.

Thus, C1

C2

= K = Distribution or Partition Coefficient.

where C1 and C2 are the concentrations of the solute in liquids 1 and 2 respectively and K is

the partition coefficient. The above equation is known as Nernst Distribution law, which

states that a substance will distribute itself between two solvents until at equilibrium the

ratio of the activities of the substance in the two layers is constant at any given temperature.

When the solutions are dilute or when they behave ideally, the activity is essentially equal to

the concentration C.

Some of the systems are as follows :

1. Benzoic acid between water and chloroform.

2. Iodine between water and carbon disulphide.

3. Hydrogen peroxide between water and ether.

4. Boric acid between water and amyl alcohol.

5. Oxalic acid between water and ether.

6. Bromine between water and carbon disulphide.

7. Phenol between water and amyl alcohol.

According to Nernst, attention must be given to the fact that the statement of the

distribution law is valid only when the solute undergoes no change such as dissociation or

association. If a solute does dissociate into ions or simpler molecules or if it is associated


into more complex molecules then the distribution law is not valid and the concentration

ratio as given in the above equation is not constant. For the dissociation or association of

the molecules, the modified form of the equation is given as follows :

Case - 1 : Normal solute molecules in solvent 1 and its association in solvent 2.

Suppose a solute is represented by a molecular formula A. Let the solute exist as normal

molecules in solvent 1 and in the association form as An in the solvent 2. If C1 is the total

concentration of normal molecules (A) in solvent 1 and C2 is the total concentration of the

solute in solvent 2, then it is observed that

C1

nC2

= constant

where 'n' is the number of molecules of the solute that associate to form a bigger molecule.

It is observed that when the equation is applied for the distribution of salicylic acid between

water and benzene, it is found that C1

C2

is not constant, but C1

C2

is constant. Thus the

association of salicylic acid molecules takes place and the value of n is two.

Case - 2 : Normal solute molecules in solvent 1 and its dissociation in solvent 2.

Suppose that the solute is represented by molecular formula A. The solute exists in the

normal form in solvent 1 and in the dissociated form in solvent 2, then it is found that

C1

C2 (1 − α) = constant

where C1 is the concentration of solute molecules in solvent 1, C2 is the concentration of

solute molecules in solvent 2 while α is the degree of dissociation of the solute in the second

solvent.

If the solute is dissociated in both the solvents, the distribution law becomes

C1 (1 − α1)

C2 (1 − α2) = constant

where α1 and α2 is the degree of dissociation of the solute in solvents 1 and 2 respectively.

The distribution law has been applied to the study of problems of both theoretical and

practical interest; such as extraction, analysis and determination of equilibrium constants.

Following are some important applications of distribution law.

(A) The Process of Extraction :

The extraction is a subject of great importance both in the laboratory and in industry.

The process of extraction is used to remove a substance from the solution by shaking it with


some suitable solvent in which the required substance is more soluble. In laboratory the

removal of dissolved substances from water solution, with solvents such as ether,

chloroform, carbon tetrachloride or benzene may be used. Again in industry extraction is

used to remove various undesirable constituents of a product, such as harmful ingredients in

petroleum oil, by treating the product with an immiscible solvent in which the impurity is

soluble.

The extraction is more economical and efficient if the given volume of the solvent is not

used in a single lot but it is used in several installments, each time separating the solvent. In

other words, it is better to extract with small volumes of solvent several times than once with

a large volume. Some conclusions apply to washing of precipitates, in which case the

process may be considered as the distribution of the impurity between the wash liquid and

precipitate.

(B) The Association and Dissociation of a Solute in a Solvent :

The distribution law has important application of deciding whether a solute is associated

or dissociated in any particular solvent. The findings are given as below :

1. C1

C2

= constant ; molecular conditions of the solute in both the solvents

are same.

2. C1

2C2

= constant ; two molecules of the solute associate in solvent two.

3. C1

3C2

= constant ; three molecules of the solute associate in solvent two.

4. C1

C2 (1 − α) = constant ; the solute molecules dissociate in solvent two.

5. C1 (1 − α1)

C2 (1 − α2) = constant ; solute molecules dissociate in both the solvents.

APPARATUS :

Reagent bottles, burette, pipette, conical flask, separating funnel, water bath etc.

CHEMICALS :

(i) Pure CCl4. (ii) Saturated solution of iodine in CCl4.

(iii) N

20 Na2S2O3. (iv)

N

100 Na2S2O3.

(v) 10% KI. (vi) Starch indicator.


PROCEDURE :

1. Clean the reagent bottles with tap water. Give number to these bottles as 1, 2, 3 and

4 with the help of marker pencil.

2. Prepare the mixtures in these bottles as shown below :

Bottle No. Volume of saturated

I2 solution in CCl4, 'ml'

Volume of pure

CCl4, 'ml'

Volume of distilled

water, 'ml'

1.

2.

3.

4.

30

25

20

15

00

05

10

15

150

150

150

150

3. Stopper each bottle and shake them thoroughly for 30 - 45 minutes. The accuracy of

the experiment depends on the efficiency of shaking.

4. After shaking is over, allow the mixture to separate into two layers. After few

minutes, two distinct layers will be observed. The upper layer is water while the lower

layer is of CCl4.

5. Transfer the mixture of say bottle No. 1 into the separating funnel. Separate the two

layers in a beaker. Estimate the iodine concentration in both layers as follows :

(A) Titration of aqueous layer :

1. Clean the burette and fill it with N

100 Na2S2O3 solution upto the mark. Clamp the

burette properly by using burette stand.

2. In a clean conical flask, pipette out 50 ml of water layer from the separated mixture

of bottle No. 1.

3. Add 5 ml of 10% KI solution.

4. Titrate this mixture against N

100 Na2S2O3 solution till a faint yellow colour appears.

5. In this solution, add 2 - 3 ml starch indicator. Blue colour is developed, continue the

titration till blue colour disappears.

6. Repeat the same procedure, take three readings and report the constant burette

reading, say 'X' ml.

(B) Titration of carbon tetrachloride layer :

1. Clean the burette and fill it upto the mark with N

20 N2S2O3 solution.


2. With the help of pipette, withdraw 5 ml of carbon tetrachloride layer from the

separated mixture of bottle No. 1 in a clean conical flask.

3. To this mixture add 25 ml of 10% KI solution.

4. Titrate the mixture against N

20 Na2S2O3 solution till faint yellow colour appears.

5. Unless and until the faint colour does not obtain do not add the indicator. Thus,

when faint colour appears, add 2 - 3 ml of starch indicator. Continue the titration till

blue colour disappears.

6. Take at least three readings and record constant burette reading, say 'Y' ml.

(C) Repeat the above procedure of separating the layers and titration of aqueous and

organic layers for remaining mixture from bottle Nos. 2, 3 and 4.

Record the observations as follows :

Bottle

No. ml of

N

100 Na2S2O3

per 50 ml

aqueous layer

ml of N

20 Na2S2O3

per 5 ml CCl4

Conc. of

aqueous

layer 'C1'

mol L−−−−1

Conc. of CCl4

layer ‘C2’

mol L−−−−1

C1

C2

1.

2.

3.

4.

CALCULATIONS :

(1) To find out the concentration C1 of aqueous layer. Let 'X' ml of N

100 Na2S2O3 solution

be required for 50 ml of aqueous layer.

Water layer =

N

100 Na2S2O3

N1 × V1 = N2 × V2 V1 = 50 ml aqueous layer taken by pipette

∴ N1 × 50 = 1

100 × V2 V2 = 'X' constant burette reading.

Thus, N1 = 1

5000 × X

N1 = 'X'

5000 = Normality of I2 in the water layer.


But gram equivalent of iodine is half of its molecular weight. Thus, concentration in

g mol−1 L−1 = Normality

2

∴ C1 = 'N1'

2 =

X

10,000 mol L−1

(2) To find out the concentration (C2) of CCl4 layer, let 'Y' ml of N

20 Na2S2O3 solution be

required for 5 ml of CCl4 layer.

N1 × V1 = N2 × V2 V1 = 5 ml of CCl4 layer taken by pipette

∴ N1 × 5 = 1

20 × Y V2 = 'Y' ml constant burette reading.

N1 = Y

100 = Normality of I2 in CCl4 layer.

∴ C2 = N1

2 =

Y

200 mol L−1

Thus, simply by substituting 'X' and 'Y' in the above final equations, obtain 'C1' and 'C2'

for all the bottles, from which C1

C2

and C1

C2

can be calculated. Record the values of C1

C2

and

C1

C2

in the result table. From these results, following conclusions are drawn.

(1) In this experiment, C1

C2

is not constant but the ratio C1

C2

is constant. Thus partition

coefficient of iodine between water and CCl4 = Mean of C1

C2

.

(2) Molecular conditions : Since C1

C2

is constant, molecular condition of iodine in both the

solvents (CCl4 and water) is the same.

Thus iodine exists as a single molecule in water and CCl4. There is no dissociation and

association of iodine in either of the solvents. Similarly, molecular weight of the iodine

remains same in both the solvents.

PRECAUTIONS :

1. Continuous and vigorous stirring/shaking is must.

2. Do not add starch until the iodine solution becomes faint yellow in colour.

3. The mixture should be separated by using separating funnel. If the line of

demarcation is not clear, shake the funnel and allow it to stand still so that clear

boundary of separation is observed.


QUESTIONS

1. Give mathematical statement of Nernst distribution law.

2. Define partition coefficient.

3. What is meant by miscible and immiscible liquid pairs ? Explain with the help of an

example.

4. State and explain distribution law. Give its limitations.

5. Explain various applications of distribution law.

6. How distribution law is helpful in determining the molecular condition of solute in

solvents ?

7. Explain the effect of association of solute molecules on partition coefficient.

✍ ✍ ✍

(39)

SECTION - B

INORGANIC CHEMISTRY PRACTICALS

(Minimum Five Mixtures)

INORGANIC QUALITATIVE ANALYSIS

The detection or identification of individual elements or ions entering into the

composition of a substance constitutes the task of qualitative analysis.

Qualitative analysis is of enormous, scientific and practical importance. It is a science

dealing with the various methods for investigating substances and their transformations.

It also plays an important role in branches of science allied to chemistry-mineralogy,

geology, physiology, microbiology, as well as in medicine, agriculture and technology.

In almost every scientific research work dealing in one way or another with chemical

phenomena, the investigator has to make use of the qualitative methods of analytical

chemistry.

According to the amount of substance used for the analytical reactions, qualitative

analysis is distinguished into four categories.

1. Macro Qualitative Analysis :

In this method, large quantities of an unknown substance (0.5 to 1.0 g) or solution

(20-50 ml) are used. The reactions are carried out in ordinary test tubes (10 - 20 ml in

capacity), beakers or flasks. Precipitates are separated by filtration through filter paper.

Analyst requires large quantities of reagents and hence it is non-economic.

2. Micro Qualitative Analysis :

Very small quantity (few mg) of substance or (about a millilitre) solution is used to carry

out this analysis. Only highly sensitive reactions are employed, which permit detection of

individual constituents by the fractional method even if they are present in small amounts.

Such reactions are carried out either by micro crystalloscopic method (using microscope

slides) or by the drop method (using spot tests).

3. Semimicro Qualitative Analysis :

It ranks between macro and micro method of qualitative analysis mentioned above. The

amount of substance used is about 50 mg of solid or 1 ml of solution. This method is more

advantageous over macro analysis, because it gives quick and reliable results and also needs

S.Y.B.Sc. Practical Chemistry 40 Inorganic Chemistry Practicals

small amounts of reagents. It is more economic than the macro method. For these reasons,

it is generally employed commonly in laboratory work.

4. Ultramicro Qualitative Analysis :

It is employed when the substance to be analysed is available in very small amount (less

than 1 mg). In this method, all the analytical operations are performed under the

microscope. The qualitative scheme of analysis given here is concerned with the methods of

qualitative analysis of inorganic substances only. Before going to the analysis, students

should refer the following important instructions.

INSTRUCTIONS :

(a) Dry Tests : These tests are performed directly with the given solid (dry) substance

and the appropriate reagent by heating them to a desired temperature.

1. Use clean and perfectly dry test tube.

2. Heat the test tube on blue flame of the burner.

3. Observe the changes upon heating very carefully.

4. Do not throw the contents of the test tube in basin, unless the test tube is cooled.

5. Make sure that all the necessary test papers are ready with you before heating the

substance. Moisten the test papers (blue and red litmus, turmeric paper, starch

paper).

6. To prepare starch iodide paper just immerse a piece of starch paper in potassium

iodide solution.

7. Do not place hot test tube or beaker directly on your table, make use of asbestos

sheet.

8. Replace the reagent bottles after use at their original place in reagent rack.

9. Read the complete test before performing it.

10. Record your observations according to the sequence given in the scheme.

(b) Wet Tests :

1. Prepare a solution of a given substance in a suitable solvent. Solution should not be

turbid one.

2. Use small fractions of this solution for every test. Large volumes need more reagents

for precipitation and also take more time for filtration.

3. Strictly follow the instructions given for each test. (e.g. instructions regarding pH,

temperature etc.)

4. If solutions are colourless, label them to avoid confusion during analysis.

5. Do not throw any solution, till the completion of analysis.

6. Once a particular group is detected, precipitate it completely, filter it and use it for

group analysis.


7. Analyse the group to find out basic radical and confirm it.

8. Use water solution only for detection and identification of group VI radicals.

SCHEME OF INORGANIC QUALITATIVE ANALYSIS :

The systematic procedure of the analysis of a given mixture involves following steps :

(a) Preliminary tests

(b) Dry tests for basic radicals

(c) Dry tests for acidic radicals

(d) Preparation of solution

(e) Wet test for basic radicals and their confirmatory tests.

(f) Wet test for acidic radicals and their confirmatory tests.

(A) PRELIMINARY TESTS :

Observation Inference

1. Colour

(a) Blue Hydrated copper salts and anhydrous cobalt

salt.

(b) Bluish green or green Chromium, nickel, copper and ferrous salts.

(c) Yellow Chromates or salts like PbO, HgO, Bi2O3,

FePO4, CdS, As2O3 etc.

(d) Orange or red Dichromates and Sb2S3.

(e) Pink or flesh Cobalt or manganese salts.

(f) Black Oxides of Cu2+, Mn2+, sulphides of Fe3+,

Sb3+, Ni2+.

(g) White/Colourless Al, Zn, Ca, Sr, Mg, Na, K, NH4 salts.

(h) Coloured Coloured radicals like Cu2+, Mn2+, Fe3+ etc.

and coloured powders like HgO, PbO.

2. Nature/Appearance

(a) Crystalline All soluble salts i.e. salts of Na, K and NH4 or

salt with Cl−, Br−, I−, SO2−

4 , NO−

2 , NO−

3 .

(b) Amorphous Insoluble salts containing CO2−

3 , S2−, PO3−

4 ,

BO3−

3 , O2−.


(c) Hygroscopic Nitrate salts.

(B) DRY TESTS FOR BASIC RADICALS (CATIONS) :

3. Heating in a Dry Test Tube :

Take a little quantity of the given mixture in a dry test tube. Heat it strongly and observe

the change which takes place during heating (more than one observation may be taken).


(a) Decrepitation, i.e. cracking noise Crystalline salts like NaCl, KBr, Pb(NO3)2,

Ba(NO3)2 etc.

(b) Substance fuses Alkali salts, nitrates of alkaline earth metals.

(c) Water vapour condenses on the cooler

part of the test tube.

The water is due to (i) hygroscopic salts,

(ii) water of crystallisation,

(iii) decomposition of hydroxides.

(d) Change in colour

Original

Colour On heating On cooling

White Yellow White Zn salts

White Brown Brown Cd salts

White Yellow Brown Pb and Bi salts

Coloured Black Black Salts of Cu, Fe, Cr, Mn, Co, Ni

White White White Salts of Ca, Ba, Sr, Al, Mg.

(e) Formation of sublimate NH4 salts and HgCl2, As2O3, Sb2O3

(f) Evolution of a gas

1. Colourless and odourless gas igniting a

glowing splinter i.e. O2 gas

Nitrates, chromates and oxides like HgO

2. Colourless and odourless gas turning

freshly prepared lime water milky

i.e. CO2 gas.

Carbonates

3. Colourless gas with pungent odour,

turning moist turmeric paper red

i.e. NH3 gas.

Ammonium salts

4. Colourless gas turning dichromate

paper green i.e. SO2 gas.

Sulphate and certain sulphides like PbS


5. Brown fumes, i.e. NO2 or Br2 gas. Nitrates or bromides (Br2 turns starch paper

yellow)

6. Violet fumes i.e. I2 gas turns starch

paper blue black.

Iodides

4. Heating in a Charcoal Cavity :

Prepare a small cavity on a piece of charcoal. Place a little quantity of mixture (Given

mixture + Na2CO3 in the ratio 2 : 1). Place a drop of water on it and heat it with the reducing

(yellow) flame using blow pipe.

White infusible residue obtained Coloured residue obtained

(Ca, Sr, Ba, Zn, Al salts) (Transition metal salts)

↓ (Cu2+, Fe2+, Fe3+, Mn2+ etc.)

Perform cobalt nitrate test as follows :

Moisten the white infusible residue with one or two drops of cobalt nitrate solution and

heat it again on oxidising flame (blue) using blow pipe.

If residue turns

Blue Al salts

Green Zn salts

Pink Mg salts

Gray Ba, Sr, Ca salts

...

[Hint : Generally, white/colourless mixtures give white infusible residue on charcoal

cavity, while coloured mixtures give coloured residue.]

5. Action of NaOH :

Take small amount of the mixture in a test tube and add a little amount of dilute NaOH

solution to it. Heat to boiling. Hold the moist turmeric paper at the mouth of the test tube

without touching the test tube. If the paper turns red, NH+

4 may be present.

(C) DRY TESTS FOR ACIDIC RADICALS (ANIONS) :

6. Action of Dilute HCl :

Take small quantity of the mixture in a test tube, add dilute HCl and heat gently.


(a) Brisk effervescence of CO2 gas turning

freshly prepared lime water milky.

CO2−

3 (Carbonate)


(b) Evolution of colourless H2S gas having

smell of rotten eggs and turning lead

acetate paper black.

S2− (Sulphide)

(c) Brownish fumes acidic to litmus,

NO2 gas. NO

−

2 (Nitrite)

7. Action of MnO2 + Conc. H2SO4 : Mix small quantity of mixture with a pinch of MnO2 in

a test tube. Add 1-2 ml of concentrated H2SO4 and heat cautiously.

(Hint : Until you don't see coloured fumes, do not test with any test paper.)


(a) Greenish yellow, Cl2 gas is evolved

which turns blue litmus paper red and

then bleaches it. It also turns starch

iodide paper blue.

Cl− (Chloride)

(b) Brown fumes of Br2 gas evolved which

turns starch iodide paper blue and

starch paper yellow and fluorescent

paper red.

Br− (Bromide)

(c) Violet vapours of I2 evolved turning

starch paper blue.

I− (Iodide)

(d) No coloured fumes. Halides may be absent.

8. Nitrate Test :

Take little quantity of mixture in a test tube, add conc. H2SO4 and heat gently, if brown

fumes are seen then add copper filings and heat. If intense brown fumes increases then

NO−

3 may be present.

9. Phosphate Test :

Take a little mixture in a dry clean test tube. Add to it concentrated HNO3 and boil well

to dissolve the mixture. To this hot solution now add excess of ammonium molybdate

solution. If canary yellow precipitate is obtained, PO3−

4 i.e. phosphate may be present.

[Hint : Many times only yellow colouration is obtained, which may not be a precipitate.

If it is a precipitate, it should settle at the bottom of test tube on keeping for few minutes.]

10. Borate Test :

Take a little mixture in a porcelain dish. Add to it drop of concentrated H2SO4 and two

drops of ethyl alcohol. Mix well with the help of glass rod and make a paste of this mixture.


Take this paste on glass rod and hold it in the flame. If a green edged flame of ethyl borate

is observed, BO3−

3 , i.e. borate may be present.

[Hint : Both phosphate and borate are never given in the mixture, therefore any one of

these two may be present in the given mixture. However, sometimes both phosphate and

borate may not be present.]

(D) PREPARATION OF A SOLUTION OF THE GIVEN MIXTURE :

Prepare only 10 - 15 ml of solution (1 test tube) :

(a) Solution in water : Take a little quantity of the mixture in a test tube and add to it

about 1/3rd test tube of water and boil. If the mixture dissolves completely, the components

of the mixture are water soluble. In such a case, the aqueous solution of the mixture is used

to detect the basic radicals.

(b) Solution in dilute HCl : If the mixture is not completely soluble in water, add about

2 - 4 ml of dilute HCl (2N) and boil. If the mixture dissolves completely, use this solution for

further analysis. If gas like CO2 or H2S is evolved, boil the solution till it is free from H2S or

CO2.

(c) Solution in conc. HCl : Take a little quantity of given mixture in a test tube and add

3 ml of concentrated HCl to it. Boil well till evolution of CO2 and H2S is completed. Dilute this

solution and use it for further analysis.

(d) If the given mixture is insoluble in dil. and conc. HCl, then use the following solvents

in the order given as dil. HNO3, conc. HNO3 and aqua regia (mixture of conc. HCl and

HNO3 3 : 1).

Mixture partly soluble in water : Take a little quantity of mixture in a test tube and add

to it about 3 - 4 ml water and boil, if mixture is partly soluble then filter or centrifuge. Collect

the filtrate and residue and analyse them separately to find the basic radicals.

Principles of Precipitation (Solubility Product) :

As quantitative analysis involves the precipitation of a radical into its insoluble salt,

qualitative analysis mainly depends on solubility product principle and chemical reaction can

be explained by it.

In a saturated solution of an insoluble binary electrolyte e.g. MA (where M = cation and

A = anion), the ionic product of M and that of A, at a given temperature remains constant

and is called the 'solubility product' and is denoted by Ksp of MA.

Ionic Product : It is the product of ionic concentration of ions.

The precipitation will occur when the ionic product exceeds the solubility product.

Hence,

1. Ionic product > Solubility product → Precipitation (supersaturated solution)

2. Ionic product < Solubility product → No precipitation (unsaturated solution)


3. Ionic product = Solubility product → No precipitation (saturated solution)

In a solution containing M+ ion, the concentration of A− is made very high, the insoluble

salt MA will be precipitated and concentration of M+ will be gently reduced, because as

[A−] increases, [M+] must decrease to bring ionic product equal to solubility product.

In a similar way, A− can be completely precipitated from its solution by increasing the

concentration of M+.


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lori

des

in

solu

ble

in

ac

ids

Wh

ite

pp

t -

Ag

Cl,

Hg

2C

l 2,

Pb

Cl 2

II

*

Hg

2+,

Pb

2+,

Bi3

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Cu

2+,

Cd

2+,

As3

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Sn

2+,

Sn

4+,

Sb

3+

Dil

. H

Cl

+ H

2S

S

ulp

hid

es

inso

lub

le

in

dil

. H

Cl

Bla

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pt

- H

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wn

pp

t -

Bi 2

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Fe3

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l (s

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d,

exce

ss)

+

NH

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SEPARATION TABLE FOR GROUP I :

Transfer the ppt. in a porcelain dish, add 5 ml water, boil it well. Filter when hot. If ppt.

dissolves, only filtrate is obtained.

Residue Filtrate

Add to the residue 5 ml of conc. NH4OH.

Warm and filter.

It may contain PbCl2

1. Filtrate + K2CrO4 solution. If yellow ppt.,

Pb2+ present.

2. Filtrate + KI solution. If yellow ppt. of

PbI2 soluble on heating and reappears

as golden crystals on cooling.

3. Filtrate + dilute H2SO4 : If white ppt.

then Pb2+ present and confirmed.

Residue

Residue + aquaregia

HCl : HNO3 (3 : 1), boil

and add SnCl2. If

black ppt., then Hg+

present and

confirmed.

Filtrate

Acidify the filtrate

with dilute HNO3. If

white ppt. of AgCl is

formed, then Ag+

present.

C.T. filtrate + KI

solution. If yellow

ppt., then Ag+

confirmed.

SEPARATION TABLE FOR GROUP II :

Transfer the ppt. into a porcelain dish and add 5 ml of yellow ammonium sulphide

(NH4)2SX and 1 - 2 ml of NaOH. Heat it to 60°C for about 5 minutes. If ppt. dissolves only

filtrate is obtained and group II B is present. Acidify the solution by adding conc. HCl, yellow

or orange ppt is obtained.

(Use separation table for group II B on page no. 50).

If ppt. does not dissolve group II A is present. [If white or dirty white ppt. is obtained,

then it is a ppt. of excess of sulphur. Therefore, it should be rejected i.e. II B group absent.]

(Use separation table for group II A as given below.)

SEPARATION TABLE FOR GROUP II A :

Transfer the ppt. of II A group metal sulphides into porcelain dish. Add 2 - 3 ml of

conc. HNO3. Boil for 2 - 3 minutes. Filter or centrifuge. If ppt. dissolves only filtrate is

obtained.


Residue Filtrate

Black ppt. of HgS

Dissolve the residue in

aquaregia (HCl + HNO3

3 : 1) by boiling.

Filtrate contains nitrates of Pb, Bi, Cu, Cd. To a small portion add

dil. H2SO4. If white ppt is obtained, it indicates the presence of

Pb++. So to the whole solution add dil. H2SO4. Filter or centrifuge.

Dilute with water. Residue Filtrate

C.T.

(i) Solution + SnCl2

solution white or grey

ppt. confirms the

presence of Hg2+.

It is white ppt. of

PbSO4. Boil the

residue with strong

solution of

ammonium acetate to

get solution.

It may contain sulphates or nitrates of

Bi, Cu, Cd. Add 1 : 1 ammonia in excess.

Filter or centrifuge.

(ii) Solution + KI if Residue Filtrate

yellow ppt. of HgI2.

Hg2+ present and

confirmed.

C.T.

(i) This solution +

acetic acid + K2CrO4.

If yellow ppt. of

PbCrO4 is obtained,

Pb2+ may be present.

(ii) Solution + KI if

yellow ppt. is

obtained, Pb2+ is

confirmed.

It is a white ppt. of

Bi(OH)3. Wash with

water. Dissolve it in

dil HCl.

(i) Solution + excess

water, if white ppt. of

BiOCl is obtained

then Bi3+ is present.

(ii) 1 ml solution +

10% Thiourea

solution if intense

yellow colour or ppt

is obtained then Bi3+

is confirmed.

If filtrate is blue,

Cu2+ may be

present. (C.T.)

(i) To a small

portion add dilute

acetic acid then

add potassium

ferrocyanide

solution. If a

chocolate red ppt.

then Cu2+ present

and confirmed.

(ii) To the other

part add K2CrO4

solution. A light

yellow ppt. of

CdCrO4 is

obtained. Cd++

present and

confirmed.


SEPARATION TABLE FOR GROUP II B :

Transfer the ppt. into porcelain dish and add 5 ml conc. HCl and boil for 5 minutes. Filter

or centrifuge.

Residue Filtrate

Residue may be yellow As2S3. Boil

the residue with conc. HNO3 and

dilute with H2O to get solution.

It may contain complexes of Sb and Sn. Neutralise it

with diluted NH4OH. Add solid oxalic acid and heat,

pass H2S gas.

C.T. Residue Filtrate

(i) Solution + ammonium molyb-

date, if yellow ppt. then As3+

present.

(ii) Solution + Mg(NO3)2, if white

ppt. is obtained, then As3+

present and confirmed.

Orange coloured ppt. of

Sb2S3. Dissolve it in

conc. HCl.

C.T. : (i) Dilute a small

portion of it with water. If

white turbidity of SbOCl,

then Sb3+ present.

(ii) Solution + Rhoda-

mine B. If red colour is

seen and (iii) Solution +

KI + pyridine. If coloured

solution, then Sb3+


It may contain a complex

of Sn and oxalic acid.

Expel H2S, add Zn and

dilute HCl Sn is

precipitated. Dissolve it

in conc. HCl to get

solution : (i) Solution +

HgCl2 solution. If white

ppt. of Hg2Cl2, then Sn2+

present and (ii) Solution

+ dilute I2 solution. If I2

decolourises, then Sn2+


REMOVAL OF INTERFERING ANIONS :

Phosphate and borate ions interfere in the detection of group of cation from III A to V

due to formation of their insoluble borates and phosphates in the alkaline medium. For

example, if given mixture contains Ni3(PO4)2 that is nickel phosphate, its original solution is

prepared in acid like HCl.

∴ Ni3(PO4)2 + 6HCl → 3NiCl2 + 2H3PO4

Thus, original solution contains Ni2+ ions along with phosphoric acid. When this solution

is tested for presence of group III A (which is not present, as Ni2+ is present in III B) by

adding NH4Cl and NH4OH, following reaction takes place.

3NiCl2 + 2H3PO4 + 6NH4OH → Ni3(PO4)2 ↑ + 6NH4Cl + 6H2O

A precipitate of nickel phosphate is obtained, which leads to conclusion of presence of

group III A which is not actually present. Similar type of misleading reaction is observed

when borate is present. Thus when phosphate or borate is present they must be removed

before the detection of group III A from the solution.


Removal of Phosphate - (If present) :

1. If group I and II are absent then use the original solution.

2. If group II is present, precipitate it completely and filter, use filtrate of group II for

the removal of phosphate after expelling H2S gas by boiling the filtrate.

3. Test for Fe2+ : Take one drop of filtrate/O.S. on porcelain tile, mix a drop of

potassium ferricyanide solution → If blue colour or precipitate is seen, Fe2+ is

present.

4. If Fe2+ is present add to the filtrate or O.S. 2 - 3 ml of concentrated HNO3 and boil so

that Fe2+ is oxidised to Fe3+.

5. Use this solution for removal of phosphate as follows :

To the solution add 1 g of NH4Cl, warm and then add 1:1 NH3 until a slight turbidity

is obtained and persist even after stirring. This means solution is alkaline and

turbidity is due to either hydroxides of III A group metals or phosphates of III B, IV

and V. Now add 10 ml of sodium acetate buffer solution and stir well.

If turbidity persist or increases, III A group present.

Stir the ppt. and filter.

If turbidity dissolves completely, phosphates of III A, IV, V group present

ppt. (Residue) contains hydroxides of Fe, Cr, Al. Use it for the analysis of group III A. Page No. 52

Filtrate contains phosphate of III B, IV, V group

Add neutral FeCl solution till phosphate is precipitated and solution acquires brownish red colour. Boil and filter.

3

Filtrate contains III B, IV, V group chlorides and excess of FeCl3 .

Add to the filtrate NH Cl solid + 1 : 1 NH3 till alkaline

4

ppt. (Residue) Discard the ppt.

of FePO4

Reddish brown ppt. of Fe(OH)3 is obtained. Boil and filter, discard the ppt. of Fe(OH)3

and concentrate the filtrate (reduce the volume) and use for detection of IIIB, IV and V group

according to the chart given above.


Removal of Borate (If present) :

1. If group II is present, precipitate it completely by passing H2S gas. Filtrate the

precipitate and boil the filtrate to expel H2S completely (test with lead acetate

paper). Then use it for removal of borate.

2. If group I and II are absent, then use original solution.

3. Removal of borate. Take O.S/filtrate from group II in a porcelain dish, add to it

3 ml conc. HCl and 5 ml ethyl alcohol and boil. Repeat this procedure till test for

borate is negative (No green edged flame on igniting solution). Then use this

solution for detection of group III A onwards according to chart given on

previous page.

SEPARATION TABLE FOR GROUP III A :

Transfer the ppt. in a evaporating dish + 4 ml NaOH + 2 ml H2O2 boil and filter. If ppt.

dissolves completely only filtrate is obtained.

Residue Filtrate

It may contain Fe(OH)2 and MnO2 ⋅ H2O.

Divide the residue into two parts.

Part I :

Dissolve 1st part of the residue in conc.

HNO3 and add to it a pinch of PbO2. Boil

gently for 2 to 3 minutes, dilute with water

and allow to settle. If violet colouration of

HMnO4 (permanganic acid) then Mn2+


Part II :

Dissolve the second part in hot dilute

HCl and divide the solution into two parts :

(a) Solution + Ammonium thiocyanate

(NH4SCN) solution. If deep blood red

colouration, then Fe3+ present.

(b) Solution + Potassium ferrocyanide

K4[Fe(CN)6], if a deep blue colouration, then

Fe3+ confirmed.

The filtrate contains NaAlO2 (colourless)

and Na2CrO4 (yellow). It is divided into two

parts.

Part I :

If the colour of the filtrate is yellow,

Cr3+ may be present :

(a) Acidify with dilute CH3COOH and then

added lead acetate solution. If yellow ppt. of

PbCrO4, then Cr3+ present.

(b) Neutralise the yellow solution by

CH3COOH and add AgNO3 solution. If a

brick red ppt. of Ag2CrO4, then Cr3+ present

and confirmed.

Part II :

(a) If filtrate is colourless acidify with HCl.

Then add NH4OH till the solution is alkaline.

Heat to boiling. If white gelatinous ppt. of

Al(OH)3 formed, then Al3+ present.

(b) Alizarin Test : A drop of Alizarin on the

test paper + 1 drop of filtrate. Hold the

paper over bottle of ammonia. If a violet red

colour is seen, then Al3+ present and

confirmed.


SEPARATION TABLE FOR GROUP III B :

Transfer the ppt. into a porcelain dish and add to it about 5 - 7 ml of dil. HCl. Stir well

and allow to stand for five minutes. Filter the solution.

Residue Filtrate

The black residue may be of NiS or

CoS. Dissolve the ppt. in conc. HCl

and add a few crystals of KClO3

It may contain MnCl2 and ZnCl2. Boil the solution to

expel H2S. Add excess of NaOH solution. Filter.

(or in aquaregia). Boil the excess Residue Filtrate

of acid and evaporate to dryness.

Cool it and extract the solid with

5 ml of water. Divide this solution

into three parts.

Tests for Co2+ :

(1) To a portion of the solution add

solid NH4Cl and ammonia till

alkaline and add K4[Fe(CN)6]. If red

solution or reddish brown ppt. on

heating, then Co2+ present.

(2) Two drops of solution on a

watch glass + one crystal of

Na2S2O3. If blue colour around the

crystal, then Co2+ present and

confirmed.

(3) To a portion of the solution add

solid NH4Cl and ammonia till

alkaline. Then add excess of

dimethyl glyoxime. If scarlet red

ppt. then Ni2+ present and

confirmed.

It is a white residue

turning brown due

to atmospheric

oxidation. Dissolve

this ppt. in conc.

HNO3 + a pinch of

PbO2, boil and

dilute with water. If

violet colouration,

then Mn2+ present

and confirmed.

It may contain Na2ZnO2.

(1) Acidify the solution by acetic

acid and then pass H2S gas.

If white ppt., then Zn2+ present.

(2) Acidify the solution by acetic

acid. Add 4 drops of K3[Fe(CN)6]

solution → Green ppt.

(3) Acidify the solution with dil.

HNO3 and add a few drops of

Co(NO3)2 solution. Heat to

dryness. If a green mass, then

Zn2+ present and confirmed.

SEPARATION TABLE FOR GROUP IV :

Transfer the ppt. of IVth group in a porcelain dish and dissolve it in dil. acetic acid. Boil to

expel CO2. The solution thus formed is called as acetate solution.

1. Take 2 ml acetate solution. Boil it + K2CrO4 if yellow ppt. then Ba2+ present.

2. If Ba2+ is present, add little excess of K2CrO4 solution to the whole solution. Boil and

filter and proceed for residue below.


3. If Ba2+ is absent, then test the acetate solution for Sr and Ca (filtrate).

Residue Filtrate

Wash yellow residue of BaCrO4

with water. Dissolve it in about

3 ml conc. HCl.

It may contain Sr and Ca acetate. Add excess of

ammonium sulphate solution and heat to boiling.

Allow to stand for 5 minutes. Filter

Solution of BaCl2 is obtained : Residue Filtrate

(1) If this solution is heated on a

platinum wire, it imparts a green

colour to the flame.

(2) To this solution add dilute

H2SO4, if white ppt. of BaSO4, then

Ba2+ present.

(3) Solution + ammonium oxalate

(NH4)2C2O4 solution, if white ppt. of

BaC2O4 then Ba2+ present and

confirmed.

(1) It is a ppt. of SrSO4.

Add 2 to 3 drops of conc.

HCl and take it on a

platinum wire. It imparts

a crimson colour to the

flame.

(2) Dissolve the residue in

conc. HCl. Take a drop of

it on a paper and add a

drop of sodium

rhodizonate solution. If

brownish red coloured

spot, then Sr2+ present

and confirmed.

(1) Concentrate the

solution. Then add

ammonium oxalate

(NH4)2C2O4 solution. If

white ppt. of CaC2O4,

then Ca2+ present.

(2) Prepare a very con-

centrated solution and

add one drop of conc.

HCl to it. This solution

imparts a red colour to

the flame, then Ca2+


ANALYSIS OF GROUP V :

Transfer the ppt. in a test tube and dissolve it in a little dil. H2SO4. Use this solution for

C.T. of Mg2+.

C.T. of Mg2+ :

1. 2 drops of solution on watch glass + 2 drops of titan yellow solution + 4 drops of

dil. NaOH solution, a rose red colour or ppt.

2. Hypoiodate reagent + 2 drops of above solution, if reddish brown ppt. of Mg(OI)2

then Mg2+ present and confirmed.

[Hypoiodate reagent : NaOH solution + equal amount of KI solution + Iodine

solution till solution becomes just yellow.]

SEPARATION TABLE FOR GROUP VI :

(i) Use water solution only.

(ii) Test the water solution for NH+

4 , Na+ and K+ as follows :


(1) Test for NH+

4 :

Water solution + NaOH boil, if evolution of NH3 gas which turns moist turmeric paper

red, NH+

4 present.

C.T. for NH+

4 : Nessler's reagent + water solution, if brown ppt. or colouration is seen,

then NH+

4 confirmed.

(Nessler's reagent : 2 drops of HgCl2 + KI solution till the scarlet red ppt. first formed

and then dissolves. Then add equal amount of NaOH.)

(2) Test for Na+ and K+ :

Water solution + Sodium cobaltinitrite : Na3[Co(NO2)6]

If yellow ppt.

then K+ present

If no yellow ppt.

then Na+ present

C.T. for K+ : (i) Water solution + Picric acid. Rub the inner side of the test tube with glass

rod, if yellow ppt. of potassium picrate then K+ confirmed.

(ii) Water solution + perchloric acid, if white ppt. then K+ confirmed.

C.T. for Na+ : Water solution + Potassium pyroantimonate (K2H2Sb2O7) if white ppt. then

Na+ confirmed.

(Hint : Generally sodium (Na+) is not given in the mixture.)

DETECTION OF ACIDIC RADICALS (ANIONS) :

Acidic radicals like CO2−

3 , S2−, PO3−

4 , BO3−

3 are easily detected during the dry tests for

anions. Halides (Cl−, Br−, I−) and nitrate (NO−

3 ) may also be detected in dry tests if the tests

for them are carried out carefully. There is no separate dry test for SO2−

4 . Presence of acidic

radicals are further confirmed by wet tests.

If the mixture is water soluble, then its water solution (WS) is used for the detection of

acidic radicals. If the mixture is insoluble in water, sodium carbonate extract and neutral

extract are used for the detection of acidic radicals. These extracts are prepared as follows :

Preparation of sodium carbonate extract : This is required only when mixture is water

insoluble.

The sodium carbonate extract is prepared to convert the water insoluble anions like

S2−, CO2−

3 to their sodium salts which are soluble in water. Take about 0.1 g of mixture and

about 0.2 g of solid pure sodium carbonate in a beaker. Add about 1/2 test tube of distilled

water and boil for about 2 minutes. Cool and filter. Discard the residue. The filtrate is called

as sodium carbonate extract.


Preparation of neutral extract : A part of sodium carbonate extract is neutralised by

adding few drops of dil HNO3 till solution becomes acidic and then NH4OH till solution

becomes just alkaline. (Use litmus paper to check acidity and alkalinity.) Boil the solution to

remove excess of NH3 and CO2 and cool. The resulting solution is called neutral extract

(NS).

(Hint : Use water solution only for the detection of acidic radicals if mixture is water soluble. Do not prepare sodium carbonate extract and neutral extract.)

If mixture is water insoluble, use neutral extract only for the detection of halides, PO3−

4 ,

BO3−

3 and SO2−

4 , while use sodium carbonate extract only for the detection of S2−, NO−

2 , NO−

3

etc.

Before you proceed for the wet tests of anions, check the results of dry tests and see which tests are positive.

Tests Observation Inference

(1) Neutral solution (N.S.)/

Water solution (W.S.) +

AgNO3 solution.

(a) Curdy white ppt insoluble

in dil. HNO3 but soluble in

NH4OH

Cl− present

(b) Pale yellow ppt. soluble in

liquor ammonia

Br− present

(c) Yellow ppt. insoluble in

dil. HNO3 and also in

NH4OH

I− present

(d) Yellow ppt. soluble in

dil. HNO3 PO

3−

4 present

(e) White ppt. soluble in

dil. HNO3 BO

3−

3 present

(2) Neutral solution or Water

solution + Ba(NO3)2

solution.

White ppt. insoluble in

dil. HNO3. SO

2−

4 present

(3) Sodium carbonate extract +

Lead acetate.

Black ppt. soluble in HNO3 S2− present

(4) Sodium carbonate extract

or W.S. + dil. acetic acid +

FeSO4 solution.

Brown coloured solution NO−

2 present

(5) Sodium carbonate extract

or W.S. + conc. H2SO4 (cool

thoroughly) + FeSO4

solution dropwise from the

sides of the test tube.

Brown ring at the junction of

two layers. NO

−

3 present

(6) Little mixture + dil. HNO3. Effervescence of CO2 gas CO2−

3 present


CONFIRMATORY TESTS FOR ANIONS :

1. Chloride (Cl−−−−) :

Chromyl Chloride Test : Little of mixture (or 2 ml water extract) + little solid K2Cr2O7

and few drops of conc. H2SO4 in a test tube. Heat gently, and pass the evolved gas

into another test tube containing lead acetate solution. If yellow ppt. of lead

chromate is formed, then Cl− is confirmed.

2. Bromide (Br−−−−) : Water solution + Chlorine water + Chloroform. Shake well. If brown

colour to the chloroform layer (at the bottom) then Br− is confirmed.

3. Iodide (I−−−−) : Water solution + Chlorine water + Chloroform. Shake well. If violet

colour to the chloroform layer (at the bottom), then I− is confirmed.

4. Sulphate ( )SO2−−−−

4 :

(a) Lead acetate test : Water solution + Lead acetate solution. If white ppt. of

PbSO4, SO2−

4 is confirmed. OR

(b) Sodium rhodizonate test : Place a drop of freshly prepared 5% aqueous

solution of sodium rhodizonate on a test paper and add to it a drop of BaCl2

solution.

A red ppt. is formed. Then add a drop of aqueous solution of the given mixture

on red spot. If red ppt. disappears, SO2−

4 is confirmed.

5. Sulphide (S2−−−−) : Sodium nitroprusside test : Take 1 ml of water extract in a test

tube, add 2 drops of NaOH solution. Then add 2 drops of sodium nitroprusside

reagent solution. If violet colouration, then S2− is confirmed.

6. Nitrite (NO−−−−

2 ) :

(a) Little of the mixture (or 2 ml water extract) + 1 drop of dil. CH3COOH + Pinch of

thiourea. Shake well and add few drops of aqueous FeCl3 solution. If red

colouration, NO−

2 is confirmed. OR

(b) Mixture + dil. H2SO4 and warm. If brown fumes of NO2 gas, then NO−

2 is

confirmed.

7. Nitrate (NO−−−−

3 ) :

Brown ring test : Take 1 ml of water solution and acidify it with conc. H2SO4. Cool

the solution. To this solution add a saturated solution of FeSO4 from the sides of the

test tube. If a brown ring at the junction of two layers is formed due to FeSO4.NO

complex, then NO−

3 is confirmed.

8. Carbonate (CO2−−−−

3 ) :

(a) Little of the mixture + 1 ml dil. H2SO4. If effervescence of CO2 (colourless gas)

which turns lime water milky, CO2−

3 is confirmed.

(Hint : No confirmatory test for PO3−

4 and BO3−

3 is required because their presence is

confirmed during dry tests for acidic radicals only.)


TEST FOR ANIONS IN PRESENCE OF EACH OTHER :

1. Nitrate in the presence of bromide : (NO−

3 and Br−). Acidify the water solution with

dil. H2SO4. Add to this solution ammonical silver sulphate to precipitate bromide as AgBr.

Filter the ppt. and test the filtrate for NO−

3 .

RESULT:

Basic Radicals (positive) Acidic Radicals (negative)

1.

2.

1.

2.

QUESTIONS

1. What is qualitative analysis ? How it differs from quantitative analysis ?

2. What are the different types of qualitative analysis ?

3. What is dry test for ammonium ?

4. What is the use of turmeric paper ?

5. For what test you use starch iodide paper ?

6. What is dry test for chloride ?

7. What is cobalt nitrate test ? When it is performed ?

8. What quantity of original solution you require ? How will you prepare it ?

9. What is solubility product and ionic product ?

10. What basic radicals are present in group III A ?

11. What are the group reagents for testing group III B ?

12. Why concentrated solution is required for testing group IV ?

13. What is C.T. for Ni2+ ?

14. What is Nessler's reagent ?

15. What is hypoiodate reagent ?

16. How will you test presence of Cl−, Br− and I− ?

17. What is the test for CO2−

3 ?

18. What is chromyl chloride test ?

19. What is C.T. for NO−

3 ?

20. Which acidic radical is not detected in dry test ?

21. What is aqua-regia ? Where it is used ?

22. When sodium carbonate extract is required ?

23. What is neutral extract ?

24. What are interfering anions ?

25. Why phosphate is to be removed before testing presence of group III A ?

26. Why borate is removed from the solution ?

27. How borate is removed from the solution ?

28. What is the test for phosphate ?

29. How we detect presence of borate ?

30. What is brown ring test ? How it is performed ?

✍ ✍ ✍

( 59 )

SECTION - C

ORGANIC CHEMISTRY PRACTICALS

(a) Separation of Binary Mixtures and

Qualitative Analysis

(Minimum Four Mixtures)

In F.Y.B.Sc., we have learnt the qualitative analysis of single organic compound. This year

we are going to learn the separation of binary mixture and individual analysis of each

component of the binary mixture.

Systematic Qualitative Analysis of a binary mixture involves following steps :

1. Determination of the type of the binary mixture.

2. Separation of the binary mixture into two components.

3. Re-crystallization of the individual components.

4. Individual analysis of the two components :

(a) Detection of saturation/unsaturation.

(b) Detection of aliphatic/aromatic character.

(c) Detection of elements.

(d) Detection of functional group.

(e) Determination of melting/boiling point.

The nature of the binary mixture can be solid-solid, solid-liquid or liquid-liquid.

(a) Solid-solid mixture [1 Mixture] :

If the nature of the given binary mixture is solid-solid, it is necessary to find out the type

of the mixture, then separate into individual components using chemical method. The

principle behind the chemical separation is to solubalize only one of the components in one

of the reagents i.e. 10% NaHCO3, 2 N NaOH or 2N HCl. The other insoluble component is

then separated by simple filtration.

(b) Solid-liquid mixture [1 Mixture]:

S.Y.B.Sc. Practical Chemistry 60 Organic Chemistry Practicals

If the nature of the binary mixture is solid-liquid, the liquid component (volatile) is first

separated by simple distillation.

(c) Liquid-liquid mixture [2 Mixtures]:

In case of liquid-liquid mixture, the low boiling (or volatile) component is separated by

distillation.

The type of the individual components is then determined.

1. Determination of Type :

To determine the type of the solid-solid binary mixture, following scheme is followed :

Binary mixture (200 mg)

+ 10% NaHCO (1 ml)

then filter3

Filtrate

+ 2N HCl

Residue

wash with waterthen add 2 N NaOH (1 ml)shake, then filter

If ppt is obtainedacid is present

Filtrate Residue

wash with waterthen add 2 N HCl(1 ml), shake, thenfilter

Filtrate Residue

+ 2 N HCl

If ppt is obtainedphenol is present

+ 2 N NaOH

If ppt is obtainedbase is present

It is neutral

Conclusion :

The type of the given mixture is _________________.

2. (A) Separation of Solid-Solid mixture :

Six types of solid-solid mixtures can be divided into three groups :

Group I : Acid-Phenol, Acid-Base and Acid-Neutral


Group II : Phenol-Base and Phenol-Neutral

Group III : Base-Neutral

Separation of Group I (Acid-Phenol, Acid-Base or Acid-Neutral) :

Take all the mixture in a beaker and add 10% NaHCO3 (20 ml), stir well till all the

effervescence of CO2 stops (add more NaHCO3 if required). Then filter it and wash the

residue (residue is either phenol, base or neutral component) with water and re-crystallize it

from aqueous ethanol. Cool the filtrate and acidify it with 2 N HCl (check with blue litmus

paper). Filter the precipitated acid on a Buchner funnel. Dry and re-crystallize it from hot

water or aqueous ethanol.

Separation of Group II (Phenol-Base or Phenol-Neutral) :

Take all the mixture in a beaker and add to 2 N NaOH (20 ml), stir well till (add more

NaOH if required). Then filter it and wash the residue (residue is base or neutral component)

with water and re-crystallize it from aqueous ethanol. Cool the filtrate and acidify it with

2 N HCl (check with blue litmus paper). Filter the precipitated phenol on a Buchner funnel.

Dry and re-crystallize it from aqueous ethanol.

Separation of Group III (Base-Neutral) :

Take all the mixture in a beaker and add 2 N HCl (20 ml), stir well till (add more HCl if

required). Then filter it and wash the residue (neutral component) with water and

re-crystallize it from aqueous ethanol. Cool the filtrate and make it alkaline with 2 N NaOH

(check with red litmus paper). Filter the precipitated base on a Buchner funnel. Dry and

re-crystallize it from aqueous ethanol.

2. (B) Separation of Solid-Liquid or Liquid-Liquid mixture :

Solid-liquid mixture could be

(a) Solid + Non-volatile liquid or (b) Solid + Volatile liquid.

Liquid-liquid mixture can be

(a) Volatile + Non-Volatile or (b) Non-volatile − Non-volatile.

The solid-liquid mixture could appear as a heterogeneous mixture (solid is immiscible in

liquid) or homogeneous mixture (the solid is miscible in liquid). To find out whether the

homogeneous mixture is a solid-liquid or liquid-liquid mixture, place few drops of the

mixture on a watch glass and blow it with mouth, if solid is obtained then the nature of the

mixture is solid-liquid otherwise it is a liquid-liquid mixture.

Volatility of one component : To find whether one liquid component is volatile or not,

place few drops of the mixture in a sodium fusion tube and place a capillary with open

mouth dipped in the fusion tube (sealed at the top). Heat the sodium fusion tube on a

boiling water bath. If continuous bubbles start appearing then one component is volatile.


Separation of solid-liquid or liquid-liquid mixture (when one component is

volatile) : Place the mixture in a round bottom flask (25 ml) with a porcelain piece. Place the

Hickman head and the water condenser as shown in Fig. 1.

Water condenser

Water

Hickman head

Pure liquid

Distillation flask

Water bath

Binary mixture

Rubber tubepipette

Fig. 1

Heat the flask on a boiling water bath till all the volatile component is collected in the

groove of the Hickman head. Collect this volatile component with the help of a Pasteur

pipette in a vial (seal it or keep it in ice-bath so that the volatile liquid does not evaporate).

Pour the residual liquid of the round bottom flask in a beaker and allow it to evaporate, so

that any volatile component still left gets evaporated. The residue could separate as a solid

or a liquid depending on the nature of the mixture.

Note : Two volatile components mixture cannot be given, as they cannot be separated

by simple distillation method.

Separation of liquid-liquid mixture (both liquids are non-volatile) : Follow the

procedure for type determination as given in (1) for solid-solid mixture. For separation,

follow the procedure given in 2 (A).

Note : The individual components do not precipitate out during regeneration from the

filtrate (as they are liquids) but they separate out as oils which in turn are separated out

using Pasteur pipette.

Individual Analysis of Acid, Phenol, Base and Neutral Component :

1. Determination of saturation/unsaturation :

An unsaturated compound is one which has multiple bonds : double or triple bond.

These multiple bonds give characteristic electrophilic addition reactions as well as oxidation

reactions.


Note : Aromatic ring, even though contains alternate double bonds, is a saturated

compound as it does not give electrophilic addition reaction. Aromatic ring gives

electrophilic substitution reactions.

The presence of unsaturation is determined by following two tests :

(a) Br2 in CCl4 test

(b) Baeyer test or KMnO4 test.

Both these tests have to be performed with care to arrive at any conclusion.

Test Observation Inference

1. Take 1 ml Br2 in CCl4 in a

small test tube, add a

pinch of given substance

with shaking and hold a

glass rod dipped in NH3

on the mouth of the test

tube.

(a) if brown colour of bromine

does not disappear.

(b) if brown colour disappears

with precipitation of solid

in the test tube and white

fumes on the glass rod.

(c) if brown colour disappears

with no white fumes on

the glass rod.

Compound is saturated


Compound is unsaturated

C C

R

R'

R

R'

Alkene

+ Br2 C C

R

R'

R

R'

CCl4

Br

Br

Browncolour Dibromo derivative

Brown colour disappears


2. Baeyer test : Take a

pinch of given substance

in a small test tube +

10% NaHCO3 (1 ml) + dil.

KMnO4 (5 drops) with

shaking.

(a) Pink colour of KMnO4

disappears

(b) No decolourization.

Compound is unsaturated


C C

R

R'

R

R'

Alkene

+ Alk. KMnO4 C C

R

R'

R

R'

OHPink

colour Dihydroxy derivative

OH

+ MnO2 2 KOH+

Brown


Conclusion :

The given organic compound is ___________. (Saturated/Unsaturated)

2. Determination of Aliphatic/Aromatic Nature :


Heat a little substance on a

clean copper gauze

(a) No sooty flame

(b) Sooty flame

Compound is aliphatic

Compound is aromatic

Conclusion :

The given organic compound is ___________. (Aliphatic/Aromatic)

3. Detection of Elements :

Organic compounds always contain C and H, while O may or may not be present, hence written inside the brackets. It is often valuable to determine the existence of other elements N, S, halogens, as the knowledge of the elemental composition is essential for the selection of appropriate classification tests for functional groups. The detection of these elements is done by sodium fusion test (Lassaigne’s test).

Sodium Fusion Test :

1. Place a small size (wheat size) of freshly cut sodium metal in a Borosil or Pyrex glass dry test tube (10 × 80 mm).

2. Heat the test tube until the sodium metal melts and turns into a silver ball. Place small amount of the given solid compound onto the sodium metal without touching the side of the test tube. [Note : The amount of compound should be less than the amount of sodium]

Note : If the compound is liquid, add one drop using capillary. If the compound is low boiling liquid then mix small amount of powdered sucrose with it prior to its addition into the test tube.

3. Heat the lower part of the test tube gently to initiate the reaction (reaction of sodium metal with the given compound) and then to red hot for 1 min.

4. Cool the test tube and add few drops of ethanol to dissolve any unreacted sodium. Repeat adding ethanol until no further bubbles of hydrogen gas are evolved.

5. Add 2 ml of distilled water to this solution, again heat and then filter using pasture pipette fitted with a cotton plug.

Use this sodium fusion filtrate to test for the presence of sulphur, nitrogen or halogens.

Test for Sulphur :


1. Place 1 drop of sodium

fusion extract on a

porcelain tile + 1 drop

acetic acid + 2 drops 1%

lead acetate solution

Black precipitate Sulphur present

Na2S + 2CH3COOH → 2CH3COσO

⊕

Na + H2S


(CH3COO)2Pb + H2S → PbS↓ + 2CH3COOH

Black ppt

2. Place 1 drop of sodium

fusion extract on a

porcelain tile + 2 drops of

2% sodium nitroprusside

solution

Deep blue-violet color Sulphur present

Na2S + Na2[Fe(CH)5]NO → Na4[Fe(CN)5]NOS

Violet colour

Test for Nitrogen :


Place 1 drop of sodium

fusion extract on a

porcelain tile + 1 drop 2 N

NaOH. To this add a pinch

of FeSO4 crystals (green

coloured) and mix it with

glass rod. Then add 1 drop

of dil. H2SO4 or HCl.

Prussian blue or bluish green

colour

Nitrogen present

FeSO4 + 6NaCN → Na4[Fe(CN)6] + Na2SO4

3Na4[Fe(CN)6] + 2Fe2(SO4)3 → Fe4[Fe(CN)6]3 + 6Na2SO4

Prussian blue

colour

Tests for Halogens :


1. Place 20 drops of sodium

fusion extract in a small

test tube + 1 drop conc.

HNO3. Then add few

drops of AgNO3 solution.

White/Yellow precipitate

Halogens present

NaX + AgNO3 dil. HNO3

→ AgX ↓ + NaNO3


ppt.

2. Place 20 drops of sodium

fusion extract + dil HNO3

(4 drops). To this add

CHCl3 (20 drops) +

freshly prepared Cl2

water (10 drops). Shake

well and allow to stand

(i) CHCl3 layer colourless

(ii) CHCl3 layer brown

(iii) CHCl3 layer violet

Chlorine is present

Bromine is present

Iodine is present

NaBr + Cl2 CCl4

→ 2NaCl + Br2 (Brown colour)

NaI + Cl2 CCl4

→ 2NaCl + I2 (Violet colour)

Conclusion :

The given organic compound has ___________ elements. (N, S, Halogens)

Once the type and elements of given organic compound is known, organic compounds

can be classified into four groups for their functional group detection.

Group I : Compounds containing C, H, (O)

Group II : Compounds containing C, H, (O) and N

Group III : Compounds containing C, H, (O), N and S

Group IV : Compounds containing C, H, (O), Halogens

4. Detection of Functional Groups :

Group I : Compounds containing C, H, (O)

If the type is Acid [Check for –COOH, −OH, −CO−, −COOR functional groups]

If the type is Phenol [Check for –OH, −CO−, −COOR functional groups]

If the compound is Neutral [Check for –CO−, −COOR, carbohydrate functional groups]

Group II : Compounds containing C, H, (O) and N

If the type is Acid [Check for –COOH, −OH, −CO−, −COOR, −NH2 (−NHR, −NR2), −NO2,

−CONH2, −NHCOCH3 functional groups]

If the type is Phenol [Check for –OH, –CO–, –COOR, –NH2 (–NHR, –NR2), –NO2, –CONH2,

–NHCOCH3 functional groups]

If the type is Base [Check for –CO–, –COOR, –NH2 (–NHR, –NR2), –NO2 functional groups]

If the compound is Neutral [Check for –CO–, –COOR, –NO2, –CONH2, –NHCOCH3

functional groups]

Group III : Compounds containing C, H, (O), N and S


If the type is Acid [Check for –COOH, –OH, –CO–, –COOR, –NH2 (–NHR, –NR2), –NO2,

–CONH2, –NHCOCH3 functional groups]

If the type is Phenol [Check for –OH, –CO–, –COOR, –NH2 (–NHR, –NR2), –NO2, –CONH2,

–NHCOCH3 functional groups]

If the type is Base [Check for –CO–, –COOR, –NH2 (–NHR, –NR2), –NO2 functional groups]

If the compound is Neutral [Check for –CO–, –COOR, –NO2, –CONH2, –NHCOCH3

functional groups].

Note : There are no separate tests for sulphur containing functional groups. Sulphur

can exist as –SO3H (sulphonic acid) or –CS–NH2 (thioamide). These functional groups

give the same test as given by –COOH and –CO–NH2 respectively.

Group IV : Compounds containing C, H, (O) and Halogens

If the type is Acid [Check for –COOH, –OH, –CO–, –COOR, –X functional groups]

If the type is Phenol [Check for –OH, –CO–, –COOR, –X functional groups]

If the compound is Neutral [Check for –CO–, –COOR, –X functional groups]

Individual functional group tests :


1. For acid (–COOH)

group :

Take 1 ml of 10%

NaHCO3 in a small test

tube, add a pinch of

given substance without

shaking

Substance dissolves with

effervescence of CO2 and

reappears by adding conc. HCl

Acid is present and

confirmed.

C

O

R OH

+

C

O

R O Na

+

Acid Water soluble salt

NaHCO3 CO2

HClNaCl+

Acid ppt.

C

O

R OH+ H O2


2. For phenolic (–OH)

group :

Take a pinch of the given

compound + 1 ml of

Blue / green / violet colour Phenol is confirmed.


alcohol + 2-3 drops of

neutral FeCl3.

Ar

O

H

Cl

Fe

Cl

Cl

O

H

Ar

O

Ar

H3 Ar-OH + FeCl3 Blue/Green/ Violetcolour


3. Carbonyl (–CO–) group :


compound + 1 ml of

alcohol + 2-3 drops of

2.4-DNP reagent

Yellow or red precipitate Carbonyl group confirmed

C

O

R R'

+

Aldehydeor Ketone

O N2 NH-NH2 O N2 N-N = C

NO2 NO2

R

R'

2, 4-DNP Yellow or Red ppt.

H

Tests for Aldehydes :

(a) Tollen’s test :

Given compound (5-6

drops) + Tollen’s reagent

(1 ml) in a small test tube.

Warm gently on water bath

for 5 min. (Note : The test

tube should be washed with

conc. HNO3 and distilled

water prior to this test).

Silver gets deposited on the

inner walls of the tube (silver

mirror formation)

Aldehyde group present


C

O

Ar H

+

Aldehyde

2 [Ag(NH )] OH3 2

OHAr-COONH4 + H O2 + 2Ag + 3 NH3

Tollen'sreagent

Blackppt.

(b) Fehling’s test :

Given compound (5-6

drops) + Fehling’s solutions

A and B (1 ml) in a small test

tube. Warm gently on water

bath for 2 min.

Red precipitate of Cu2O

Aldehyde group present

(Ar/R–CHO)

C

O

Ar H

+

Aldehyde

2 Cu(OH) citrate2 Ar-COO Na citrate + Cu O2

Fehling'sreagent

Redppt.

-3 H O2

2 NaOH

(c) Schiff’s reagent test :

Given compound (5-6

drops) + Schiff’s reagent

(0.5 ml) in a small test tube.

Pink colour develops in the

organic layer

Aromatic aldehyde present

(Ar–CHO)

HN

NH2

SO H3C O

R

H

Aldehyde

+ 2

Schiff's reagent Pink adduct

SOHO

SOHO

NH

NH2

S

OH

NH

+HSO3

O HR

O

S

OH

HN

OH

R

O

If aldehyde group is absent assume that the compound has ketone group.

(d) Iodoform test :

Given compound + NaOH

(2 ml) + I2 solution (5 ml)

and then heat

Yellow precipitate

Methyl ketone present

(CH3–CO–)


C

O

H C3 R

+

C

O

R O Na

Methyl ketone Iodoform

NaOHI2

CHI3 +


4. Ester (–COOR) group :

(a) Phenolphthalein test : To a very dilute NaOH solution add 1 drop of phenolphthalein to obtain very faint pink color. (Note: If the pink colour is very dark, dilute it with water till faint pink colour is obtained). To this add 2-3 drops of the given compound and heat it.

Pink colour disappears

Ester group is present

C

O

OR'

+

C

O

R O Na

Ester

Colourless

[NaOH + Phenolphthalein] + R'-OH + Phenolphthalein

R Pink colour

D

(b) Hydroxamic test :

Substance + alcoholic

NH2OH.HCl + 10% NaOH

(till alkaline). Heat to boil,

cool and neutralize with

dil. HCl. Then add 1-2 drops

of FeCl3.

Red-violet colour develops

Ester group is confirmed

(i) 3RCOOR' + 3 NH2OH.HCl NaOH

→ 3 R−CO−NH−OH + R'−OH + NaCl + H2O

3 R CO NHOH + FeCl3

RC

O

HNO

Fe + 3HCl

Red colour

(ii)

3


5. Carbohydrate group :

(a) Molish test :

Violet red colouration at the

Carbohydrate present


Compound + 3-4 drops of

α-naphthol in alcohol. Shake

and then add 5 drops of conc.

H2SO4 from the walls of the

tube.

junction of the two layers

OH

CHO

(CHOH)4

CH OH2

+H SO2 4

Glucose

OHC

(CHOH)4

CH OH2

a-Naphthol Violet-red colour

(b) Compound +

conc. H2SO4 :

Charring of the substance Carbohydrate present

Cn(H2O)n + H2SO4 → nC + nH2O

Charcoal

Hydrocarbons (Water insoluble liquid or solid)

If none of the above functional group is present then the given compound is

hydrocarbon.


6. Nitro (–NO2) group :

(a) Neutral reduction

test : Compound + 1 ml of

alcohol + 6 drops CaCl2 +

pinch of Zn dust. Heat to

boiling for at least 4 mins.

and then filter into Tollen’s

reagent.

Black or grey precipitate

Nitro group present.

NO2

CaCl /Zn2

NHOH

Tollen's

NO

reagent+ Ag + NH3 + H O2

Blackppt.

Hydroxylamine

(b) Ferrous hydroxide

test : Compound + freshly

prepared 5% ferrous

ammonium sulphate

solution + 1 drop of dil.

Red-brown precipitate

Nitro group is confirmed


H2SO4 + excess of KOH.

Shake the tube quickly

Ar-NO2 + 6 Fe(OH)2 + 4 H2O Ar-NH2 + 6 Fe(OH)3

BrownBlue

colour


7. Test for Primary/

Secondary/Tertiary

amines :


compound + few drops of

conc. H2SO4. Add excess of

NaNO2 solution.

(i) Yellow solid appears.

(ii) Red colouration which

turns green solid on

addition of NaOH

(iii) No colouration but on

addition of β-naphthol in

NaOH gives orange

dyestuff

Secondary amine present

Tertiary amine present and

confirmed

Primary amine present and

confirmed

(i) C H NHR6 5

Secondaryamine

HONOC H — N — N == O6 5

RYellow colour oil

or solid

C H N6 5

R

R

(ii)(i) HONO

(ii) NaOH

R

R

N C H N O6 4

Green ppt. (para-nitroso compound)

C H NH6 5 2

HONOC H N N6 5

+ b - Naphthol

in NaOH

OH

N N C H6 5

Orange dye

(iii)

Diazonium ion


8. Anilide (−−−−NHCOCH3)

group :

Given compound + few

drops of conc. HCl. Boil for

3 min, then cool and add

excess of NaNO2 solution.

Then add β-naphthol in

NaOH

Orange dyestuff

Anilide present

(i) Ar – NH – CO – R + HClD

Ar — NH + R — COONa2

Primary amine

– +


Ar NH2HONO

Ar N N+ b - Naphthol

in NaOH

OH

N N Ar

Orange dye

(ii)

Primary amine Diazonium ion

9. Amide/Thioamide

(–CO–NH2/–CSNH2)

group :

Compound + NaOH (1 ml)

and boil well.

Evolution of NH3 which turns

moist turmeric paper red

Amide present

RC

NH2

+ NaOHD

X

RC

O Na+ NH3

X = O AmideX = S Thioamide

X = OX = S

+-

X


10. Halogen (–X) group :

Compound + NaOH (1 ml)

and boil well. Neutralize

with HNO3 and add AgNO3

solution.

(i) Precipitate

(ii) No precipitate

Aliphatic halogen

compound

Aromatic halogen

compound

(i) R-X + NaOH → R-OH + NaX

(ii) NaX + AgNO3 HNO3→ AgX↓ + NaNO3

X = Cl White

= Br Yellowish white

= I Yellow/Violet

5. Determination of Melting/Boiling Point :

(A) Melting Point :

Melting point is defined as the temperature at which the thermal energy of the particles

overcomes the inter crystalline forces of attraction which hold them together.

Procedure for determining the melting point :

1. Fill the capillary, sealed at one end, with powdered sample by gently tapping the

capillary on the table.


2. Repeat this procedure till sufficient quantity is present in the capillary.

3. Attach the capillary to the thermometer with the help of a thread such that the lower

end of the capillary and the thermometer bulb are at the same level.

4. Insert the thermometer in the Thiele’s tube such that only the thermometer bulb is

dipped inside the paraffin oil.

5. Heat the lower side arm of the Thiele’s tube with the help of the burner (blue flame)

slowly back and forth. If the heating is too fast remove the burner for few seconds,

and then resume heating.

6. The rate of heating should be low near the melting point (about 1°C/min.).

7. Record the exact melting point of the compound.

Sealed capillary(i)

Sodium fusiontube with liquid

compound(ii)

Capillary insertedwith open end

dipped in the liquid(iii)

Sodium fusiontube attached to the

thermometer(iv)

Fig. 2

(B) Boiling Point :

Boiling point is defined as the temperature at which the vapour pressure of the liquid

becomes equal to the atmospheric pressure.

Procedure for determining the boiling point :

1. Take a sodium fusion tube and add few drops of the liquid sample into it.

2. Seal one end of the capillary and insert it in the fusion tube with its open end dipped

in the liquid.

3. Tie this assembly to the thermometer with the help of a thread such that the lower

end of the sodium fusion tube and the thermometer bulb are at the same level.

4. Insert the thermometer in the Thiele’s tube such that only the thermometer bulb is

dipped inside the paraffin oil.


5. Heat the lower side arm of the Thiele’s tube with the help of the burner (blue flame)

slowly back and forth. If the heating is too fast remove the burner for few seconds,

and then resume heating.

6. Stop heating and record the temperature when vigorous, continuous bubbling

starts. (If the bubbling is not continuous then continue heating). Continue looking at

the fusion tube till the last bubble comes out. At this point the liquid rises in the

capillary. Record this temperature as the boiling point.

7. Repeat this procedure till constant boiling point is recorded.

Thermometerattached with capillary

Thiele's tube

Paraffin oil

Burner

Sealed capillary(ii)

Sealed capillarywith compound

(iii)

Sealed capillaryattached to thethermometer

(iv)

Capillary(i)

Fig. 3

Result Table :

Sr. No. Description Component 1 Component 2


1. Nature of the given binary mixture

2. Type of the given substance

3. Aliphatic/Aromatic

4. Saturation/Unsaturation

5. Elements

6. Functional groups

7. Melting / boiling point

VIVA VOCE

1. What is organic qualitative analysis ?

2. What are the four main types of organic compounds ?

3. In solid - solid binary mixture, Neutral-neutral type is not given. Why ?

4. What is the principle of binary mixture separation ?

5. How will you separate the following mixtures ?

(a) Acid-phenol (b) Acid-base

(c) Acid-neutral (d) Phenol-base

(e) Phenol-base (f) Base-neutral

6. Why acid-phenol mixture is not separated by using 10% NaOH solution ?

7. After separation of mixture why is it essential to purify the individual components ?

8. What do you mean by volatile liquid ? How is it detected ?

9. NaCl and CHCl3 both contain Cl but AgNO3 gives precipitate with NaCl but not with

CHCl3. Why ?

10. Why do aromatic compounds burn with sooty flame ?

11. How will you detect the presence of S, N and halogen in an organic compound ?

12. What happens to the organic compound when it is fused with sodium metal ?

13. How will you separate solid-liquid mixture ?

14. Why sodium metal is kept under kerosene ?

15. Can potassium, lithium or magnesium be used instead of sodium in making

Lassaigne’s reagent ?

16. What are the functional group tests for the following groups :

(a) Carboxylic (b) Phenolic


(c) Primary amine (d) Secondary amine

(e) Tertiary amine (f) Ester

(g) Anilide (h) Amide

(i) Nitro (j) Carbonyl

(k) Aldehyde (l) Ketone

(m) Methyl ketone.

− − −

(b) Organic Preparations

(Any Two)

Experiment No. 1

PHTHALIC ANHYDRIDE TO PHTHALIMIDE

AIM :

To prepare phthalimide from phthalic anhydride.

PRINCIPLE :

This is an example of conversion of cyclic anhydride to cyclic amide (imide). When the

cyclic anhydride such as phthalic anhydride is treated with ammonia or urea, it undergoes

the substitution of O by N−H to form the cyclic imide i.e. phthalimide.

REACTIONS :

O

O

O

NH CONH2 2

Phthalic anhydride

NH

O

O

Phthalimide

CHEMICALS :

(i) Phthalic anhydride : 1 gm

(ii) Urea : 0.5 gm

(iii) Methanol : 5 ml

(iv) Water : 10 ml

PROCEDURE :

Weigh 1 gm of phthalic anhydride and 0.5 gm of urea crystals in a dry conical flask

(100 ml). Heat this mixture on a sand bath. When the temperature reaches to 140°C, the


reaction mixture melts and effervescence commences and gradually increases. After

10 minutes the reaction mixture solidifies. Now stop heating and cool the flask to room

temperature. Add 10 ml of distilled water, stir vigorously. The white precipitate of

phthalimide separates out. Filter it at suction pump. Wash the crystals with cold water. Dry

the crystals and take the weight of the crystals.

Recrystallisation : Take all crude phthalimide in a conical flask, add 5 ml of methanol

and heat on a water bath to dissolve the crystals. Filter the hot solution through fluted filter

paper. Cool the filtrate in ice. White crystals of phthalimide are formed. Filter these crystals

on the suction pump. Wash with cold methanol and dry the crystals well. Find out the m.p.

The m.p. of crystalline phthalimide is 234°C.

TLC : Dissolve small quantities of phthalic anhydride and phthalimide in acetone in

sodium fusion tubes separately. Use this solution for spotting on TLC plate. Run the TLC in

toluene as solvent and calculate the Rf values for phthalic anhydride and phthalimide

(product).

CALCULATIONS :

1. Theoretical yield :

Phthalic anhydride = Phthalimide

C8H4O3 = C8H5O2N

148 gm = 147 gm

1 gm = 0.999 gm (1 gm)

2. Practical yield = 'A' gm

3. % Practical yield, 1 gm of product = 100% yield

A gm of product = A × 100

1.0 = % of phthalimide

RESULTS :

1. M.P. of phthalimide = ………°C

2. Practical yield of phthalimide = ……… gm

3. % Practical yield of phthalimide = ……… %

4. Rf value of phthalic anhydride = ………

5. Rf value of phthalimide = ………

6. Solvent used for TLC = Toluene

✍ ✍ ✍

Experiment No. 2

GLUCOSE TO GLUCOSAZONE


AIM :

To prepare glucosazone from glucose.

PRINCIPLE :

Osazone is a solid derivative for reducing sugars (carbohydrate). The sugars i.e.

carbohydrates are the polyhydroxy aldehydes or ketones. When these sugars e.g. D-Glucose,

D-Fructose and D-Mannose are treated with one mole of phenyl hydrazine then it forms

phenyl hydrazone derivatives. When these sugars are treated with excess of phenyl

hydrazine, it forms the osazone derivatives.

This is one of the tests for detection of reducing sugars.

REACTIONS :

(1) Formation of glucose phenyl hydrazone :

(2) Oxidation of CHOH to C = O group :

C H NHNH6 5 2

H C N NHC H6 5

CHOH

(CHOH)3

CH OH2

Glucose phenyl hydrazone

H C NNHC H6 5

C O

(CHOH)3

CH OH2

C H NH + NH6 5 2 3+

(3)

C H NHNH6 5 2

H C NNHC H6 5

C O

(CHOH)3

CH OH2

HC N•NHC H6 5

C NNHC H6 5

(CHOH)3

CH OH2

Glucosazone

CHEMICALS :

(i) D-glucose : 0.5 gm

(ii) Phenyl hydrazine : 1.5 ml or phenyl hydrazine hydrochloride : 1.0 gm


(iii) Sodium acetate : 1.0 gm

(iv) Glacial acetic acid : 2.0 ml

(v) Ethanol : 5 ml

(vi) Distilled water

PROCEDURE :

Method - I : Dissolve 0.5 gm of sugar (D-glucose) in about 5 ml of water in hard glass

tube. Add to it 1.5 ml phenyl hydrazine dissolved in 2.0 ml glacial acetic acid. Shake the

mixture well. Loosely cork the tube and stand it in a boiling water bath for about 10 minutes.

Yellowish orange precipitate of osazone separates out. Filter it on suction pump, wash with

water and recrystallize it from water or ethanol and record the yield and m.p. The m.p. of

glucosazone is 204 − 5°C.

Method - II : Dissolve 0.5 gm of sugar (D-glucose) in about 10 ml water in hard glass

tube, add 1.0 gm of phenyl hydrazine hydrochloride and 1.0 gm of sodium acetate. Warm

gently to dissolve. Filter if necessary over a cotton plug. Place the content in boiling water

bath. Crystals of osazone appear within 10 -15 minutes.

(D-Mannose and D-Fructose give the same osazone.)

TLC : Dissolve small amount of sugar (D-glucose), crystalline osazone (product) in small

sodium fusion tubes separately. Use them for spotting on TLC plates. Run the TLC in toluene

as a solvent and record the Rf values for glucose (starting material) and glucosazone

(product).

CALCULATIONS :

1. Theoretical yield : Sugar = Osazone

(D-Glucose, D-Mannose or D-Fructose)

C6H12O6 = C18H22O4N4

180 gm = 328 gm

0.5 gm = 0.994 gm

2. Practical yield of glucosazone = 'A' gm

3. % Practical yield : 0.994 gm of osazone = 100% yield

A gm of osazone = A × 100

0.994 gm

= 'B' gm %

RESULTS :


1. M.P. of crystalline osazone = ……… °C

2. Practical yield of osazone = ……… gm

3. % Practical yield of osazone = ……… %

4. Rf value of sugar = ………

5. Rf value of osazone = ………


− − −

Experiment No. 3

ACETANILIDE TO p-BROMO ACETANILIDE

AIM :

To prepare p-bromo acetanilide from acetanilide.

PRINCIPLE :

Acetanilide undergoes electrophilic substitution on reaction with bromine in acetic acid.

The anilide group being activating group directs the incoming electrophile at o- and

p-positions. However in case of acetanilide, the substitution takes place only at para position

as the ortho positions are sterically crowded due to large anilide group.

REACTION :

O

CCH3N

H

Acetanilide

Br2

AcOH O

CCH3N

H

p-Bromoacetanilide

Br

CHEMICALS :

(i) Acetanilide : 1.0 gm

(ii) Bromine in AcOH (25% w/v) : 5.0 ml

(iii) Acetic acid : 1 ml

PROCEDURE :

Dissolve 1.0 gm of acetanilide in 1 ml acetic acid in a round bottom flask (50 ml). Add to

it 5.0 ml of bromine solution in acetic acid (25% w/v). Shake the mixture for one hour and

then pour it in water of 20 ml. Filter the solid product obtained on suction pump. Wash with

cold water. Dry the crude product and record the practical yield. The practical yield is

approximately 1.4 gm.


Crystallisation : Take all the crude product of p-bromo acetanilide in a conical flask.

Add to it 5 ml of ethanol. Add 1 ml porcelain piece and boil it in water bath to dissolve the

crude product. Filter this hot solution through a fluted filter paper in a beaker. Cool the

filtrate at room temperature. Fine crystals of p-bromo acetanilide are formed. Filter these

crystals on a buchner funnel, wash with cold water. Dry the crystals and take the m.p. of

p-bromo acetanilide. The m.p. of pure p-bromo acetanilide is 167°C.

TLC : Dissolve small quantities of acetanilide and p-bromo acetanilide in sodium fusion

tubes separately in suitable solvents and use these solutions for TLC. Spot the two spots of

these solutions on a TLC plate and run the TLC in toluene as a solvent. Find out Rf values for

acetanilide and p-bromo acetanilide.

CALCULATIONS :

1. Theoretical yield : Acetanilide = p-Bromo acetanilide

C8H9NO = C8H8BrNO

135 gm = 214 gm

1 gm = 1.58 gm

2. Practical yield of p-Bromo acetanilide = 'A' gm

3. % Practical yield :

1.305 gm of p-Bromo acetanilide = 100% yield

'A' gm of p-Bromo acetanilide = A × 100

1.58

= 'B' % of practical yield

RESULTS :

1. M.P. of crystalline p-Bromo acetanilide = ……°C

2. Practical yield of p-Bromo acetanilide 'A' = …… gm

3. % Practical yield of p-Bromo acetanilide 'B' = …… gm

4. Rf value of acetanilide = ……

5. Rf value of p-Bromo acetanilide = ……


✍ ✍ ✍

Experiment No. 4

BENZALDEHYDE TO DIBENZYLIDENE ACETONE


AIM:

To prepare dibenzylidene acetone from benzaldehyde.

PRINCIPLE:

Aromatic aldehydes condense with aliphatic or mixed alkyl aryl ketones in the presence

of aqueous alkali to form α, β-unsaturated ketones. This reaction is called as Claisen-Schmidt

reaction. The first step is a condensation of the aldol type involving the nucleophilic addition

of the carbanion derived from the methyl ketone to the carbonyl-carbon of the aromatic

aldehyde. In the second step the dehydration of hydroxy ketone results in the formation of

conjugated α, β-unsaturated ketone spontaneously. When two moles of aromatic aldehydes

are used, condensations occur on both sides of the ketone.

REACTION:

PhC

H

aq. NaOH

Benzaldehyde

O

H C3

CCH3

Acetone

O

+

PhC

C

H

HCH

C

O

CH

Ph

Dibenzylidene acetone

2

CHEMICALS:

(i) NaOH : 1.0 gm

(ii) Ethanol : 0.8 ml

(iii) Pure benzaldehyde : 1.0 ml

(iv) Acetone : 0.4 ml

(v) Distilled water : 10 ml

PROCEDURE:

Take 1.0 gm of NaOH in distilled water (10 ml) and ethanol (8 ml) in a round bottom

flask (50 ml). Add to it one half the quantity of previously prepared mixture of pure

benzaldehyde (1.0 ml) and acetone (0.4 ml) and stir vigorously. A flocculant precipitate

forms in 2-3 minutes. After fifteen minutes add the remaining benzaldehyde-acetone

mixture and continue stirring for further 30 min. Filter the product on suction pump and

wash with cold water. Dry the crude product and record the practical yield. The practical

yield is approximately 1.0 gm.

Crystallization : Take all the crude product of dibenzylidene acetone in a conical flask.

Add to it 10 ml of ethanol or ethyl acetate. Add 1 porcelain piece and boil it in water bath to

dissolve the crude product. Filter this hot solution through a fluted filter paper in a beaker.

Cool the filtrate at room temperature. Fine crystals of dibenzylidene acetone are formed.

Filter these crystals on a buchner funnel, wash with cold ethanol. Dry the crystals and take

the m.p. of dibenzylidene acetone. The m.p. of pure product is 122°C.


TLC : Dissolve small quantities of benzaldehyde and dibenzylidene acetone in a sodium

fusion tubes separately in suitable solvents and use these solutions for TLC. Spot the two

spots of these solutions on a TLC plate and run the TLC in toluene as a solvent. Find out Rf

values for benzaldehyde and dibenzylidene acetone.

(Follow the general procedure given to run the TLC.)

CALCULATIONS :

1. Theoretical yield :

2 equivalent Benzaldehyde = Acetone = Dibenzylidene acetone

2 C7H6O = C3H6O = C17H14O

2 × 106 gm = 58 gm = 234 gm

1 gm benzaldehyde = 1.1 gm dibenzylidene acetone

2. Practical yield of dibenzylidene acetone = 'A' gm

3. % Practical yield :

1.1 gm of dibenzylidene acetone = 100% yield

'A' gm of dibenzylidene acetone = A × 100

1.1

= 'B' % of practical yield

RESULTS :

1. M.P. of crystalline dibenzylidene acetone = …… °C

2. Practical yield of dibenzylidene acetone 'A' = …… gm

3. % Practical yield of dibenzylidene acetone 'B' = …… gm

4. Rf value of benzaldehyde = ……

5. Rf value of dibenzylidene acetone = ……


− − −

(85)

SECTION - D

ANALYTICAL CHEMISTRY PRACTICALS

(Any Five)

Experiment No. 1

ESTIMATION OF SODIUM CARBONATE

AIM :

(a) Preparation of standard solution of sodium carbonate (0.1 M) and standardization of

approximate 0.1 M hydrochloric acid solution.

(b) Determination of the Na2CO3 content of the washing soda.

THEORY :

Na2CO3 content of commercial sample of washing soda can be determined by

volumetric method. The process of determining the concentration of unknown sample by

allowing it to react with a solution of known concentration is called volumetric analysis.

The solution whose concentration is accurately known is called standard solution. The

concentration of the solution used in volumetric analysis is expressed either in terms of

(i) Normality or (ii) Molarity.

Na2CO3 content of commercial sample of washing soda can be found out by titrating the

solution of sample with standard solution of hydrochloric acid. However hydrochloric acid is

not a primary standard substance and hence its standard solution cannot be prepared. Exact

concentration (Normality/Molarity) of hydrochloric acid thus can be found out by

standardization with a primary standard substance like sodium carbonate.

Two molecules of HCl solution supply two atoms of hydrogen for neutralization purpose

as shown below.

2HCl + Na2CO3 → 2NaCl + CO2↑ + H2O

This means that one molecule of Na2CO3 can react with two molecules of HCl, so its

equivalent weight is equal to molecular weight divided by two. Thus, by taking the

equivalent quantity of standard sodium carbonate solution and titrating it against HCl

solution using methyl orange indicator, exact normality/molarity of HCl solution can be

found out.

S.Y.B.Sc. Practical Chemistry 86 Analytical Chemistry Practicals

Washing soda is mainly comprised of Na2CO3. The quantity of Na2CO3 present in

washing soda can be found out by dissolving the sample in water and fixing it to a known

volume and a aliquat of resulting solution is then titrated against standardised solution of

hydrochloric acid. The quantity of Na2CO3 is then calculated using equivalent system.

CHEMICALS :

(a) A sample of washing soda.

(b) HCl solution (0.1 N, approximately).

(c) Pure AR grade Na2CO3

(d) Methyl orange indicator solution.

PROCEDURE :

This experiment is performed in following three parts.

Part - (A) : Preparation of standard solution of sodium carbonate (0.1 N) :

1000 ml 1 N sodium carbonate solution contains 53.0 g Na2CO3

1000 ml 0.1 N sodium carbonate solution contains 5.30 g Na2CO3

100 ml 0.1 N sodium carbonate solution contains 0.53 g Na2CO3.

1. Weigh accurately 0.53 g of AR grade sodium carbonate on watch glass.

2. Transfer it in a clean beaker and rinse the watch glass with distilled water.

3. Add 25 - 30 ml distilled water and stir the solution with glass rod to dissolve the

crystals. Now pour this clear solution into 100 ml volumetric flask.

4. Take little distilled water into the same beaker, rinse the glass rod with small quantity

of distilled water and transfer the contents of beaker into the volumetric flask.

Finally, dilute the solution with distilled water upto the mark.

5. Take out all the solution in a clean beaker to make homogeneous solution. This is

0.1 N sodium carbonate solution.

6. Use this solution for the standardisation of given HCl solution.

Part - (B) : Standardisation of HCl solution :

1. Fill the burette No. 1 with 0.1 N (approx.) HCl solution.

2. Fill the burette No. 2 with standard 0.1 N sodium carbonate solution.

3. Take 9.0 ml of sodium carbonate solution from burette No. 2 in a 100 ml conical

flask. Call this volume as V2 ml.

4. Add to this solution 3 - 4 drops of methyl orange indicator solution. The solution will

have yellow colour.


5. Titrate this solution of sodium carbonate against HCl solution added slowly (drop by

drop) from burette No. 1 with constant shaking of the flask.

6. Continue the addition of HCl solution till the solution in the conical flask becomes

permanently orange. Record the burette reading and call this as x1 ml.

7. Add to the same conical flask 1.0 ml solution of 0.1 N sodium carbonate from

burette No. 2. Yellow colour reappears.

8. Add HCl solution from burette No. 1 with constant shaking of the flask till the

solution becomes permanently orange. Record the burette reading and call this as x2

ml.

9. To the same solution in the conical flask add 1.0 ml solution of 0.1 N sodium

carbonate from burette No. 2. Yellow colour reappears.


solution becomes permanently orange. Record the burette reading and call this as x3

ml.

Part - (C) : Estimation of sodium carbonate :

1. Accurately weigh 1.3 to 1.4 g of sample of washing soda on a watch glass. Call this

weight as w grams.


3. Add about 50 ml distilled water and stir the solution with glass rod to dissolve the

solid. If solid dissolves completely pour this clear solution into 100 ml volumetric

flask. If sample do not dissolve completely, filter it through glass funnel with cotton

plug to remove the undissolved solid.


of distilled water and transfer the contents of beaker into the volumetric flask.


5. Take out all the solution into a clean beaker to make homogeneous solution. This is

solution of sample of washing soda.

6. Fill the burette No. 1 with standardised HCl solution.

7. Clean the burette No. 2 and fill it with diluted solution of sample of washing soda.

8. Take 9.0 ml of diluted solution of washing soda from burette No. 2 in a 100 ml

conical flask. Call this volume as V2 ml.

9. Add to this solution 3 - 4 drops of methyl orange indicator solution. The solution will

have yellow colour.

10. Titrate this solution of washing soda against standardised solution of HCl added

slowly (drop by drop) from burette No. 1 with constant shaking of flask.


11. Continue the addition of HCl solution from burette No. 1 till the solution in the

conical flask becomes permanently orange. Record the burette reading and call this

as Y1 ml.

12. Add to the same conical flask 1.0 ml solution of washing soda from burette No. 2.

Yellow colour reappears.


solution becomes permanently orange. Record the burette reading and call this as

Y2 ml.

14. To the same solution in the conical flask add 1.0 ml solution of washing soda from

burette No. 2. Yellow colour reappears.


solution becomes permanently orange. Record the burette reading and call this as

Y3 ml.

Observation Table :

Part - (B) : Standardisation of HCl solution

To Find : Exact normality of HCl solution

Burette No. 1 : HCl solution (approx. 0.1 N)

Burette No. 2 : Na2CO3 solution (exact 0.1 N)

Indicator : Methyl orange

End point : Yellow to orange

Reaction : 2HCl + Na2CO3 → 2NaCl + CO2↑ + H2O

Obs. No. Burette - 1

x ml

Burette - 2

V2 ml Normality N =

0.1 ×××× V2

x

1. x1 = 9.0 N1 =

0.1 × 9

x1

2. x2 = 9 + 1 = 10.0 N2 =

0.1 × 10

x2

3. x3 = 10 + 1 = 11.0 N3 =

0.1 × 11

x3

Mean Normality = N4 = N1 + N2 + N3

3 = ……… N


Part - (C) : Estimation of sodium carbonate

To find : Quantity of Na2CO3 in the washing soda sample

Burette No. 1 : Standardised HCl solution

Burette No. 2 : Washing soda solution

Indicator : Methyl orange

End point : Yellow to orange

Reaction : Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

Obs. No. Burette - 1, y ml Burette - 2, V2 ml

1. y1 = 9.0 ml

2. y2 = 9 + 1 = 10.0 ml

3. y3 = 10 + 1 = 11.0 ml

Mean burette - 1 reading

= y1 + y2 + y3

3 = y = … ml

Mean burette - 2 reading

= 9 + 10 + 11

3 = 10.0 ml

CALCULATIONS :

From the knowledge of chemical reaction between HCl and Na2CO3, it is observed that

one molecule of Na2CO3 require 2 molecules of HCl for the complete neutralisation. Hence,

equivalent weight of Na2CO3 is 1

2 of its molecular weight, that is,

106

2 = 53.0 g. Therefore,

1000 ml 1 N HCl = 53.0 g of Na2CO3

∴ Y ml N4 HCl = ?

= Y × N4 × 53.0

1000 = A g of Na2CO3

Here, Y = Mean burette - 1 reading, N4 = Mean normality of HCl solution. Hence, 10 ml

diluted solution of washing soda contains Ag of Na2CO3.

∴ 100 ml diluted solution of washing soda would contain A × 10 = B g of Na2CO3.

Now, w g of washing soda contains B g of Na2CO3

∴ 100 g of washing soda would contain

100 × B

w = C … % of Na2CO3

Here w = weight of washing soda taken for the analysis.


RESULT :

1. 10 ml 0.1 N sodium carbonate solution required = X2 ml of HCl

2. Exact normality of HCl solution = N4 = … N

3. 10 ml diluted washing soda solution required = Y2 ml of HCl

4. Amount of Na2CO3 in the washing soda sample = B …… g

5. % of Na2CO3 in the washing soda = C …… %

QUESTIONS

1. What is volumetric analysis ?

2. What is normal and molar solution ?

3. What do you mean by equivalent weight ?

4. What is standard solution ?

5. What are primary and secondary standard substances ?

6. What do you mean by standardisation ?

7. What is the composition of washing soda ?

8. Which other indicators can be used for this titration ?

− − −

Experiment No. 2

DETERMINATION OF CALCIUM IN THE PRESENCE OF MAGNESIUM

AIM :

(a) Preparation of standard solution of zinc sulphate (0.01 M) and standardization of

approximate 0.01 M EDTA solution.

(b) Determination of calcium in the presence of magnesium.

THEORY :

Many natural samples like hard water, limestone, dolomite etc. contain both calcium and

magnesium. Determination of one in the presence of another is then always a problem.

Ethylene Diamine Tetra Acetic acid, EDTA, is the best complex forming agent that form

complexes with 40 different metal ions, but only a particular pH is required for the complex

formation. Calcium can be directly titrated against EDTA solution at pH = 12 using methyl

thymol blue indicator. Above pH = 10 magnesium also form complex with EDTA and thus

interferes in the titration. To estimate calcium in the presence of magnesium, masking or

blocking of magnesium is required. There are different methods of blocking magnesium that

allow estimation of calcium in the presence of magnesium.


Disodium dihydrogen ethylene diamine tetra acetate of AR quality is available

commercially, but this may contain a trace of moisture hence it should not be used as a

primary standard. Solutions of EDTA of the concentration like 0.1M, 0.05M and 0.01M are

suitable for most titrations. The solution must be standardised by titration with solution of

zinc sulphate at pH = 10 using eriochrome black T indicator.

CHEMICALS :

(a) EDTA solution (0.01 M, approximately)

(b) Pure AR grade ZnSO4 ⋅ 7H2O

(c) Eriochrome black T indicator

(d) Sample solution containing calcium and magnesium

(e) Patton and Reeder’s indicator powder

(f) Buffer pH = 10

(g) Solid NaOH pellets.

PROCEDURE :

This experiment is performed in following three parts.

Part - (A) : Preparation of standard solution of zinc sulphate (0.01 M) :

1000 ml 1.0 M zinc sulphate solution contains 287.54 g ZnSO4 ⋅ 7H2O




1. Weigh accurately 0.287 g of AR grade zinc sulphate on watch glass.

2. Transfer the solid in a clean beaker and rinse the watch glass with distilled water.

3. Add 25-30 ml distilled water and stir the solution with glass rod to dissolve the

crystals. Now pour this clear solution into 100 ml volumetric flask.


of distilled water and transfer the contents of the beaker into the volumetric flask.



0.01 M zinc sulphate solution.

6. Use this solution for the standardisation of given EDTA solution.

Part - (B) : Standardisation of EDTA solution :

1. Fill the burette No. 1 with 0.01 M (approx.) EDTA solution.

2. Fill the burette No. 2 with standard 0.01 M zinc sulphate solution.


3. Take 10.0 ml of zinc sulphate solution from burette No. 2 in a 100 ml conical flask.

Call this volume as V2 ml.

4. Add to this solution 5.0 ml buffer solution of pH = 10, 3 - 4 drops of eriochrome

black T indicator solution. The solution will have wine red colour.

5. Titrate this solution against EDTA solution added slowly from burette No. 1 with

constant shaking of the flask.

6. Continue the addition of EDTA solution till the solution in the conical flask becomes

permanently blue. Record the burette reading and call this as V1 ml.

7. Take two more readings following steps 3 to 6 and find constant burette reading.

Call this reading as V1 ml.

8. From the knowledge of volumes of zinc sulphate and EDTA solution, that is, V2 and

V1 ml, find exact molarity of the EDTA solution.

Part - (C) : Estimation of calcium :

1. Dilute the given solution of calcium and magnesium upto the mark, that is, 100 ml

with distilled water.

2. Take out whole solution into a clean beaker to make homogeneous solution.

3. Fill the burette No. 1 with standardised EDTA solution.

4. Clean the burette No. 2 and fill it with diluted solution of calcium and magnesium.

5. Take 10.0 ml of diluted solution of calcium containing magnesium from burette

No. 2 in a 100 ml conical flask.

6. Add to this solution 5.0 ml buffer solution of pH = 10 and 1 - 2 pellets of solid

NaOH. Stir the flask to dissolve NaOH and check the pH of the solution using small

piece of pH paper. The pH of the solution must be in the range of 12 - 14.

7. Now add a pinch of (50 mg) Patton and Reeder’s indicator powder. Shake the flask

and see the solution has wine red colour.

8. Titrate this solution of calcium containing magnesium against standardised solution

of EDTA added slowly from burette No. 1 with constant shaking of the flask.

9. Continue the addition of EDTA solution from burette No. 1 till the solution in the

conical flask becomes permanently blue. Record the burette reading and call this as

x ml.

10. Take two more readings following steps 5 to 9 and find constant burette reading.

Call this reading as x ml.

11. From the knowledge of volume of EDTA solution required, that is, x ml and exact

molarity of EDTA solution (M1) calculate the amount of calcium in the given solution.


OBSERVATION TABLE :

Part - (B) : Standardisation of EDTA solution

To find : Exact molarity of EDTA solution

Burette No. 1 : EDTA solution (approx. 0.01 M)

Burette No. 2 : Zinc sulphate solution (exact 0.01 M)

Indicator : Eriochrome black T.

End point : Wine red to blue

Reaction : ZnSO4 + Na2EDTA → Na2SO4 + Zn − EDTA


V1 ml

Burette - 2

V2 ml

Constant burette - 1

reading

1. ……… 10.0

2. ……… 10.0 V1 = …… ml

3. ……… 10.0

Molarity of EDTA solution

EDTA = ZnSO4

M1V1 = M2V2

∴ M1 = M2V2

V1

M1 = 0.01 × 10

V1 = …… M

Exact molarity of EDTA solution = M1 = …… M

Part - (C) : Estimation of calcium in the presence of magnesium :

To find : Quantity of calcium in the given solution

Burette No. 1 : Standardised EDTA solution

Burette No. 2 : Diluted calcium solution

Indicator : Patton and Reeder (solid)

End point : Wine red to pure blue

Reaction : Ca++ + Na2EDTA → Ca − EDTA + 2Na+



V1 ml

Burette - 1

x ml

Constant burette - 1

reading

1. 10.0 ……

2. 10.0 …… x = …… ml

3. 10.0 ……

CALCULATIONS :

From the knowledge of chemical reaction between EDTA and calcium, it is observed that

one molecule of calcium require 1 molecule of EDTA for the complete complexation.

Therefore,

1000 ml 1 M EDTA = 40.08 g of calcium

∴ x ml M1 EDTA = ?

= x × M1 × 40.08

1000 = A g of calcium

Here, x = Mean burette-1 reading

M1 = Exact molarity of EDTA solution

Hence, 10 ml diluted solution of calcium (containing magnesium) contains A g of

calcium.

∴ 100 ml diluted solution of calcium would contain A × 10 = B g of calcium.

RESULTS :

1. 10 ml 0.01 M zinc sulphate solution required V1 = …… ml

2. Exact molarity of EDTA solution M1 = …… M

3. 10 ml diluted solution of calcium solution required x = …… ml

4. Amount of calcium in the given solution B = …… g

QUESTIONS

1. What do you mean by complexometric titrations ?

2. What is EDTA ?

3. What is the speciality of EDTA titrations ?

4. What is buffer solution ?

5. What pH is required to estimate calcium in the presence of magnesium ?

6. Which indicators are used in this titration ?

− − −


Experiment No. 3

ESTIMATION OF H2O2

AIM :

(a) Preparation of standard solution of oxalic acid (0.05 N) and standardisation of

approximate 0.05 N KMnO4 solution.

(b) Determination of the strength of given H2O2 solution with standardised 0.05 N KMnO4

solution.

THEORY :

Volumetric Analysis : The process of determining the concentration of unknown

solution by allowing it to react with a solution of known concentration is called volumetric

analysis. The measurement of the solutions required for volumetric analysis is done in terms

of volumes and hence it is known as volumetric analysis.

Standard solution : A standard solution is one which contains a known weight of

reagent in a definite volume of solution or solution with known concentration is called

standard solution.

The concentration of the solution used in volumetric analysis is expressed either in

(i) Normal terms or (ii) Molar terms.

Normal solution : A solution which contains one gram equivalent of a solute per litre of

the solution is called as normal solution. A normal solution is designated by N, where

N stands for normality.

Normality = Number of gram equivalents

Volume in litres

= Weight in grams/equivalent weight

Volume in litres

Molar solution : A solution which contains one gram mole of solute in one litre of

solution is called as molar solution and designated by M, where M stands for molarity.

Molarity = Number of gram moles

Volume in litres

= Weight in grams/molecular weight

Volume in litres

Volumetric methods are generally classified into four main classes :

1. Neutralisation reactions.

2. Complex formation reactions.

3. Precipitation reactions.

4. Oxidation - reduction reactions.


Oxalic acid has formula H2C2O4.2H2O. Oxalic acid does not undergo change in

composition after exposed to air for long time. So it is called a primary standard substance.

When known weight of oxalic acid is dissolved in known volume of solution, it gives a

standard solution of oxalic acid. One molecule of oxalic acid takes one atom of oxygen for

oxidation purpose.

H2C2O4 + (O) 2CO2 + H2O.

This means that one molecule of oxalic acid can take 2 electrons, so its equivalent weight

is half of its molecular weight. Therefore 126

2 = 63. Carbon is in +3 oxidation state in H2SO4

while carbon is in +4 oxidation state in CO2. In the reaction, 1 molecule of H2C2O4 can be

converted to two molecules of CO2. When 63 g oxalic acid is dissolved in solvent and final

volume is made one litre, 1 N solution is obtained.

Potassium permanganate has the formula KMnO4. It undergoes a change in composition

when exposed to air and sunlight, so it is called a secondary standard substance. Its standard

solution cannot be prepared and thus its exact normality is found out by standardisation

with a primary standard substance like oxalic acid. Two molecules of KMnO4 in acid solution

supply five atoms or ten gram equivalents of oxygen for oxidation purpose.

2KMn+7O4 + 3H2SO4 K2SO4 + 2Mn+7SO4 + 3H2O + 5(O)

This means that one molecule of KMnO4 can supply 5 electrons, so its equivalent weight

is equal to molecular weight divided by 5. Thus, 158

5 = 31.6 is the equivalent weight of

KMnO4. Thus when 31.6 g of KMnO4 is dissolved in solvent to make one litre of its solution,

it becomes 1 N.

The reaction between oxalic acid and KMnO4 is a redox reaction. Here oxalic acid is

oxidised to CO2, while KMnO4 is reduced to MnSO4, in the presence of dil. H2SO4. H2O2 can

act as an oxidising as well as a reducing agent. But when H2O2 reacts with KMnO4 it acts as a

reducing agent. Thus, titration between KMnO4 and H2O2 is a redox titration. In this titration,

hydrogen peroxide is oxidised to water and oxygen, while KMnO4 is reduced to MnSO4 in

the presence of dil. H2SO4. In this titration, KMnO4 itself acts as an indicator. Hydrogen

peroxide molecule takes up one oxygen atom in its reaction with KMnO4.

H2O2 + [O] H2O + O2

Commercially available hydrogen peroxide is usually in the form of aqueous solution

containing about 6%, 12%, 30% hydrogen peroxide and it is referred as 20 volume,

40 volume and 100 volume hydrogen peroxide respectively. This volume strength means

one volume of H2O2 solution evolves so many volumes of oxygen after complete

decomposition at N.T.P.


CHEMICALS :

(i) A sample of H2O2 solution in a volumetric flask.

(ii) KMnO4 (approx. 0.05 N) solution.

(iii) Oxalic acid solid (A.R. grade).

(iv) 2 N H2SO4.

PROCEDURE :

This experiment is performed in following three parts :

Part - (A) : Preparation of standard solution of oxalic acid (0.05 N) :

1000 ml 1 N oxalic acid solution contains 63 g H2C2O4.2H2O.

1000 ml 0.1 N oxalic acid solution contains 6.3 g H2C2O4.2H2O.

100 ml 0.05 N oxalic acid solution contains 0.315 g H2C2O4.2H2O.

1. Weigh accurately 0.315 g of A.R. grade oxalic acid crystals on watch glass.


3. Stir the solution with glass rod and dissolve the crystals. Pour the solution into

100 ml volumetric flask.

4. Dilute the solution with distilled water upto the mark.


0.05 N oxalic acid solution.

6. Use this solution for standardisation of given KMnO4 solution.

Part - (B) : Standardisation of KMnO4 solution :

1. Fill the burette No. 1 with 0.05 N (approx.) KMnO4 solution.

2. Fill the burette No. 2 with standard 0.05 N oxalic acid solution.

3. Take 9 ml of oxalic acid solution from burette No. 2 in a 100 ml conical flask. Call this

volume as V2 ml.

4. Add to it 1/2 test tube (10 ml) of 2 N (dilute) H2SO4.

5. Heat the solution on a wire gauze to about 60°C to 70°C (as indicated by

condensation of water vapours on cooler part of conical flask).

6. Titrate this hot solution of oxalic acid against KMnO4 solution, added slowly (drop by

drop) from burette No. 1 with constant shaking of the flask.

7. Continue the addition of KMnO4 till the solution in the conical flask becomes

permanently faint pink. Record the burette reading and call this as X1 ml.

8. Add to the same conical flask 1 ml solution of 0.05 N oxalic acid from burette No. 2.

Heat the flask again so that solution in the flask acquires a temperature of about

60 - 70°C and the solution becomes colourless.


9. Add KMnO4 solution from burette No. 1 till a permanent faint pink colour appears to

the solution. Record the burette reading and call this as X2 ml.

10. To the same solution in the conical flask add 1 ml solution of 0.05 N oxalic acid from

burette No. 2 (see that the solution in the flask is hot, if not, heat it again to

60 - 70°C). The pink colour disappears.

11. Add KMnO4 solution from burette No. 1 till a faint permanent pink colour appears to

the solution. Call this reading as X3 ml.

Part - (C) : Determination of strength of H2O2 :

1. Dilute the given solution of hydrogen peroxide (H2O2) to 100 ml with distilled water.

Take out whole solution in a beaker and stir it well.

2. Fill the burette No. 1 with standardised KMnO4 solution.

3. Clean the burette No. 2 and fill it with diluted H2O2 solution.

4. Take 9 ml of diluted H2O2 solution from burette No. 2 in a 100 ml conical flask. Call

this volume as V2 ml.

5. Add to it 1/2 test tube (10 ml) of 2 N (dilute) H2SO4.

6. Do not heat the solution.

7. Titrate this solution of H2O2 against KMnO4 solution, added slowly (drop by drop)

from burette No. 1 with constant shaking of the flask.

8. Continue the addition of KMnO4 till the solution in the conical flask becomes

permanently faint pink. Record the burette reading and call this as Y1 ml.

9. Add to the same conical flask 1 ml solution of H2O2 from burette No. 2. The pink

colour disappears. Do not heat the solution.

10. Add KMnO4 solution from burette No. 1 till a permanent faint pink colour appears to

the solution. Record the burette reading and call this as Y2 ml.

11. To the same solution in the conical flask add 1 ml solution of H2O2 from burette No.

2. The pink colour disappears.

12. Add KMnO4 solution from burette No. 1 till a faint permanent pink colour appears to

the solution. Call this reading as Y3 ml.

OBSERVATION TABLE :

Part - (B) : Standardisation of KMnO4 solution :

To find : Exact normality of KMnO4 solution

Burette No. 1 : KMnO4 solution (approx. 0.05 N)

Burette No. 2 : Oxalic acid solution (0.05 N exact)

Indicator : KMnO4 (self indicator)

End point : Colourless to pink


Reaction :

2KMnO4 + 3H2SO4 + 5H2C2O4 K2SO4 + 2MnSO4 + 8H2O + 10CO2


X ml

Burette - 2

V2 ml Normality N =

0.05 ×××× V2

X

1. X1 = 9.0 N1 =

0.05 × 9

X1

2. X2 = 9 + 1 = 10.0 N2 =

0.05 × 10

X2

3. X3 = 10 + 1 = 11.0 N3 =

0.05 × 11

X3

Mean Normality N4 = N1 + N2 + N3

3 = ……… N

Part - (C) : Determination of strength of H2O2 in the given solution :

To find : Strength of H2O2 in the given solution

Burette No. 1 : Standardised KMnO4

Burette No. 2 : H2O2 solution (given)

Indicator : KMnO4 (self indicator)

End point : Colourless to pink

Reaction :

2KMnO4 + 3H2SO4 + 5H2O2 K2SO4 + 2MnSO4 + 8H2O + 5O2


Y ml

Burette - 2

V2 ml

1. Y1 = 9.0

2. Y2 = 9 + 1 = 10.0

3. Y3 = 10 + 1 = 11.0

Mean Burette Reading 1

= Y1 + Y2 + Y3

3 = Y …… ml

Mean Burette Reading 2

= 9 + 10 + 11

3 = 10 ml

CALCULATIONS :

From above equation,

(Oxidising agent) = (Reducing agent)


2KMnO4 = 5H2O2

1

5 KMnO4 =

1

2 H2O2

1

5 (158) =

1

2 (34)

31.6 = 17 (Equivalent weight)

But one fifth of the formula weight (Mol. wt.) of KMnO4 is its equivalent weight i.e. 31.6.

Equivalent weight of hydrogen peroxide (H2O2) when it acts as a reducing agent in acidic

medium is half of its molecular weight. Therefore equivalent weight of hydrogen peroxide

(H2O2) is 17.

(a) Strength of H2O2 in normal terms :

(Oxidising agent) = (Reducing agent)

KMnO4 = H2O2

Since 10 ml (average) H2O2 solution required Y ml of N4 KMnO4 therefore, normality of

H2O2 solution can be calculated as follows :

Y × N4 = N5 × 10

∴ N5 = N4 × Y

10 = Exact normality of H2O2

(b) Strength of H2O2 in grams per litre :

Strength = Normality × Equivalent weight

= N5 × Equivalent weight of H2O2

= N5 × 17

= A g/litre

Amount of H2O2 in the given solution = …………… (A) g/litre.

(c) Strength of H2O2 in volume unit :

Hydrogen peroxide solution when heated alone, it is decomposed as

2H2O2 2H2O + O2

68 g 36 gm or 22.4 lits.

∴ 68 g of H2O2 liberates = 22.4 litres of oxygen at N.T.P.

'A' g of H2O2 will liberate = 22.4 × A

68 = B litres of oxygen at N.T.P.


Thus one litre of H2O2 solution contains 'A' g of H2O2 which produces B litres of oxygen

measured at N.T.P.

Hence given solution of H2O2 is 'B' volume.

Note : 68 g of H2O2 produces 22.4 litres of oxygen at N.T.P.

17 g of H2O2 will produce 5.6 litres of oxygen at N.T.P. Hence 1 N solution of H2O2 is

5.6 volume.

RESULTS :

1. 10 ml 0.05 N oxalic acid solution required = 'X2' ml of KMnO4

2. Exact normality of KMnO4 solution = N4 ………… N

3. 10 ml diluted H2O2 solution required = Y2 ml of KMnO4

4. Normality of H2O2 solution = N5 ………… N

5. Strength of H2O2 = A ………… g/litre

6. Strength of H2O2 in volume unit = 'B' ………… volume

QUESTIONS

1. What do you mean by volumetric analysis ?

2. What is meant by normal solution and molar solution ?

3. Define the term equivalent weight of a substance.

4. What is equivalent weight of KMnO4 ?

5. What is standard solution ?

6. How will you prepare 100 ml 0.05 N oxalic acid solution ?

7. Which solutions are called standard solutions ?

8. What is a primary standard substance ?

9. Why is standardisation of KMnO4 required ?

10. Why KMnO4 solution should not be added rapidly during a titration ?

11. What do you understand by volume strength of H2O2 ?

12. What is the role of dilute sulphuric acid in this titration ?

13. List the oxidising and reducing agents that you know.

✍ ✍ ✍


Experiment No. 4

ESTIMATION OF ASPIRIN

AIM :

Estimation of aspirin from a given tablet and to find errors in quantitative analysis.

THEORY :

APC tablet contains aspirin, phenacetin and caffine out of which the percentage of

aspirin is more. Aspirin is acetyl salicylic acid (o-acetoxy-benzoic acid) and has analgesic and

antipyretic action. It can relieve headache and also can be used to control rheumatic fever.

Aspirin on hydrolysis gives salicylic acid which can be detected by ferric chloride test and

estimated by back titration with 0.1 M HCl solution using phenol red or some suitable

indicator.

In any quantitative analysis, some errors are usually associated. Therefore, it is necessary

to take more number of readings and do various statistical calculations to find errors.

In order to obtain reliable results from an analytical method, sources of error must be

identified and either it should be eliminated completely or it should be atleast minimized.

Errors may be expressed in absolute terms as the difference between an analytical result,

Y and the known true value (or mean value) T.

∴ D = (T − Y)

When this difference (D) is expressed as an unsigned number, it is known as an absolute

error.

The relative error, Erel is used to determine the accuracy of measurement and is typically

expressed as the percentage of the known true value :

Erel = D

T ; %Erel =

D

T × 100

Since the relative error is a dimensionless number, it can be used to determine the

accuracy of results as well as to compare the accuracies of results expressed in different

units.

Apart from this mean deviation, standard deviation, relative mean deviation and relative

standard deviation are also important. The mean deviation or the relative mean deviation is

a measure of precision (reproducibility) while, the spread of the values is measured most

efficiently by the standard deviation 'S'.


APPARATUS :

Burettes (preferably a match pair), burette stand, conical flask, beaker, volumetric flask,

glass rod, droppers, stem-cut funnel, etc.

CHEMICALS :

(i) APC tablets.

(ii) 1 M NaOH solution.

(iii) 0.1 M HCl (exact).

(iv) Phenol red/phenolphthalein indicator.

PROCEDURE :

This experiment is performed in following two parts :

Part - I : Back Titration :

1. Powder 2 APC tablets into fine powder and weigh accurately ('W' g) which contains

about 0.200 g of aspirin.

2. Transfer all powder ('W' g) in a clean conical flask and add 10 ml of 1 M NaOH

solution with the help of a common burette.

3. Place a stem-cut funnel on the mouth of a conical flask (to avoid loss due to

evaporation) and boil the reaction mixture on a water bath for 15 minutes. Aspirin

on alkaline hydrolysis gives sodium salicylate. Cool the solution and transfer it to a

clean 100 ml volumetric flask.

4. Dilute the solution to 100 ml with distilled water. Take out all the solution in a clean

beaker, stir with glass rod to make the solution homogeneous.

5. Fill Burette-1 with this diluted solution. Take 10 ml diluted solution in a conical flask.

6. Titrate this solution against 0.1 M (exact) HCl solution from Burette-2 using

phenolphthalein (or phenol red) as an indicator.

7. The end point is noted when colour changes from pink to colourless (red to yellow

if phenol red indicator is used).

8. In the similar way, take two more readings.

9. Note down the burette reading, call it as Y1, Y2 and Y3 ml.

10. From the weight of sample taken and actual amount of aspirin present in the

sample, find out the known true value or ask for the known true value of % aspirin in

the sample to the practical incharge or examiner.

11. Calculate the practical percentage of aspirin from three readings taken and use the

data to calculate absolute error, relative error, standard deviation etc.


Part - II : Blank Titration :

1. Take 10 ml of 1 M NaOH solution in a 100 ml volumetric flask with the help of a

common burette and dilute it upto the mark with distilled water.

2. Take out all the solution in a clean beaker, stir it with glass rod to make the solution

homogeneous.

3. Fill Burette-1 with this diluted solution. Take 10 ml diluted solution in a conical flask.

4. Add 2 drops of phenolphthalein (or phenol red) indicator and titrate it against 0.1 M

(exact) HCl solution from Burette-2.

5. The end point is pink to colourless (red to yellow if phenol red is used).

6. Take two more readings in a similar way and note the constant burette reading as

'X' ml or call it as blank reading.

7. (X − Y) ml or (blank-back) ml gives the quantity of NaOH required for hydrolysis of

aspirin, from which the % of aspirin in given tablet can be calculated.

Hint : If weight of powdered sample i.e. 'W' g is between 0.150 to 0.200 g, then 10 ml,

1 M NaOH solution is sufficient for hydrolysis. In that case dilute to 100 ml. But, if the

amount of sample taken is more i.e. 'W' g is more than 0.300 g, then it may require 25 ml,

1 M NaOH solution for hydrolysis, in that case dilute to 250 ml.

OBSERVATIONS

Part - I : Blank Titration

Given 0.1 M (exact) HCl

To find : Amount of NaOH consumed

Burette 1 : Diluted solution of hydrolysed aspirin (10 ml)

Burette 2 : 0.1 M (exact) HCl

Indicator : Phenolphthalein (or phenol red)

End point : Pink to colourless (or red to yellow if phenol red)

Reaction :

Readings Pilot First Second Third

Burette - 1

(Diluted aspirin)

10 ml 10 ml 10 ml 10 ml

Burette - 2

(0.1 M HCl)

One ml range close to end

point (in whole number)

Y1 Y2 Y3


Part - II : Blank Titration

Given 0.1 M HCl (exact)

To find : Amount of NaOH

Burette No. 1 : Diluted solution of NaOH (10 ml)

Burette No. 2 : 0.1 M HCl (exact)

Indicator : Phenolphthalein (or phenol red)

End point : Pink to colourless (or red to yellow if phenol red)

Reaction :

NaOH + HCl → NaCl + H2O

Readings Pilot First Second Third Constant B.R.

Burette - 1

(Diluted NaOH)

10 ml

10 ml

10 ml

10 ml

'X' ml

Burette - 2

(0.1 M HCl)

One ml range close

to end point

(in whole number)

X1

X2

X3

(Blank)

CALCULATIONS :

1. To find out amount and hence % of aspirin :

We know that, 2 moles of NaOH = 1 mole of aspirin = 180 g

∴ 1 mole of NaOH = 1

2 mole of aspirin = 90 g

Now, 1 mole of NaOH = 1 mole of HCl

∴ 1 mole of HCl = 90 g of aspirin

∴ 1000 ml of 1 M HCl = 90 g of aspirin

∴ X − Y1 ml 1 M HCl = X − Y1 × 1 × 90

1000 × 1 =

X − Y1 × 90

1000

= A g of aspirin

∴ 10 ml diluted solution contains = A g of aspirin

∴ 100 ml diluted solution contains = A × 10 = B g of aspirin

Now, W g of APC sample = B g of aspirin

∴ 100 g of APC sample = B × 100

W = Z1

i.e. % of aspirin in the APC sample = Z1 % for the first reading i.e. Y1 ml


In the same way, find out % of aspirin using Y2 and Y3 i.e. second and third readings and

call them as Z2 and Z3 %.

Find the average % of aspirin as Z1 + Z2 + Z3

3 = Z%

2. To calculate errors in aspirin estimation :

(a) Known true value : APC tablet contains aspirin, phenacetin and caffine. In addition

certain binder material is also present. Actual or true % of aspirin can be calculated using

knowledge of weight of sample taken and actual amount of aspirin present in the sample

(as given on the wrapper of the tablet). Known true value can be asked to the practical

incharge or examiner during examination. Call this known true value as T.

(b) Absolute error 'D' :

Absolute error (D) = (Known true value) − (Practical value)

i.e. D = T − Z1 For 1st reading

D = T − Z2 For 2nd reading

and D = T − Z3 For 3rd reading

(c) Relative error, Erel :

Erel = Absolute error

Known true value =

D

T

% Erel = D

T × 100

For the three values of D, three values of % relative error may be calculated.

(d) Standard deviation 'S' :

S = Σ (Y1 − T)2

n − 1 =

(Y1 − T)2 + (Y2 − T)2 + (Y3 − T)2

n − 1

Here n = Number of measurements which is 3.

(e) Relative standard deviation, Sm :

Relative standard deviation = Standard deviation × 100

True value

i.e. Sm = S × 100

T = CV

where, CV = Coefficient of variance or variation.


RESULTS :

Sr. No. Description Unit

1. % of aspirin in the given APC sample 'Z' %

2. Absolute error value D

3. Relative error value Erel

4. Standard deviation S

5. Relative standard deviation Sm

QUESTIONS

1. What are the components of APC tablet ?

2. What is meant by hydrolysis ?

3. What are the products of alkaline hydrolysis of aspirin ?

4. How salicylic acid is qualitatively detected ?

5. Explain the principle behind the estimation of aspirin.

6. Which indicators are available for this titration ? What is the colour change at the

end point ?

7. Distinguish between back and blank titration.

8. What are errors ? How they are found ?

9. Define the terms : (a) Absolute error and (b) Relative error.

10. Explain how standard deviation and relative standard deviation are calculated.

11. What is the use of APC tablet ?

✍ ✍ ✍

Experiment No. 5

ESTIMATION OF ALUMINIUM

AIM :

To estimate amount of Al (III) from the given solution of aluminium by using Eriochrome

Black - T indicator.

THEORY :

To estimate aluminium quantity volumetrically, sodium salt of ethylene diamine tetra-

acetic acid (EDTA) should be used. However, some complex - forming reactions of EDTA are

quite slow. This is particularly true with trivalent ions like Al+3, Cr+3, Co+3, Fe+3 etc. When the

reactions are very slow the elements are determined by back titration method. These

trivalent metals can be determined in neutral solution by adding excess of EDTA and back

titrating with a suitable metal sulphate solution like zinc sulphate.


Aluminium can be quantitatively estimated by using Eriochrome Black-T indicator at

neutral pH and at about 60°C if titrated rapidly by this back titration method. The metal-

indicator complex (i.e. Al-Erio T complex) formation is slow and therefore the method is

suitable. The end point is detected with the help of metal indicator such as Eriochrome

black-T which respond to the zinc ions introduced in the back titration rather than

aluminium ions.

APPARATUS :

Burettes (preferably a match pair), conical flask, beaker, volumetric flask, glass rod,

droppers etc.

CHEMICALS :

(i) Aluminium solution - supplied in 100 ml volumetric flask.

(ii) Zinc sulphate - 0.01 M (exact).

(iii) EDTA solution - 0.01 M (approx.)

(iv) Ammonia solution - (to adjust pH).

(v) Eriochrome Black-T indicator.

(vi) Phenol red paper/pH paper/pH meter.

PROCEDURE :

Perform this experiment in two parts :


1. Dilute the given solution of aluminium with distilled water upto the mark in a

100 ml volumetric flask and take out all the solution in a beaker, stir with glass rod to

make the solution homogeneous.

2. Fill Burette-1 with this diluted aluminium solution.

3. Take 10 ml of the diluted aluminium solution in conical flask. Add one test tube

water.

4. Now add 10 ml 0.01 M EDTA solution by means of a common burette in a conical

flask containing aluminium solution.

5. Adjust the pH of solution between 7 and 8 by adding ammonia solution dropwise

with constant stirring (Check the pH with the help of phenol red paper/pH paper/pH

meter).

6. Boil the solution for few minutes (till 60°C) to ensure complete complexation of the

aluminium with EDTA. Then cool the solution to room temperature and adjust the

pH to 7-8 by adding ammonia.


7. Now add 2-3 drops of Eriochrome Black-T indicator, the colour of the solution

becomes blue.

8. Titrate this solution rapidly against standard 0.01 M zinc sulphate solution from

Burette-2 with constant shaking till the colour changes to wine red.

9. The end point of titration is obtained when the blue colour changes to wine red.

10. In the similar way take two more readings.

11. Note down the constant burette reading as 'Y' ml or call it as back reading. (Do not

take the mean reading).


1. Take 10 ml of 0.01 M EDTA solution by means of common burette in conical flask.

Add one test tube water.

2. Now add 3 ml of ammonia solution to adjust the pH.

3. Then add 2-3 drops of Eriochrome Black-T indicator, the colour of the solution

becomes blue.

4. Titrate this solution against standard 0.01 M zinc sulphate solution from Burette-2

with constant shaking till the colour changes to wine red.

5. The end point of the titration is obtained when the blue colour changes to wine red.

6. In the similar way, take two more readings.

7. Note down the constant burette reading as 'X' ml or call it as blank reading. (Do not

take the mean reading.)

(X − Y) ml or (blank-back) ml gives the quantity of EDTA consumed by 10 ml of the

diluted aluminium solution in terms of 0.01 M ZnSO4 solution. Or, in other words, every

millilitre difference between the volume of 0.01 M EDTA added and the 0.01 M zinc sulphate

solution used in the back titration corresponds to 0.2698 mg of Al.

PRECAUTIONS :

1. pH of the solution should be adjusted to desired value strictly.

2. To ensure complete complexation in back titration, heating to 60°C is must.

3. Titration should be performed rapidly with constant shaking.

4. After standing for few minutes, the fully titrated solution acquires a reddish violet

colour; this change is irreversible, so that over-titrated solutions are lost.


OBSERVATIONS :


Given : 0.01 M ZnSO4 solution

To find : Unreacted (excess) EDTA in terms of 0.01 M ZnSO4

Burette No. 1 : Diluted aluminium solution (10 ml)

Burette No. 2 : 0.01 M ZnSO4 solution

Indicator : Eriochrome Black-T

End point : Blue to wine red

Reactions :

(i) Al+3 + EDTA (excess) Al-EDTA complex + EDTA (unused)

(ii) Zn2+ + EDTA (unused) Zn-EDTA complex


Burette - 1

(Aluminium)

solution

10.0 ml

10 ml

10 ml

10 ml

'Y' ml

Burette - 2

(0.01 M ZnSO4)

solution

One ml range close

to end point

(in whole number)

Y1

Y2

Y3

Back


Given : 0.01 M ZnSO4 solution

To find : Total quantity of EDTA in terms of 0.01 M ZnSO4

Common burette : 10 ml EDTA solution

Burette No. 2 : 0.01 M ZnSO4 solution

Indicator : Eriochrome Black-T

End point : Blue to wine red

Reaction :

Zn2+ + EDTA Zn − EDTA complex


Burette (common)

(EDTA solution)

10.0 ml

10 ml

10 ml

10 ml

'X' ml

Burette-2

(0.01 M ZnSO4)

solution

One ml range

close to end point

(in whole number)

X1

X2

X3

Blank


CALCULATIONS :

Since every ml difference between the volume of 0.01 M EDTA and the 0.01 M zinc

sulphate solution used in the back titration corresponds to 0.2698 mg of Al, we have the

relation as

1 ml 0.01 M ZnSO4 = 0.2698 mg of Al

∴ (X − Y) ml 0.01 M ZnSO4 = (X − Y) × 0.2698 × 0.01

1 × 0.01 = (X − Y) × 0.2698

= 'A' mg of Al

This amount of aluminium is present in 10 ml diluted solution.

∴ Total amount of aluminium in 100 ml = 'A' × 10 mg

= A × 10

1000 = B g

RESULTS :

Sr. No. Description Unit

1. Back titration reading 'Y' ml

2. Blank titration reading 'X' ml

3. Amount of EDTA consumed in terms of 0.01 M

ZnSO4 solution by 10 ml of aluminium solution

(X − Y) ml

4. Amount of aluminium in the given solution 'B' g

QUESTIONS

1. How Al (III) is estimated using EDTA ?

2. What is the full form of EDTA ?

3. Why aluminium is estimated by back titration method ?

4. Which indicator is used in Al-EDTA titration ? What is the change in colour at the end

point ?

5. What are the conditions to be maintained in this titration ?

6. Why heating is required in back titration of this experiment ?

7. The results are good if rapid stirring is done during titration. Explain.

8. Explain how the amount of aluminium is calculated in this experiment.

9. What is the role of ZnSO4 solution in this titration ?

✍ ✍ ✍


Experiment No. 6

ESTIMATION OF COPPER

AIM :

To determine the amount of copper from the given solution by iodometric method.

THEORY :

The direct iodometric titration method called iodimetry is the titration in which standard

iodine solution is used. The indirect iodometric titration method called iodometry deals with

the titration of liberated iodine in a chemical reaction. In the iodometric determination of

copper, Cu++ ions oxidise I− from KI and iodine is liberated. The reaction taking place is,

2CuSO4 + 4KI 2CuI + I2 + 2K2SO4

i.e. 2Cu++ + 4I− 2CuI + I2

The liberated iodine is estimated by titrating it against standardised sodium thiosulphate

solution using starch solution as an indicator.

Strong reducing agent such as sodium thiosulphate reacts completely and rapidly with

iodine even in the acid solution. The reaction taking place is,

I2 + 2S2O2−

3 2I− + S4O2−

6

From which it follows that, 2CuSO4 + I2 + 2Na2S2O3.

Two important sources of error in titration involving iodine are loss of iodine owing to its

appreciable volatility and oxidation of iodide in acid solutions by oxygen from the air :

4I− + O2 + 4H+ 2I2 + 2H2O

In the presence of excess iodide, the volatility is decreased remarkably because of

formation of the tri-iodide ion.

I2 + I− I−

3

At room temperature the loss of iodine by volatilisation from a solution containing

atleast 4% of potassium iodide is negligible, provided the titration is not prolonged unduly.

Titrations must be carried out in cold solutions in conical flask only. If a solution is to stand,

it should be kept in a glass stoppered vessel.

The atmospheric oxidation of iodide is negligible in neutral solution in the absence of

catalysts. The reaction is catalysed by certain metal ions particularly copper and also by

strong light. For this reason, titration should not be performed in direct sunlight and

solutions containing iodide should be stored in ambar glass bottles.


A solution of iodine in aqueous iodide has an intense yellow to brown colour. The test

for the presence of iodine is made much more sensitive by using a solution of starch as an

indicator. Starch reacts with iodine in the presence of iodide to form an intensely blue

coloured complex which is visible at very low concentration of iodine. Only freshly prepared

starch solution should be used. Two millilitres of a 0.1% solution per 100 ml of the solution

to be titrated is a satisfactory amount. In the titration of iodine, starch must be added until

just before the end point is reached. Apart from the fact that the fading of the iodine colour

is a good indication of the approaching end point, if the starch is added when the iodine

concentration is high i.e. at the beginning of titration, some iodine may remain adsorbed

even at the end point.

Sodium thiosulphate (Na2S2O3 ⋅ 5H2O) is readily available in a state of high purity, but

there is always some uncertainty regarding exact water content because of the efflorescent

nature of the salt. The substance is therefore not a primary standard. Its sub-standard

solution is made and is standardised using K2Cr2O7 solution of known concentration. Sodium

thiosulphate is a reducing agent by virtue of the following half cell reaction.

2S2O2−

3 S4O2−

6 + 2e−

CHEMICALS :

(i) A sample of CuSO4 solution given in a volumetric flask.

(ii) Sodium thiosulphate solution (approx. 0.025 N) solution.

(iii) Solid K2Cr2O7 crystals.

(iv) 10% KI solution.

(v) 2 N sodium hydroxide.

(vi) 2 N acetic acid.

PROCEDURE :

This experiment is performed in following three parts :

Part - I : Preparation of standard solution of potassium dichromate (0.025 N) :

Equivalent weight of K2Cr2O7 is 49 (1/6 of molecular weight).

∴ 1000 ml 1 N K2Cr2O7 solution = 49 g of K2Cr2O7

∴ 100 ml 0.025 N K2Cr2O7 solution = 0.1225 g of K2Cr2O7

1. Weigh accurately 0.1225 g of A.R. grade K2Cr2O7 crystals on a watch glass.


2. Transfer the weighed crystals carefully in a clean 100 ml beaker and rinse the watch

glass with distilled water. Transfer the washing to the same beaker. Stir the solution

with glass rod and dissolve the crystals.

3. Pour this solution in a clean 100 ml volumetric flask. Rinse the beaker with distilled

water and transfer washings into the volumetric flask. Add distilled water and dilute

upto the mark.

4. Take out whole solution in a clean beaker.

5. This becomes 0.025 N K2Cr2O7 solution. Use this solution for standardisation of given

sodium thiosulphate solution.

Part - II : Standardisation of sodium thiosulphate solution :

1. Fill the burette No. 1 with 0.025 N (approx) sodium thiosulphate solution.

2. Fill the burette No. 2 with standard 0.025 N potassium dichromate solution.

3. Take 9.0 ml of potassium dichromate solution from burette No. 2 in a 100 ml conical

flask. Call this volume as V2 ml.

4. Add 3 ml concentrated HCl and add 3 ml 10% KI solution. Shake the flask well and

keep it for about five minutes.

5. Titrate the liberated iodine with Na2S2O3 solution added from burette No. 1. Initially

solution in the conical flask is dark brown in colour. Continue addition of Na2S2O3

solution till colour becomes pale yellow. Then add 1 - 2 ml starch indicator. The

solution will become blue in colour. Continue the titration.

6. The end point of titration is recorded when colour changes from dark blue to green.

There is sudden change in colour. Call this burette reading as 'X1' ml.

7. To the same conical flask add 1.0 ml of 0.025 N K2Cr2O7 solution from burette No. 2

(Do not add KI solution and starch indicator.) The colour of the solution turns blue.

8. Titrate this solution against Na2S2O3 solution added from burette No. 1 till the colour

of solution changes from blue to green. Record this reading as 'X2' ml.

9. To the same conical flask add 1.0 ml of 0.025 N K2Cr2O7 solution from burette No. 2.

(Do not add KI or starch solution.) The colour of solution again turns blue.

10. Titrate this solution against Na2S2O3 solution added from burette No. 1 till the colour

of solution changes from blue to green. Record this reading as 'X3' ml.

11. From the burette readings X1, X2 and X3 find out the exact normality of Na2S2O3

solution.


Part - III : Iodometric determination of copper :

1. Dilute the given solution of copper sulphate to 100 ml with distilled water. Take out

whole solution in a beaker and stir it well.

2. Fill the burette No. 1 with standardised Na2S2O3 solution.

3. Clean the burette No. 2 and fill it with diluted copper sulphate solution. Take 10.0 ml

of this solution in a 100 ml conical flask.

4. Add to it 2 N NaOH solution till slight turbidity is obtained. Dissolve this turbidity by

adding drop by drop 2 N acetic acid. Shake the flask well. Add 2 − 3 drops of acetic

acid in excess.

5. Add 3 ml 10% KI solution by means of measuring cylinder. Shake the flask well and

keep it for about five minutes.

6. Titrate the liberated iodine with Na2S2O3 solution from burette No. 1 till the colour of

the solution becomes pale yellow.

7. Then add 1 − 2 ml starch indicator. The solution becomes blue.

8. Continue the titration. The end point of the titration is recorded when colour

changes from blue to colourless. Record this reading as Y1 ml.

9. Take two more readings by repeating Step No. 3 to 8 and record the constant

burette reading.

10. From this reading determine the amount of copper in the given solution.

OBSERVATIONS :

Part II : Standardisation of Na2S2O3 solution :

To find : Exact normality of Na2S2O3 solution

Burette No. 1 : 0.025 N (approx.) Na2S2O3 solution

Burette No. 2 : 0.025 N (exact) K2Cr2O7 solution

Indicator : Starch solution (freshly prepared)

End point : Blue to green

Reactions :

1. K2Cr2O7 + 6KI + 14 HCl 8KCl + 2CrCl3 + 7H2O + 3I2

2. 2Na2S2O3 + I2 Na2S4O6 + 2NaI



X ml

Burette - 2

V2 ml Normality N =

0.025 ×××× V2

X

1. X1 = 9.0 N1 = 0.025 × 9.0

X1

2. X2 = 9.0 + 1.0 = 10.0 N2 = 0.025 × 10.0

X2

3. X3 = 10.0 + 1.0 = 11.00 N3 = 0.025 × 11.0

X3

Mean normality N4 = N1 + N2 + N3

3 = ……… N

Exact normality of Na2S2O3, N4 = ………… N

Part III : Estimation of copper :

To find : Amount of copper in the given solution

Burette No. 1 : Standardised Na2S2O3 solution

Burette No. 2 : Diluted copper sulphate solution

Indicator : Starch solution (freshly prepared)

End point : Blue to colourless

Reactions :

1. 2CuSO4 + 4KI 2CuI + I2 + 2K2SO4

2. I2 + 2Na2S2O3 2NaI + Na2S4O6


Burette-2

Diluted copper

solution

10.0 ml

10 ml

10 ml

10 ml

'Y' ml

Burette-1

Na2S2O3 solution

One ml range close

to end point

(in whole number)

Y1

Y2

Y3


CALCULATIONS :

1. To calculate amount of copper in the given solution :

We know that, 2CuSO4 ⋅ 5H2O + I2 + 2Na2S2O3 ⋅ 5H2O + 2Cu

∴ Na2S2O3 ⋅ 5H2O + Cu

∴ 248 g = 63.5

The relation therefore becomes,

1000 ml 1 N Na2S2O3 = 63.5 g of copper

∴ Y ml N4 Na2S2O3 = Y × N4 × 63.5

1000 × 1 = ……… A g of Cu

∴ 10 ml diluted solution contains A g of Cu.

∴ 100 ml diluted solution contains A × 10 = B g of Cu.

RESULTS :

1. Exact normality of Na2S2O3 solution N4 = …… N

2. Volume of Na2S2O3 solution required to

titrate 10 ml diluted solution

Y = …… ml

3. Amount of copper in the given solution B = …… g

QUESTIONS

1. What is the difference between Iodimetry and Iodometry ?

2. How standard solution of K2Cr2O7 is made ?

3. Why Na2S2O3 ⋅ 5H2O is not a primary standard substance ?

4. How sodium thiosulphate solution is standardised ?

5. Which indicator is used in this titration ? When it is added ?

6. Why starch is added near the end point of titration ?

7. How volatilization of iodine is minimized ?

8. Why excess of KI is added in this titration ?

9. This titration is not performed in direct sun light. Why ?

10. Iodine solutions are generally stored in amber glass bottles. Why ?

11. Iodometric titrations are carried out in cold condition and in conical flask or

stoppered bottle only. Why ?

✍ ✍ ✍


Experiment No. 7

INDUSTRIAL EDUCATIONAL VISIT

AIM :

To submit a report on industrial educational visit.

(A) Why such industrial visit(s) is necessary ?

In our routine syllabus, we study about various industries like sugar, cement, glass,

fertilizers, dyes, drugs and pharmaceuticals, soaps and detergents, ceramics, fermentations,

explosives, paper, distilleries etc. Chemistry plays a vital role in these industries. Many of the

industries have their own chemistry laboratories. They may have their own Research and

Development and quality control units.

The knowledge about the Chemical Industry learnt from the book is helpful upto

certain extent only. By visiting the actual chemical plant operation one can study about what

goes on in a typical chemical industry. One can get the knowledge about actual plant

operations. Because of such industrial visit(s), interest and understanding of chemistry is

increased in the mind of students.

Therefore, such industrial visits must be arranged. Discussion on such visits must be

arranged before and after visits.

(B) Planning before visit(s) :

1. Selection of proper industry for visit.

2. Get maximum information about the industry before visit.

3. Get familiar with safety precautions those are to be taken during visit.

4. Prepare questionnaries regarding the industry you visit.

5. Students should have some ideas regarding the various processes, operations in that

industry.

6. Take few flow-charts regarding such processes with you.

7. Take a diary and required stationary with you.

(C) Submission of the Report :

Prepare a neat report on your visit to industry in a systematic way and submit it to your

practical incharge within four days. You can include following points in your report :

1. Name of the industry :

2. Address and location :


3. Distance from your college : ……………… km (approx)

4. Contact numbers (telephone, fax, e-mail address etc.) :

5. Date of visit :

6. Organiser of your visit :

7. Number and name(s) of the teachers participated :

8. Number of students participated : M F

9. Person(s) met during your visit with their designation.

10. Detail report (about 40 lines) :

Your report should include following important points :

1. Raw materials used in industry (source of raw material, cost, way of transport,

hazardous, non-hazardous etc.)

2. Safety precautions taken during transport of raw material.

3. Outline of processes/operations in the industry (flow sheet).

4. Chemical reactions in the plant.

5. Quality control.

6. Research and development unit in the industry.

7. Products from the industry, by-products during the process. Disposal of by-

products, students are expected to report on different properties, uses and

applications of the product and by-products. Try to note the purchaser of these

products. Also note the cost of product and by-products.

8. Safety precautions taken in the industry. Note whether the workers are informed

regarding toxicity of chemicals used in the plant.

9. Pollution control.

10. Number of labours in the industry.

11. Shifts in industry, weekly holiday.

12. Various machineries/equipments used in the industry.

QUESTIONS

1. Does the industry you visited have chemical laboratory ?

2. How many staff members are there in the laboratory ? Give their names.

3. What type of analysis work is done in the laboratory ?

4. Have you studied any such technique in your laboratory ?


5. List the equipments those were used/or in the laboratory.

6. Name the incharge of the laboratory along with his qualifications. Also name the

other staff members, their qualification and about their duties in the laboratory.

7. What types of chemicals are used in the laboratory ? (A.R., L.R., C.R. etc.)

8. What types of glasswares are used ?

9. In which respect you compare your college laboratory with the laboratory in the

industry ?

10. Do you realise the importance of chemistry in various industries ? Justify your

answer.

11. Do you like to become a chemist ?

12. Are you benefited from the industrial visit ? If yes ? In what respect ?

13. Which industry you have visited ?

14. What type of industry is it ?

15. How care is taken regarding pollution ?

16. Whether that industry is taking precautions regarding safety wares ?

✍ ✍ ✍

(121 )

APPENDIX - I

(A) PHYSICAL CHEMISTRY EXPERIMENTS

1. Partition Coefficient :

(i) 0.05 M Na2S2O3 solution : Dissolve 62.0 g of Na2S2O3.5H2O in distilled water and

dilute to 5 litres.

(ii) 0.01 M Na2S2O3 solution : Dissolve 12.4 g of Na2S2O3.5H2O in distilled water and

dilute to 5 litres.

(iii) 10% KI solution : Dissolve 50 g of KI in distilled water and dilute to 500 ml.

2. Heat of Neutralisation :

(i) 1 M HCl : 90 ml conc. HCl in 1 litre distilled water.

(ii) 1 M H2SO4 : 56 ml conc. H2SO4 in 1 litre distilled water.

(iii) 1 M CH3COOH : 60 ml acetic acid in 1 litre distilled water.

(iv) 1 M NaOH : 40 g NaOH, dissolve and dilute to 1 litre ≡ 1 N NaOH.

(B) ANALYTICAL CHEMISTRY EXPERIMENTS

1. Estimation of Na2CO3 :

(i) 0.1 N HCl solution : Accurately measure 44 ml of concentrated AR HCl solution and

dilute this solution to make the final volume of 5000 ml.

(ii) Methyl orange indicator solution : Dissolve 0.5 g of the sodium salt of methyl

orange in 1 litre of water, add 15.2 ml of 0.1 M HCl and filter if necessary.

2. Estimation of Calcium in the Presence of Magnesium :

(i) Patton and Reeder’s Indicator : 0.5 g indicator powder + 50 g sodium sulphate,

grind well to make a homogeneous mixture.

(ii) Stock Solution containing Calcium and Magnesium : Take 10.0 g CaCO3 and 6.0 g

MgCO3 in a clean beaker. Add dilute HCl solution till effervescence of CO2 gas stops

completely and solution becomes clear. Transfer this solution to 1000 ml volumetric

flask and dilute upto the mark. Give 8 - 10 ml of this solution for the estimation.

S.Y.B.Sc. Practical Chemistry 122 Appendix

3. Estimation of H2O2 :

(i) H2O2 solution (Eq. wt. 17) : Dissolve 17 ml of 100 volume H2O2 in 1 litre distilled

water (distribute 8 to 12 ml of this stock solution in 100 ml volumetric flask).

(ii) 0.05 N KMnO4 solution (Eq. wt. 31.6) : Dissolve 158.0 g KMnO4 in water and dilute

to 10 litres with distilled water.

4. Estimation of Aspirin :

(i) 0.1 M HCl : 45 ml HCl, dissolve and dilute to 5 litres.

(ii) 1 M NaOH : 40 g NaOH, dissolve and dilute to 1 litre.

5. Estimation of Aluminium :

(i) 0.01 M EDTA solution : Dissolve 3.72 g of pure EDTA and dilute it to 1 litre.

(ii) Buffer solution of pH 10 : Take 17.5 g of ammonium chloride, add 143 ml of

conc. ammonia and dilute the solution to 250 ml with distilled water.

(iii) Al (III) stock solution : Dissolve 50 g aluminium in about 200 ml of water and add

10 ml conc. H2SO4 and dilute it to 1 litre with distilled water.

(iv) Eriochrome Black T : Dissolve 0.2 g of indicator powder in a mixture of 15 ml

triethanolamine and 5 ml ethyl alcohol.

(v) Stock solution of aluminium : Dissolve 48 gm of hydrated potassium aluminium

sulphate in minimum amount of water and finally dilute to one litre. The

concentration of the resulting solution is 0.1 M. Give any volume between 8 to 12 ml

for titration.

6. Estimation of Copper :

(i) 0.025 N Na2S2O3 solution : Dissolve 39.0 g of Na2S2O3 and dilute to 5 litres with

distilled water.

(ii) Copper stock solution : Dissolve 62.375 g of CuSO4 ⋅ 5H2O and dilute to 1 litre with

distilled water. Give 8 - 10 ml of this solution in 100 ml volumetric flask.

(iii) 10% KI solution : Dissolve 50 g of KI in 500 ml distilled water.

(iv) 2 N NaOH solution : Dissolve 40 g of NaOH in 500 ml distilled water.

(v) 2 N Acetic acid solution : Dissolve and dilute 57.5 ml of acetic acid to 500 ml in

distilled water.

(vi) Starch solution : Dissolve about 1.0 g soluble starch powder in 250 ml of hot distilled

water.


(C) INORGANIC CHEMISTRY PRACTICALS

Inorganic Qualitative Analysis

List of Inorganic Mixtures :

• ZnSO4 + MnSO4 + Borax

• CuSO4 + AlPO4

• Co3(PO4)2 + NiCO3

• CdCl2 + NH4Br

• (NH4)3PO4 + KBr

• MnCl2 + AlPO4

• ZnCO3 + Ni3(PO4)2

• BiCl3 + BaCl2 + Borax

• CoCO3 + NiCl2

• CuSO4 + (NH4)3PO4

• CuCO3 + FeSO4

• MgCO3 + MnCl2

• Al2O3 + Mg3(PO4)2 + MgCO3

• SrBr2 + KCl

• NH4Cl + KBr

• ZnCO3 + Ni3(PO4)2

• KCl + NH4Cl + Borax

• CO3(PO4)2 + KCl

• FeSO4 + K2SO4 + Borax

• ZnS + MgSO4

• MnCl2 + BaCl2 + Borax

• CaCl2 + Ba3(PO4)2

• Cd(NO3)2 + KNO2

• MgBr2 + SrBr2 + Borax

• KI + NiCl2

• CrPO4 + MnCl2

• SbCl3 + CuCl2 + Borax

• KBr + NH4Br + Borax

• CoCO3 + NH4Br

• ZnCO3 + MnCO3 + Borax

• FeSO4 + AlPO4

• SrBr2 + KI

• CaCl2 + Ba3(PO4)2

• MgCl2 + KBr

(D) ORGANIC CHEMISTRY EXPERIMENTS

1. Qualitative Analysis :

(i) Lucas reagent : Take 13.65 g of zinc powder in 10 ml conc. HCl and dilute to 100 ml

with distilled water (add more conc. HCl if required). Filter the solution and use it

freshly for detection of alcohols.


(ii) 2, 4 DNP reagent : Take 1 g of 2 : 4 dinitro phenyl hydrazine reagent and 30 ml

methyl alcohol. Stir well and add cautiously (dropwise) 4 ml of conc. H2SO4. Cool the

solution in an ice bath. (If clear solution is not obtained, add little conc. H2SO4.)

(iii) Schiff's reagent : Take 0.1 g of rosaniline hydrochloride in 200 ml water and pass SO2

gas into it, until the magenta colour disappears.

(iv) Fehling's solution A : Take 17.3 g of crystalline copper sulphate, add 250 ml of water

and few drops of conc. H2SO4 to make clear solution.

(v) Fehling's solution B : Take 8.65 g of Rochelle salt (sodium potassium tartarate), add

35 g of caustic soda (NaOH), dissolve in 250 ml of water. Mix equal volumes of

solutions A and B before use.

✍ ✍ ✍


APPENDIX - II

Table I : International Atomic Weights

Name of

element

Symbol Atomic

No.

Atomic

weight

Name of

element

Symbol Atomic

No.

Atomic

weight

Aluminium

Barium

Bromine

Calcium

Carbon

Chlorine

Chromium

Copper

Hydrogen

Iodine

Iron

Lead

Magnesium

Al

Ba

Br

Ca

C

Cl

Cr

Cu

H

I

Fe

Pb

Mg

13

56

35

20

6

17

24

29

1

53

26

82

12

26.98

137.36

79.916

40.08

12.011

35.457

52.01

63.64

1.008

126.91

55.85

207.21

24.32

Manganese

Mercury

Nickel

Nitrogen

Oxygen

Phosphorous

Potassium

Silver

Sodium

Strontium

Sulphur

Tin

Zinc

Mn

Hg

Ni

N

O

P

K

Ag

Na

Sr

S

Sn

Zn

25

80

28

7

8

15

19

47

11

38

16

50

30

54.94

200.61

58.69

14.008

16.000

30.975

39.100

107.880

22.997

87.63

32.066

118.70

65.38

Table II : Molecular Weights and Equivalent Weights of Some Common Substances

Name of the Substance Formula Mol. Wt. Eq. Wt.

Acids : Sp. Gr.

Hydrochloric acid (11.3 N) [1.18]

Nitric acid (16 N) [1.42]

Sulphuric acid (36 N) [1.84]

Acetic acid (16 N) [1.05]

Oxalic acid

Bases :

Sodium hydroxide

Potassium hydroxide

Sodium carbonate (anhydrous)

Potassium carbonate (anhydrous)

Sodium bicarbonate

HCl

HNO3

H2SO4

CH3COOH

H2C2O4 : 2H2O

NaOH

KOH

Na2CO3

K2CO3

NaHCO3

36.5

63.0

98.0

60.0

126.0

40.0

56.0

106.0

138.0

84.0

36.5

63.0

49.0

60.0

63.0

40.0

56.0

53.0

69.0

84.0


Oxidising and Reducing Agents

Name of the Substance Formula Mol. Wt. Eq. Wt.

Bases :

Potassium permanganate

Potassium dichromate

Iodine

Hydrogen peroxide

Ferrous sulphate

Ferrous ammonium

sulphate (Mohr's salt)

Sodium thiosulphate

Some common compounds :

Ammonium chloride

Ammonium sulphocyanide or

Ammonium thiocyanate copper

sulphate (in iodometric titration)

Potassium hydrogen phthalate

Potassium oxalate

Sodium oxalate

KMnO4

K2Cr2O7

I2

H2O2

FeSO47H2O

FeSO4(NH4)2SO4

6H2O

Na2S2O3.5H2O

NH4Cl

NH4CNS

CuSO4.5H2O

C8H5O4K

K2C2O4.H2O

Na2C2O4H2O

158.0

294.0

254.0

34.0

278.0

329.0

248.0

53.7

76.0

249.7

204.0

184.0

134

31.6

49.0

127.0

17.0

278.0

392.0

248.0

53.7

76.0

249.7

204.0

92.0

92.0

Approximate Compositions of Some Common Acids and Bases

Reagent Mol. wt. % by weight Specific

gravity

Normality Volume per litre

for 1 N solution

Hydrochloric acid

Nitric acid

Sulphuric acid

Acetic acid (glacial)

Perchloric acid

Ammonia (aq.)

36.46

63.00

98.00

60.05

100.45

17.03

35

70

96

99.5

70

27

1.18

1.42

1.84

1.05

1.66

0.9

11.3

16.0

36.0

17.4

11.6

14.3

89

63

28

58

86

70


Some Combinations for Acetic Acid - Sodium Acetate Buffers

Volume of 0.2 M

Acetic Acid in ml

Volume of 0.2 M

Sodium Acetate in ml

pH at 25°°°°C

9.5

9.0

8.0

7.0

6.0

5.0

4.0

3.0

2.0

1.0

0.5

0.5

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

9.5

3.42

3.70

4.05

4.27

4.45

4.63

4.80

4.99

5.23

5.57

5.89

− − −

UNIVERSITY OF PUNEUNIVERSITY OF PUNEUNIVERSITY OF PUNEUNIVERSITY OF PUNE

SCHEMESCHEMESCHEMESCHEME OF CHEMISTRY PRACTICOF CHEMISTRY PRACTICOF CHEMISTRY PRACTICOF CHEMISTRY PRACTICAL (CHAL (CHAL (CHAL (CH----223)223)223)223) EXAMINATIONEXAMINATIONEXAMINATIONEXAMINATION

1. Chemistry Practical Examination will be of six hours duration and of 100 marks.

2. Candidate has to perform two experiments as follows :

Questions Marks

Q. 1 Physical Experiment …………… 35

OR

Analytical Experiment …………… 35

Q. 2 Inorganic Qualitative Analysis …………… 35

OR

(a) Organic Qualitative Analysis …………… 20

(b) Organic Preparation …………… 15

Q. 3 Oral Examination (Compulsory) …………… 10

Total 80

Instructions :

1. Chemistry practical examination will be of six hours duration and of 80 Marks.

2. Candidates have to perform two experiments as above.


3. For Q.1 half of the students in the batch should be given Physical Chemistry experiments,

whereas half of the students in the same batch should be given Analytical Chemistry

experiments.

4. For Q.2 half the students in the batch should be given Organic Chemistry experiments

whereas half of the students in the same batch should be given Inorganic Chemistry

experiments.

Note :

Books/Type written/cyclostyled/printed material will be allowed during the examination.

RECORD EXPERIMENT IN JOURNAL AS

(Left Hand Page) (Right Hand Page)

Pg. No. Pg. No.

Date : Expt. No.

Title

Diagram :

Equations :

Observations :

Calculations :

* If this space is not sufficient, continue on next

left hand side page.

Aim :

Apparatus :

Chemicals :

Procedure :

(1)

(2)

(3)

Result Table :

* If this space is not sufficient, continue on next

right hand side page.

✍ ✍ ✍

practical chemistry (ch - 223) b. sc g. s. gugale a. v. nagawade r. a. pawar s. s. jadhav v. d....

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