PRACTICAL CHEMISTRY B. Sc G. S. Gugale A. V. Nagawade R. A. Pawar S. S. Jadhav V. D. Bobade A. D. Natu D. R. Thube P. C. Mhaske L. K. Nikam Nirali Prakashan

A Book Of

PPRRAACCTTIICCAALL CCHHEEMMIISSTTRRYY T.Y.B.Sc. Chemistry : CH-347, CH-348, CH-349 As Per New Revised Syllabus of Savitribai Phule Pune University With Effect from June 2015

Dr. G. S. Gugale Head & Associate Professor,

Dept. of Chemistry, H. V. Desai College, PUNE

Dr. A. V. Nagawade Dr. R. A. Pawar Associate Professor, Dept. of Chemistry, Associate Professor, Dept. of Chemistry Ahmednagar College, Baburaoji Gholap College, AHMEDNAGAR. Sangvi, PUNE.

Dr. S. S. Jadhav Dr. V. D. Bobade Vice-Principal, Head & Associate Professor, Associate Professor, Dept. of Chemistry, New Arts, Commerce and Science College, H.P.T. Arts, R.Y.K. Science College, AHMEDNAGAR. NASHIK.

Dr. A. D. Natu Dr. D. R. Thube Ex. Head, Dept. of Chemistry, Head, PG Dept. of Chemistry, Abasaheb Garware College, New Arts, Commerce & Science College, PUNE. Parner, AHMEDNAGAR.

Dr. P. C. Mhaske Dr. L. K. Nikam Associate Professor, Associate Professor, Dept. of Chemistry, Dept. of Chemistry Baburaoji Gholap College, S. P. College, PUNE. Sangvi, PUNE.

Price ` 240.00

N1848

T.Y.B.Sc. : Practical Chemistry ISBN 978-93-51645-84-9 Second Edition : July 2017 © : Authors The text of this publication, or any part thereof, should not be reproduced or transmitted in any form or stored in any computer storage system or device for distribution including photocopy, recording, taping or information retrieval system or reproduced on any disc, tape, perforated media or other information storage device etc., without the written permission of Authors with whom the rights are reserved. Breach of this condition is liable for legal action. Every effort has been made to avoid errors or omissions in this publication. In spite of this, errors may have crept in. Any mistake, error or discrepancy so noted and shall be brought to our notice shall be taken care of in the next edition. It is notified that neither the publisher nor the authors or seller shall be responsible for any damage or loss of action to any one, of any kind, in any manner, therefrom.

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Preface …

It gives us an immense pleasure to place this “Text Book of Practical Chemistry” in

the hands of T.Y.B.Sc. students. The Book has been written according to the revised

semester pattern of Savitribai Phule Pune University implemented from June 2015.

The book is divided into three sections : (i) Physical, (ii) Inorganic, (iii) Organic

Chemistry Practicals. An attempt is made to give appropriate theoretical background of

each experiment. Procedures, observation tables and calculations are made more simple.

At the end of each experiment, viva-voce questions are given for practice.

All the experiments are written keeping in view the specification of learning

objectives. Micro techniques, double burette methods are adopted wherever possible.

Sincere efforts are made to make the practicals more simple.

We feel that the book will be very useful to the students during their regular

practicals as well as at the time of examination.

We are thankful to Mr. Dineshbhai Furia and Mr. Jignesh Furia for taking keen

interest in publishing this book.

We are also thankful to staff of Nirali Prakashan especially, Mr. Ilyas Shaikh, Mr. Kiran

Velankar and Mrs. Anjali Muley for completing this book on time.

We are also thankful to Marketing staff especially, Mr. Nilesh Deshmukh and others

for coordinating the matter well in time.

Suggestions for improving the book will be highly appreciated.

Authors

Contents …

Physical Chemistry Practicals CH - 347 Group A

1. Chemical Kinetics (Any Five) 1. To study the effect of concentration of the reactants on the rate of

hydrolysis of an ester. … 7

2. To compare the relative strength of HCl and H2SO4 by studying the kinetics of hydrolysis of an ester.

… 9

3. To compare the relative strength of HCl and H2SO4 by studying the kinetics of inversion of cane sugar using Polarimeter.

… 12

4. To study the kinetics of iodination of acetone. … 17 5. To determine the first order velocity constant of the decomposition of

hydrogen peroxide by volume determination of oxygen. … 20

6. To determine the energy of activation of the reaction between potassium iodide and potassium persulphate.

… 23

7. To determine the order of reaction between K2S2O8 and KI by half-life method.

… 27

2. Viscosity 8. To determine the molecular weight of a high polymer by using solutions of

different concentrations. … 32

3. Adsorption : 9. To investigate the adsorption of oxalic acid/acetic acid by activated

charcoal and test the validity of Freundlich/Langmuir isotherm. … 39

4. Phenol-Water System 10. To study the effect of addition of salt on critical solution temperature of

phenol-water system. … 43

5. Transport Number 11. To determine the transport number of cation by moving boundary method. … 476. Refractometry (Any Two) 1. To determine the specific refractivities of the given liquids A and B and

their mixture and hence determine the percentage composition of their mixture C.

… 55

2. To determine the molecular refractivity of the given liquids A, B, C and D. … 57 3. To determine the molar refraction of homologues methyl, ethyl and propyl

alcohol and show the constancy contribution to the molar refraction by −CH2 group.

… 59

Group B 1. Colorimetry (Any Two)

1. Determination of λmax and concentration of unknown solution of KMnO4 in 2 N H2SO4.

… 63

2. Determination of λmax and concentration of unknown solution of CuSO4. … 64

3. To titrate Cu2+ ions with EDTA photometrically. … 68

4. To determine the indicator constant of methyl red indicator. … 70

2. Potentiometry (Any Three) 1. To prepare standard 0.2 M Na2HPO4 and 0.1 M citric acid solution, hence

prepare four different buffer solutions using them. Determine the pKa value of these and unknown solutions.

… 80

2. To determine the concentrations of strong acid and weak acid present in the mixture by titrating with strong base.

… 83

3. To determine the formal redox potential of Fe2+/Fe3+ system potentiometrically.

… 86

4. To determine the amount of NaCl in the given solution by potentiometric titration against silver nitrate.

… 90

5. To determine the amount of Cl− and Br− from the given halide mixture by titrating with silver nitrate solution.

… 93

3. pH Metry (Any Two) 1. To determine the degree of hydrolysis of aniline hydrochloride. … 101 2. To determine the pKa value of given weak acid by pH-metric titration with

strong base. … 103

3. To determine the dissociation constant of oxalic acid by pH-metric titration with strong base.

… 105

4. To determine pH of various mixtures of sodium acetate and acetic acid in aqueous solution and hence to find the dissociation of acetic acid.

… 108

4. Radioactivity (Any One)

1. To determine plateau voltage of the given GM counter. … 113

2. To determine the resolving time of GM counter. … 115

3. To determine Emax of beta particles. … 1165. Conductometry (Any Two) 1. To determine the cell constant of the given cell using 0.01 M KCl solution

and hence determine dissociation constant of a given monobasic weak acid. … 124

2. To estimate the amount of lead present in the given solution of lead nitrate by conductometric titration with sodium sulphate.

… 128

3. To investigate the conductometric titration of any one of the following : (a) Strong acid against strong base. (b) Strong acid against weak base. (c) Strong base against weak acid. (d) Weak acid against weak base.

… 131 … 131 … 132 … 134 … 135

Appendices … 139Inorganic Chemistry Practicals

CH - 348 (A) Gravimetric Estimations (Any Three) 1. Fe as Fe2O3. … 143 2. Nickel as Ni - DMG. … 145 3. Al as Aluminium oxide. … 148 4. Gravimetric estimation of Ba as BaSO4 using homogeneous precipitation

method. … 151

(B) Volumetric Estimations (Any Four) 1. Mn by Volhard’s method. … 154 2. Estimation of NO−

2 by using KMnO4. … 159 3. Estimation of % purity of given sample of Sodium chloride. … 163 4. Analysis of Brass - Estimation of copper by Iodometry. … 166 5. Fertilizer analysis (PO4)3−. … 171(C) Inorganic Preparations (Any Four) 1. Preparation of Hexamminenickel(II) chloride, [Ni(NH3)6]Cl2. … 176 2. Preparation of Potassium trioxalatoferrate(III) trihydrate,

K3[Fe(C2O4)3].3H2O. … 179

3. Preparation of Tetraamminecopper(II) sulphate monohydrate, [Cu(NH3)4]SO4.H2O.

… 182

4. Preparation of Manganese(III) acetylacetone [Mn(acac)3]. … 184 5. Preparation of Tris (Thiourea) Copper(I) chloride [Cu(Thiourea)3]Cl. … 186(D) Colorimetric Estimations (Any Two) 1. Iron by thiocyanate method. … 188 2. Cobalt by using R-nitroso salt method. … 192 3. Titanium by H2O2. … 195(E) Separation of Binary Mixture of Cations by Column Chromatography … 202

(3 Mixtures) (One Mixture should be colourless, Zn + Al, Zn + Mg)

OR (E) Flame Photometry Analysis (Any Three) 1. Estimation of Na by flame photometry by calibration curve method. … 208 2. Estimation by Na by flame photometry by regression method. … 211

3. Estimation of K by flame photometry by calibration curve method. … 212 4. Estimation of K by flame photometry by regression method. … 215 (F) Inorganic Qualitative Analysis (4 Mixtures including Borates and Phosphates) … 217 Appendices … 238

Organic Chemistry Practicals CH - 349

(A) Separation of Binary Mixtures and Qualitative Analysis (Minimum 8 Mixtures) (a) Solid-Solid (4 Mixtures). … 247 (b) Solid-Liquid (2 Mixtures). … 247 (c) Liquid-Liquid (2 Mixtures). … 249 Atleast one mixture from each of the following should be given : Acid-Base, Acid-Phenol,

Acid-Neutral, Phenol-Base, Phenol-Neutral, Base-Neutral, Neutral-Neutral. (i) Separation and qualitative analysis of the binary mixtures should be carried out on

micro scale using micro scale bits. (ii) Name and structure of the separated components of the binary mixture is not

necessary. (iii) Students are expected to record the Type, Preliminary tests, Physical constants,

Elements and Functional groups only. (iv) The purified samples of the separated components should be submitted. (B) Organic Estimations (Any Four) (1) Estimation of Acetamide. … 265 (2) Estimation of Glucose. … 268 (3) Estimation of Ethyl benzoate. … 271 (4) Determination of molecular weight of monobasic/dibasic organic acids by

volumetric methods. … 274 (C) Organic Preparations (Any Eight) (1) Preparation of Adipic acid from Cyclohexanone (Oxidation by conc. HNO3) … 279 (2) Benzoquinone from Hydroquinone (Oxidation by KBrO3/K2CrO3) … 280 (3) p-nitroacetanilide from Acetanilide (Nitration) … 282 (4) 2-Naphthyl methyl ether from 2-naphthol (Methylation by DMS, NaOH) … 283 (5) Hippuric acid by Glycine (Benzoylation) … 285 (6) p-iodonitrobenzene from p-nitroaniline (Sandmeyer reaction) … 286 (7) Benzoic acid from Ethyl benzoate (Ester hydrolysis) … 288 (8) p-Bromoacetanilide from Acetanilide (Bromination) … 289 (9) Paracetamol from p-Hydroxyaniline (Acetylation) … 290 (10) Ethyl benzene from Acetophenone (Wolff Kishner reduction) … 292 Appendices … 295

− − −

1

Course - I

CH-347

PHYSICAL CHEMISTRY

PRACTICALS

T.Y.B.Sc. Practical Chemistry (Physical Practicals) Chemical Kinetics

2

STRUCTURE OF PRACTICAL EXAMINATION

COURSE – I CH – 347

Physical Chemistry Practicals

Marks Q.1 One Experiment from Group – A 35 Q.2 One Experiment from Group – B 35 Q.3 Oral 10


3

(GROUP - A)

1…

Chemical Kinetics (Any Five)

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

− dCdt = k ⋅ Cn1

1 ⋅ Cn22 ⋅ Cn3

3 …

where, k = constant, and C1, C2, C3 etc. are concentrations of reactants 1, 2, 3, etc. then order of reaction = n1 + n2 + n3 + …


4

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 reaction. The rate expression is Rate = k where k is a 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

− dCdt = 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 = − dCdt =

d(a − x)dt

= − dadt +

dxdt

but − dadt = 0 since ‘a’ is constant

∴ Rate = dxdt

According to law of mass action the rate of a reaction is directly proportional to the concentration of reactant ‘A’ at an instance. ∴ Rate ∝ (a − x) Combining above two equations,

dxdt = 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


5

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


6

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 = dxdt = 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 concentration of ‘A’ and ‘b’ moles/lit be the initial concentration 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 = dxdt = 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 ba + log ⎝⎜

⎛⎠⎟⎞(a − x)

(b − x)

∴ log (a − x)(b − x) = ⎝⎜

⎛⎠⎟⎞k (a − b)

2.303 t − log ba

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

∴ By plot the graph of log (a − x)(b − x) against t.

The nature of graph is a straight line with intercept log ba

Slope = k (a − b)

2.303

∴ k = 2.303 × slope

(a − b)

Units of k are conc−1 time−1.


7

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

Experiment No. 1 Aim : To study the effect of concentration of reactants on the rate of hydrolysis of an ester. 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 N H2SO4), 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

dxdt = 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 in two sets. Set I : 5 ml methyl acetate + 100 ml 0.5 N HCl 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.


8

4. After 10 to 15 minutes the liquids will have assumed 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. Similarly carry out the experiment for set II in the same manner as in the set I. The only difference in set two is the use of 10 ml of methyl acetate in place of 5 ml.

9. Infinite time readings : Take 25 ml of the reaction mixture in a flask. Cork it and keep in water bath at around 500°C temperature, for 90 minutes to complete the hydrolysis. Finally, titrate 5 ml of it against 0.1 N NaOH and record the readings as T∞.

Set II : 10 ml methyl acetate + 100 ml 0.5 N HCl 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

10 20 30 40 50

Prepare similar observation table for set II. 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 for first and second set 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. 1: log (a − x) vs. t

(ii) Plot the graph of log a

(a − x) against t.

Slope = k

2.303

∴ k = 2.303 × slope

log

time/min

a

(a x)�

Fig. 2: log a

Q − x vs. t

Straight line nature of both the graphs indicates that the reaction of hydrolysis of methyl acetate is first order one. Result Table :

Set I k (min–1) Set II k (min–1)

By calculations By graph By calculations By graph

Conclusion : k value for first and second set are nearly constant. This shows that the rate constant is independent of initial concentration of reactant.

Experiment No. 2 Aim :

9


10

To compare the relative strength of HCl and H2SO4 by studying the kinetics of hydrolysis of an ester. Chemicals : Methyl acetate 0.5 N HCl, 0.5 N H2SO4, 0.1 N NaOH, phenolphthalein indicator, ice cold water etc. Apparatus : Stoppered bottles, pipettes, burette, measuring cylinder, beakers, stop watch, water bath etc. Theory : The hydrolysis of ester is catalyzed by H+ ions.

CH3COOCH3 + H2O H+

⎯→ CH3COOH + CH3OH Therefore, the rate of hydrolysis increases with increase in concentration of H+ ions. If the hydrolysis is carried out in presence of hydrochloric acid and of sulphuric acid independently, with their equal concentrations, the rate constants for hydrolysis, kHCl and kH2SO4, depend on the degree of dissociation of HCl and H2SO4. The rate of hydrolysis using either of the catalyst is proportional to the concentration of methyl acetate (reactant) and in turn to the concentration of acetic acid (product). Experimental measurement of the rate of reaction (kHCl and kH2SO4) can be studied by determining the amount of acetic acid formed in the course of reaction. Amount or concentration of acetic acid formed at different time intervals can be easily determined by titrating the definite amount of reaction mixture with standard alkali. Hence, the relative strength can be calculated as

Relative strength = kHCl

kH2SO4

Thus the strength of the two acids can be compared by kinetic measurements. Procedure : Perform the experiment in two sets. Set I : 5 ml methyl acetate + 100 ml 0.5 N HCl : 1. Take 100 ml 0.5 N HCl in a clean and dry stoppered bottle (bottle 1). In another

stoppered bottle take 5 ml methyl acetate (bottle 2). 2. Keep the bottles in a water bath for 10-15 minutes to attain the constant

temperature. 3. Fill the burette with 0.1 N NaOH solution upto the zero mark. 4. Take few pieces of crushed ice or around 30 ml ice cold water in a conical flask and

add to it 2-3 drops of phenolphthalein indicator. 5. Transfer the 100 ml 0.5 N HCl in bottle 1 to bottle 2 containing 5 ml methyl acetate.

Shake the mixture and start the stop watch. Note the time of mixing as zero time.


11

6. Immediately pipette out 5 ml reaction mixture in titration flask containing ice and phenolphthalein indicator. (Place the bottle in water bath).

7. Shake it well and titrate with 0.1 N NaOH till faint pink colour appears and persists for 30 seconds to the solution.

8. Record the titration reading as T0.

9. Shake the reaction mixture from time to time and titrate 5 ml of reaction mixture similarly at the successive intervals of 10, 20, 30, 40 and 50 minutes. Record the titration readings as Tt.

10. Infinity reading (T∞) : Take 25 ml of the reaction mixture in a flask. Cork it and keep in water bath at around 50°C temperature, for 10 minutes to complete the hydrolysis. Finally, titrate 5 ml of it against 0.1 N NaOH and record the readings as T∞.

11. Similarly carry out the experiment for set II in the same manner as the set I. The only difference in set two is the use of H2SO4 in place of HCl.

Set II : 5 ml methyl acetate + 100 ml 0.5 N HCl :

Observations :

Set I : 5 ml methyl acetate + 100 ml 0.5 N HCl

Initial reading (T0) = ……...ml

Infinity 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

aa − x k (min–1)

10

20

30

40

50

Prepare similar observation table for set II.

Calculations :

1. 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 for first and second set at 10, 20, 30, 40 and 50 min. intervals. Calculate mean value of k for each set.

∴ Relative strength = kHCl

kH2SO4

2. Rate constant (k) by graphical method : (i) Plot the graph of log (a – x) against time for both the sets, straight line nature of graph shows that reaction is first order.

Slope = −k

2.303

∴ −k = 2.303 × slope log (a x)�

time/min

Fig. 1: log (a − x) vs. t

(ii) Plot the graph of log [a/(a – x)] against time, the nature of graph is a straight line passing through the origin.

Slope = k

2.303

∴ k = 2.303 × slope

Find out kHCl and kH2SO4 from both the graphs. time/min

loga

(a x)�

Fig. 2: log a

a − x vs. t

Result Table :

Sr. No. Description By calculations By graph 1. 2.

3.

Rate constant (kHCl) Rate constant (kH2SO4)

Relative strength = ⎝⎜⎛

⎠⎟⎞kHCl

kH2SO4

……… min–1

……… min–1

………

……… min–1

……… min–1

………

Conclusion : For equal concentration of HCl and H2SO4, hydrochloric acid is more dissociated than sulphuric acid.

∴ kHCl

kH2SO4 > 1

12


13

Experiment No. 3 Aim : To determine the relative strength of HCl and H2SO4 by studying the kinetics of inversion of cane sugar using polarimetry. Apparatus : Polarimeter, sodium lamp, stop watch, conical flask, etc. Chemicals : Hydrochloric acid (1 N), sulphuric acid (1 N), distilled water, sugar etc. Theory : Substances like cane sugar that are capable of rotating the plane polarized light are optically active and the phenomenon is called Optical Activity. Substances that rotate the plane of polarized light to the right are called dextro rotatory and those that rotate it to the left are called laevo rotatory. In fact, one name for glucose, dextrose, refers to the fact that it causes plane of polarization to rotate to the right or dexter side (clockwise). Similarly, levulose, commonly known as fructose, causes the plane of polarization to rotate to the left (counter-clockwise). Fructose is even more strongly laevo rotatory than glucose is dextro rotatory. Hydrolysis of cane sugar is termed as “inversion of cane sugar” since with the progress in hydrolysis, the sign of rotation changes from positive to negative (the conversion causes the direction of rotation to "invert" from right to left or dextro to laevo). The amount by which the light is rotated is known as the angle of rotation. The fall in optical rotation is proportional to the amount of fructose formed and the amount of cane sugar hydrolyzed. Polarimeter is the instrument used to measure the angle of rotation of a plane polarized light. Sucrose molecule has number of asymmetric carbon atoms and can be hydrolyzed to give glucose and fructose.

C12H22O11 + H2O H+

⎯⎯→ C6H12O6 + C6H12O6 Cane sugar Glucose Fructose The hydrolysis of cane sugar is catalyzed by H+ ions. Therefore, the rate of hydrolysis increases with increase in concentration of H+ ions. If the hydrolysis is carried out in presence of hydrochloric acid and of sulphuric acid independently, with their equal concentration, the rate constants for hydrolysis, kHCl and kH2SO4, depend on degree of dissociation of HCl and H2SO4. Experimental measurement of change in optical rotation with the inversion of cane sugar can be used to determine the rate constants kHCl and kH2SO4. Hence, the relative strength can be calculated as

Relative strength = kHCl

kH2SO4


Polarimeter :

A simple polarimeter consists of a long tube with flat glass ends, into which the sample is placed. At each end of the tube is a Nicol prism, one is called polarizer and the other analyzer. Polarizer is kept fixed and converts the monochromatic light into plane polarized beam. The angle of rotation of plane polarized beam by the sample is measured by passing monochromatic light through polarizer. This beam gets rotated as it passes through the sample already filled in the polarimeter tube. The sample is usually prepared as a solution where the optically active substance is dissolved in an optically inactive chemical such as distilled water or methanol. After passing through the sample the analyzer is rotated manually. When the analyzer is rotated to the proper angle, the maximum amount of light will pass through and shine onto a detector (telescope).

0o

90o

180o

+90o

Analyzer, A

Polarimeter tube, T

Fixed polarizer, P

Light source, S

Unpolarimeter light

, E

Polarizedlight

Fig. 1

The principle of the set-up of the polarimeter consists of the light source (S), polarizer (P), analyzer (A) with angle measuring scale and eyepiece (E). Two Nicol prisms are placed between the source of monochromatic light, S (usually sodium lamp) and the eyepiece. Light passes through the polarizer and gets plane polarized (vibrations are restricted to one plane only). This plane polarized light then passes through the second Nicol prism, A, the analyzer and reaches the eye. The field of view is observed through the telescope eyepiece (E). However, when the axis of the analyzer is perpendicular to that of the polarizer, no light reaches the eye. The polarizer and analyzer are arranged in such a way that no light reaches the eye. A tube T filled with cane sugar solution is placed between the two prisms. Some light

14


15

can be now seen through the telescope eyepiece. The analyzer should be rotated so that once again no light reaches the eye.

Note that some light could be seen when sugar solution was placed between the prisms whose axes were perpendicular to each other and no light was visible originally. So what made this light visible on the other end when neither of the two prisms had been moved ? Light is visible only if the plane of polarization has been rotated by some way. Thus the plane of polarization must have been rotated by cane sugar solution. Procedure : 1. Prepare 20% sugar solution by dissolving 20 gram of sugar in about 40 ml distilled

water dilute it to 100 ml using volumetric flask. Filter the solution. 2. Take two different clean and dry conical flasks. Place 25 ml of sugar solution in one

and 25 ml of 1N HCl in other. Keep them in a thermostat at room temperature. 3. Clean and dry the polarimeter tube. 4. Fill the polarimeter tube with distilled water. Ensure that no air bubble is trapped

inside the tube. 5. Observe the field (the light passed through analyzer) through the eyepiece. The

circular field is seen either with uniform illumination or with two halves illuminated distinctly. Rotate the analyzer and adjust it for equally illuminated halves. This reading is added or subtracted subsequently as the case may be.

6. Make polarimeter tube empty by removing water from it. 7. Mix the two solutions in the bottles and shake well. Transfer immediately the reaction

mixture in the polarimeter tube. Keep the tube in the polarimeter jacket and close it. Record the mixing time.

8. Rotate the analyzer and adjust it for equally illuminated halves. Note this reading as initial or zero time reading, α0.

9. Take similar readings at the intervals of 10, 20, 30, 40 and 50 minutes and record them as αt.

10. Infinity reading : The reaction mixture is heated at about 60 to 70°C for half an hour for reaction to complete. It is then cooled and poured into the polarimeter tube to measure the angle of rotation. This reading is α∞.

11. Repeat the experiment in the same manner by using 1 N H2SO4 in place of HCl. Observation Table : Set I = 25 ml 20% sugar solution + 25 ml 1N HCl Initial reading (α0) = Infinity reading (α∞) = ∴ Initial concentration (a) = α∞ – α0 = ……….


Time in min.

Angle of rotation ‘αt’ α = α0 – αt a – x = αt – α∞ log (αt – α∞) log

α0 – α∞αt – α∞

10 20 30 40 50

Set II : 25 ml sugar solution + 25 ml 1 N H2SO4 Prepare similar observation table for Set II. Calculations : (a) k by calculations :

k = 2.303

t log a

a − x

k = 2.303

t log α0 − α∞

αt − α∞

where, a, the initial concentration of sugar = α0 – α∞ a – x, the concentration of sugar at time t = αt – α∞ angle of rotation at zero time t = αt − α∞ angle of rotation at time t = αt angle of rotation at infinite time = α∞ Calculate the rate constant, kHCl, for Set I. Similarly calculate the rate constant, kH2SO4 for Set II.


kH2SO4

(b) k by graph : (i) Plot the graph of log (a − x) against t. The nature of the graph is a straight line with negative slope.

Slope = −k

2.303

∴ k = −2.303 × slope

log ( )� �t ��

Time (min) Fig. 2

(ii) Plot the graph of log α0 − α∞

αt − α∞ against time. Nature of graph is a straight line passing

through origin.

16


17

Slope = k

2.303

∴ k = 2.303 × slope

Calculate kHCl and kH2SO4 from slope of both the graphs.


kH2SO4

log� �0 �

�

� �t ��

Time (min) Fig. 3

Result Table :

Sr. No. Description By calculations By graph

1. 2.

(kHCl) (kH2SO4)

……… min–1 ……… min–1

……… min–1 ……… min–1

3. Relative strength =

kHCl

kH2SO4

……… ………

Conclusion : Since experimentally it is verified that relative strength is greater than one, i.e.

kHCl

kH2SO4 > 1

it indicates that hydrochloric and dissociates more than sulphuric acid under the identical set of conditions.

Experiment No. 4

Aim :

To study the kinetics of iodination of acetone.

Apparatus :

Stoppered bottles (five), burette, pipettes, stop watch, conical flask, etc.

Chemicals :

1 M H2SO4, acetone, 0.1 M I2 in 10% KI, 1 M sodium acetate, 0.01 M sodium thiosulphate.

Theory :

Iodine reacts with acetone in the presence of acid to form iodoacetone and hydrogen iodide.

C

O

H C3 CH3

+ I2

H+

C

O

H C3 CH2I

+ I�

H+

+

acetone iodoacetone


The reaction is catalyzed by H+ ions and the rate of reaction is found to be independent of iodine concentration.

Rate = −d[I2]

dt = k' [CH3COCH3]a [H+]b [I2]c

Since, a very important characteristic of this reaction is that it turns out to be zero order with respect to iodine (i.e. c = 0). This means that the rate of the reaction does not depend on [I2] at all as [I2]0 = 1, no matter what the value of [I2] is as long as it is NOT zero. This is consistent with the mechanism.

+

18

C

O

C

HO

H C3 CH3

Acetone

H+

+ H C3 CH3

Ion

C

OH+

+ H O2

C

HO

H C3 CH2

Enol

H O+

3+

H C3 CH3

Ion

C

O

H C3 CH2

Iodoacetone

H+

+CH C3 CH2

Enol

OH I

+ I�

I2+

Since the reaction is said to be zero order with respect to iodine concentration the rate law simplifies to the equation Rate = k' [(CH3)2C=O]a [H+]b If acetone and acid is taken in large excess relative to [I2], then change in their concentration will be minimal during the course of reaction. This means [CH3COCH3] and [H+] will remain (essentially) constant. Therefore, the rate law can be written as :

Rate = −d[I2]

dt = k' [(CH3)2C=O]a [H+]b = constant = k

A plot of [I2] versus time is relatively a straight line whose slope is the reaction rate equal to k.

T.Y.B.Sc. Practical Chemistry Chemical Kinetics (Physical Practicals)

0.1

0.2

0.3

0.4

0.5

0.6

00 10 20 30

[ ]I o2

[ ]I2

Slope = k

Slope = k

Time (min) Fig. 1

Decrease in concentration of iodine in the course of reaction can be used for determination of the order with respect to iodine. Experimentally unreacted iodine is titrated against standard sodium thiosulphate solution using starch as a indicator. Sodium acetate or sodium bicarbonate can be used to stop the reaction while performing the titration. Procedure : 1. Prepare four sets of mixture solutions from stock in four stoppered bottles as given

below.

Set I II III IV

Acetone (ml) Water (ml)

Sulphuric acid (ml)

10 70 10

10 65 10

15 55 10

15 50 10

2. Take about 100 ml iodine solution in another bottle say bottle V. Keep all these bottles in water bath to attain uniform temperature.

3. Fill a burette with 0.01 M sodium thiosulphate solution. 4. Pipette out 10 ml of sodium acetate solution each in 4 to 5 conical flasks and keep

them ready for titrating reaction mixture. 5. After around 10 minutes add 10 ml of iodine solution from bottle by means of

pipette to bottle no. 1 and start the stop watch. 6. After three minutes, pipette out 10 ml of the reaction mixture in a conical flask

containing 10 ml of sodium acetate solution.

7. Titrate the reaction mixture against 0.01 M sodium thiosulphate solution. Add 2 ml of starch indicator just before the solution becomes faint yellow. At the end point solution turns blue to colourless.

8. Shake the reaction mixture from time to time and similarly titrate 10 ml of it at successive intervals of 6, 9, 12, 15, 18 and 21 minutes. Note the titration readings in the observation table.

19


9. Similarly carry out the reaction between acetone solution in bottle no. II, III and IV with 15 ml, 20 ml and 25 ml of iodine solution in bottle no. V respectively and repeat the titration procedure as mentioned in steps 4 to 8 above.

Observation Table :

Time in min. Burette reading in ml

Set I Set II Set III Set IV

3

6

9

12

15

18

21

Calculations : Plot the graph of titration reading against time for all four sets.

Burette reading(ml)

Time (min)

IV

III

II

I

Fig. 2

Straight line graph with negative slope indicates that the rate of reaction is independent of the concentration of iodine. ∴ k = − slope Conclusion : All the graphs are straight (and parallel) lines with same slopes (k). It proves that order of this reaction is zero with respect to reactant iodine.


20


To determine the first order velocity constant of decomposition of hydrogen peroxide by volume determination of liberated oxygen. Apparatus : Gas burette, conical flask, magnetic stirrer, rubber tubings, etc. Chemicals : 0.1 M KI, 3% H2O2 solution etc. Theory : Hydrogen peroxide decomposes into oxygen and water spontaneously. It is an exothermic reaction with heat of reaction ΔH° = −98.2 kJ mol−1. Rate of decomposition is dependent on the temperature and concentration of the peroxide as well as on the pH of the reaction medium. Decomposition occurs more rapidly in alkali. In acidic medium this reaction follows second order kinetics, whereas in neutral medium it exhibits first order kinetics.

In neutral medium, decomposition is represented as 2H2O2 ⎯→ 2H2O2 + O2 ↑

Besides transition metals, potassium iodide catalyzes this reaction greatly. Since the reaction is catalyzed by I− ions, the rate of reaction increases with increase in concentration of I− ions. Elementary steps for this reaction are : (i) H2O2 + I− ⎯→ H2O2 + IO− (slow step) (ii) IO− ⎯→ I− + 1/2 O2 (fast step) H2O2 ⎯→ H2O + 1/2 O2 (overall reaction) The progress and hence the kinetics of the reaction can be studied by measuring the volume of oxygen liberated at different time intervals.

B

CE

D

A

F

Fig. 1: Experimental set up for measurement of volume of oxygen

liberated in the reaction Procedure : Perform the experiment in two sets. Set I : 10 ml, 0.1 M KI + 15 ml water + 5 ml 3% H2O2

21


22

1. Take 10 ml of KI (0.1N) solution in the reaction flask D. Add 15 ml distilled water to it, insert a magnetic needle and cork gently. Mount the flask on the stirrer.

2. Connect the gas burette B and burette A using a rubber tube. 3. Fill the burette with tap water and adjust the water level in gas burette B to zero level

mark by moving burette A upward down. 4. Connect open end of gas burette to reaction flask D using a glass and a rubber tube. 5. Add 5 ml 3% H2O2 solution to reaction flask and immediately cork it. Start the stop

watch. Note this time as zero time. 6. Start the stirrer. Care should be taken that oxygen leakage does not occur during the

experiment. 7. Equalize the two levels of water in two burettes after 3 minutes and measure the

volume of oxygen liberated in gas burette B. Record this volume in observation table. 8. Similarly record the volume of oxygen liberated as Vt at different time intervals

6, 9, 12, 15, 18, 21 and 24 minutes. 9. No more oxygen will be liberated when the reaction is completed, this volume of

oxygen is called as V∞. Observations : (i) Initial reading (V0) = ……ml (ii) Infinity reading (V∞) = ……ml a = V∞ – V0 = ……ml

Obs. No.

Time t (min) Volume of oxygen collected

Vt (ml)

(a – x) = V∞ – Vt

log (V∞ – Vt) k (min–1)

1 2 3 4 5 6 7 8

3 6 9 12 15 18 21 24

Set II : 20 ml, 0.1M KI + 5 ml distilled water + 5 ml 3% H2O2 Repeat the same procedure as in Set I for the Set II and write the observations in observation table for Set II. Calculations :


23

1. Rate constant (k) by calculations : The first order velocity constant is given by equation-

k = 2.303

t log V∞ − V0

V∞ − Vt

k = 2.303

t log a

a − x

log ( )� �t ��

Time (min) Fig. 2

Calculate the value of k for various time intervals for both the sets. Also calculate average rate constant (k) for each set. 2. Rate constant by graph : Plot the graph of log (V∞ − Vt) against time t. Nature of the graph is a straight line with negative slope.

∴ Slope = −k

2.303

∴ k = −2.303 × slope Calculate k from the slope for both the sets. Result Table :

Description Set By calculations By graph

Velocity constant k (min–1)

I II

Conclusion : Constant values of k and straight line nature of the graph proves that the reaction is of first order.

Experiment No. 6 Aim : To determine the energy of activation of the reaction between potassium iodide and potassium persulphate. Apparatus : Stoppered bottles, burette, pipettes, stop-watch, measuring cylinder, thermostat, etc. Chemicals : 0.1N K2S2O8, 0.1N KI, 0.002 N Na2S2O3, starch indicator, crushed ice or ice cold water etc. Theory : Activation energy and the pre-exponential factor A are two quantities of great importance in kinetics. Most chemical reactions are thermally activated; their rate increases


24

with temperature. Theories of reaction rates predict that not all the colliding molecules of reacting substances undergo chemical change. An energy barrier must be overcome in order to transform reactants into products. This barrier is referred to as the activation energy for the reaction. The fraction of reacting molecules that has an energy that exceeds the activation energy is increased by increasing the temperature. This allows the reaction to proceed faster. Thus, energy of activation is the minimum amount of energy required by the reactant molecules to undergo chemical reaction and form the product. The reaction between potassium persulphate and potassium iodide is a second order reaction. The overall reaction is K2S2O8 + 2KI ⎯→ 2K2SO4 + I2

or S2O−28 + 2I− ⎯→ 2SO

−24 + I2

The reaction is first order with respect to each reactant and hence second order overall. The reason is that the reaction occurs in two steps as represented below.

(i) S2O−28 + I− ⎯→ S2O8I3−

(ii) S2O8I3− + I− ⎯→ 2SO−24 + I

2−2

The progress of reaction can be studied by titrating the liberated iodine against the 0.002 N sodium thiosulphate using starch as an indicator. The rate constant for the second order reaction with unequal initial concentration can be calculated by using equation-

k = 2.303

t (a − b) log b (a − x)a (b − x)

where a and b are initial concentrations of reactants K2S2O8 and KI, (a – x) and (b – x) are their concentrations at time t. The above expression can be written as

k = 2.303

t (a − b) ⎝⎜⎛

⎠⎟⎞log

(a − x)(b − x) + log

ba

∴ k × t (a − b)

2.303 = log (a − x)(b − x) + log

ba

∴ log (a − x)(b − x) = ⎝⎜

⎛⎠⎟⎞k (a − b)

2.303 t − log ba

This equation is equivalent to equation y = mx + c and thus represents a straight line.

If the graph of log (a − x)(b − x) against time ‘t’ is plotted, a straight line graph with negative

slope is obtained.

∴ Slope = k (a − b)

2.303


25

∴ k = 2.303 × slope

(a −b)

Thus rate constant k can be obtained graphically. Rate of reaction depends on temperature. Rate constants can be determined at two different temperatures experimentally. The relationship between rate constants k1 and k2 at temperatures T1 and T2 is given by Arrhenius equation as –

log k2

k1 =

Ea

2.303 R ⎝⎜⎛

⎠⎟⎞T2 − T1

T1T2

where, Ea = Energy of activation R = Molar gas constant Procedure : Perform the experiment in two sets for two different temperatures. Set I : Reaction at lower temperature (T1 K) : 1. Take 20 ml 0.1 N KI and 20 ml distilled water in a clean and stoppered bottle and

10 ml 0.1 N K2S2O8 solution and 30 ml distilled water in another stoppered bottle. Keep these two bottles in water bath for 10-15 minutes to attain constant temperature. Record the temperature (T°C) of the water bath.

2. Fill a clean burette with 0.002 N sodium thiosulphate.

3. When the two solutions have attained the same temperature, mix them to start the reaction. Start the stop watch and note the mixing time as zero time.

4. Keep the reaction mixture in thermostat and shake it from time to time.

5. Keep a titration flask ready with around 30 ml crushed ice or ice-cold water.

6. Withdraw 10 ml of reaction mixture after 10 minutes from the start by pipette and transfer it to conical flask containing ice. Titrate the liberated iodine against sodium thiosulphate solution using starch indicator. Starch should be added near the equivalence point i.e. when the reaction mixture becomes faint yellow during titration.

7. Similarly titrate the reaction mixture at the interval of every 5 minutes.

8. Take seven such readings.

Observation Table :

Temperature of thermostat (t1) = …….. °C

Temperature (T1) = 273 + t1 = …… K

Time ‘t’ (min)

Volume of 0.002 N Na2S2O3 (ml)

(a – x) (ml)

(b – x) (ml)

log a − xb − x

k1 (min–1)


26

10

15

20

25

30

35

40

Set II : Reaction at higher temperature (t1 + 10°C) :

1. Maintain the temperature of the water bath at t1 + 10°C. Record the temperature as t2°C.

2. Repeat all the steps given in Set I in this set also. The only difference in two sets is the temperature of water bath.

3. Prepare the separate observation table for this set.

Calculations :

1. Calculation of initial concentration of KI, (‘a’) :

Total volume of reaction mixture = 80 ml.

[(20 ml 0.1 N KI + 20 ml water) + (10 ml 0.1N K2S2O8 + 30 ml water)] The resultant normality of KI in reaction mixture can be calculated as - KI ≡ Reaction mixture N1V1 = N2V2 0.1 × 2.0 = N2 × 80 ∴ N2 = 0.025 N Therefore initial concentration of KI in 10 ml of reaction mixture (in terms of 0.002 N Na2S2O3) is as,

a = 0.25 × 10

0.002

∴ a = 125 ml 2. Calculation of initial concentration of K2S2O8, (b) : Total volume of reaction mixture = 80 ml The resultant normality of K2S2O8 in reaction mixture can be calculated as- K2S2O8 ≡ Reaction mixture N1V1 = N2V2 ∴ 10 × 0.1 = N2 × 80

∴ N2 = 10 × 0.1

80


∴ N2 = 0.0125 N Therefore initial concentration of K2S2O8 in 10 ml of reaction mixture (in terms of 0.002 N Na2S2O3) is

b = 0.0125 × 10

0.002

∴ b = 62.5 ml The rate constant k, for second order reaction with unequal initial concentration is given by

k = 2.303

t (a − b) log b (a − x)a (b − x)

Calculate the value of k for both the sets at the instances 10, 15, 20, 25, 30, 35 and 40 minutes. Determine mean value of kI (at temperature T1 K) and kII (at temperature T2 K). 3. Rate constant (k) by graphical method :

Plot the graph of log (a − x)(b − x) against time (t).

The nature of graph is a straight line with slope

Slope = k (a − b)

2.303

∴ k = 2.303 × slope

(a − b)

log

Time (min)

Set II

Set I(a x)�

(b x)�

Fig. 1

Thus, find kI at temperature T1 K and kII at temperature T2 K. 4. Calculation of energy of activation : Energy of activation can be calculated by using Arrhenius equation –

log kII

kI =

Ea

2.303 R ⎣⎢⎡

⎦⎥⎤T2 − T1

T1T2 where T2 > T1

∴ Ea = 2.303 × R × T1 × T2

(T2 − T1) log

kII

kI

where, kI is rate constant at temperature T1 and kII is rate constant at temperature T2 °K. R = Molar gas constant = 1.987 cal degree−1 mole−1 ∴ Ea = …….. cal mole−1 = …….. kcal mole−1 Thus, find energy of activation by calculation as well as by graph. Result Table :

Description By calculations By graph Energy of activation

Rate constant (kI) at temperature TI K

…. dm3 mol–1 min–1 ….. dm3 mol–1 min–1 …… k ⋅ cal mole–1

Rate constant (kII) …. dm3 mol–1 min–1 …. dm3 mol–1 min–1 …… k ⋅ cal mole–1

27


28

at temperature TII K

Conclusions :

1. The value of k at a temperature T remains fairly constant and the graph of a − xb − x

against time t is a straight line. It indicates that reaction follows second order kinetics. 2. Activation energy of this reaction is …….. k⋅cal mole–1.

Experiment No. 7 Aim : To determine the order of reaction between K2S2O8 and KI by half life method. Apparatus : Stoppered bottles, burette, pipette, conical flask, water bath, stop watch etc. Chemicals : 0.05 N K2S2O8, 0.05 N KI, 0.005 N Na2S8O3, starch indicator, crushed ice etc. Theory : The reaction between potassium persulphate and potassium iodide is a second order reaction. The overall reaction is K2S2O8 + 2KI ⎯→ 2K2SO4 + I2

or S2O−28 + 2I− ⎯→ 2SO

2−4 + I2

The reaction is first order with respect to each reactant and hence second order overall. It is a two step reaction as represented below.

(i) S2O2−8 + I− ⎯→ S2O8I3−

(ii) S2O8I3− + I− ⎯→ 2SO2−4 + I2

The progress of reaction can be studied by titrating the liberated iodine against 0.002 N sodium thiosulphate using starch as a indicator. The rate constant for the second order reaction with equal initial concentration can be calculated by using equation-

k2 = 1t

xa (a − x) min−1 conc−1

Half-life period (t1/2) : The time required for completion of half of the reaction is called the half-life period. At half-life period t = t1/2, x = a/2.

∴ k2 = 1

t1/2 ×

1/2a (a − a/2)


29

∴ k2 = 1

t1/2 ×

1a =

1a × t1/2

For third order reaction,

k3 = 1

a2 × (t1/2)

∴ In general for nth order reaction,

kn = 1

t1/2 × an−1

∴ Half-life period (t1/2) = 1

kn ⋅ an−1

∴ t1/2 = 1

kn ⋅ an−1

For nth order reaction if (t1/2)I and (t1/2)II are the times for half change with initial concentrations, a1 and a2 then –

(t1/2)I = 1

k2 ⋅ an−11

(t1/2)II = 1

k2 ⋅ an−12

∴ (t1/2)I

(t1/2)II = ⎝⎜

⎛⎠⎟⎞a2

a1

n−1

∴ log (t1/2)I

(t1/2)II = log ⎝⎜

⎛⎠⎟⎞a2

a1

n−1

∴ log (t1/2)I

(t1/2)II = (n − 1) log

a2

a1

∴ (n − 1) = log (t1/2)I − log (t1/2)II

log a2 − log a1

∴ Order, n = 1 + log (t1/2)I − log (t1/2)II

log a2 − log a1

This equation enables the experimental determination of order of a reaction. Two sets of experiment are performed by taking equal initial concentrations of two reactants as a1 in Set I and a2 in Set II. Half times (t1/2)I and (t1/2)II can be calculated from the rate constants for the two sets Hence, order can be determined using the above formula. Procedure : Set I : 25 ml 0.05 N KI + 25 ml 0.05 N K2S2O8 + 50 ml distilled water :


30

1. Take 25 ml 0.05 N KI and 20 ml distilled water in clean stoppered bottle and 25 ml 0.05 N K2S2O8 and 25 ml distilled water in another stoppered bottle. Keep the bottles in thermostat or water bath for 10 – 15 minutes to attain constant temperature.

2. Fill the burette with 0.05 N Na2S8O3 solution. 3. When the two solutions have attained the same temperature, mix them to start the

reaction. Start the stop watch and note the mixing time as zero time. 4. Keep the mixture in water bath and shake it from time to time. 5. Keep a titration flask ready with around 30 ml crushed ice or ice-cold water. 6. Withdraw 10 ml of reaction mixture after 10 minutes from the start by pipette and

transfer it to the conical flask containing ice. Titrate the liberated iodine against sodium thiosulphate solution using starch indicator. Starch should be added near the equivalence point i.e. when reaction mixture becomes faint yellow during titration.

7. Similarly titrate the reaction mixture at the interval of every 5 minutes. 8. Take seven such readings. Observation Table :

Time ‘t’ (min) Volume of Na2S8O3 x (ml)

10 15 20 25 30 35 40

Set II : 50 ml 0.05 N KI + 50 ml 0.05 N K2S2O8 : Repeat the experiment in the same manner as in set I except with the concentrations of reactants mentioned above. Prepare the separate observation table for Set II. Calculations :

1. To determine initial concentration (a) :

Set I : 25 ml 0.05 N KI + 25 ml 0.05 N K2S2O8 + 50 ml distilled water :

The two reactants KI and K2S2O8 are having equal initial concentrations. Resultant volume of the reaction mixture is 100 ml. The resultant molarity of KI and K2S2O8 in the reaction mixture can be calculated as –

KI (or K2S2O8) ≡ Reaction mixture

N1V1 = N2N2

∴ 0.05 × 25 = N2 × 100


N2 = 0.05 × 25

100

∴ N2 = 0.0125 N

∴ Initial concentration of KI or K2S2O8 in 10 ml of reaction mixture in terms of 0.005 N Na2S8O3 would be

a1 = 0.0125 × 10

0.005

∴ a1 = 25 ml

Set II : 50 ml 0.05 N KI + 50 ml 0.05 N K2S2O8 :

The two reactants KI and K2S2O8 are having equal initial concentrations. Resultant volume of the reaction mixture is 100 ml. The resultant molarity of KI and K2S2O8 in the reaction mixture can be calculated as –

KI (or K2S2O8) ≡ Reaction mixture

N1V1 = N2N2

∴ 0.05 × 50 = N2 × 100

N2 = 0.05 × 50

100

∴ N2 = 0.025 N

∴ Initial concentration of KI or K2S2O8 in 10 ml of reaction mixture in terms of 0.005 N Na2S8O3 would be

a1 = 0.025 × 10

0.005

∴ a1 = 50 ml 2. To determine the half-life period of the reaction : Plot the graph of x (burette reading) against time t for both the sets.

Set II

Set Ix

a0/2

a0/2

t1/2 t1/2 time (t)

time

31


32

Fig. 1 : x vs. t, showing determination of half lives from initial concentrations From graph calculate half-life periods (t1/2)I and (t1/2)II for Set I and Set II from concentration x = a0/2. Thus order can be calculated using the equation –

Order (n) = 1 + log (t1/2)I log (t1/2)II

log a2 − log a1

Result Table :

Order of the reaction (n ) = ..........

- - -

32

2…

Viscosity

Experiment No. 8 Aim: To determine the molecular weight of high polymer using it’s solution of different concentrations by viscometry. Apparatus : Ostwald’s viscometer, density bottle, stop watch, measuring cylinder, etc. Chemicals : Pure solvent, 0.1, 0.2, 0.3, 0.4 and 0.5% solution of polymer in given solvent. Theory : Viscosity is the quantity that describes a fluid's resistance to flow. Fluids resist the motion of layers with differing velocities within them as well as to the relative motion of immersed objects through them. Definition of Viscosity : “It is the property of a fluid that resists the force tending to cause the fluid to flow.” This resistance produces a frictional force known as viscous force proportional to the area

of contact, A, between the flowing regions and to the velocity gradient, dvdx .

F ∝ A ⋅ dvdx

∴ F = η A ⋅ dvdx … (1)

The constant of proportionality is called the coefficient of viscosity, η. The coefficient of viscosity η, is the quantitative measure of viscosity and is often called the viscosity. Presence of short-range attractive intermolecular forces and the momentum transfer between layers of moving molecules are the cause of the force of viscosity. The first one dominates in flow of liquids and second one in gases. The kinetic theory of gases is successfully applied to calculate η that manifests in the formula :

η = ⎝⎜⎛

⎠⎟⎞MRT

π3σ4N2

1/2

… (2)

T.Y.B.Sc. Practical Chemistry (Physical Practicals) Viscosity

33

where, M is the molar mass, R is the gas constant, σ is the molecular diameter, and N is the Avogadro’s number.

Viscosity co-efficient is defined as the force per unit area required to maintain the unit difference of velocity between the two different layers of liquid which are one centimeter apart. It is measured in terms of “poise”. Viscosity of liquid is said to be one poise, when a force of one dyne maintains a velocity difference of one centimeter per second between two parallel layers of liquids, which are one centimeter apart and having area of contact one square centimeter. It is also measured in terms of milli poise, centipoise etc.

The SI unit of viscosity is the pascal second (1Pa s = 1 kg/m-s). The most common unit of viscosity is dyne s/cm2, which is given the name poise [P] in the honor of the French physiologist Jean Louis Poiseuille (1799-1869). Ten poise equals one pascal second [Pa s] or a centipoise [cP].

Thus, 1 g/cm-s = dyne-seconds/cm2 = 1 pascal second = 10 poise.

Poiseuille derived the following formula for the volume, V, of a fluid of viscosity, η, moved through a tube of length l and radius r in time t at a pressure P,

Vt = ⎝⎜

⎛⎠⎟⎞πr4

8ηl ΔP … (3)

From this equation, one see that a measure of the time that a specific volume of a liquid takes to move through a tube, with a very small diameter under a constant pressure drop is a direct measure of the viscosity. The above equation is directly applicable to a device such as the Ostwald viscometer to measure the viscosity. With Ostwald viscometer, a liquid between two marks (i.e. a definite volume) is allowed to flow through a small bore capillary and the time of flow is measured.

Factors affecting Viscosity of Liquids :

1. Strong intermolecular force of attraction resists the flow of liquids and show higher viscosity.

2. Polar compounds have more viscosity due to stronger attractions.

3. Increase in molecular weight of liquid, results in increase in viscosity.

4. Increase in temperature increases the kinetic energy of molecules and decreases the viscosity of the liquid. (In case of gases the trend is reversed).

5. The increase in pressure increases viscosity to small extent.

Viscosity of Polymer Solutions : Eyring’s theory shows that solution properties depend not only on the type of constituents, but on the amount of each present. The basis for explaining the concentration dependence of the viscosity of a polymer solution is Einstein’s relationship for the viscosity of dilute solution of spherical particles that follows :


34

ηη0

= 1 + 2.5 φ … (4)

where, η is the viscosity of a polymer solution with molecular volume φ, η0 is the viscosity of the pure solvent. The specific viscosity, ηsp, is the increase in the viscosity over that of the pure solvent caused by the addition of the polymer. Specific viscosity of a polymer solution is thus expressed as

ηsp = (η − η0)

η0 … (5)

According to the Einstein relation, Eq. (4), the specific viscosity can be related to the concentration by equation :

ηsp = ⎝⎜⎛

⎠⎟⎞10πr3

3 m ⋅ cm … (6)

where, r is the radius of the “spherical” molecule, m is its mass, and cm is the concentration of polymer solution. The concentration dependence of specific viscosity is expressed in terms of intrinsic viscosity, [η], as :

[η] = lim

cm → 0 (ηsp/cm) … (7)

Like other parameters extrapolated to infinite dilution (zero concentration), the intrinsic viscosity describes the interaction of a single average polymer molecule with a bulk of solvent. For a spherical molecule, [η] should be proportional to the molecular volume. For many synthetic polymers in dilute solution, the chains tend to adopt a tight, ball-like configuration in solution, the average structure of which is defined by a radius of gyration, RG. For such molecules, one predicts that :

[η] = K' R

3G

M … (8)

where, M is the molar mass and K' is a constant. For a polymer that adopts a random-coil conformation, dependence of intrinsic viscosity on molar mass as is found to vary as –

[η] = KM1/2 … (9)

This generally applies to polymer solutions under certain conditions.

In practice, the dependence of intrinsic viscosity on molar mass is more accurately given by the empirical Mark-Houwink equation :

[η] = kMα … (10)

where, k and α are two parameters that depend on the solvent and the polymer. The values of k and α for some polymer solvent systems at 25°C are given in Table 1. Once k and α are known for a combination of polymer and solvent and the intrinsic viscosity determined experimentally, one may use to calculate the viscosity-average molar mass Mv of a polymer


material. The values of k and α for some polymer solvent systems at 25°C are given in Table 1.

Table 1 : Mark-Houwink parameters, k and α, for various polymers and solvents Sr. No.

Polymer Solvent Temperature/°C k/(cm3/gm) α

1. Polystyrene Toluene 25 3.7 × 10−3 0.62 2. Cellulose acetate Acetone 25 1.49 × 10−3 0.82 3. Polymethyl methacrylate

(Perspex) Benzene 25 0.94 × 10−3 0.76

4. Poly (isobutylene) Benzene 25 0.083 0.53 5. Poly (isobutylene) Toluene 25 0.087 0.56 6. Poly (isobutylene) Cyclohexane 25 0.040 0.72 7. Crape rubber Benzene 25 2.02 × 10−3 0.89 8. Polypropylene Toluene 30 0.0218 0.7259. Polypropylene Cyclohexane 25 0.016 0.80

10. Poly (vinyl alcohol) Water 25 0.0002 0.76 Measurement of Viscosity : The common method used for the measurement of viscosity is to measure the time of flow of liquid through a capillary tube. According to Poiseuille equation, co-efficient of viscosity is given by equation,

η = πpr4t8Vl

where p = hydrostatic pressure on the liquid.

r = radius of capillary t = time of flow of liquid V = volume of liquid flowing

through the capillary in time t l = the distance travelled by the

liquid in time t (or length of capillary)

A

a

b

V

C

l

B

Fig. 1: Viscosity is measured using

Ostwald’s viscometer Ostwald’s viscometer is a U-shape tube with two bulbs A and B and a capillary C. The volume V of liquid between marks ‘a’ and ‘b’ of bulb ‘A’ is allowed to pass through capillary B (having length l) and time of flow is measured.

35


36

In practice a liquid of known viscosity and density is used to calibrate the Ostwald viscometer. Then for any other liquid

η1

η2 =

p1 ⋅ t1

p2 ⋅ t2 … (11)

Same viscometer being used in the experiment, equation (10) reduces to equation (11). As pressure of liquid is directly proportional to density (ρ) of liquid,

η1

η2 =

ρ1 ⋅ t1

ρ2 ⋅ t2

Molecular weight of polymer is very high and ranges from 5,000 to 2,00,000. The molecular weight of each polymer molecule depends on the degree of polymerization i.e. the number of repeating units in the polymer molecule. By comparing the viscosity coefficients of solvent and polymer solution, the molecular weight of high polymer can be determined using equation :

(η/η0 − 1)

C = k ⋅ Mα

where, M = molecular weight of polymer C = concentration of polymer in gram per 100 ml of solution η = viscosity of solution η0 = viscosity of pure solvent k = constant for a given polymer for particular solvent and temperature α = a constant that is a function of geometry of polymer molecule

But, ⎣⎢⎡

⎦⎥⎤η

η0 − 1 = ηsp = specific viscosity

∴ ηsp

C = kMα

This equation is valid for dilute solution only (concentration less than 1%). Since ηsp is directly proportional to the concentration (C) of the polymer in the solution, the plot of ηsp/C against C is a straight line. The intercept of this straight line on y-axis gives the intrinsic viscosity [η].

Similarly plot of 2.303

C log ηη0

against C gives a straight line with y intercept [η].

The intrinsic viscosity [η] is defined by the relations

[η] = lim

C → 0 ηsp

C = lim

C → 0 2.303

C log ηη0

Knowing the intrinsic viscosity the molecular weight is determined by Mark-Houwink equation [η] = kMα ∴ log [η] = log k + α log M


37

∴ (log [η] – log k)

α = log M

∴ Mol. wt. (M) = Antilog ⎝⎜⎛

⎠⎟⎞log [η] – log k

α

Procedure :

(A) Measurement of Flow Time for Solvent :

1. Take a clean dry Ostwald viscometer and clamp it in a perfectly vertical position to an iron stand. (Clean and dry the viscometer by running a few millilitres of acetone through it. Drain the acetone and aspirate for about a minute to evaporate all the acetone.)

2. Suspend the viscometer into a large beaker of water.

3. Pipette sufficient quantity of solvent of known density into the viscometer (bulb B) and allow time for the liquid to equilibrate to the temperature of the bath.

4. Then suck the liquid level up above the upper mark ‘a’ on the bulb A. (A pipette bulb may be used to pull the liquid up).

5. Allow the solvent to run back down and start the stop watch/timer exactly as the meniscus passes the upper mark “a”.

6. Stop the stop watch just as the meniscus passes the lower mark “b”. Repeat at least twice. Your flow times should agree to within ± 0.4 seconds.

7. Record the observations as shown in the observation table.

(B) Determination of Time of Flow for the Solution of Polymer :

1. Introduce sufficient quantity of 0.1% polymer solution in viscometer.

2. Determine the flow times as measured for solvent above.

3. Similarly, find the flow time for 0.2, 0.3, 0.4 and 0.5% polymer solutions.

Record the observations in observation table.

Observations :

Part A : Solvent

Solvent Time of flow (seconds) Mean time of flow

t0 (seconds) 1 2 3


Part B : Polymer Solution

Conc. of Polymer

%

Flow time t tt0

= ηη0

ηη0

= 1 = ηsp ηsp

C2.303

C log ηη01 2 3

Mean Time

0.1

0.2

0.3

0.4

0.5

Calculations : Plot the graph of ηsp/C against concentration C. Extrapolate to find intercept on y axis as intrinsic viscosity [η]. Graph :

38

�sp/C

Concentration

[n]

�

�o

Concentration

[n]2.303

Clog

Fig. 2

Similarly plot the graph of 2.303

C log ηη0

against concentration C.

The intrinsic viscosity is determined by extrapolating the straight line to zero concentration. Thus, knowing [η], k and α, the molecular weight (M) is calculated using equation :

M = antilog ⎝⎜⎛

⎠⎟⎞log [η] – log k

α

Result Table :

1. Intrinsic viscosity [η] ………poise

2. Molecular weight of given polymer ………


39

- - -

39

3…

Adsorption

Experiment No. 9 Aim: To investigate the adsorption of oxalic acid/acetic acid by activated charcoal and test the validity of Freundlich and Langmuir isotherm. Apparatus : Five stoppered bottles, burette, pipettes, conical flasks etc. Chemicals : 0.5 N oxalic acid, 0.5 N acetic acid, 0.1 N sodium hydroxide solution, phenolphthalein indicator, powdered activated charcoal, etc. Theory : The molecular forces at the surface of liquid as well as solid are in a state of imbalance. As a result exposed solid and liquid surfaces tend to satisfy their residual forces by attracting gases or dissolved substances from an adjacent gas or liquid and retaining them on surfaces. This phenomenon of concentration of substances on the surface of liquid or solid is called as ‘adsorption’. Thus adsorption is a surface phenomenon. The extent of adsorption mainly depends on temperature, pressure, area of adsorbent etc. A substance that has been or is to be adsorbed on a surface is said to be adsorbate. A substance having capacity or tendency to adsorb another substance is called as ‘adsorbent’. Adsorption Isotherm : Adsorption is usually described through isotherms. The relation between the amount of substance adsorbed by an adsorbent and the equilibrium concentration (or pressure in case of gas) at constant temperature is called adsorption isotherm. Adsorption phenomenon is divided into two types : (1) Physical adsorption and (2) Chemical adsorption. In 1909, Freundlich gave an empirical expression representing the isothermal variation of pressure with adsorption of a quantity of gas adsorbed by unit mass of solid adsorbent. This equation is known as Freundlich Adsorption Isotherm or simply Freundlich Isotherm.

xm = kP1/n

For solutions, equation takes the form

xm = KC1/n (for solution)


where, x = mass of gas or solute adsorbed m = mass of adsorbent P = equilibrium pressure of adsorbate C = concentration of the solute in gram moles per litre. K and n are constants whose values depend upon adsorbent and gas or solute at particular temperature. Taking logarithm on both sides, above equation becomes

log xm =

1n log C + log K

This equation represents the straight line. The graph of log x/m against log C is a straight line, with slope = 1/n and y intercept log K. Slope can give the value of n since, n = 1/slope

xm

P

(PS)

Adsorption Isotherm

Saturation pressure

log x/m

log K

log C

(a) (b)

Fig. 1: Freundlich Isotherms The straight line nature is observed at lower concentrations. Deviations are marked at higher concentration. Freundlich Isotherms failed to predict value of adsorption at higher concentration /pressure. Langmuir Adsorption Isotherm : In 1916, Langmuir proposed another adsorption isotherm known as Langmuir Adsorption Isotherm. For the adsorption of solute from solution on solids Langmuir isotherm is

xm =

aC1 + bC

where, a and b are constants. Rearranging above equation

C

x/m = ba ⋅ C +

1a

Graph of C

x/m against concentration is a straight line with slope = ba and y intercept equal

to 1a .

40

T.Y.B.Sc. Practical Chemistry (Physica Chemical Kinetics l Practicals)

Cx/m

Concentration

Slope = b/a

y intercept = 1/a

Fig. 2

Procedure :

1. Take five clean stoppered bottles and label them as 1 to 5.

2. Add oxalic acid solution and distilled water to these bottles as per the following chart.

Bottle No. ml. of 0.5 N oxalic acid ml. of water Amount of

activated charcoal

1 50 00 1 gm

2 40 10 1 gm

3 30 20 1 gm

4 20 30 1 gm

5 10 40 1 gm

3. Weigh and add 1.00 gram of the activated charcoal to each of the bottle.

4. Stopper the bottles, place them in water bath and shake well for 30 minutes from time to time.

5. Mean while titrate 10 ml oxalic acid against 0.1 N NaOH solution using phenolphthalein indicator to determine the initial concentration of acid. Record it as ‘V’.

6. After 30 minutes filter the solution from each bottle in separate flask. Titrate 10 ml of the filtrate from each bottle with 0.1 N NaOH. Take three readings for each bottle and note the mean as Ce, the equilibrium concentration.

Prepare observation table as given below :

(i) Weight of charcoal, (m) = 1.00 gm.

(ii) Volume of 0.1 N NaOH required (burette reading) for 10 ml. of oxalic acid, (V) =.…..ml.

41


42

Obs No.

ml. of NaOH required for10 ml of acid in bottles

∴ ml. of NaOH required for 50 ml of acid in bottles

C log C x x/m log

(x/m) Before adsorption

After adsorption

(B.R.)

Before adsorption

(C0)

After adsorption

(Ce)

1 V =…….. V1 =…..… 5V =…..… 5V1 =…..…

2 4/5V =…….. V2 =…..… 4V =…..… 5V2 =…..…

3 3/5V =…….. V3 =…..… 3V =…..… 5V3 =…..…

4 2/5V =…..… V4 =…..… 2V =…..… 5V4 =…..…

5 1/5V =…….. V5 =…..… V =…….. 5V5 =…..…

Calculations : Equilibrium concentration of oxalic acid in gram equivalent per litre,

C = Ce × Total volume of acid solution × Eq. wt. of acid

1000 = Ce × 50 × 63

1000

Amount of acid adsorbed, x = (C0 − Ce) × 50 × 63

1000

Plot the graph of log x/m against log Ce. A straight line graph indicates the validity of Freundlich equation in the adsorption of oxalic acid by activated charcoal. Find the value of K and n.

Validity of Langmuir adsorption isotherm can be tested by plotting Ce

x/m against Ce.

A straight line graph is in agreement with Langmuir adsorption isotherm. Calculate the constants a and b. Result Table :

Obs. No. Symbol Values

1 K

2 N

3 a

4 b

Conclusion : Straight line nature of both the graphs indicates the validity of Freundlich and Langmuir adsorption isotherm.

- - -

43

4…

Phenol-Water System


To study the effect of addition of salt on critical solution temperature of phenol-water system.

Apparatus :

Hard glass test tube, stirrer, thermometer, water bath, 10 ml measuring cylinder etc.

Chemicals :

Pure phenol, distilled water, 0.1 M KCl.

Theory :

The solubility of liquid in liquid is an important part of solution chemistry. Miscibility is the property of liquids to mix in all proportions, forming a homogeneous solution. Water and ethanol, for example, are miscible since they mix in all proportions. By contrast, substances are said to be immiscible if they do not form a solution, in all the proportions. For example, diethyl ether is fairly soluble in water. These solvents are not soluble in all the proportions. Hence they are immiscible. Liquids tend to be immiscible when attractions between like molecules are much stronger than attractions between mixed pairs.

When two liquids are mixed together, one of the following cases may arise :

(a) the two liquids are completely miscible at all proportions forming one homogeneous solution e.g. water and ethanol.

(b) the two liquids are partially miscible forming either one or two liquid phases, depending on their composition in the mixture. e.g. phenol and water.

(c) the two liquids are immiscible forming always two distinct phases irrespective of composition e.g. water and carbon tetrachloride (CCl4).

The mutual solubility of partially miscible liquids usually increases with temperature. This miscibility temperature is different for different compositions of the mixture. The highest miscibility temperature is called the “critical solution temperature”. Above this temperature, all compositions of this mixture are completely miscible. For some liquid pairs such as ether and water, the mutual solubility decreases with temperature. For such systems the lowest miscibility temperature is called the “critical solution temperature”. Below this temperature, all compositions of this mixture are completely miscible.

T.Y.B.Sc. Practical Chemistry (Physical Practicals) Phenol-Water System

44

There are certain liquid pairs like nicotine and water, which have upper as well as lower critical solution temperature. Liquid pairs like ethyl acetate and water or chloroform and water does not have upper or/and lower critical solution temperature. Such systems are called as systems without critical solution temperature or incomplete systems.

A solute soluble in one of the two liquids increases the critical solution temperature if present in partially miscible liquids in equilibrium. For example, addition of small amount of KCl, increases critical solution temperature of phenol-water system. KCl is soluble in water only and not in phenol. If the solute is soluble in both the liquids the critical solution temperature decreases. For example, presence of succinic acid in phenol-water system decreases the critical solution temperature. The change in critical solution temperature is found to be linearly proportional to the percentage of solute.

N.B. : Phenol is highly corrosive. Avoid its contact with skin. Phenol sticks to measuring cylinder and bottle from outside and may come in contact with skin.

Procedure :

Set I :

1. Measure 9.5 ml of phenol into the hard glass tube (boiling tube). Fill a burette with distilled water and add 6 ml water into the tube containing phenol.

2. Fix the thermometer and stirrer and mount the tube in a water bath (water in a beaker) placed on tripod stand. Turbidity (cloudiness) appears on stirring the solution.

3. Heat it with mild stirring until the turbidity disappears. Note the temperature as mixing temperature at which turbidity just disappears.

4. Now cool the tube in a second water bath with mild stirring until the turbidity reappears. Once again note the temperature.

5. Similarly prepare phenol-water mixtures of various compositions in the hard glass tube as given in the observation table below. Record their mixing and separation temperatures in the same way as given in steps 3 and 4.

6. Prepare 0.1 M KCl salt solution mixture of various compositions as before and find the mutual solubility of each mixture.

Set II :

1. Repeat the procedure in Set I except with 0.1 M KCl salt solution in place of water. Note the mutual solubility temperatures for mixtures of various compositions.

2. Tabulate the result as follows.


45

Observation Table : Density of salt solution = 1.09 g/ml. Set I : Phenol - Water system

Amount of Phenol (g)

Amount of Water (g)

% Phenol % Water

Miscibility temperature (°C)

disappear appear average

10 (9.5 ml) 10 10 10

6 8 12 15

5 (4.8 ml) 5

5 10

2.5 (2.4 ml) 2.5 2.5

15 20 25

Set II : Phenol - Salt solution

Amount of Phenol (g)

Amount of 0.1 M KCl solution (g)

% Phenol

% Water

Miscibility temperature (°C)

disappear appear average

10 (9.5 ml) 10 10 10

6 8 12 15

5 (4.8 ml) 5

5 10

2.5 (2.4 ml) 2.5 2.5

15 20 25

Calculations :

% phenol = Weight of phenol

Weight of phenol + Weight of water

% phenol = Weight of water

Weight of water + Weight of phenol

% phenol = Weight of salt

Weight of salt + Weight of phenol

Plot the graph of mutual solubility temperature against % of phenol (composition) for two sets. Use the same graph paper.


0% % Phenol

P = 1

P = 2

tCII

tCI

Te

mp

era

ture

Critical composition

Fig. 1: Mixing temperature vs. Composition curves (P → phases)

Temperature corresponding to the maxima on the parabolic curve gives the upper critical solution temperature. Find the increase in critical solution temperature due to addition of salt.

Increase in critical solution temperature = (tCII – tCI

) °C.

Result Table :

1. Critical solution temperature of phenol-salt system (tCII) = ……... °C.

2. Critical solution temperature of phenol-water system (tCI) = ……. °C.

3. Increase in critical solution temperature (tCII – tCI

) = ……….. °C.

4. Percentage of phenol at critical solution temperature = ..….... %. Conclusion : Liquids tend to be immiscible when attractions between like molecules are much stronger than attractions between mixed pairs. Presence of a salt modifies these interactions. Presence of KCl in phenol-water system delays the miscibility. Increase in critical solution temperature in this system indicates that the phenol-water attractions are weakened due to presence of salt.

- - -

46

5…

Transport Number

Experiment No. 11 Aim : To determine the transport number of cation by moving boundary method. Theory : In a solution of an electrolyte the current is carried by the ions. As a result the concentration of the electrolyte changes. The equivalent amounts of positive and negative ions are discharged at respective electrodes. But the velocity with which the positive and negative ion move is different. Consequently the amount of electricity carried by cation in one direction is different from that carried by anions in opposite direction. These two amounts are in the ratio of the mobilities of cations (u+) and anion (u−) respectively. The total current passing through the solution is proportional to the sum of ionic mobilities (u+ + u−). The fraction carried by cation and anion is given by the equations

t+ = u+

u+ + u− and t− =

u−

u+ + u−

These fractions are called as transport number of cation (t+) and transport number of anion (t−). These numbers can be determined by Hittorf’s method and moving boundary method.

Moving Boundary Method : A sharp boundary is formed in a tube between solutions of two salts having a common ion say H+Cl− and Li+Cl−. The position of the boundary is observed by the colour of the methyl orange. Now a potential gradient is set up along the tube by applying voltage across the electrodes situated at the ends of the tube. The current will flow by the migration of cation

_

+

b

a

H+

Cl�

Li+

Cl�

b

a

Fig. 1: Basis of the moving boundary method

for determination of transport number

towards the cathode and anions towards the anode. At the boundary the H+ and Li+ ions will move the same direction and hence the boundary will be observed to move. The tube is graduated so that the volume V swept through in time t(s) during the passage of QC of

47

T.Y.B.Sc. Practical Chemistry (Physical Practicals) Transport Number

electricity can be determined. The amount of charge carried by H+ ions across any section bb in HCl solution is equal to the number of moles of H+ ions, which would have passed that section (CV) multiplied by charge carried by each mole (96500 C). Hence, the fraction of the current carried by H ions (t+) is given by

t+ = C × V × 96500

Q

where, C = Concentration V = Volume swept Q = Number of coulombs passed through the solution If current passing through the tube is kept constant, then above equation becomes

t+ = CF

1000 × I × dvdt

where, C = Concentration of H+ ion F = Faraday I = Current

dvdt = the slope of the volume versus time curve

Apparatus : Glass assembly as shown in Fig. 2. Chemicals : 0.1 N HCl solution, methyl orange indicator, LiCl solution.

_+Cathode Anode

48

D

G E

I

A

LiCl

B

F

CH


Fig. 2: Apparatus for determination of transport number of HCl by moving boundary method

Procedure : 1. Rinse the cleaned apparatus twice with HCl solution and fill it with the same acid

solution containing methyl orange indicator. 2. Insert the stopper G and close the stopcock A. 3. Pour off HCl solution from E and clean this part with distilled water. Fill E with LiCl

solution. 4. Insert the anode assembly as shown in figure above. 5. Keep the whole apparatus in glass sided water bath. 6. Connect the cathode and anode to the source. 7. Open fully the stopcock very slowly. 8. Switch on the current of 10 mA at once and bring the boundary in the calibrated

volume rapidly. 9. Reduce the current and keep it constant between 2-5 mA by adjusting the

potentiometer. 10. Start the stop-watch as soon as the boundary reaches the first graduation on the

tube BC. 11. Record the volume swept by the boundary and time till the boundary reaches the

bottom of the tube. Observations :

Obs. No. Volume on tube BC Volume swept/ml Time/second

1. 2. 3. 4. 5.

Graph :

Vo

lum

esw

ep

t/

ml

time / s

Plot the graph of volume swept (ml) against time (s). If the current is kept constant, the graph is a straight line as shown in Fig. 3.

Find the slope of the line i.e. dvdt .

49


50

Fig. 3: Plot of volume swept in ml against time in second

Calculations : Calculate transport number of cation by using the equation

t+ = 96500 × C1000 × I ×

dvdt

Result :

t+ = ……….

VIVA VOCE 1. Define transport number of the cation (t+). 2. Define transport number of the anion (t−). 3. State different methods used for the measurement of transport number. 4. State the equation for transport number determination by moving boundary method.

- - -

6…

Refractometry (Any Two)

Refraction : When a ray of light travels from one medium into another its velocity changes. This causes the wave to change its direction. This change of direction as waves pass from one material into another is called refraction. The Refractive Index : The refractive index 'n' of a substance is defined as

n = Vv

V

where, Vv is the velocity of light in vacuum and V is the velocity of light in the substance. Assuming that velocity of light in air is essentially same as that in vacuum, then refractive index of air is one. Light slows down when it enters form air into a substance so that the refractive index of the substance is always greater than one. In other words, when light passes from the rarer medium into the denser medium it bends towards the normal. This bending of light is called refraction. Snell’s Law :

Snell's law is the simple formula used to calculate the refraction of light when travelling between two media of different refractive indices. In the diagram, two media of refractive indices n1 and n2, meet at an interface (horizontal line) and n2 > n1 and the light has a low velocity within the second medium. A light ray PO in the upper less denser medium strikes the interface or point ‘O’. A vertical line drawn from O at right angles to the interface is called normal.

�1

�2

n1

P

InterfaceO

Q

Normal

n2

Fig. 1

The angle between the incident ray PO and the normal is called angle of incidence θ1. The light ray continues through the interface into the denser medium on the lower side

51

T.Y.B.Sc. Practical Chemistry (Physical Practicals) Refractometry

52

shown by the ray OQ. The angle between the ray OQ and the normal is θ2 and is called angle of refraction.

Snell"s law gives the relation between θ1 and θ2 as

n1 sin θ1 = n2 sin θ2

or sin θ1

sin θ2 =

n2

n1

It should be noted that for the case of θ1 = 0 (a ray perpendicular to the interface, the value of θ2 = 0 irrespective of values of n1 and n2). That is a ray hitting the interface at right angles and passing into another medium never bent. The above law is also valid for light travelling from a dense to a less dense medium. A qualitative rule to determine the direction of refraction is that the ray in the denser medium is always closer to the normal.

When a ray of light passes from a dense to a less dense medium i.e. n1 > n2, the angle of refraction becomes 90° for a particular angle of incidence (called critical angle, θcrit). If the angle of incidence is farther increased there will be no refraction but the light is completely reflected into the first medium, this phenomenon is known as total internal reflection.

Critical angle for a pair of media can be defined as the angle of incidence for which the angle of refraction is 90° when light travels from a dense to a less dense medium.

By Snell's law,

sin θ1

sin θ2 =

n2

n1

When θ1 = θcrit, θ2 = 90° and sin 90° = 1.

Therefore, sin θcrit

sin 90 = n2

n1 or sin−1

n2

n1 =

sin θcrit

1 = Acrit

Therefore, Acrit = sin−1 n2

n1 or arc sin ⎝⎜

⎛⎠⎟⎞n2

n1

The equation θcrit = sin−1 n2

n1 is frequently used to find out the refractive indices of liquids

using refractometers.

Measurement of Refractive Index :

Following are the different types of refractometers used for determination of refractive indices of liquids.

(i) Pulfrich refractometer – accuracy of measurement : 0.00002

(ii) Immersion refractometer − accuracy of measurement : 0.00004

(iii) Abbe refractometer – accuracy of measurement : 0.00002.


The most commonly used refractometer is the Abbe refractometer. The schematic of which is shown in Fig. 2.

S

F

P

K

T

E

M

X

2

1

Fig. 2: Abbe refractometer-schematic

E = Eye piece

S = Refractive index scale

T = Telescope

P = Prism box (with split prisms)

M = Adjustable mirror

F = Adjustment screw

K = Compensating system screw

X = Field view when setting is complete

The Abbe refractometer measures refractive indices from 1.30 to 1.70 with accuracy of measurement to 0.0002. It requires only a few drops of liquids for measurement. The optical parts of Abbe refractometer are shown in the labelled diagram. For measurement of refractive index of sodium-D line, light is used.

The liquid whose index is to be measured is spread on the lower split prism. The prism box is closed and tightly locked. The mirror is directed to reflect light from lamp on to the prism. The light emerging from the prism is received by the fixed telescope after moving the arc to which index scale is attached. The field of view may initially show blue or red fringes dividing a grey or dark zone on one side and a white zone on the other side. The dispersed light is neutralised by rotating the compensative prisms (prism box) and line demarking the grey zone and white zone is made sharp and made to coincide with the crossing of the lines in the field view. This adjustment indicates the critical angle condition and the scale reads the refractive index. The refractive index is read from the scale.

Specific and Molar Refraction :

(i) The specific refraction r, is defined by the equation

r = n2 − 1n2 + 1 ×

1ρ

n = refractive index, ρ = density of liquid

(ii) The molar refraction, RM is defined by the equation

53


54

RM = n2 − 1n2 + 1 ×

Mρ = r × M

where, M is molecular weight of the liquid. The molar refraction R is constitutive and additive property. The molar refraction [R]AB of a mixture of A and B liquids is the sum of the contributions of the separate components A and B i.e.

[R]A, B = XA[R]A + XB[R]B

where, XA and XB are mole fractions of A and B respectively, and [R]A and [R]B are molar refractions of A and B respectively.

Also, RA, B = n

2A,B − 1

n2A,B + 2

⎣⎢⎡

⎦⎥⎤XAMA + XBMB

ρA,B

where, nA, B and ρA, B are refractive index and density of mixture AB.

Mixture Law :

The percentage of A and B in their mixture is given by the mixture law as

rA, B × 100 = rAPA + rB (100 − PA)

where, rA, rB and rA,B are specific refractivities of liquids A, B and their mixture AB respectively. PA is percentage of A, PB = (100 − PA) is percentage of B in the mixture.

The molar refraction is an additive as well as a constitutive property. Every constituent atom of the molecule makes a definite contribution to the molar refraction. The double bonds, triple bonds etc. also contribute to the total molar refraction.

Contribution of −CH2 group to the molar refraction of successive members of homologous series :

The successive members of homologous series like (i) methanol, ethanol, n-propanol, n-butanol and (ii) methyl acetate, ethyl acetate, propyl acetate differ by a CHO2 group. Therefore, the difference in molar refraction of successive members of the series must be same and should give the contribution of −CH2 group and it is found to be so. The following table of experimental values proves this.

Table 1

Homologous series Molar refraction R (Na-D line)

Contribution of −CH2 group, ΔR

Methanol, CH3OH 8.213 −

Ethanol CH3CH2OH 12.74 4.522


55

n-propanol, CH3CH2CH2OH 17.415 4.675

n-Butanol CH3CH2CH2CH2OH 22.130 4.615

Density and refractive indices of some liquids are given below : Table 2

Liquid Refractive index ‘n’ Density ρ, gm ml−1

Acetone Aniline Benzene n-Butyl acetate Carbon tetrachloride Chlorobenzene Chloroform Cyclohexane Ethanol Ethyl acetate Ethyl ether Hexane Methanol n-propyl acetate Toluene Methyl acetate Water

1.3588 1.5861 1.5011 1.3940 1.4601 1.5241 1.4460 1.4266 1.3611 1.3723 1.3526 1.3751 1.3288 1.3820 1.4961 1.3610 1.330

0.785 1.022 0.874 0.882 1.584 1.106 1.480 0.714 0.785 0.900 0.714 0.655 0.787 0.889 0.774 0.932 1.00

Experiment No. 1 Aim : To determine the specific refractivities of given liquids A and B and hence to determine the percentage composition of liquid mixture C containing A and B. Apparatus : Abbe refractometer, specific gravity bottles, droppers etc. Chemicals : Given pure liquids A and B and the following mixtures of A and B prepared to weight to weight ratio. Mixture C : Unknown Mixture D : 20 %A + 80 %B Mixture E : 40 %A + 60 %B Mixture F : 60 %A + 40 %B Mixture G : 80 %A + 20 %B Procedure :


56

As described in the previous experiment of refractometry, determine densities and specific refractivities of pure liquids A and B as well as their different mixtures. Record your observations as follows. Observation Table : (I) Determination of densities of liquids : Weight of empty specific gravity bottle (W1) = ………. g

Obs. No.

Liquid mixture

Weight of sp. gr. bottle + liquid W2 g

Weight of liquid W2 − W1 = W3 gm

Density ρ = W3/W4 g ml−1

1. 2. 3. 4. 5. 6. 7. 8.

A B C D E F G

Water

W4

(II) Refractive index measurement :

Obs. No.

Liquid Composition of

mixture Refractive index

Density ρ gm l−1

Specific refractivity

r = n2 − 1n2 + 2 ×

1ρ

I II III Mean 1. 2. 3. 4. 5. 6. 7.

A B C D E F G

100 %A + 0 %B 0 %A + 100 %B Unknown 20 %A + 80 %B 40 %A + 60 %B 60 %A + 40 %B 80 %A + 20 %B

Calculations :

Calculate specific refractivity, r = n2 − 1n2 + 2 ×

1ρ

Graph :


57

Plot a graph of specific refractivity against percentage of A. From the graph find the % of A corresponding to the specific refractivity of unknown C. %B = 100 − %A Mixture Law :

PA = 100 (rC − rB)

(rA − rB) and PB = 100 − PA 20 40 60 80 100

r rc

Unknowncomposition

% of A

0

Fig. 1

where, rA, rB and rC are specific refractivities of %A, 100%B and unknown composition respectively. Result Table : Percentage composition of mixture C.

Graphical Method Mixture Law % A = % B =

% A = % B =

Experiment No. 2 Aim : To determine molecular refractivity of given liquids A, B, C and D. Apparatus : Abbe refractometer, specific gravity bottle, beakers etc. Chemicals : Given pure liquids A, B, C and D. [The liquids can be carbon tetrachloride, chloroform, methyl acetate, ethyl acetate, benzene, methanol, ethanol, n-hexane, propanol etc.] Procedure : (A) Measurement of Refractive Index of liquids A, B, C and D : 1. Open the prism box of the refractometer. Clean the prism surfaces by wiping gently

with cotton moistened with alcohol. Then wipe with dry cotton plug. 2. Spread the given liquid A by putting few drops of liquid on the surface of the lower

prism and immediately close the prism box and tighten it with the locking screw. 3. Adjust the mirror below the prism box so that the light from the light source enters

the prism box. The illumination can be seen through eyepiece 1. 4. Move the arc to which index scale is attached, till light emerging from the prism is

received by the telescope.


5. The field of view is divided by a grey or dark zone on one side and white zone on the other side with blue or red fringes. Fringes are due to dispersion of light.

6. To neutralise the dispersion of light rotate the compensator prism using compensating system screw till the line demarking grey zone and white zone become sharp.

7. Rotate the prism box till the sharp line dividing grey and white zone coincides crossing of lines in the field view.

8. Read the refractive index from the scale through eyepiece 2. 9. In a similar way find refractive indices of liquids B, C and D.

Field view Fig. 1

(B) Measurement of Density of liquids A, B, C and D : 1. Take a specific gravity bottle of 10.0 ml or 15.0 ml capacity. Wash the bottle with

water and then with acetone and dry the bottle. 2. Weigh accurately the empty specific gravity bottle with the stopper (W1 gms). 3. Fill the bottle with the liquid A to the rim and place the stopper, weigh the bottle

again. 4. Remove the liquid A, wash the bottle again and rinse it with a small quantity of

acetone and dry it. Fill it with liquid B and find the weight of bottle again. 5. In a similar way weigh the bottle with liquids C and D. Find the weights of liquids

A, B, C, D. 6. Now fill the dry bottle with water and find the weight of water. 7. Find the density of liquids.

Density of liquid = ρ = Weight of liquid

Weight of same volume of water

Observation Table :

Obs. No.

Liquid Weight of sp. gr. bottle + liquid (W2)

Weight of liquid W3 = W2 − W1 g

Density = Wt. of liquid

Wt. of same vol. of water

ρ = W3/W4

1. 2. 3. 4. 5.

A B C D

Water

W4

(I) Density Measurements : Weight of empty specific gravity bottle = (W1) = ………. g. (II) Refractive Index Measurements :

58


59

Obs. No. Liquid Refractive Index ‘n’

Mean ‘n’ I II III

1. 2. 3. 4.

A B C D

Calculations : 1. Specific Refraction :

Specific refraction r = n2 − 1n2 + 2 ×

1ρ

2. Molar Refraction :

R = n2 − 1n2 + 2 ×

Mρ

Ask for the molecular weights of liquids A, B, C and D. Result Table :

Liquids A B C D Molar Refraction RM

Experiment No. 3 Aim : To determine the molar refraction of methyl, ethyl, propyl alcohol and to show the constancy contribution to the molar refraction made by −CH2 group. Apparatus : Abbe refractometer, light source, beaker etc. Chemicals : Methyl, ethyl and n-propyl alcohol. Procedure : As in the previous experiment of refractometry find the densities and refractive indices of liquids methyl, ethyl and n-propyl alcohol. Record the observations in the table given below. Find the specific and molar refraction values of the liquids. Find the contribution made by −CH2 group to the molar refraction of the liquids. Observation Table : (I) Measurement of Density :

Obs. No.

Liquid Weight of sp. gr. bottle + liquid, W2 g

Weight of liquid W3 = W2 − W1

gm

Density ρ = W3/W4

gm l−1

1. Methyl alcohol (CH3OH) 2. Ethyl alcohol

(CH3CH2OH)


60

3. n-propyl alcohol CH3CH2CH2OH

4. Water W4 (II) Refractivity Measurements : Obs. No.

Liquid Mole-cular wt. M

Refractive index ‘n’

Density ρ gm l−1

Specific refractivity

r = n2 − 1n2 + 2 ×

1ρ

Molar refraction RM = r.M I II III Mean

1. 2. 3.

CH3OH CH3CH2OH

CH3CH2CH2OH

32 46 60

Weight of empty specific gravity bottle = (W1) = ………. g (III) Calculations :

1. Specific refraction r = n2 − 1n2 + 2 ×

1ρ

2. Molar refraction = r × molecular weight (M) 3. ΔR, the contribution of CH2 group to R. Find : ΔR1 = RCH3CH2OH − RCH3OH and ΔR2 = RCH3CH2CH2OH − RCH3CH2OH ΔR1 = ΔR2 = Contribution of CH2 group to the molar refraction of

homologous alcohol. Result Table :

Liquid Molar Refraction RM

ΔR Contribution of −CH2 group

1. Methyl alcohol (CH3OH) 2. Ethyl alcohol (CH3CH2OH) 3. n-propyl alcohol (CH3CH2CH2OH)

VIVA VOCE 1. What is refraction ? 2. Define refractive index. 3. State and explain Snell's law. 4. What is total internal reflection ? 5. What happens when angle of incidence exceeds the critical angle ? 6. What is critical angle ? 7. Define specific refractivity and molar refraction. 8. Give the relation between specific refraction and refractive index ? 9. What is the effect of temperature on specific and molar refraction ? 10. Explain how the molar refraction is useful in elucidating structure of organic

compounds. 11. What is an additive and a constitutive property ?


61

12. Explain with examples how molar refraction is an additive as well as a constitutive property.

13. Define density and specific gravity of a substance. 14. How will you determine the density of a liquid ? 15. What is mixture law in terms of percentage of individual components ? 16. What is the expression for molar refraction of a mixture of two liquids in terms of

molar refractions of the two liquids ? 17. Give Gladstone and Dale equation of molar refraction of a liquid ? 18. What is Lorenz and Lorentz equation of molar refraction ? 19. How you will determine contribution of methylene group in homologous alcohol and

esters ? 20. Will the contribution of methylene group is same for both homologous series of

alcohol and esters ? - - -

(GROUP – B)

1…

Colorimetry (Any Two)

Theory : The variation of intensity of colour of a solution with the change in the concentration of the coloured solute forms the basis of colorimetric analysis. The intensity of the colour of a substance can be measured by determining the fraction of particular wavelength of light absorbed by it. The advantage of colorimetric determination is that, this method is much more rapid than those involving volumetric or gravimetric methods. Also it is applicable to the determination of small amounts of substances in a material, i.e. less than 1 percent. Fundamental Laws of Colorimetric Measurement : When light of intensity Io, passes through a solution, some of the radiation is scattered or reflected by the walls of the container and by the suspended particles in the solution (Is), some is absorbed (Ia) and remaining is transmitted (It), i.e. Io = Is + Ia + It. Since, the colorimetric methods are based on the relation between the amount of l ight absorbed and the concentration of the absorbing material. Therefore, it is necessary to eliminate Is, so that a measurement of Io and It will be sufficient to determine Ia. In colorimetric

Io It

Is

Is

Ia

Fig. 1: The distribution of incident light

passing through a cell containing a solution

determinations, the amount of light scattered is not eliminated, but it is kept constant by using cells that have identical optical properties while measuring Io and It. The important laws of photochemistry are the Lambert’s law, Beer’s law and combination of these two, and is called as Lambert - Beer’s law. Lambert’s Law : The rate of decrease in the intensity of a monochromatic light passing through a transparent medium, with the thickness or optical path-length of the medium is proportional to the intensity of the incident light.

61

T.Y.B.Sc. Practical Chemistry (Physical Practicals) Colorimetry

62

Thus, − dIdl = kI

where, I = Intensity of the light l = Thickness of the medium k = Constant and is characteristic of the medium By solving the above differential equation we obtain the following relation

ln I0It

= kl, Intensity of incident light = I0

It = I0e−kl, Intensity of transmitted light = It In general, this law have the simple relation i.e. optical density or absorbance ∝ thickness of the light absorbing medium. But this law is not sufficient to find out the concentration of solute in the solution. Beer’s Law : In a path of constant length, equal fractions of incident radiation are absorbed by equal changes in the concentration of the light absorbing material. Thus,

− dIdC = kI

ln I0It

= kC

Thus, the short form of this law is given as Optical density or absorbance ∝ concentration. In order to find out the concentration of coloured solute in given solutions, it is necessary to consider both laws simultaneously. Thus, the combination of Beer’s law and the Lambert’s law gives suitable form of the equation where both i.e. thickness of the absorbing material and its concentration is considered.

Thus, ln I0I = alC

A or D = ∈.C.l = log I0I = − log T

where ‘C’ is the concentration of the solute and ∈ is a constant characteristic of the solute. When ‘C’ is expressed as mol dm−3, the constant ∈ is called the molar absorption coefficient, or molar extinction coefficient or molar absorptivity. The important terms involved in it are as follows : 1. Radiant Power (P) : It is the rate at which energy in a beam of radiation arrives at some fixed point. It is also called as intensity (I). 2. Transmittance (T) : It is the ratio of intensity of transmitted radiation (It) to that of intensity of incident radiation (Io).

T = ItI0

It does not have any unit.


3. Absorbance (A) : It is also called optical density. It is the log to the base 10 of the intensity of incident radiation to that of the intensity of transmitted radiation. Thus,

A = log I0It

It does not have any unit. 4. Molar absorptivity (∈) : It is the absorbance of a solution divided by the product of its optical path l (in centimeter) and molar concentration, C, of the absorbing molecules or ions.

∈ = A

l ⋅ C lit mole−1 cm−1

This property is also known as molar extinction coefficient.

Monochromaticlamp

+ ++++

Concavemirror

Collimatinglens

Slit(S )1

Filter Cuvette Slit(S )2

Photocell and recorder

Fig. 2: Single beam photoelectric colorimeter A simple single beam photocolorimeter consists of the components as shown in Fig. 2. Now-a-days digital colorimeters are used in laboratories, which are very easy to operate. But with these colorimeters we read only optical density. The transmittance is found out by using the equation O.D. = − log T

Experiment No. 1 Aim : Determine the λmax and concentration of unknown solution of KMnO4 in 2 N H2SO4 solution. Apparatus : Colorimeter, filters, 50 ml volumetric flasks, micropipette, beakers, measuring cylinder etc. Chemicals : 0.01 N KMnO4 stock solution, 2 N H2SO4, distilled water, unknown KMnO4 solution.

63


64

This experiment is also performed in three parts : Part A : Preparation of different concentrations of KMnO4 solutions. First clean the glasswares properly. Label the volumetric flask from 1 to 8. The different concentrations of KMnO4 are prepared by using the relation N1V1 = N2V2. Preparation of solutions :

Flask No.

Conc. of KMnO4

solution (N)

Amount of 0.01 N

KMnO4 taken for dilution

(ml)

2 N H2SO4 added (ml)

Distilled water added

(ml)

Total final volume

(ml)

1. 0.0005 2.5 25 22.5 50 2. 0.0010 5.0 25 20.0 50 3. 0.0015 7.5 25 17.5 50 4. 0.0020 10.0 25 15.0 50 5. 0.0025 12.5 25 12.5 50 6. 0.0040 20.0 25 5.0 50 7. Unknown given volume 25 upto the mark 50 8. Reference − 25 25 50

Part B & Part C : Procedure, graphs, calculations and results refer the previous experiment.

Experiment No. 2 Aim : Determine the λmax and concentration of unknown solution of copper sulphate. Chemicals : 1. 0.01 M CuSO4 stock solution. 2. Unknown CuSO4 solution. 3. Liquor ammonia. 4. Distilled water. Apparatus : 1. Colorimeter 2. 50 ml capacity calibrated eight volumetric flasks. 3. Micropipette. 4. Measuring cylinder. Procedure : The experiment is carried out in three parts as follows :


Part A :

Preparation of different concentration of CuSO−4 − NH3 complex solutions from the given

0.01 M CuSO4 stock solution. The glasswares i.e. volumetric flask, beaker, measuring cylinder, which are required for the experiment were properly cleaned with tap water. The volumetric flasks should be labelled with numbers one to eight. By using the 0.01 M CuSO4 stock solution the lower concentrations i.e. 0.006 M, 0.005 M, 0.004 M, 0.003 M, 0.002 M and 0.001 M were prepared by using the equation N1V1 = N2V2. Thus, in order to prepare 0.006 M and 50 ml solution from 0.01 M CuSO4 stock solution, the volume to be taken is calculated as follows :

Old (Stock) New (Stock) N1V1 = N2V2 0.01 × V1 = 0.006 × 50

N1 = 0.01 M V1 = ? N2 = 0.006 M V2 = 50 ml

∴ V1 = 0.006 × 50

0.01 = 30 ml.

Thus, to prepare 0.006 M and 50 ml solution, 30 ml of stock solution should be withdrawn and diluted to 50 ml by adding distilled water. i.e. (30 ml 0.01 M stock solution + 20 ml distilled water). Thus, all the solutions were prepared by using same method and are summarised in the following tabular form. Preparation of solutions :

Flask No.

Conc. of CuSO4 solution

M

0.01 M CuSO4 solution taken

for dilution (ml)

Liquor NH3 added (ml)

Distilled water (ml)

Total volume after dilution

1. 0.006 30 12 08 50

2. 0.005 25 12 13 50

3. 0.004 20 12 18 50

4. 0.003 15 12 23 50

5. 0.002 10 12 28 50

6. 0.001 05 12 33 50

7. Unknown given volume 12 upto the mark

50

8. Reference − 12 38 50

These solutions are prepared in calibrated 50 ml capacity volumetric flask. The solutions should be shaken well to make them homogeneous. For the preparation of solution, same measuring cylinder and micropipette should be used throughout the experiment. Thus the stock solution, liquor ammonia and distilled water should be taken by using same measuring cylinder and micropipette.

65


66

Part B : Determination of λmax : The photoelectric colorimeter is having inbuilt 6 to 8 filters. From these filters, one proper filter is to be selected. To perform this experiment a blank solution (reference) is prepared by diluting 12 ml of liquor ammonia to 50 ml using distilled water. From the prepared solutions the one having highest concentration i.e. 0.006 M is used for this part. Suppose filter no. 1 is inserted at the proper place. The reference solution is placed in the cell and the instrument is made on. With the help of zero adjustment knob, the optical density is made zero. Then the reference solution is removed and the cuvette having 0.006 M solution is inserted at that place. Waiting for few minutes constant optical density is recorded. Thus, same procedure is repeated for all the filters and wavelength and optical density is recorded. The wavelength is plotted against the optical density and from the graph, the filter which is showing maximum optical density is called as the λmax and is used for the next part. Observation Table : For λmax

Filter No. Wavelength λ in nm Optical density 1. 2. 3. 4. 5. 6. 7. 8.

450 470 520 540 570 600 620 670

Part C : To find out the O.D. and %T for the known and the unknown solution : (Using Digital Colorimeter) 1. Insert the filter having λmax in proper place. If inbuilt filter system is there, then adjust

the λmax filter (wavelength) in the light path with the help of filter knob. 2. Rinse the cuvette with reference solution and fill it with the same solution upto the

mark. The outer surface of the cuvette is cleaned with filter paper and is inserted at the sample cuvette.

3. The instrument is made on. On the display it shows reading. With the help of zero adjustment knob it is adjusted to read zero optical density. Then this zero adjustment knob is kept undisturbed throughout the experiment. Now the reference solution is removed. The cuvette is washed with distilled water and rinsed with the 0.006 M solution. Then it is filled upto the mark with 0.006 M solution. Clean it from outside and keep it in the sample cuvette. Make the instrument on wait for few minutes and record the constant optical density. Repeat the same procedure for remaining solutions as well as unknown. Report the observations in the observation table as follow.


Observation Table :

Flask No. Concentration in moles/lit.

Absorbance or Optical density

% Transmission

1.

2.

3.

4.

5.

6.

7.

0.006

0.005

0.004

0.003

0.002

0.001

Unknown

Graphs :

From the observations, plot the following graphs :

(a) Wavelength against optical density.

(b) Optical density against concentration.

(c) % Transmission against concentration.

0.2

0.4

0.6

0.8

1.0

O.D.

�max

O.D.

0.2

0.4

0.6

0.8

1.0

100 200 300 400 500 600 700

Wavelength (nm)(a)

0.002 0.004 0.006

Unknown

% T

20

40

60

80

100

0.002 0.004 0.006

Unknown

Concentration (M)(b)

Concentration (M)(c)

Fig. 1

Knowing the optical density and % T of unknown solution, the corresponding concentration is easily obtained from graphs (b) and (c).

Result Table :

Sr. No. Description Value with units

1. The λmax of the test solution = ………. nm.

2. Conc. of unknown solution from graph (b) = ………. moles/l

3. Conc. of unknown solution from graph (c) = ………. moles/l

67


Experiment No. 3 Aim : To determine the amount of copper present in the given solution of copper sulphate by colorimetric titration method using standard solution of EDTA. Apparatus : Colorimeter and accessories, volumetric flasks, burette etc. Theory :

HOOCH C2

�

OOCH C2

HN CH2 CH N H+

2

CH COO2

�

CH COOH2 Ethylenediamine tetraacetic acid (H2A) is a strong complexing agent and forms complexes with various metal ions. It is usually used as its disodium salt, Na2H2A. It forms complexes with copper (II) and many other ions in 1 : 1 mole ratio. The quantitative estimation of Cu(II) is possible with photometric measurements at a particular wavelength. The copper – EDTA complex absorbs in the visible region more strongly at wavelength of its λmax. Known amount of copper solution is taken for estimation and solution of EDTA is added with the help of burette. The absorbance is noted at 660 nm for each addition of EDTA. The absorbance goes on increasing with increase in concentration of Cu-EDTA complex upto the equivalence point. The total Cu ions turn into equivalent amount of Cu-EDTA complex at the equivalence point giving maximum absorbance. After the equivalence point further addition of EDTA gives limiting value of absorbance. Chemicals : 0.05 M (approximate) CuSO4, 0.1 M EDTA, acetate buffer of pH 2.2 − 2.4, (add 1 molar solution of hydrochloric acid to 350 ml of a solution of sodium acetate of 1 molar concentration until the pH of the mixture is 2.2 measured by pH meter). Procedure : The experiment is carried out in two parts : Part A : Determination of λmax of Cu-EDTA Complex : Take 10 ml of 0.05 M (approximate) copper sulphate solution and 20 ml acetate buffer solution and add 8 ml of 0.1 M EDTA solution to it. Shake the solution and use this for the determination of λmax of Cu-EDTA complex (use water as blank). Repeat the same procedure as in previous Experiment No. 2 and find out the λmax. Part B : Estimation of Copper : 1. Pipette out 10 ml of given solution of copper sulphate in a conical flask and add

20 ml acetate buffer solution and 50 ml water in it. 2. Fill the burette with 0.1 M solution of EDTA. From the burette add 0.5 ml of EDTA into

the conical flask and shake the solution. From this solution take sufficient quantity into the properly cleaned cuvette and find out its absorbance at λmax, which is adjusted to read zero optical density by taking water as a blank.

68


3. Transfer the solution from the cuvette back into the conical flask and further add into it again 0.5 ml of EDTA solution from burette. Shake well and again take it in the same cuvette upto the mark and find the absorbance of the solution as above. Thus continue the same procedure upto 2 ml of the burette reading.

4. After 2 ml of burette reading carry on the titration by adding 0.2 ml of EDTA solution at a time. After the equivalence point constant readings of absorbance are obtained. Continue the addition till 5-6 fairly constant readings are recorded.

Observation Table : The filter used for the experiment. i.e. λmax = …… nm. Observation No. ml of 0.1 M EDTA added Absorbance

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

0.5 1.0 1.5 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6

Graph : Plot a graph of absorbance against ml of EDTA added. The intersection of two lines gives equivalence point of the titration. Let it be ‘X’ ml.

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

O.D.

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8

X ml

ml of 0.1 M EDTA added Fig. 1

69


70

Calculations : 1000 ml 1 M EDTA = 63.54 g of Cu2+ 1000 ml 0.1 M EDTA = 6.354 g of Cu2+

‘X’ ml 0.1 M EDTA = 6.354 × 'X'

1000 g of Cu2+

= ‘W’ g. 10 ml of given solution of CuSO4 contains W g of Cu2+. ∴ 1000 ml of given solution of CuSO4 contains (100 × W) g of Cu2+. Result Table :

Sr. No. Description Value with units 1. λmax of Cu-EDTA complex solution = ………. nm. 2. Concentration of Cu2+ in the given solution = ………. g l−1

Experiment No. 4 Aim : To determine the indicator constant of methyl red indicator. Theory : The acid-base indicator is a substance which varies in colour with the pH of the solution. It possesses the properties of weak acid or weak base, in some form or the other. According to the Bronsted concept of acids and bases, the acidic and basic forms of indictor will exist in the following equilibrium in aqueous solution.

InA + H2O ≡ H3O+ + InB (acid form) (basic form) The forms InA and InB have different colours because one of them has a tautomeric form of different colour existing in equilibrium with it. The equilibrium constant of the above equilibrium can be written as −

KIn = [InB] [H3O+]

[InA]

neglecting the activity coefficient terms, KIn is called the indicator constant. Thus, the above equation can be written as −

pH = pKIn + log [InB][InA] … (1)

Since the intensity of the colour of a solution is a proportional to the concentration of the substance responsible for the colour. The above equation can be written as

pH = pKIn + log Colour due to alkaline (basic) form

Colour due to acidic form … (2)

When colour due to the alkaline form predominates over the colour due to the acidic form ([InB] >> [InA]), the solution has the colour of the alkaline form of the indicator and vice-versa. Over a certain range of pH, neither of the colours predominates. This range of pH is called the indicator range and generally lies between pKIn − 1 and pKIn + 1.


Now to evaluate the indicator constant, consider the acid-base indicator methyl red. The methyl red have the acidic form (HMR) and basic form (MR−) as given below.

HMR ≡ MR− + H+

N(CH )3 2

N

+NH

COO�

N(CH )3 2

N

NH

COO�

N(CH )3 2

N

N

COO�

+

+ H+

Acid formHMR (Red)

Basic formM (Yellow)R

The dissociation constant K, of this equilibrium between acid form and base form is given by using equation (1).

pH = pK + log [MR−][HMR]

In this above equation if [MR−] = [HMR] the equation reduces to pK = pH From the above discussion, the determination of pK can be done as follows. The [HMR] and [MR−] have strong absorption peaks in the visible region. Series of solutions having known pH (between 2 to 10) are prepared and equal amount of methyl red is added to each of these solutions. The colour intensity of these solutions depends upon the degree of dissociation which in turn depends on pH of the solution. The solution having low pH values have red colour while those having higher pH values shows yellow colour. The solution having pH in between shows intermediate colour between red and vellow. The absorbances of solutions are determined as a function of pH at proper wavelength. The point of inflection represents [MR−] = [HMR] and the pH at this situation is pK. Apparatus : Colorimeter with filters, pH meter, burettes, 50 ml volumetric flasks (12), calibrated micro pipette, hard glass test tubes (12), etc. Chemicals : Methyl red solution, (Dissolve 20 mg of methyl red in 60 ml of ethanol and dilute to 100 ml by water), 0.2 M disodium hydrogen phosphate, 0.1 M citric acid. Procedure : The experiment is performed in three parts as follows :

71


72

Part A : Preparation of solutions : 1. Clean all the 50 ml capacity volumetric flasks and label them from 1 to 12. 2. Clean two burettes of 50 ml capacity and fill them by 0.2 M Na2HPO4 and 0.1 M citric

acid solution. 3. To prepare the buffer solutions from Na2HPO4 and citric acid in the labelled

volumetric flasks, use the following table : Flask No. Expected pH 0.2 M Na2HPO4 (ml) 0.1 M citric acid (ml)

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

3.4 3.8 4.2 4.6 5.0 5.4 5.8 6.2 6.6 7.0 7.4 8.0

11.40 14.20 16.56 18.70 20.60 22.30 24.18 26.44 29.10 32.94 36.34 38.90

28.60 25.80 23.44 21.30 19.40 17.70 15.82 13.56 10.90 7.06 3.66 1.10

4. The hard glass test tubes should be properly cleaned and labelled as 1 to 12. From the above solutions withdraw 10 ml from each flask and place into the hard glass test tubes respectively.

5. With the help of micro pipette add one ml of indicator in each of the test tubes and shake well.

Part B : Determination of pH of buffer solutions : The buffer solutions prepared in volumetric flask are labelled 1 to 12, their pH is determined with the help of pH meter as follows : 1. Standardize the pH meter by using standard 0.05 M potassium hydrogen phthalate

solution (as explained in pH metre experiment). 2. Take sufficient amount of buffer solution from flask No.1 in a beaker and dip in it the

combined glass and calomel electrode. Wait for few minutes and record the constant pH of the solution.

3. Repeat the same procedure for remaining solutions and report their readings as observed pH of solutions.

Part C : Determination of absorbance : 1. First find out the λmax by using the procedure given in the previous experiment.

Usually for methyl red, blue green filter (490-520 nm) is used. 2. In digital colorimeter inbuilt filters are there, so adjust the filter. In a clean cuvette

take buffer solution from flask no.1 as blank upto the mark. Make cuvette dry and clean from outside with the help of filter paper and insert it into the sample compartment.


3. Make the instrument on and adjust zero optical density. 4. Remove the cuvette, wash it with distilled water. Now take the solution from hard

glass test tube No. 1 into the cuvette upto the mark. Clean it from outside and insert it into the sample compartment, make the instrument on, wait for few minutes and record the constant optical density.

5. Repeat the same procedure for remaining solutions of the hard glass test tube. Report the observations in the tabular form as follows :

Observation Table : Solution

from Test Tube

Observed pH

Absorbance A

% T ΔA ΔpH ΔAΔpH

Mean pH

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Graphs :

(i) Absorbance against observed pH, (ii) ΔAΔpH against mean pH.

0.9

Absorbance(A)

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

1 2 3 4 5 6 7 8 9 10

pH = pKa

[MR ]�

[HMR]= 1

Observed pH

Mean pH

�

�

A

pH

Fig. 1

73


74

Calculations : From the graph, the pH value at the inflection or at peak position is equal to pKa. Knowing the pKa, Ka is obtained as Ka = − log pKa. Thus, the value of dissociation constant Ka of the indicator is obtained. Result Table :

Sr. No. Description Value with units 1. The λmax of indicator solution. = ………. nm. 2. pKa of the indicator methyl red. = ……….

3. Ka of the indicator methyl red. = ……….

VIVA VOCE 1. Define Beer's and Lambert's law. 2. What is optical density ? What is its unit ? 3. Define the terms : (i) Radiant power, (ii) Transmittance, (iii) Molar extinction coefficient. 4. What are the different components of single beam colorimeter ? 5. What is the relationship between O.D. and concentration and why ? 6. What is meant by λmax ? Why is it necsessary to determine it ? 7. 'Beer-Lambert's law is applicable to dilute solutions only, explain. 8. Which equation is used to prepare the solution having lower concentration than that

of given stock solution ? 9. In colorimetric estimation of CuSO4 solution, liquor ammonia is added in the

preparation of solution. Why ? 10. How the percent T is obtained from measured O.D. ? 11. Why 2 N H2SO4 is added in the preparation of dilute solutions of KMnO4 ? 12. What is the structural formula of EDTA ? 13. Why the acetate buffer of pH 2.2 - 2.4 is added ? 14. Why there is no change in O.D. after the equivalence point ? 15. Define the indicator constant. 16. Define the indicator range. Give the limiting value of indicator range. 17. What is the nature of absorbance against pH graph and why ?

- - -

75

2…

Potentiometry (Any Three)

Theory :

In potentiometry, electromotive force (E.M.F.) of cells and electrode potentials of single electrodes are measured. Potentiometry has wide range of applications in analytical, inorganic and physical chemistry.

Electrochemical cells :

Electrochemical cells are used to convert (i) electrical energy to chemical energy and (ii) chemical energy to electrical energy. A cell is basically an arrangement of two electrodes dipping in an electrolyte solution, capable of generating electricity due to chemical change within the cell, or of producing chemical change due to passage of electricity through the cell.

A cell in which electrical energy is converted into chemical energy is called electrolytic cell. In such a cell, a non-spontaneous reaction is made to occur by passage of electricity through the electrolyte. Oxidation and reduction reactions are forced to occur.

A cell in which chemical change generates electricity is called a galvanic or voltaic cell. In such a cell, spontaneous oxidation and reduction reactions occur producing electricity. These cells are formed by combining two single electrodes through a porous partition or a salt bridge.

An electrode in an electrolyte is called a single electrode. At every single electrode there exists a potential difference called single electrode potential. When two such electrodes differing in their potential are combined and connected, a spontaneous flow of electrons takes place generating electricity. This difference in electrode potential of the two electrodes is called electromotive force (E.M.F.).

Measurement of Electrode Potential :

Electrode potential of different electrodes are measured with reference to standard hydrogen electrode which is a primary standard electrode. Due to difficulties in construction of hydrogen electrode, secondary reference electrodes such as calomel electrodes are commonly used.

T.Y.B.Sc. Practical Chemistry (Physical Practicals) Potentiometry

Calomel Electrode :

This secondary reference electrode essentially consists a paste of mercury-mercurous chloride (Hg2Cl2) and potassium chloride solution of specified concentration as shown in Fig. 1.

The electrode is represented as

Pt⏐Hg⏐Hg2Cl2(s)⏐KCl(satd.)

The reaction at the electrode is

Hg2Cl2(s) + 2eσ ≡ 2Hg(l) + 2Cl−(satd.)

This electrode is reversible to Cl− ions. The potential of the electrode depends on concentration of KCl solution used. The electrode potential at any given temperature ‘t’, with different concentrations of KCl are given below :

0.1 N KCl E = 0.3338 − 7 × 10−5 (t − 25)

1.0 N KCl E = 0.2800 − 2.4 × 10−4

Saturated KCl l2 = 0.2415 − 7.6 × 10−4 Calomel electrode with saturated KCl solution is most convenient to use because of ease of replacing the solution. Nernst Equation : The single electrode potential of an electrode where the reaction Mn+ + ne− → M (metal ions in equilibrium with the metal) occurs is given by the Nernst equation

Ered = Eored −

2.303 RTF log

aM

aMn+

Pt

KCl solution (saturated)

Paste of Hg + Hg Cl2(s)2

KCl crystals

Ceramic disc orsintered glass disc

Hg

Fig. 1

where, Ered and Eored are the reduction potential and standard reduction potential of the metal,

respectively; R is the molar gas constant, n is the number of electrons involved in the reaction at the electrode, F is faraday the unit of electricity, aM and aMn+ are the activities of metal and metal ion respectively. Activity of all metals is taken as one and therefore Nernst equation becomes

Ered = Eored −

2.303 RTnF log

1aMn+

= Eored +

2.303 RTnF log aMn+

Measurement of Cell E.M.F. using Potentiometer : The potentiometer used for e.m.f. measurements operate on Poggendorff’s compensation principle. In this method, an unknown e.m.f. is opposed by another e.m.f. till they are balanced. The circuit diagram of potentiometer is as shown below.

76


S

A

X

_+

K

G

C C'B

_ +

P

Fig. 2

AB is a stretched wire of uniform cross-section usually one meter in length. P, S and X are the storage cell, standard cell and unknown cell respectively. At any given time only two sources of E.M.F. are in circuit P and either S or X. P, S and X are connected in such a way that, their e.m.f.’s oppose each other. There is a provision to take either S or X in the circuit. The E.M.F. of P is greater than E.M.F. of S or X. K is the tap key. G, the galvanometer is connected through a sliding contact which can be moved along AB. The e.m.f. of P is uniformly distributed over the entire length of wire AB. First the standard cell is brought into circuit and the sliding contact is moved along AB till at a point C' no current flows through the circuit (indicated by no deflection in the galvanometer). At this point the bridge is said to be balanced. The potential drop across AC' is balanced by E.M.F. of standard cell. i.e. ES ∝ length AC' Then standard cell is taken out of the circuit and unknown cell whose e.m.f. is to be measured is brought in the circuit. The sliding contact is again moved along AB till the bridge is balanced at point C. The potential drop across AC is balanced by E.M.F. of unknown cell X. i.e. EX ∝ length AC

Therefore, EX

ES =

ACAC'

or EX = ES × ACAC'

Knowing ES and measuring lengths AC and AC’, EX, E.M.F. of unknown cell can be determined. Standard Cell : The standard cell used in measurement of e.m.f. of unknown cells by potentiometry is the saturated Weston standard cell. The requirements to be met by a standard cell are (i) its e.m.f. must be reproducible and constant with time, (ii) it should be reversible and should not be subject to permanent damage due to passage of current through it and (iii) its temperature coefficient of e.m.f. should be small. The cadmium Weston standard cell satisfies

77


all above requirements. The cell is shown in following figure, when the cell operates the following reversible reaction takes place.

Cd(s) + Hg2SO4(s) + 8/3 H2O(l) ≡ CdSO4 ⋅ 8/3 H2O(s) + 2Hg(l) The potential of the cell at 20°C is 1.01830 volt and at 25°C is 1.01807.

Sealing wax

Cork

Saturated CdSO solution4

Hg + Hg SO2 4

Hg

CdSO 8/3 H O crystals4 2

Cd Amalgam

Fig. 3

Potentiometric Titrations : The variation of electrode potential with concentration of ions is used as an indicator in potentiometric titrations. The potentiometric method is applicable to a wide range of titrations including titrations of highly coloured solutions where visual indicators are useless. Indicator Electrodes :

For the electrode, a metal and a solution of its own ions Mn+ + neσ ≡ M, the electrode potential is given by,

E = E° + 2.303 RT

nF log aM+n

For dilute solutions the activity can be replaced by molar concentration i.e.

E = E° + 2.303 RT

nF log Cm+n … (1)

The electrode potential of oxidation-reduction system (where platinum is used as electrode), the electrode reaction is

Oxidized form + ne ≡ Reduced form and the electrode potential is given by,

E = E° + 2.303 RT

nF log Cox

Cred … (2)

where, Cox and Cred are molar concentrations of oxidized and reduced forms respectively. The electrode potential depends on concentration of both the oxidized form and reduced form.

78


From the above two equations it is clear that the change in electrode potential reflects a

change in Cm+ or Cox

Cred .

This forms the basis of potentiometric titration. It is impossible to determine E on its own, the indicating electrode is combined with a reference electrode such as calomel electrode. The potential of reference electrode does not change during titration. Salt Bridge : Salt bridge is used to connect the reference electrode with the indicator electrode in the test solution. The salt bridge is a inverted U shaped tube as shown in Fig. 4.

Stop cock

Sintered glass disc

Agar Agar + KNO� 3

Fig. 4 The two ends of U tube are fitted with sintered glass discs of fine porosity. The bridge is filled with jelly of agar-agar and an electrolyte such as KCl, KNO3, NaNO3 etc. The positive and negative ions of these salts have nearly same mobility, and hence the junction potential is minimized. Titration Cell : The cell used in potentiometric titration is prepared by combining a reference electrode and an indicator electrode through a salt bridge. Typical cells in acid-base titration are – 1. Pt⏐Hg⏐Hg2Cl2(s)⏐Sat. KCl⏐Salt Bridge⏐quinhydrone⏐Pt. 2. Pt⏐H2⏐H+⏐Salt Bridge⏐Sat. KCl⏐Hg2Cl2(s)⏐Hg⏐Pt. Some of the commonly employed indicator electrodes are : Ag⏐AgCl⏐Cl−, Pt⏐Fe2+, Fe3+ and Pt⏐quinhydrone, H+.

79


Quinhydrone Electrode : Quinhydrone is a equimolar mixture of quinine and hydroquinone. The reaction that occurs is

OH

OH

O

O

+ 2H + 2e+

The above reaction can be represented by general equation

QH2 ≡ Q + 2H+ + e The platinum wire is dipped into a saturated solution of quinhydrone to obtain this electrode. The electrode potential of quinhydrone electrode depends on concentration of H+ ions.

Its standard potential at any given temperature is given by Eored = − 0.6994 + 0.0007 (t − 25)

volt and at 25°C, Eored = − 0.6994 volt.

Experiment No. 1 Aim : To prepare standard 0.2 M Na2HPO4 and standard 0.1 M citric acid solution and to prepare four different buffer solutions using them. Determine the pKa value of these and unknown solution. Apparatus : Potentiometer, calomel electrode, platinum electrode, cadmium Weston standard cell, 250 ml standard flask, burette etc. Chemicals : Disodium hydrogen phosphate (Na2HPO4.12H2O), citric acid, quinhydrone powder, 0.05 M potassium hydrogen phthalate (pH = 4) solution (solution A) and two given unknown buffer solutions F and G. Procedure : (A) Preparation of standard 0.2 M Na2HPO4 and 0.1 M citric acid solutions : (i) 0.2 M Na2HPO4 : Na2HPO4.12H2O molecular weight is 358.2. Therefore, to prepare 250 ml of 0.2 M

Na2HPO4 solution, the quantity required is 17.91 gms. Accurately weigh 17.91 g of Na2HPO4.12H2O and prepare 250 ml solution in standard

250 ml flask, using distilled water. (ii) 0.1 M citric acid solution, molecular weight is 192.1 (anhydrous) : Therefore to prepare 250 ml 0.1 M citric acid solution accurately weigh 4.8025 g of

citric acid and prepare 250 ml solution of it in a standard 250 ml flask, using distilled water.

80


81

(iii) Preparation of buffer solutions : Using above 0.2 M Na2HPO4 and 0.1 M citric acid solutions, prepare four buffer

solutions B, C, D and E as given in the following table. Buffer Solution Table : 0.2 M Na2HPO4.12H2O (17.91 g/250 ml) + 0.1 M citric acid (anhydrous) (4.8025 g/250 ml).

Buffer solution Volume of 0.2 M Na2HPO4.12H2O in ml

Volume of 0.1 M citric acid in ml

B C D E

10.3 25.8 41.2 48.6

39.7 24.2 08.8 01.4

(B) Standardization of Potentiometer : (i) Connect the Weston standard cell (E = 1.0185 V) to the potentiometer. (ii) Standardize the potentiometer. After doing this do not disturb the position of

standardization knobs on your potentiometer throughout the experiment. (C) Determination of ECalomel : Set up the cell Pt⏐Hg⏐Hg2Cl2(s)⏐Cl−⏐⏐ H+, quinhydrone⏐Pt+ calomel (Satd) pH = 4 indicator electrode 1. Take about 50 ml of 0.05 M (pH = 4) solution of potassium hydrogen phthalate in

small beaker. Add quinhydrone powder into it till the solution is saturated with it. Stir well.

2. Place platinum and calomel electrodes into the solution. Connect platinum electrode to the positively marked terminal on the potentiometer and calomel to the negatively marked terminal.

3. Record the e.m.f. of this cell as Eobs. (D) Determination of pH of buffer solutions : 1. Take about 50 ml of the buffer solution B prepared in a small beaker. Add

quinhydrone powder till it is saturated with it. Stir well. 2. Place platinum and calomel electrodes into the solution. Connect platinum electrode

to the positively marked terminal and calomel electrode to the negatively marked terminal on the potentiometer.

3. Determine the e.m.f. of the cell. 4. In a similar way find e.m.f. of the buffer solutions C, D, E and given unknown solutions

F and G. 5. Record the room temperature as t°C.


82

Observation Table : 1. Room temperature = (t) = ……….. °C. 2. E.M.F. of 0.05 M potassium hydrogen phthalate (solution A) = ……….. volt.

Buffer E.M.F., volts

B C D E F G

(E) Calculations : Calculation of Ecalomel :

pH = E

oQ − Ecal − Eobs

2.303 RTF

= E


0.0001984 × T

EoQ = 0.699 volts, T = t + 273

∴ pH = 0.699 − Ecal − Eobs

0.0001984 × (t + 273)

Using pH of 0.05 M potassium hydrogen phthalate (pH = 4) and its E.M.F. = Eobs find Ecal.

∴ 4 = 0.699 − Ecal − Eobs of solution A

0.0001984 × (t + 273)

Find Ecalomel, ∴ Ecalomel = ……….. (F) Calculation of pH of buffer solutions :

pH = 0.699 − Ecalomel − Eobs

0.0001984 × (t + 273)

Substitute the value of Ecal calculated above and Eobs = E.M.F. of buffer solutions A, B, C, D, E, F and G and calculate pH of the respective solution. Result Table :

Buffer pH

B C D E F G


83

Experiment No. 2 Aim : To determine the pKa value of given monobasic weak acid by potentiometric titration. Theory : A weak monobasic acid dissociates as

HA ≡ H+ + A−

and dissociation constant Ka = [H+] [A−]

[HA]

∴ [H+] = Ka [HA][A−]

where [H+], [HA] and [A−] are all equilibrium concentrations. Taking − log of both sides, we get

−log [H+] = − log Ka − log [HA][A−]

∴ pH = pKa − log [HA][A−] = pKa + log

[A−][HA] when acid is half neutralised, [A−] = [HA]

∴ log [A−][HA] = 0 and hence pH = pKa

Therefore, during the course of titration, when the acid is half neutralised, its pH = pKa. From pKa value, Ka dissociation constant can be calculated. Therefore, if we are able to find pH or [H+] ion concentration, when the acid is half neutralised we can determine dissociation constant of the weak acid. The quinhydrone electrode can be used as an indicator electrode in acid-base titration to find H+ ion concentration. Apparatus : Potentiometer, Standard cell, Platinum electrode, Calomel electrode etc. Chemicals : 0.1 N (approximate) acetic acid solution. 0.5 N (exact) NaOH solution, quinhydrone powder etc. Procedure : (A) Standardization of Potentiometer : Connect the standard cell (E = 1.0185 volt, Weston cell) to the potentiometer and standardize the potentiometer to 1.0185 volts. After standardization do not disturb the positions of standardization knobs of your potentiometer. (B) Potentiometric Titration : For the acid-base titration set up the cell as follows : Pt⏐Hg⏐Hg2Cl2(s)⏐Cl−⏐⏐ H+, quinhydrone⏐Pt Calomel Acetic acid 10 ml


1. In a small beaker (100 ml capacity) take 10.0 ml 0.1 N acetic acid and add about 40.0 ml distilled water. To this solution add quinhydrone powder till a saturated solution of it is obtained. Stir the solution well.

2. Dip a calomel and a platinum electrode into the solution, connect the calomel electrode to negatively marked terminal and platinum to positively marked terminal of potentiometer.

3. Determine the e.m.f. of the cell. 4. Add 0.2 ml of 0.5 N NaOH from the burette, stir the solution. And measure the e.m.f.

of the cell.

Microburette

Magnetic stirrer

0.1 N aceticacid + 40 ml watersaturated withquinhydrone powder

Iron stand

To potentiometer

+_

Pt

Calomel

0.5 NNaOH

Fig. 1

5. In a similar way add 0.2 ml NaOH each time and measure the e.m.f. of the cell every time and record the observations in the observation table.

6. Measure cell e.m.f. till the end point exceeds by about 1.0 ml (i.e. one ml after you notice a sharp change in e.m.f.). The potentiometer shows negative e.m.f. Take 2-3 negative readings and stop the titration.

7. Plot graph of : (i) E.M.F. against ml of NaOH added and

(ii) ΔEΔV against mean V of NaOH

Plot both graphs on the same graph paper. ΔE is the difference in two successive e.m.f. readings. ΔV is difference in two successive additions i.e. 0.2 ml of NaOH and mean V is the

mean of two successive additions of NaOH.

84


8. From the graph of e.m.f. against ml of NaOH added, find the neutralization point of the titration.

9. From the graph ΔEΔV against mean V of NaOH, find e.m.f. at half neutralization point.

(C) Determination of Ecalomel : 1. Wash the calomel and platinum electrodes. 2. In a small beaker take about 50 ml of 0.05 M potassium hydrogen phthalate (pH = 4)

solution. Add quinhydrone powder till the solution is saturated with it. Stir well. 3. Dip calomel and platinum electrodes into the solution and connect them to

negatively and positively marked terminals respectively. 4. Determine the e.m.f. of the cell, Eobs. Observation Table :

Obs. No.

ml of 0.5 N NaOH added

E.M.F. (volts)

ΔE (volts)

ΔV (ml)

ΔEΔV

(volt/ml)

Mean V (ml)

1 2 3 4 − − − − − −

3.0

0 0.2 0.4 0.6 − − − − − −

E1 E2 E3 E4 E5 − − − − −

−E1 − E2 E2 − E3 E3 − E4 E4 − E5

− − − − −

−0.2 0.2 0.2 − − − − − −

− − 0.1 0.3 0.5 − − − − − −

Graphs :

Half eq.pt.

Mean V

�

�

E

V

Equivalencepoint x

0.5 N ml ofNaOH added

+_

E.M.F.

E1/2

Mean V

x2

Fig. 2

85


86

Calculations : Calculation of Ecalomel : Using Eobs of 0.05 M potassium hydrogen phthalate (pH = 4) solution find Ecalomel using the expression

pH = E


0.0001984 × (t + 273), EoQ = 0.699 volts given

∴ 4 = 0.699 − Ecal − Eobs pot. hy. phthalate

0.0001984 × (t + 273)

∴ Ecal = ……….. volts Calculation of pKa of acetic acid :

1. From graph of e.m.f. vs. ml of 0.5 N NaOH added and ΔEΔV vs. mean V

(i) Neutralization point = x = ……….. ml

(ii) Half neutralization point = x2 = ……….. ml

(iii) E1/2 e.m.f. at 12 neutralization point = ……….. volts.

pKa = pH at 12 neutralization point

pH = E

oQ − Ecal − E1/2

0.0001984 × (t + 273)

pKa = ………..

1. E.M.F. of calomel electrode e.g. Ecal = ………. volt 2. Neutralization point = ………. ml

3. E1/2 e.m.f. at 12 neutralization point = ………. volt

4. pKa = ……….

Experiment No. 3 Aim : To determine the formal redox potential of Fe2+, Fe3+ system potentiometrically. Theory : Consider the following cell involving redox electrode Pt⏐Fe2+, Fe3+,

Pt⏐Hg⏐Hg2Cl2(s)⏐Cl−⏐⏐ Fe2+, Fe3+⏐Pt Calomel The cell e.m.f. is given by ECell = EFe2+, Fe3+ − ECal


87

Now, EFe2+, Fe3+ = Eo

Fe2+, Fe3+ + 2.303 RT

F log Fe3+, Fe2+

∴ ECell = ⎣⎢⎡

⎦⎥⎤ E

o

Fe2+, Fe3+ + 2.303 RT

F log Fe3+, Fe2+ − ECal

The electrode potential of Fe2+, Fe3+ system depends on concentration of Fe2+ and Fe3+.

If we add KMnO4 in Pt/Fe2+, Fe3 half cell, concentration of Fe2+ decreases and Fe3+ increases by the reaction

5Fe2+ + MnO−4 + 8H+ ⎯→ 5Fe3+ + Mn2+ + 4H2O

and the ratio ⎣⎢⎡

⎦⎥⎤Fe3+

Fe2+ increases and cell e.m.f. increases.

Now near end point there is a sudden decrease of Fe2+ ion and therefore e.m.f. sharply increases. The above reaction involves H+ and electrode potential also depends on H+. Therefore, to avoid the change in H+, the titration is carried out with large excess of H2SO4 (2M). At half equivalence point, [Fe2+] = [Fe3+] then,

[Ecell]1/2 neutralization point = EFe2+, Fe3+ − Ecal

∴ Eo

Fe2+, Fe3+ = (Ecell)1/2 neutralization point + Ecalomel

Eformal redox = Ecell at 1/2 neutralization point + Ecalomel

Apparatus :

Potentiometer, calomel electrode, platinum electrode, standard cell etc.

Chemicals :

0.05 M potassium hydrogen phthalate, 0.01 N ferrous ammonium sulphate, 0.05 N KMnO4.

Procedure :

(A) Standardization of Potentiometer :

Connect the Weston standard cell (E = 1.0185 V) to your potentiometer, standardize the potentiometer to 1.0185 volts using fine and course knobs on your potentiometer. After doing this do not disturb these knobs throughout the experiment.

(B) Determination of Ecalomel :

1. Take 50 ml of 0.05 N potassium hydrogen phthalate in a small beaker. Add quinhydrone powder till saturated with it. Stir the solution using a magnetic stirrer.

2. Dip a platinum and a calomel electrode. Connect calomel and platinum electrodes to the terminal marked as positive and negative respectively on the potentiometer.

3. Find e.m.f. of the cell Eobs for 0.05 N potassium hydrogen phthalate solution.


(C) Redox Titration : Set up the cell as shown in Fig. 1.

Microburette0.05 N KMnO4

Ironstand

Calomelelectrode

Magnetic stirrer

10 ml 0.01 N FerrousAmmonium Sulphate+ 40 ml 2N H SO2 4

Platinum electrode

To potentiometer

+_

Fig. 1

Pt⏐Hg⏐Hg2Cl2(s)⏐KCl⏐⏐ Fe2+, Fe3+⏐Pt calomel (satd) 1. In a small beaker take 10 ml 0.01 N ferrous ammonium sulphate and add about 40 ml

2 N H2SO4 solution to it. 2. Dip calomel and platinum electrodes in the solution. 3. Connect platinum electrode and calomel electrode to the terminals marked positive

and negative respectively on the potentiometer. Stir the solution. Measure e.m.f. of the cell.

4. Add 0.2 ml 0.05 N KMnO4 solution from the microburette in the ferrous ammonium sulphate solution, stir the solution. Measure the e.m.f. of the cell.

5. In a similar way take e.m.f. measurements each time adding 0.2 ml of KMnO4 solution till the end point exceeds by about one ml (indicated by sharp change in e.m.f.).

6. Plot the graphs

(i) E.M.F. against ml of KMnO4 added and (ii) ΔEΔV against mean V of KMnO4 added.

7. From the graph of ΔEΔV against mean V, find equivalence point.

8. From the graph of e.m.f. against ml of KMnO4 added, find E1/2 e.m.f. at half equivalence point.

88


Observation Table : (i) E.M.F. of 0.05 N potassium hydrogen phthalate cell = ………. volts. (ii) Room temperature = ………. °C.

Obs. No.

ml of 0.05 N KMnO4 added

E.M.F. (volts)

ΔE (volts)

ΔV (ml)

ΔEΔV

(volt/ml)

Mean V (ml)

1 2 3 4 − − − − − − 16

0.0 0.2 0.4 0.6 − − − − − −

3.0

E1 E2 E3 E4 − − − − − −

− E2 − E1 E3 − E2 E4 − E3

− − − − − −

− 0.2 0.2 0.2 − − − − − −

− − 0.1 0.3 0.5 − − − − − −

Graphs :

E.M.F.

E1/2

ml of KMnO4

Half eq. pt. x/2

Mean V

�

�

E

V

Equivalencepoint x

Fig. 2 ∴ Equivalence point = ………. and E1/2 e.m.f. at half equivalence point = ………. Calculations : 1. Determination of Ecalomel :

pH = Eo

Q − Ecal − Eobs

0.0001984 (t + 273)

EoQ = 0.699 volt and t = Room temperature

89


90

Using pH = 4 for solution of 0.05 M potassium hydrogen phthalate and its Eobs, find Ecal.

∴ 4 = 0.699 − Ecal − Eobs

0.0001984 (t + 273)

∴ Ecalomel = ………… volts. 2. Formal Redox Potential :

Eo

Fe2+, Fe3+ = E1/2 + Ecal

Result Table : 1. Ecalomel = ………. volt 2. Equivalence point = ………. ml 3. E1/2 e.m.f. at 1/2 equivalence point = ………. volt 4. E

o

formal = E

o

Fe2+, Fe3+

= ………. volt

Experiment No. 4 Aim : To determine the amount of sodium chloride in the given solution by potentiometric titration against silver nitrate solution. Theory : Consider the following concentration cell :

−Ag⏐AgNO3⏐Salt Bridge⏐AgNO3⏐Ag+

C1 C2 The cell e.m.f. is due to difference in concentration of silver nitrate in the two half cells. If we add sodium chloride in say right hand side half cell the following reaction takes place.

AgNO3 + NaCl → AgCl ↓ + NaNO3 A precipitate of silver chloride is obtained. As a result the concentration C2 decreases and the electrode potential undergoes a change and as a consequence the cell e.m.f. changes. This change in cell e.m.f. corresponds to the amount of sodium chloride added. This provides a basis for the potentiometric titration of AgNO3 against NaCl. Apparatus : Potentiometer, silver electrodes, KNO3 or NaNO3 or NaNO2 salt bridge, standard cell etc. Chemicals : 0.01 N AgNO3, 0.05 N (approximate) NaCl. Procedure : (A) Standardization of Potentiometer : Connect the Weston standard cell (E = 1.01850 volts) to the potentiometer and standardize the potentiometer using knobs on your potentiometer to 1.0185 volts. After this do not disturb the knobs till the experiment is over.


(B) Titration of AgNO3 with NaCl :

0.05 M NaCl

Ironstand

Magnetic stirrer

20 ml.0.01 MAgNO3

MicroburetteTo potentiometer

_ +

Salt bridge

Ag

10 ml. 0.01 MAgNO + 40 ml.

distilled water3

Ag

Fig. 1

Set up the following cell : −Ag AgNO3

20 ml 0.01 M

NaNO2 Salt bridge

AgNO3 10 ml 0.01 M

+ 40 ml distilled water

Ag+

1. In a small beaker take 10 ml 0.01 M AgNO3 and add about 40 ml distilled water and this is the half cell in which titration is carried out.

2. In another small beaker take about 20 ml 0.01 M AgNO3 solution. 3. Place two silver electrodes in the two beakers. 4. Connect the two half cells (beakers) using a NaNO2 or KNO3 or NaNO3 bridge.

(Do not use KCl salt bridge because diffusion of Cl− ions from salt bridge will precipitate silver chloride. NaNO2 salt bridge is preferable because it quickly forms a gel with agar-agar or gelatine.]

5. Connect the silver electrode in the titration flask to the positive terminal and the other silver electrode to the negative terminal of the potentiometer.

6. Determine the e.m.f. of the cell. 7. Add 0.2 ml of 0.05 M NaCl solution into the titration flask. Stir the solution using a

magnetic stirrer. A precipitate of silver chloride is obtained. Find e.m.f. of the cell. 8. In a similar way add 0.2 ml of 0.05 M NaCl every time and find the e.m.f. of the cell. 9. Take e.m.f. measurement till the equivalent point exceeds by about one ml (upto

3.0 ml addition of NaCl). Report the observations in the observation table.

10. Plot the graphs – (i) E.M.F. vs. ml of NaCl added and (ii) ΔEΔV against mean V of NaCl

added.

91


11. From the graph ΔEΔV against mean V, find the neutralization point.

Observation Table :

Obs. No.

ml of 0.05 N NaCl added

E.M.F. (volts)

ΔE (volts)

ΔV (ml)

ΔEΔV

(volts/ml)

Mean V (ml)

1

2

3

4

−

−

−

−

−

−

16

0

0.2

0.4

0.6

−

−

−

−

−

−

3.0

−

−

−

−

−

−

−

−

−

−

−

−

−

−

0.2

0.2

0.2

−

−

−

−

−

−

− −

0.1

0.3

0.5

−

−

−

−

−

−

2.9

ΔE is the difference between two successive e.m.f. values and ΔV is the difference between two successive additions of NaCl. Mean V is the mean of two successive additions of NaCl. Graphs :

Mean V

�

�

E

V

E

ml of AgNO added3

E

ml of AgNO added3

( )I ( )II ( )III

Equivalencepoint (V )1

Fig. 2 Depending on the concentrations C1 and C2 in the two cell you may get graph of type I or type II.

92


93

Calculations :

From the graph ΔEΔV versus mean V find the equivalence point, that is amount of NaCl

required to react with 10.0 ml 0.01 M AgNO3 (V1). (i) Normality of NaCl : NaCl = AgNO3 N1V1 = N2V2 N1 × V1 = 0.01 × 10

∴ N1 = 0.01 × 10

V1 g. equi. l−1

(ii) Strength of NaCl : Strength = Normality × Equivalent weight = N1 × 58.5 g l−1 (iii) Amount of NaCl in 10.0 ml given solution

= N1 × 58.5 × 10

1000 g

Result Table :

1. Equivalence point or amount of NaCl (0.05 N) = required to precipitate AgCl completely.

= ………. ml

2. Exact normality of NaCl = ………. g equi. l−1

3. Strength of NaCl solution = ………. g l−1

4. Amount of NaCl in 10 ml of given NaCl solution = ………. g.

Experiment No. 5 Aim : To determine the amount of chloride and bromide from their mixture by potentiometric titration with silver nitrate solution. Theory : When silver nitrate is added to a solution of mixture of chloride and bromide, silver ions react with solution of Cl− and Br− ions to form AgCl and AgBr. Though both are sparingly soluble salts the solubility of AgBr is less than solubility of AgCl. Therefore, first precipitation of AgBr occurs. After the complete precipitation of AgBr, the precipitation of AgCl begins. This forms the basis for titration of halide mixture against silver nitrate solution potentiometrically. The cell used in this titration is

−Pt⏐Hg⏐Hg2Cl2(s)⏐(satd.) KCl ⏐⏐ Halide solution (Br−, Cl−)⏐Ag+ calomel electrode


94

The plots of e.m.f. vs. ml of AgNO3 added and ΔEΔV against mean V show two equivalence

points corresponding to complete precipitation of AgBr and AgCl respectively. From these amounts of chloride and bromide present in the mixture can be determined.

Apparatus :

Potentiometer, standard cell, calomel and silver electrodes, microburette etc.

Chemicals :

Given halide mixture (0.01 approximate equinormal mixture of Cl− and Br− ions), 0.01 N AgNO3 solution.

Procedure :

(A) Standardization of Potentiometer :

Connect the standard cell (Weston) to the potentiometer and standardize the potentiometer to 1.0185 volts using the fine and course adjustment knobs on the potentiometer. After this do not disturb the standardization knobs throughout the experiment.

(B) Titration :

1. In a small beaker take 10 ml of the given halide mixture containing Cl− and Br−. Add about 40 ml distilled water. Stir the solution using a magnetic stirrer.

2. Dip a silver and a calomel electrode into the solution. Connect the calomel electrode to negatively marked terminal and silver electrode to positively marked terminal of the potentiometer.

3. Measure the e.m.f. of the cell.

4. Add 0.5 ml of 0.01 N AgNO3 solution into the halide mixture, stir the solution well and measure e.m.f. of the cell.

5. In a similar way take measurement of e.m.f. values every time adding 0.5 ml AgNO3 till the total equivalence point exceeds by about 4.0 ml (Take about 25 measurements). Report the readings in the observation table as below.

6. Plot the graphs :

(i) E.M.F. vs. ml of AgNO3 added.

(ii) ΔEΔV against mean V of AgNO3 added.

7. From the graphs find the amount of AgNO3 required for complete precipitation of AgBr and AgCl.


Observation Table : Obs. No.

ml of 0.01 N AgNO3 added

E.M.F. E

(volts)

ΔE (volts)

ΔV (ml)

ΔEΔV

(volt/ml)

Mean V (ml)

1 2 3 4 5 − − − − −

0 0.5 1.0 1.5 2.0 − − − − −

14.0

E1 E2 E3 E4 E5 − − − − −

− E1 − E2 E2 − E3 E3 − E4

E4 − E5

− − − − −

− 0.5 0.5 0.5 0.5 − − − − −

0.5

− − 0.25 0.75 1.25 1.75 − − − − −

Graphs :

95

E.M.F.

ml of AgNO3

Br

Cl

V1 V2

�

�

E

V

+_

Mean V

Fig. 1

From the graph ΔEΔV against mean V :

1. Amount of AgNO3 0.01 N required for complete precipitation of Br− = V1 ml. 2. Amount of AgNO3 0.01 N required for complete precipitation of Cl− = (V2 − V1) ml. Calculations : 1. Calculation of bromide present in the mixture : Atomic weight of bromide = 79.9

AgNO3 + Br− = AgBr ↓ + NO−3

1 mole 1 mole ppt. ∴ 1000 ml 1 N AgNO3 ≡ 1 N Br− = 79.9 g Br−


96

∴ 1000 ml 0.01 N AgNO3 = 0.799 g Br− ∴ V1 ml of 0.01 N AgNO3 = ?

= 0.799 × V1

1000 g Br−

= X (say) X gms of Br− present in 10 ml of given solution

∴ Br− present in 1 litre solution = 1000 × X

10 = 100 X g l−1

2. Amount of Cl− : Atomic weight of chloride = 35.5

AgNO3 + Cl− ⎯→ AgCl ↓ + NO−3

1 mole 1 mole 1000 ml 1 N AgNO3 ≡ 1 N Cl− = 35.5 g Cl− ∴ 1000 ml 0.01 N AgNO3 = 0.355 g Cl−

∴ (V2 − V1) ml 0.01 N AgNO3 = 0.355 × (V2 − V1)

1000 g Cl−

= Y say Amount of Cl− per litre 10 ml solution contains Y gms of Cl− ∴ 1000 ml solution contains 100 × Y g Cl−

Result Table : 1. Equivalence point of Br− (V1) = ………. ml 2. Equivalence point of Cl− (V2 − V1) = ………. ml 3. Amount of Br− in g l−1 = ………. g l−1 4. Amount of Cl− in g l− = ………. g l−1

VIVA VOCE

1. What is a cell ?

2. Define electrolytic and galvanic cells.

3. What is a single electrode ?

4. What is single electrode potential ?

5. What is half cell ? How two half cells are combined to form a cell ?

6. What is solution pressure and osmotic pressure ?

7. Single electrode potential depends on which factors ?

8. Give the Nernst equation for single electrode potential.


97

9. Define oxidation and reduction potential.

10. How the cells are represented ?

11. How will you represent a silver electrode in a silver nitrate solution ?

12. Mention the different types of electrodes.

13. What are the conventions used in the representation of a cell ?

14. A cell is formed by combining calomel electrode with standard hydrogen electrode using KCl salt bridge. How you will represent it ?

15. What is a primary standard and a secondary standard electrode ?

16. What is a reference electrode ?

17. What is an indicator electrode ?

18. Explain the construction and working of standard hydrogen electrode.

19. Mention the different types of cells.

20. What is a quinhydrone electrode ? Explain how it is useful in acid-base titrations.

21. What is a concentration cell ?

22. What is a cell with and without transference ?

23. What is junction potential ? How it is minimized ?

24. What is salt bridge ? What are its advantages ?

25. Which electrolytes are used in salt bridges ?

26. Explain the construction and working of calomel electrode.

27. What is a standard cell ?

28. Why a potentiometer is standardized ?

29. What is a potentiometer ?

30. Why a voltmeter is not used instead of potentiometer, in e.m.f. measurements of a cell ?

31. Explain the working of Cadmium Weston cell. What is its e.m.f. at 25°C ?

32. What is e.m.f. series and what are its applications ?

33. What is standard electrode potential ?

34. What is e.m.f. of a cell ?

35. Give the dependance of e.m.f. on temperature.

36. What is a potentiometric titration ?

37. What are the different types of potentiometric titration ?


98

38. The following cell is used in an acid-base titration.

Pt⏐H2⏐H+(1M)⏐⏐ H

+, quinhydrone⏐Pt

(1 atm)

Identify the indicator electrode.

39. What is a redox electrode ?

40. What are the advantages of potentiometric titrations ?

41. What is the accuracy of determination of equivalence point in a potentiometric titration compared to a conventional volumetric titration ?

42. Explain the potentiometric titration curve of an acid against a base ?

43. Why a peak is obtained at the equivalence point in the plot of ΔEΔV against mean V ?

44. Why pH = pKa at half neutralization point ?

45. How will you determine half neutralization point ?

46. How will you find pH at half neutralization point ?

47. What is the sign convention of e.m.f. ?

48. Which expression you use in calculating Ecal in a potentiometric titration of pKa determination ?

49. Explain now pH of a solution is determined by using potentiometric method ?

50. What is buffer solution ?

51. How pH of buffer solutions are determined using potentiometer ?

52. Give Henderson equation of pH of a buffer solution.

53. Which electrodes you will use in precipitation titration of AgNO3 against Na2SO4 ?

54. Can you determine pK1 and pK2 of a dibasic weak acid using potentiometric titration ?

55. pK1 and pK2 should differ by what amount so that two peaks are obtained in the plot

of ΔEΔV against mean V ?

56. A tribasic acid H3PO4 is titrated against NaOH potentiometrically. What is the nature of the graph of e.m.f. against ml of NaOH added ?

- - -

99

3…

pH Metry (Any Two)

Theory : The acidity and basicity of a solution is measured in terms of pH and pOH. For dilute solutions the concentration of acids or bases such as 0.1 or 10−1 M, 0.01 or 10−2 M or 10−3 M solution involve negative exponents. Sorenson suggested a convenient method to express hydrogen ion concentration without involving negative exponents. He defined the terms pH and pH scale. Mathematically pH is defined as –

pH = log10 1

[H+] = − log [H+]

and pOH = log 1

[OH−] = − log [OH−]

For any aqueous solution [H+] [OH−] = Kw ionic product of water. Taking negative logarithm to base 10 of both sides, we get − log10 [H+] − log10 [OH−] = − log10 Kw or pH + pOH = − log10 Kw at 25°C Kw ionic product of water = 10−14 Therefore, at 25°C, pH + pOH = − log 10−14 or pH + pOH = 14 In water, [H−] = [OH−] ∴ [H+] [OH−] = 10−14 at 25° C or [H+] = 10−7 = [OH−] ∴ pH of water = − log 10−7 = 7 = pOH ∴ For neutral solutions, pH = pOH = 7 For acidic solution, [H+] = 10−7 ∴ pH is less than 7 from above definition For bases, pOH is greater than 7 or pH is less than 7. ∴ Thus, (i) pH < 7, pOH > 7 Solution is acidic (ii) pH > 7, pOH < 7 Solution is basic

T.Y.B.Sc. Practical Chemistry (Physical Practicals) pH Metry

(iii) pH = pOH = 7 Solution is neutral

Now by above definition when [H+] = 1 gm mole l−1, then pH = 0 and hence pOH = 14. Similarly, when [OH−] = 1 gm mole l−1, pOH = 0 and pH = 14.

∴ pH range from 0 to 14 is taken as a pH scale for dilute solutions.

pH ⎯→ 0 7 14

Scale Acidic Neutral Basic

Measurement of pH : The pH values of acidic or basic solutions are conveniently measured by using pH meter. The electrodes used in pH metry are glass electrode and calomel electrodes.

Glass Electrode : When a glass surface is in contact with acidic solution, some potential difference exists between glass surface and the solution. The magnitude of this potential difference depends on the nature of the glass and H+ ion concentration of the solution. If two solutions differing in their [H+] ion concentration i.e. of different pH are put on the two sides of the glass surface a potential difference is established. If the pH of one of the two solutions is kept constant then potential developed depends only on the pH of the other solution. This is the principle used in glass electrode. In the simplest form, the glass electrode consists of a tube connected to a thin walled glass bulb with high conductivity.

Ag or Pt wire

Thin walledglass bulb

Ag-AgCl electrode(s)

0.1 M HClsolution

Fig. 1: Schematic of glass electrode

The bulb contains a solution of constant hydrogen ion concentration and an electrode of definite potential such as silver chloride electrode in 0.1 M HCl solution. The bulb is inserted in the experimental solution whose pH− is to be measured. This electrode can be expressed as

Ag⏐AgCl(s), 0.1 M HCl⏐glass⏐experimental solution The commonly used reference electrode in combination with glass electrode is calomel electrode. The cell used for pH measurements is represented as

Ag⏐AgCl(s)⏐Glass⏐Experimental solution⏐Cl−⏐Hg2Cl2⏐Hg 0.1 M HCl Glass electrode Calomel electrode

100


101

The potential of such a cell can be measured using potentiometer and hence the pH of the solution.

Experiment No. 1 Aim : To determine the degree of hydrolysis of aniline hydrochloride by pH metric method. Theory : Aniline hydrochloride is a salt of a strong acid and a weak base. It hydrolyses as

C6H5NH3Cl ≡ C6H5NH+3 + Cl−

C6H5NH+3 + H2O ≡ C6H5NH2 + H3O+

The hydrolysis constant Kh of the reaction is given by,

Kh = [C6H5NH2] [H3O+]

[C6H5NH+3]

= h2C

1 − h

Kh ≈ h2C where, ‘h’ is degree of hydrolysis and C is concentration in moles/lit. At equilibrium, if [H3O+] = [H+] = hC. ∴ If [H3O+] = hC is measured then h and Kh can be determined. Apparatus : pH meter, glass electrode, calomel electrode, burette etc. Chemicals : Aniline hydrochloride powder, 0.05 M potassium hydrogen phthalate solution (pH = 4) etc. [A] Standardization of pH meter : Wash the glass and calomel electrodes with distilled water. 1. In a small beaker take about 50 ml. of 0.05 M potassium hydrogen phthalate solution. 2. Dip a calomel and a glass electrode into the solution. Ensure that the bulb of glass

electrode is completely dipped into the solution. Connect the two electrodes to the proper terminals of pH meter.

3. Adjust the temperature knob to room temperature. Put the pH range selector switch on 0-7 pH range side, if it is there with pH meter.

4. Adjust the course and fine standardization knobs so that pH meter reads the pH = 4. After this adjustment do not disturb the standardization knobs throughout the experiment.

[B] Preparation of aniline hydrochloride solutions :


102

1. Prepare aniline hydrochloride solution by dissolving accurately weighed 1.295 g of aniline hydrochloride in water and diluting it to 100 ml. using a 100 ml. volumetric flask.

2. From 0.1 M aniline hydrochloride solution prepare the following solutions : (i) Take 50 ml of 0.1 M aniline hydrochloride solution using 25 ml pipette and dilute

to 100 ml by distilled water, which is 0.05 M aniline hydrochloride. (ii) 0.025 M Aniline hydrochloride : Take 50 ml of 0.05 M aniline hydrochloride and

dilute to 100 ml by distilled water. (iii) 0.0125 M Aniline hydrochloride : Take 50.0 ml of 0.025 M aniline hydrochloride

and dilute to 100 ml by distilled water. (iv) 0.00625 M Aniline hydrochloride : Take 50 ml of 0.0125 M aniline

hydrochloride and dilute to 100 ml by distilled water. [C] Determination of pH of aniline hydrochloride solutions : 1. Take 50 ml of 0.1 M aniline hydrochloride in a small beaker. 2. Wash the glass and calomel electrodes with distilled water and then dip into aniline

hydrochloride solution. Ensure that the bulb of glass electrode is completely immersed in the solution.

3. Keep the temperature knob at room temperature value. 4. Put pH selector switch at 0-7 pH range. 5. Find the pH of the solution. 6. In a similar way find pH of remaining aniline hydrochloride solutions. Record the

reading of pH of all solutions in the observation table as below. Observation Table :

Obs. No.

Conc. of aniline

hydrochloride mol l−1

pH Degree of hydrolysis ‘h’

Hydrolysis constant

Kh

Average value of Kh

1. 2. 3. 4. 5.

0.1 0.05 0.025 0.0125 0.00625

Calculations :

Hydrolysis constant Kh = h2C

1 − h = h2C ( 1 − h 1)

∴ h = ⎝⎜⎛

⎠⎟⎞Kh

C

1/2


103

Now, [H+] = hC = C × Kh

C = KhC = (KhC)1/2

pH = − log [H+] = − log hC = − log (KhC)1/2

∴ pH = − 12 log Kh −

12 log C

∴ 2 × pH = − log Kh − log C or log Kh = − 2 × pH − log C or Kh = antilog [−2 pH − log C] Substituting the value of pH and C of the solutions of aniline hydrochloride, find Kh. Find average Kh. Now, Kh = h2C Substituting the values of Kh and C, find h of each of the solution of aniline hydrochloride. Result Table :

Hydrolysis constant of aniline hydrochloride = ….

Experiment No. 2 Aim : To determine the pKa value of weak acid by pH metric titration with a strong base. Theory :

A weak acid HA ionizes as HA ≡ H+ + A− and its dissociation constant Ka is given by

Ka = [H+] [A−]

[HA] .

[H+], [A−] and [HA] are equilibrium concentrations of the respective species.

∴ [H+] = Ka [HA][A−]

∴ log [H+] = log Ka + log [HA][A−]

or − log [H+] = − log Ka + log [A−][HA]

or pH = pKa + log [A−][HA]

at half neutralization point, [HA] = [A−] and hence at half neutralization point, pH = pKa. Therefore, by finding pH of the solution when the solution in the cell is half neutralized pKa can be calculated. This forms the basis for determination of pKa by pH metry. By performing pH metric titration neutralization point is determined graphically and hence the half neutralization point.


104

Apparatus : pH meter, glass electrode, calomel electrode, burette, beaker etc. Chemicals : 0.1 N acetic acid, 0.5 M NaOH, 0.05 M potassium hydrogen phthalate (pH = 4) solution.

Procedure :

1. Using 0.05 M potassium hydrogen phthalate (pH = 4) solution standardize the pH meter as described in the previous experiment of pH metry.

2. Take 10.0 ml of 0.1 M acetic acid in a small beaker. Add about 40 ml distilled water. Stir the solution well.

3. Dip calomel and glass electrodes into the solution. Ensure that the bulb of the glass electrode dips completely in the solution. Connect the glass and calomel electrodes to the proper terminals of the pH meter.

4. Adjust the temperature selector switch to room temperature value.

5. Turn the pH selector switch to pH 0 - 7 range.

6. Find the pH of the solution.

7. Add 0.2 ml of 0.5 M NaOH into the acid solution. Stir the solution and find the pH.

8. In a similar way find pH values of solution by adding 0.2 ml alliquotes of 0.5 M NaOH each till the neutralization point exceeds by about one ml (upto 3.0 ml addition of NaOH). Report the readings in the observation table.

9. Plot the graphs : (i) pH against ml of NaOH added and (ii) ΔpHΔV against mean V of

NaOH added. Find neutralization point and pH at half neutralization point which is equal to pKa.

Observation Table :

Obs. No.

0.5 M NaOH added (ml)

pH ΔpH ΔV (ml) ΔpHΔV

Mean V (ml)

1

2

3

4

5

−

−

0

0.2

0.4

0.6

−

−

−

pH1

pH2

pH3

pH4

−

−

−

−

pH2 − pH1

pH3 − pH2

pH4 − pH3

−

−

−

−

0.2

0.2

0.2

−

−

−

− −

0.1

0.3

0.5

0.7

−

−


−

−

16

−

−

3.0

−

−

−

−

−

−

−

−

−

Graphs :

�

�

pH

V

x Neutralization point

Mean V

Half neutralization point

x/2ml of NaOH added

point

pH

pH at halfneutralization

Fig. 1

Calculations :

From the graph of ΔpHΔV against mean V, find the neutralization point. From the graph of

pH against ml of NaOH added, find pH at half neutralization point which is equal to pKa. pH at half neutralization point = pKa Result Table :

pKa of given weak acid = ……….

Experiment No. 3 Aim : To determine the dissociation constant of oxalic acid by pH metric titration with a strong base. Theory : Oxalic acid HOOC − COOH is a dibasic acid and ionizes in two steps as

1.

COOH⏐COOH

K1⎯→ COO−

⏐COOH

+ H+

K1 is dissociation constant of first ionization.

2.

COO−

⏐COOH

K2⎯→ COO−

⏐COO−

+ H+

105


106

K2 is dissociation constant of second ionization. The ionization of oxalic acid is characterized by two different ionization constants K1 and K2. It is observed that for oxalic

acid K1

K2 > 104 and when the solution of oxalic acid is titrated against NaOH solution, the

titration curve will show two pronounced inflections.

If K1

K2 < 104 the titration curve shows only one inflection. This is true for all dibasic acids.

For tribasic acids which ionize in three steps there are three ionization constants K1, K2 and K3.

1. If K1

K2 < 104 and

K2

K3 < 104 only one inflection is observed in the titration curve.

2. If K1

K2 > 104 and

K2

K3 < 104, two inflections are observed.

3. If K1

K2 > 104 and

K2

K3 > 104, three pronounced inflections are observed in the titration

curve. Thus, in order to find values of ionization constants in which different steps are involved in polybasic acids, the ionization constants should differ by large amount (104 times).

Now, in oxalic acid, K1

K2 > 104.

Therefore, two pronounced inflections are obtained, one when COOH⏐COOH

⎯→ COO−

⏐COOH

+ H+

the ionized species COO−

⏐COOH

itself behaves as weak acid and undergoes further ionization with

ionization constant K2. pH-metry offers a very convenient method to obtain K1 and K2 of oxalic acid.

Apparatus : pH-meter, glass electrode, calomel electrode etc. Chemicals : 0.05 N oxalic acid, 0.2 N NaOH, 0.05 M potassium hydrogen phthalate (pH = 4 solution). Procedure : (A) Standardization of pH meter : Standardize the pH-meter using 0.05 M potassium hydrogen phthalate (pH = 4) solution as described in the first experiment of pH-metry. (B) Titration : 1. Take 10.0 ml of 0.05 N oxalic acid in a beaker and add about 40 ml distilled water

into it, stir the solution and as described in previous experiment, measure pH of the solution using pH meter.


2. Add 0.2 ml of 0.2 N NaOH into the oxalic acid solution, stir the solution and measure pH of the solution.

3. In a similar way go on adding 0.2 ml of 0.2 N NaOH each time and measure pH of the solution every time upto 4.0 ml total addition of NaOH (i.e. end point exceeds by about 1.5 ml). Report the readings in the observation table as shown below.

4. Plot the graphs : (i) pH against ml of NaOH added, (ii) ΔpHΔV against mean V.

5. From the graph, find two neutralization points corresponding to two ionization constants K1 and K2.

Observation Table : Obs. No.

0.2 N NaOH added (ml)

pH ΔpH ΔV (ml) ΔpHΔV

Mean V (ml)

1 2 3 4 − − − −

0 0.2 0.4 0.6 − − − −

4.0

pH1 pH2 pH3 pH4 − − − − −

− pH2 − pH1 pH3 − pH2 pH4 − pH3

− − − − −

− 0.2 0.2 0.2 − − − −

0.2

− − 0.1 0.3 0.5 − − − −

3.9

Graphs :

pH

(pH )1/22

(pH )1/21

1/2V1

ml of NaOH added

1/2(V V )�2 1

�

�

pH

V

Mean V

V1 V2

Fig. 1

From the graph of ΔpHΔV versus mean V, find :

107


108

1. V1 the amount of NaOH for the complete dissociation of the first −COOH group K1. 2. (V2 − V1) the amount of NaOH required for complete ionization of second −COOH

group. 3. From the graph of pH against ml of NaOH added find (pH1)1/2 = pK1 for ionization

and (pH2)1/2 = pK2 for the second ionization as shown in the above plot. Result Table :

1. pK1 of oxalic acid = ………. 2. pK2 of oxalic acid = ……….

Experiment No. 4 Aim : To determine the pH of various mixtures of sodium acetate and acetic acid in aqueous solutions and hence to find the dissociation constant of acetic acid. Theory : The mixture of sodium acetate and acetic acid in aqueous solution is a buffer solution. Buffer solutions are the solutions which resist a sudden change in pH due to addition of small amounts of strong acid or base. An acidic buffer is a mixture of a weak acid and its salt with a strong base. A basic buffer is a mixture of a weak base and its salt with strong acid. For example, CH3COOH + CH3COONa is an acidic buffer and NH4Cl + NH4OH is a basic buffer. The pH of an acidic buffer HA + BA is given by the Henderson equation which is

pH = pKa + log [Salt][Acid]

According to the equation, the pH of an acidic buffer depends on the ratio [Salt][Acid]

because pKa is constant. pH of any solution can be accurately determined by pH meter and

if the ratio [Salt][Acid] is known the dissociation constant can be calculated.

Apparatus : pH meter, glass electrode, calomel electrode, 50 ml volumetric flasks etc. Chemicals : 0.1 M acetic acid, 0.1 M sodium acetate, 0.05 M potassium hydrogen phthalate solution. Procedure : (A) Prepare the following buffer solutions using 0.1 M acetic acid and 0.1 M sodium acetate solutions :

Buffer solution ml of 0.1 M Acetic Acid

ml of 0.1 M Sodium Acetate

A 47.5 2.5


109

B C D E F G H I J K L

45 40 35 30 25 20 15 10 7.5 5.0 2.5

5.0 10.0 15.0 20.0 25.0 30 35 40

42.5 45.0 47.5

(B) Standardization of pH meter, as explained in the previous experiment. Standardize the pH meter using 0.05 M potassium hydrogen phthalate (pH = 4) solution : (C) Measurement of pH of buffer solutions : 1. In a small beaker take the buffer solution A. Dip the calomel and glass electrode after

washing and ensure that the bulb of the glass electrode is immersed completely in the solution.

2. Adjust the temperature selector switch at room temperature value. 3. Put the pH-range selector switch at 0 - 7 pH-range. 4. Determine pH of the solution A. 5. In a similar way find pH of remaining buffer solutions. Observation Table :

Sol. No.

ml of 0.1 M Acetic Acid

ml of 0.1 M Sodium Acetate

Conc. of Acid

[Acid]

Conc. of Salt

[Salt]

log [Salt][Acid]

pH

A B C D E F G H I J K L

47.5 45 40 35 30 25 20 15 10 7.5 5.0 2.5

2.5 5.0 10.0 15.0 20.0 25 30 35 40

42.5 45.5 47.5

0.95 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.015 0.01 0.005

0.005 0.01 0.02 0.03 0.04 0.05 0.07 0.07 0.08 0.085 0.09 0.095

− 1.2787 − 0.9542 − 0.6020 − 0.368 − 0.1761

0.00 0.3680 0.3680 0.6021 0.7533 0.9542 1.2787

Calculations : 1. Calculation of concentration of acid in buffers : 47.5 ml 0.1 M acetic acid is diluted to 50 ml (Total volume).


∴ [Acid] = 47.5 × 0.1

50 = 0.95

2.5 ml 0.1 M sodium acetate is diluted to 50 ml (Total volume).

∴ [Salt] = 2.5 × 0.1

50 = 0.05

In a similar way, calculate the value of [Acid] and [Salt] for the remaining solutions. Calculated values are given in the observation table. Graph :

Plot a graph of pH against log [Salt][Acid] .

pH = pKa + log [Salt][Acid]

when, [Salt] = [Acid], log [Salt][Acid] = 0

∴ pH = pKa Find the intercept of y-axis which gives the value of pKa. − log Ka= pKa

pH

pH = pKa

log[salt]

[acid]

+_

Fig. 1 ∴ log Ka = − pKa ∴ Ka = antilog [−pKa] Result Table :

Dissociation constant of acetic acid (Ka) = ……….

VIVA VOCE 1. Define pH and pOH. 2. Why pH of 0.1 N HCl solution and 0.1 N acetic acid are not same ? 3. What expression you will use to find pH of a weak monobasic acid ? 4. Prove the relation pH + pOH = 14. 5. What is ionic product of water ? 6. What is the pH range of acidic and basic solutions ? 7. What is the pH scale ? 8. What is the pH of 0.001 N HCl ? 9. What is the pH of 0.01 N H2SO4 and 0.01 M H2SO4 ? 10. What is Ostwald’s dilution law ? 11. What is the expression for [H+] ions of a weak acid in terms of degree of ionization

and dissociation constant ? 12. What is basicity of an acid ? Explain with examples. 13. What is the expression of pH of a buffer solution ?

110


111

14. What different methods you will employ to determine pH of a solution ? 15. What is the pH of 0.1 N NaOH ?

- - -

4…

Radioactivity (Any One)

Theory : Certain elements which emit α, β or γ radiations spontaneously at any temperature and pressure are known as radioactive elements and the phenomenon is called as Radioactivity. During the radioactive disintegration process, the original radioactive element transforms into another element which may or may not be radioactive. The initial radioactive element is called as parent while the new element that is formed is called daughter. Usually the heavy particles like α are emitted by radio elements having mass number greater than 210. While the lighter particles alike β+ or β− are emitted by lighter as well as heavier elements. From the behaviour of ion pairs in the electric field different regions have been observed. Thus in the ionization chamber region and proportional region the counters which are working are used to measure α particle. In the Geiger-Muller region in which G.M. counter works and used to measure the β particles. Geiger-Muller Counter :

To high voltagesource and counter

_

+

Argon 80 mm pressure+ quenching agent 20 mm

pressure

Glass bead

Tungsten wire (anode)

Copper cylinder (cathode)

Glass or quartz outer jacket

Thin mica window

(2-3 mg cm thick)�2

Radioactive source

Source holder

Stand

Lead castle

Fig. 1

Principle : Gases get ionized when α, β or γ radiations are allowed to pass through them.

111

T.Y.B.Sc. Practical Chemistry (Physical Practicals) Radioactivity

112

Construction : It consists of a copper tube which acts as cathode and fused inside the glass or quartz tube. The diameter of this tube is 2 cm, while is 10 cm in length. At the centre of tube is a thin tungsten wire which acts as anode. The tube is completely evacuated and filled with argon gas 80 mm of Hg pressure and quenchers like ethyl iodide or halogens at 20 mm of Hg pressure. At the other end of the tube there is a thin mica window through which the β particles are allowed to enter inside the tube. The G.M. tube is kept inside the thick lead castle to minimize the background counts. Usually the G.M. tube is mounted on a stand having number of groves to keep radioactive sample infront of the G.M. tube. The experimental set up of G.M. counter is shown in Fig. 1. Working : When the β particle enters inside the G.M. tube through the window, it produces ionization in the gas resulting in the production of ion pairs. Due to the application of high potential the electrons are quickly attracted towards the anode. The electrons produce secondary ionisation which is called as Townsend avalanches. The electrons in the avalanches will be collected by the anode, living behind the positive ions sheath which has a low mobility. The discharge takes place at the electrodes and electric pulse is registered. Under the influence of high applied voltage, the electrons reach the central anode in about 5 × 10−7 sec, after their generation. The positive ions move very slowly and cluster around the central wire. These positive ions reduce the potential gradient much below the one required for gas multiplication, the counter therefore becomes insensitive. The positive ions react with cathode in about 100 to 500 µsec. after their generation. Till they reach the cathode, the counter is insensitive and is not in a position to register a pulse of current. The period of time, after registering the ionization event, during which the counter is not in a position to register another such event, is called the dead time of the counter. During dead time, the counter recovers and becomes ready to register another ionization event. The positive ion sheath in non self quenching tubes gradually travels towards the cathode and the positive ions combine with electrons from the cathode. The neutralized positive ions come to the ground state emitting photons. The liberated photons strike the filling gas molecules like that of argon and ionise the filling gas molecules to form ion pairs. Thus unless these photoelectrons are stopped, they again give rise to Townsend avalanche process by acceleration towards the anode and the discharge process goes into cycle, which is unwanted for efficient Geiger counting. The multiple discharge in the G.M. counter can be prevented by addition of filling gas as quenching agent and are, the ethyl alcohol, ethyl iodide or halogens. A positive ion of the filling gas in trying to reach the cathode strikes quenching gas molecules. Because of the difference in ionization potentials the charge is transferred to quench the gas molecules. The neutralised ion emits the excess energy photon. The quenching gas molecules absorb these radiations getting themselves dissociated. The quenching gas ions which are formed in the process of exchange of charge with the filling gas, positive ions go to the cathode surface and get neutralized by capturing an electron from the cathode surface. Its excess energy is utilized to dissociate the quenching molecule.


The radioactive gases present in the atmosphere as well as due to the cosmic radiations the G.M. counter will show some counts. Thus the G.M. counter without keeping any radioactive source infront of the G.M. tube shows some counts. These counts are called as background counts. In order to minimize the background counts the G.M. tube is kept inside the thick lead castle. In order to get the actual counts the background counts are to be subtracted from the observed counts.

Thus, ⎣⎢⎡

⎦⎥⎤Corrected counts

per min = ⎣⎢⎡

⎦⎥⎤Observed counts

per min − ⎣⎢⎡

⎦⎥⎤Background counts

per min

Experiment No. 1 Aim : To determine the plateau voltage of the given G.M. counter. Theory : The behaviour of ion pairs in an electric field shows that as the applied potential increases the counts per minute go on increasing. (Refer Fig. 1). For the counting of β particles in the Geiger region it is observed that in a particular voltage region the number of counts per minute are found to be constant. Thus in the graph of CPM against the applied potential the curve has nearly flat portion and is known as plateau region. This region is small for α particles because of their great specific ionization

C.P.M.

Applied voltage

Geiger plateau

Working potential

Fig. 1: Plot of applied voltage Vs. C.P.M.

while for β particles it is greater because of their lower specific ionization. The pulse rate increases with the applied voltage. After a particular voltage the pulse rate becomes independent of the applied voltage. This voltage is being called Geiger threshold. The range of voltage over which the pulse rate is independent of the applied voltage is called Geiger plateau. The mid point of this Geiger plateau region and the corresponding applied voltage is taken as the working potential of the. G.M. tube. The main advantage of G.M. counter is large size of current pulses. The amplification is therefore unnecessary. Apparatus and Instruments : Geiger-Muller tube, Geiger counting system, radioactive source, lead castle, stand, radioactive source holder etc. Procedure : The G.M. tube is mounted on a stand and is kept inside the thick lead castle. The G.M. tube is connected to the counting system properly. The voltage knob of the counter is kept on the minimum and the counter is made on. The system is allowed to warm up at least five minutes.

113


114

1. Set up the timer for 3 minutes and the paralysis time knob on 250 μsec. position. 2. With the help of source holder keep the radioactive source infront of the G.M. tube at

a fixed position in a definite groove of the stand. 3. Now apply the minimum voltage say 100 volts and press the reset button and then

press the start button. Record the counts for three minutes. Take at least three trials and record it in the observation table.

4. Increase the voltage by 30 volts. Press the reset and then start button. Again record the counts for three minutes. Take three trials. This process is repeated till the plateau and a small part of the rising portion of characteristic curve is reached. Do not increase the potential above 1500 volts for old G.M. tubes (now-a-days G.M. tubes are having working potential upto 700 volts). For these tubes do not increase the potential above 900 volts otherwise G.M. tube will be damaged. Slowly bring the high voltage to the minimum, then switch off high voltage as well as mains.

In this experiment the counts for three minutes are recorded and it is found that there may be large variation in counts for the same applied potential. Therefore mean counts per minute are calculated. Observation Table :

Obs. No. Voltage

applied (V)

Counts for three minutes Counts per

minute I II III Average

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Graph : Plot the graph of CPM against applied voltage. From the graph note the plateau region. Find out the mid-point of plateau region and the voltage corresponding to it, is the working potential of G.M. counter. Result Table :

Sr. No. Description Value with unit 1. Plateau voltage of the G.M.

counter ………. volts


Experiment No. 2 Aim : To determine the resolving time of the given G.M. counter. Instruments and Apparatus : G.M. counter system, two nearly equal and strong radioactive sources, lead castle, stand etc. Procedure : Connect the G.M. counter system to the high voltage source. Keep the voltage knob of the counter on minimum. Make the mains on and allow the system to warm up for at least 5 minutes. Mean while set up the timer as well as paralysis time (250 µs, 500 µs) to any desired value with the help of respective knobs. After five minutes of warming period set up the voltage knob equal to the working potential of G.M. tube as noted in the previous experiment. 1. Now first find out the background counts for 5 minutes and hence obtain the

background counts per minute. 2. From the two selected radioactive sources call any one as S1 and the remaining as S2.

Take the source S1 on a sample holder the position 1 and keep it infront of the G.M. tube on the definite groove of the stand in the lead castle. Record the counts for two minutes. Take three trials and find out the counts per minute. Let it be R1.

3. Remove the source S1, keep the source S2, at the position number 2, on the sample holder (planchet). Now keep the holder having source S2 in the same position as S1 in the definite groove in the lead castle. Measure the counts for two minutes. Take two trials and report counts per minute. Let it be R2.

4. Keep the source S1 at position number 1 and source S2 at position number 2 on the planchet. Keep the planchet along with the sources in the same groove in the lead castle. Measure the counts for two minutes. Take two trials and find out the counts per minute (R1,2).

S1 2 S21 S1 S2

Planchet Fig. 1

Observation Table : Sr. No. Description Value

1. Background counts per minute = RB = ………. 2. Counts per minute for source S1' = R1 = ………. 3. Counts per minute for source S2' = R2 = ………. 4. Counts per minute for S1 and S2 together = R1, 2 = ……….

115


116

Calculations :

Resolving time T = R1 + R2 − R1, 2 − RB

(R1, 2)2 − (R1) − (R2)2

Express the resolving time T in microseconds. Result :

Sr. No. Description Value with units

1. Resolving time of the G.M. counter … microseconds

Experiment No. 3 Aim : To determine the Emax of β particles. Theory : Majority of radioactive elements emit β particles. β particles are the electrons having negative charge. The particles are very tiny having negligible mass and so high penetration power and low specific ionization. The β particles emitted by some radioactive source have various energies i.e. from few millivolt to 3.15 MeV. Thus the maximum energy associated with β particles can be find out by performing this simple experiment. In this experiment, the absorption of β particles is studied by placing radioactive source infront of G.M. tube and recording the counting rate with suitable thickness of absorbing material inserted between the source and the counter. The aluminium foils are used most frequently as an absorber. The maximum range of β particles can be obtained in terms of mg cm−2 of aluminium using the following relation Rmax = 543 Emax − 160

Emax = Rmax + 160

543 = Thickness range + 160

543 MeV

Apparatus : G.M. counter, β emitting source, Aluminium absorbers of various thickness. Procedure : Do the connection and all settings as explained in the previous experiment and then apply the plateau voltage (or the working potential of G.M. tube) as obtained in the earlier experiment to the G.M. counter. 1. With the help of planchet keep the β emitting radioactive source in a suitable groove

in the lead castle. Record the counts for two minutes. Take at least two trials and then convert into counts per minute.

2. With the help of absorber holder keep the thinnest aluminium absorber in the groove above the source. Again measure counts for two minutes. Take two trials and convert into counts per minute.


3. Repeat the same procedure for all the aluminium absorbers, as well as make some combinations of the absorbers. Report the observations in the observation table as follows :

Observation Table :

Source = ………. Working Plateau Voltage = ………. volts

Sr. No.

Filter or Absorber

No.

Thickness of the filter

(mg cm−2)

Counts for two minutes Counts per min. C.P.M.

log C.P.M. I Trial II Trial Average

1.

2.

3.

4.

5.

6.

7.

8.

Graph :

Plot the graph of log CPM against the thickness of the absorber.

log CPM

R�

Rmax

Absorber thickness

(mg cm )�2

Fig. 1 From the graph obtained find Rmax and then calculate Emax by using the equation

Emax = Rmax + 160

543 MeV

117


118

Result : Sr. No. Description Value with

units 1. Range of β particles, Rmax ….. mg cm−2 2. Emax of β particles ….. MeV

VIVA VOCE 1. Define the term Radioactivity. 2. Which particles are emitted by radioactive elements ? 3. What is the principle of G.M. counter ? 4. Define specific ionization. 5. What is the use of thick lead castle ? 6. Why is it necessary to measure the background counts ? 7. Define dead time and resolving time of G.M counter. 8. Explain the terms plateau region and plateau voltage. 9. The plateau region of β is larger than α particles, explain. 10. Which is the equation to determine the resolving time ? 11. Explain the meaning of the terms Emax and Rmax. 12. What is the relation to find out Emax of β particles ?

- - -

119

5…

Conductometry (Any Two)

Conductivity of Electrolytes : The solutions of electrolytes conduct electricity through ions. In the solution, the electrons are carried by migration of positive and negative ions toward the electrodes under the influence of an applied potential. The solvents which are poor conductors of electricity are used for the preparation of solutions of electrolytes so that the conductivity of electrolyte solution is almost entirely due to ions obtained from the ionization of the electrolytes. Water is the best solvent for conductivity measurements. Like metallic conductors, the electrolyte solutions obey Ohm’s law. Ohm’s Law : The resistance of an uniform column of electrolyte placed between two parallel electrodes of cross-sectional area, ‘A’ square centimeter and separated by a distance of ‘l’ centimeter is given by

R ∝ lA

or R = ρ × lA

The constant of proportionality ρ is called as specific resistance. When l = 1 cm and A = 1 cm2, then R = ρ. Therefore, the specific resistance is defined as the resistance of one centimeter cube of conducting medium. The specific resistance is measured in unit ohm cm. Conductance : The reciprocal of R is called conductance and is measured in ohm−1. Conductivity or Specific Conductance (σ) : The reciprocal of ρ, the specific resistance is called specific conductance or simply the conductivity.

Conductivity or specific conductance σ = 1ρ

Using Ohm’s law R = ρ lA we get,

Conductance = 1R =

1ρ ×

Al

T.Y.B.Sc. Practical Chemistry (Physical Practicals) Conductometry

120

Conductance = σ Al

∴ Conductance = Specific conductance × Al

For electrolytic conductance, the ratio lA i.e. the ratio of distance between the electrodes,

l and area of cross-section, A is called cell constant.

Therefore, Conductance = Specific conductance × 1

Cell constant .

Specific conductance or conductivity is measured in ohm−1 cm−1, the specific conductance of strong and weak electrolytes increases with concentration of the electrolytes. This is due to increase in number of ions per unit volume of the solution. Concentration of solutions are measured in terms of moles per litre or gram equivalent per litre. Taking this into account two more quantities called molar conductivity and equivalent conductivity are defined. Both of these are properties of the conducting medium. Molecular or Molar Conductivity μ : It is defined as the conductivity of a volume of solution containing one gram mole of electrolyte placed between two electrodes one centimeter apart and large enough to contain all the solution. Molecular conductivity is given by the expression

μ = 1000 × σ

C = σ × V

where, C is concentration of electrolyte in moles per litre and V is the volume of solution containing one mole of electrolyte. Molecular conductivity is measured in ohm−1 cm2. Equivalent Conductivity, ^ :

It is defined as the conductivity of a volume of solution containing one gram equivalent of electrolyte placed between two electrodes one centimeter apart and large enough to contain all the solution. Equivalent conductivity is given by the expression :

^ = 1000 × σ

C = σ × V

where, C is the concentration of electrolyte in gram equivalents per litre of solution and V is the volume of solution containing one gram equivalent of electrolyte. Equivalent conductivity is measured in ohm−1 cm2. According to Faraday’s laws in electrolytic conductance, the amount of substances liberated in equivalent conductivity is more commonly used. Equivalent conductance decreases with concentration for both strong and weak

electrolytes because decrease in ^ due to the term 1C is more than increase in ^ due to

increase in specific conductivity.


121

The equivalent conductance is given by the equation,

^ = 1000 × Specific conductance (σ)

C

where, C is concentration of electrolytes in gram equivalents per litre of the solution.

Equivalent conductance decreases with concentration for both strong and weak electrolytes or in other words increases with dilution. This is because the increase in σ with

concentration is more than compensated by the decrease in term 1C with concentration.

Equivalent Conductance at Infinite Dilution :

For both strong and weak electrolytes, equivalent conductance increases with dilution and reaches a limiting value at zero concentration of electrolytes. This limiting value is called as equivalent conductance at infinite dilution represented by the symbol ^0.

For strong electrolytes the ^ is calculated by graphical method using the Kohlrausch equation for dilute solutions.

^ = ^0 − b C

For weak electrolyte the ^0 value is calculated by using Kohlrausch law of independent

migration of ions. This law states that at infinite dilution, the dissociation of electrolyte is complete and each ion moves independently of the other ion and makes a definite contribution to the equivalent conductance at infinite dilution. This definite contribution by each ion is called equivalent ionic conductance at infinity dilution. Therefore, Kohlrausch law of independent migration of ions can be stated mathematically as

^0 = l0+ + l

0−

where, l0+ and l

0− are equivalent ionic conductances at infinite dilution of positive and negative

ions respectively.

The Degree of Dissociation and Conductance : The equivalent conductance ^ at any dilution is due to dissociation of electrolyte at that dilution and ^0 value at infinite dilution is due to complete dissociation of electrolyte.

Therefore, the ratio ^^0

represents the degree of dissociation.

Further, for weak electrolyte, the dissociation constant is given by the expression

K = α2 C1 − α

By measuring ^ and ^0, the degree of dissociation α and hence dissociation constant of

weak electrolyte can be determined.


Measurement of Conductance :

Resistance and conductance of electrolytic solution are measured by using Wheatstone Bridge method.

Wheatstone Bridge Circuit :

R1 R2

GRx

c

a

Rs

b

I1

I2

d

EI

Fig. 1

ab - A wire of uniform cross-section

Rx - The unknown resistance

Rs - Variable known resistance

d - Sliding contact

E - Battery

G - Galvanometer

At point b, the current I from the battery is divided into two currents I1 and I2. The sliding contact ‘d’ is moved along ab till the point d is found when no current flows through the galvanometer. The bridge is said to be balanced. The points d and c are at the same potential when the bridge is balanced. The potential drops from b to d and b to c are equal and so also the potential drops from d to a and c to a are equal i.e.

RSI1 = R2I2 and RXI1 = R1I2

From these two equations, we get RX = RS × ⎝⎜⎛

⎠⎟⎞R1

R2 .

R1 and R2 are proportional to lengths ad and db and can be measured. Hence, RX unknown resistance can be measured.

For measurement of resistance of electrolyte solutions, the following modifications are necessary in the Wheatstone bridge method.

(i) Direct current cannot be used because it causes electrolysis and concentration changes occur at the electrodes. To avoid this alternating current (A.C.) is used. The products of electrolysis obtained during the passage of first half cycle of the alternating current are almost completely eliminated by the subsequent second half cycle of the alternating current. That is alternating current eliminates polarisation.

(ii) With the use of alternating current, the balancing of the bridge is sharpened by putting a variable condenser (capacity) across the standard resistance.

(iii) As a.c. is used instead of galvanometer as detector, a cathode ray oscilloscope or any other suitable device is employed.

122


Pt PtPtPt

(a) (b)

Pt

Pt

(c) Dipping type (d) Dipping type Fig. 2

The use of alternating current eliminates the polarisation. But following two factors must be taken into account to completely eliminate polarisation. (a) The electrode must not dissolve. Therefore inert metal like platinum as electrode

should be used. (b) The surface area of the electrode should be large. For this purpose, platinised

platinum electrodes are used and are called platinum black electrodes. By electrolysis method platinum in the form of fine platinum particles are adsorbed over the surface of platinum electrodes, thereby increasing the surface area to a large extent.

Preparation of Solutions : All theoretical relations developed in electrolytic conductance are for dilute solutions of electrolytes. Therefore while doing conductivity measurements, dilute solutions are to be used. Therefore all solutions are to be prepared carefully using accurately weighed quantities of electrolytes. All glass apparatus used in conductivity experiments must be made of resistance glass such as jena, pyrex or quartz. Conductivity cells : Depending on the type and accuracy of measurement conductivity cell of different shapes and sizes are used. Basically all conductivity cells are made of resistance glass such as pyrex and contain two parallel platinised platinum electrodes. The distance between the electrodes is measured by measuring resistance of a solution of known specific conductance. The different types of cells used are shown in Fig. 2.

123


124

For solutions of high conductance the cell with large distance between the two electrodes, cell a, is used; for solution of low conductance, cell with small distance between the electrodes, cell b, c and d are used. Conductivity water : Water, which is usually used as solvent for conductivity measurement must be of good purity. The specific conductance of electrolyte cannot always be equal to (specific conductance of solution) − (specific conductance of water), the specially prepared water called conductivity water should be used. The conductivity water should not have specific conductance greater than 3 × 10−6 ohm−1 cm−1. The tap water and deionised water show specific conductance more than 10−6 and hence should not be used. Conductivity water is obtained by mixing double glass distilled water with small amount of KMnO4 and NaOH and then the mixture is redistilled with a tall reflux, alkali trap, and condenser. The first and last fraction of distillate is rejected. The middle fraction shows conductance of less than 1 × 10−6 ohm−1 cm−1. The middle fraction is collected and stoppered immediately to avoid contamination of carbon dioxide. Temperature Control : Conductance depends on temperature, therefore, conductance measurements are made keeping temperature constant using a thermostat.

Experiment No. 1 Aim : To determine the cell constant of the given cell using 0.01 M KCl solution and hence determine the dissociation constant of given weak monobasic acid. Apparatus : Conductivity bridge, magnetic stirrer, beaker (100 ml), standard flasks 50 ml or 100 ml etc. Procedure : (I) Determination of cell constant of the given cell : 1. Wash the electrodes of given conductivity cell first with conductivity water and then

with 0.01 M KCl. 2. In a small beaker (100 ml) take about 50 ml of 0.01 M KCl solution. Place the cell into

the solution. The electrodes of the cell should completely dip into the solution. 3. Place the conductivity cell setup into a thermostat at a constant temperature either at

25°C or at 30°C. 4. Connect the electrodes of the cell to the conductivity bridge and find the

conductance or resistance of the solution. (II) Determination of conductance of 0.1 N acetic acid : 1. Wash the electrodes of the conductivity cell first with conductivity water and then

with 0.1 N acetic acid. During both washing take care that the position of the electrodes is not distributed.


125

2. Place the cell in 0.1 N acetic acid solution (about 50 ml), see that the electrodes immerse completely into the solution. Place this setup in a thermostat at a constant temperature either at 25°C or at 30°C.

3. Connect the conductivity cell to the conductivity bridge and find conductance or resistance of 0.1 N acetic acid.

4. Now prepare 0.05 N, 0.025 N, 0.0125 N acetic acid solutions and measure conductance of each of these solutions.

Preparation of 0.05 N, 0.025 N and 0.0125 N acetic acid solution : 1. 0.05 N Acetic acid solution : Take 50 ml of 0.1 N acetic acid solution and dilute to

100 ml in a 100 ml volumetric flask using conductivity water. 2. 0.025 N Acetic acid solution : Take 50 ml of 0.05 N acetic acid prepared above and

dilute to 100 ml using conductivity water. 3. 0.0125 N Acetic acid solution : Take 50 ml of 0.025 N acetic acid and dilute to

100 ml using conductivity water. Observations : 1. Temperature of water bath = ………… °C 2. Conductance of 0.01 M KCl solution = ………… ohm−1

OR 2. Resistance of 0.01 M KCl solution = ………… ohm 3. Given that specific conductance of 0.01 M KCl solution = 0.001412 ohm−1 cm−1 at 25°C = 0.001552 ohm−1 cm−1 at 30°C 4. Given that equivalent conductivity of acetic acid at infinite dilution ^0 = 390.8 ohm−1 cm2

Observation Table :

Sr. No. Concentration of acetic acid

Conductance in ohm−1

1. 2. 3. 4.

0.1 N 0.05 N 0.025 N 0.0125 N

Calculations : (I) Cell constant of given cell :

Cell constant = Specific conductivity of 0.01 M KCl

Conductance of 0.01 M KCl


126

(II) Determination of dissociation constant Ka of acetic acid : (i) Specific conductivity of acetic acid = Cell constant × Conductance

(ii) Equivalent conductance ^ = 1000 × Specific conductance

C

C = 0.1 for 0.1 N acetic acid = 0.05 for 0.05 N acetic acid and so on

(iii) Degree of dissociation, α = ^^0

(iv) Dissociation constant Ka = α2C

1 − α

Using the above four steps calculate dissociation constant of acetic acid. Tabulate the results as below. Result Table : (i) Cell constant of the given cell = ……….. cm−1.

Conc. of acetic acid

Conductance ohm−1

Specific conductance ohm−1 cm−1

Equivalent conductance ^ ohm−1 cm2

Degree of dissociation

α

Dissociation constant

Ka

0.1 N

0.05 N

0.025 N

0.0125 N

CONDUCTOMETRIC TITRATION End point of various titrations can be determined using conductivity measurements. A titration where end point is determined by conductance measurement is called conductometric titration.

Addition of one electrolyte solution into another electrolyte solution will result in change of conductance because of change in volume and possible ionic reactions.

Now consider the addition of dilute solution of electrolyte (A+ B−) to another dilute solution of electrolyte (C+ D−). If the change in volume is negligible and temperature is kept constant then

(a) If there is no chemical reaction, then conductance of CD gradually increases on the addition of AB due to increase in number of ions.

(b) If there is an ionic reaction to produce either only slightly ionised or insoluble substance represented as

A+ B− + C+ D− ⎯→ A+ D− + CB


Then there is marked change in conductance at the equivalence point. From the above equation it is clear that when solution of AB is added to CD, C+ ion is replaced by A+ ion. The ionic conductances of different ions are different and conductance may increase or decrease depending on the mobility values of A+ and C+. If the ionic mobility of A+ is greater than ionic mobility of C+ then conductance increases and if ionic mobility of A+ is less than C+ then conductance decreases. This is the basic principle of any conductometric titration. As an illustration let us consider the following titrations : (1) Titration of a strong acid against a strong base and (2) A precipitation titration of a solution of lead nitrate against sodium sulphate solution. 1. Titration of a strong acid against a strong base : NaOH + HCl → NaCl + H2O Na+ + OH− + H+ + Cl− → Na+ + Cl− + H2O When NaOH solution is added to HCl solution, H+ ion in solution is replaced by Na+ ion. The mobility or ion conductance of H+ ion is more than mobility or ion conductance of Na+ ion. A high mobile cation is replaced by a less mobile cation and conductance of the solution decreases. This decrease in conductance takes place with every addition of NaOH solution till HCl solution or H+ ions are present in the solution. When neutralisation is complete, that is all the H+ ions are replaced by Na+ ions, further addition of NaOH increases conductance due to presence of excess of OH− ions. If a graph of volume of NaOH added against conductance is plotted, we get two straight lines in V shape intersecting at equivalence point as shown below.

ml of NaOH added

Co

nd

ucta

nce

Equivalence point

Fig. 1

2. Precipitation titration between Pb(NO3)2 and Na2SO4 solution : When a dilute solution of Na2SO4 is added to another dilution of Pb(NO3)2, the following reaction occurs : Na2SO4 + Pb(NO3)2 → PbSO4 ↓ + 2NaNO3 or

2Na+ + SO−24 + Pb+2 + 2NO

−23 → PbSO4 ↓ + 2Na+ + 2NO

−3

ppt

127


128

As Na2SO4 solution is added high mobile Pb+2 ion is replaced by two Na+ less mobile ions, resulting in slight decrease of conductance. When the reaction is complete, that is all Pb+2 ions are replaced by Na+ ions, the further addition of Na2SO4 results in increase in

conductance owing to the presence of excess of SO−24 ions. Therefore, before equivalence

point there is slight decrease in conductance and after equivalence point there is a sharp increase in conductance. A graph of ml of Na2SO4 against conductance consists of two lines intersecting at equivalence point. During conductometric titrations, to avoid large volume changes due to addition of one electrolyte solution into another solution, the titrant (the solution in burette) must be at least 5 times stronger than the other solution.

Experiment No. 2 Aim : To estimate the amount of lead present in the given solution of lead nitrate by conductometric titration with sodium sulphate. Apparatus : Conductivity bridge, conductivity cell, microburette of capacity 5 or 10.0 ml, magnetic stirrer, breaker, pipette etc. Chemicals : 0.1 N (approximate) Pb(NO3)2 solution, 0.5 N Na2SO4 solution, conductivity water. Procedure : 1. Wash the electrodes of conductivity cell with conductivity water and then rinse them

with 0.1 N Pb(NO3)2 solution. 2. Take 10.0 ml of 0.1 N Pb(NO3)2 in a small beaker of capacity not more than 100 ml.

Add about 40 to 50 ml of conductivity water and dip a conductivity cell into the solution. The electrodes should be completely immersed into the solution. Put a magnetic needle.

3. Place the beaker with the solution with conductivity cell on a magnetic stirrer and stir the solution for few minutes.

4. Connect the conductivity cell to the conductivity bridge and determine the conductance or resistance of 0.1 N Pb(NO3)2 solution. Set up the apparatus as shown in Fig. 1.

5. Fill a microburette with 0.5 N Na2SO4 solution. 6. From burette add 0.2 ml of Na2SO4 solution into the lead nitrate solution and stir the

solution. Observe the formation of lead sulphate precipitate. Determine the conductance or resistance of the solution.

T.Y.B.Sc. Pract nductometry ical Chemistry (Physical Practicals) Co

Conductometer

0.5 N Na SO2 4 Microburette

Ironstand

Conductivitycell

10 ml 0.1 N Pb(NO ) + 50 ml

conductivity water3 2

Magnetic stirrer

Magnetic needle

Fig. 1

7. In a similar way add 0.2 ml of sodium sulphate solution each time and determine conductance or resistance of the solution after sufficient stirring of the solution every time. Determine the conductance or resistance till the equivalence point exceeds by about one ml. (That is take readings of conductance upto 3.0 ml addition of Na2SO4 solution).

8. Plot a graph of conductance against ml of Na2SO4 added.

9. From the graph find the equivalence point.

Observation Table :

Sr. No. ml of 0.5 N Na2SO4 added Conductance in mho

1.

2.

3.

:

:

:

:

0

0.2

0.4

…

…

…

3.0

129


Graph :

Co

nd

ucta

nce

Equivalence point

ml of Na SO added2 4

Fig. 2 Calculations : 1. Calculate exact normality of Pb(NO3)2 Pb(NO3)2 ≡ Na2SO4 N1V1 = N2V2 N1 × 10 = 0.5 × V2 (V2 from graph)

∴ N1 = 0.5 × V2

10

2. Strength of Pb(NO3)2 solution : Strength = Normality × Equivalent weight = N1 × 165.6 gL−1 3. Amount of lead in 10 ml 0.1 N solution Atomic weight of Pb = 207.2 1000 ml 1 N Pb(NO3)2 solution ≡ 207.2 g Pb

10 ml 0.1 N Pb(NO3)2 solution = 10 × N1 × 207.2

1000 g Pb

Result Table :

1. Equivalence point = ………. ml

2. Exact normality of Pb(NO3)2 solution = ………. N

3. Strength of Pb(NO3)2 solution = ………. g l−1

4. Amount of lead in 10 ml 0.1 N Pb(NO3)2 = ………. gm

130


131

Experiment No. 3 (A)

Aim :

To investigate the conductometric titration of a strong acid against a strong base.

Apparatus :

Conductivity bridge, conductivity cell, magnetic stirrer, microburette etc.

Chemicals :

0.01 N (approximate) HCl solution, 0.05 N NaOH solution, conductivity water.

Procedure :

1. Take 10.0 ml of 0.1 N HCl in a small beaker, add about 40 ml of conductivity water.

2. Dip a conductivity cell into the solution and ensure that the electrodes of the cell are completely immersed into the solution. Put a magnetic needle into it and place the electrode assembly on a magnetic stirrer. Stir the solution well.

3. Connect the electrodes to the conductivity bridge and measure conductance or resistance of 0.1 N HCl solution.

4. From the burette add 0.2 ml. of 0.05 N NaOH into the HCl solution and stir the solution. Measure the conductance of the solution.

5. In a similar way add 0.2 ml of 0.05 N NaOH solution each time and every time measure the conductance. Take conductance readings till the equivalence point exceeds by one ml i.e. take readings upto addition of 3.0 ml NaOH.

6. Plot a graph of conductance against ml of 0.05 N NaOH added. From the graph find neutralization point and hence find exact normality of HCl solution.

Observation Table :

Sr. No. ml. of 0.05 N NaOH added Conductance, ohm−1

1.

2.

3.

4.

:

16

0

0.2

0.4

0.6


Graph :

ml of NaOH added

Co

nd

ucta

nce

Neutralization point (V )2

Fig. 1

Calculations : 1. Exact normality of HCl solution HCl ≡ NaOH N1V1 = N2V2 (from graph) N1 × 10 = 0.05 × V2

N1 = 0.05 × V2

10

∴ Exact normality of given HCl = ………… 2. Strength of HCl = Normality × Equivalent weight = N1 × 36.5 g l−1 Result Table :

1. Neutralization or Equivalence point = ………. ml 2. Exact normality of given HCl solution = ………. g equi−1

3. Strength of given HCl solution = ………. g/litre.

Experiment No. 3 (B) Aim : To investigate the conductometric titration of a strong acid against a weak base. Apparatus : Conductivity bridge, conductivity cell, microburette, magnetic stirrer, burette, beaker etc. Chemicals : 0.1 N (approx.) HCl solution, 0.5 N NH4OH, conductivity water. Procedure : 1. In a small beaker take 10.0 ml of 0.1 N HCl solution and add about 40 ml conductivity

water to it. Stir the solution by keeping the beaker on a magnetic stirrer.

132


2. Dip the given conductivity cell into the solution of HCl. Ensure that the electrodes immerse completely into the solution.

3. Connect the cell to the conductivity bridge and measure the conductance or resistance of the solution.

4. Add 0.2 ml of NH4OH from a microburette into the HCl solution. Stir the solution and measure conductance or resistance of the solution.

5. In a similar way add 0.2 ml of 0.5 N NH4OH each time and every time measure the conductance of the solution.

6. Plot a graph of conductance against ml of NH4OH added. 7. From the graph find the neutralization point of the titration. Observation Table :

Obs. No. ml of 0.5 N NH4OH added Conductivity, ohm−1

1. 2. 3. 4. :

16

0 0.2 0.4 0.6 :

3.0

Graph :

ml of NH OH added4

Conducta

nce

Equivalence point (V )2

Fig. 1

Calculations : 1. Exact normality of HCl HCl = NH4OH N1V1 = N2V2 N1 × 10 = 0.5 × V2 (from graph)

∴ N1 = 0.5 × V2

10

2. Strength of HCl solution = Normality × Equivalent weight = N1 × 36.5 g l−1

133


134

Result Table : 1. Equivalence point from graph = ………. ml 2. Exact normality of the given HCl solution = ………. g equi−1

3. Strength of given HCl solution = ………. g/litre.

Experiment No. 3 (C) Aim : To investigate the conductometric titration of a strong base against a weak acid. Apparatus : Conductivity bridge, dip type conductivity cell, microburette, beaker etc. Chemicals : 0.01 N (approx.) acetic acid, 0.05 N NaOH. Procedure : 1. Take 10 ml of 0.01 N acetic acid in a small beaker and add about 40 ml conductivity

water. Put a magnetic needle and place the beaker on a magnetic stirrer. 2. Dip a conductivity cell into the solution, ensure that the electrodes of the cell are

completely immersed in the solution. Stir the solution. 3. Connect the cell to conductivity bridge and measure conductance of acetic acid

solution. 4. Add 0.2 ml of 0.05 N NaOH into the acetic acid solution. Stir the solution and find

conductance of the solution. 5. In a similar way take titration readings adding 0.2 ml of 0.05 N NaOH each time till

the end point exceed by about one ml. 6. Plot a graph of conductance against ml of NaOH added. 7. From the graph find the equivalence point and hence find the normality and strength

of given acetic acid solution. Reactions : CH3COOH + NaOH → CH3COONa + H2O CH3COOH + Na+ + OH− → CH3COO− + Na+ + H2O Observation Table :

Obs. No.

ml. of 0.05 N NaOH added Conductance, ohm−1

1. 2. 3. 4. :

16

0 0.2 0.4 0.6 :

3.0


Graph :

ml of 0.05 N NaOH added

Co

nd

ucta

nce


Fig. 1

Calculations : 1. Exact normality of acetic acid Acetic acid ≡ NaOH N1V1 = N2V2 N1 × 10 = 0.05 × V2

N1 = 0.05 × V2

10

2. Strength of given acetic acid = Normality × Equivalent weight = N1 × 60 g l−1 Result Table :

1. Equivalence point = ………. ml 2. Exact normality = ………. g equi−1

3. Strength of acetic acid = ………. g/litre.

Experiment No. 3 (D) Aim : To investigate the conductometric titration of a weak acid against a weak base. Apparatus : Conductivity bridge, dip type conductivity cell, microburette, beaker etc. Chemicals : 0.1 N (approx.) acetic acid, 0.5 N NH4OH, conductivity water. Procedure : 1. Take 10 ml of 0.1 N (approximate) acetic acid in a small beaker. Add about 40 ml

conductivity water. Put a magnetic needle. Place the beaker on a magnetic stirrer.

135


2. Dip the given conductivity cell into the acetic acid solution ensuring that the electrodes are completely immersed into the solution. Stir the solution.

3. Connect the conductivity cell to the conductivity bridge and measure resistance or conductance of acetic acid solution.

4. Add 0.2 ml of 0.5 N NH4OH into the acid solution, stir the solution and measure the conductivity or resistance of the solution.

5. In a similar way, measure conductivity by adding 0.2 ml NH4OH each time till the equivalent point exceeds by about one ml.

6. Plot a graph of conductance against ml of NH4OH added. 7. From the graph find the equivalence point and hence find the exact normality and

strength of given acetic acid solution. Reaction :

CH3COOH + NH+4 + OH− → CH3COO− + NH

+4 + H2O

Observation Table :

Obs. No. ml. of 0.5 N NH4OH added Conductance, ohm−1

1. 2. 3. 4. :

16

0 0.2 0.4 0.6 :

3.0

Graph :

ml of NH OH added4

Conducta

nce


Fig. 1

136


137

Calculations : 1. Exact normality of given acid solution Acetic acid = NH4OH

N1V1 = N2V2 N1 × 10 = 0.5 × V2

∴ N1 = 0.5 × V2

10

2. Strength of given acid solution = Normality × Equivalent weight = N1 × 60 g l−1 Result Table :

1. Equivalence point = ………. ml 2. Exact normality of given acetic acid solution = ………. g equi−1

3. Strength of given acetic acid solution = ………. g l−1

VIVA VOCE 1. What is a strong and a weak electrolyte ? 2. What is Ohm’s law ? 3. What is conductance ? 4. Why solutions of electrolytes show conductance ? 5. Define specific resistance, specific conductance. 6. Define Ohm. 7. In which unit conductance is measured ? 8. Explain how resistance of a conductor is measured using Wheatstone bridge. 9. Why direct current is not used in determination of conductance ? 10. Explain the variation of specific conductance with concentration. 11. What is conductivity cell ? 12. What is cell constant ? 13. Why inert electrodes like platinum are used for conductance measurements ? 14. What is a null point ? 15. What is equivalent conductance ? 16. Explain the variation of equivalent conductance with concentration. 17. What is the effect of temperature on conductance ? 18. How equivalent conductance of strong and weak electrolytes is measured ? 19. Can you use the Kohlrausch equation for the determination of equivalent

conductance at infinite dilution of a weak electrolyte ?


138

20. What is equivalent conductance at infinite dilution ? 21. Explain how you will find ^0 of weak electrolytes.

22. State Kohlrausch law of independent migration of ions. 23. Give the expressions to find (i) specific conductance, (ii) equivalent conductance and

degree of dissociation. 24. What is conductivity water ? And why it is required for conductivity measurements ? 25. What are the different types of cells used in conductivity measurements ? 26. Which type of conductivity cell will you use (i) for solution of low conductance and

(ii) for solution of high conductance ? 27. How conductivity water is obtained ? 28. State Ostwald’s dilution law. 29. What is conductometric titration ? 30. What are the advantages of conductometric titrations ? 31. Is equivalence point obtained by a conductometric titration is more accurate as

compared to a conventional volumetric titration ? True or False. 32. Explain the nature of titration curves of following conductometric titrations : (a) Strong acid and strong base, (b) Strong acid and weak base, (c) Weak acid and strong base, (d) Weak acid and strong base and precipitation titration of Pb(NO3)2 vs. Na2SO4 ? 33. What is ionic mobility ? 34. Can you titrate a mixture of a strong and a weak acid with a strong base ? If yes, how

you will find the equivalent points for strong acid and weak acid neutralization ? 35. How will you determine dissociation constant of a weak acid by conductivity

measurement ? 36. Give the factors that affect conductance of a solution. 37. Why platinum electrode used in conductivity cells is coated with platinum black ? 38. How platinum electrode is platinised ?

- - -

( 139 )

AppendicesAppendicesAppendicesAppendices

Appendix - I : K and αααα values of polymer :

Polymer Solvent K at 25°°°°C αααα at 25°°°°C

1. Polyvinyl alcohol Water 2.0 × 10−4 0.76

2. Crape rubber Benzene 9.40 × 10−3 0.76

3. Cellulose acetate (Table Tennis Ball) Acetone 1.49 × 10−4 0.82

4. Polystyrene (yellow rubber

used in shoes)

Toluene 3.7 × 10−1 0.62

Appendix - II : Densities and refractive index of liquids :

Liquid Refractive index n at 15°°°°C Density, g ml−−−−1 at 15°°°°C

Benzene

Chloroform

Carbon tetrachloride

Methyl alcohol

Ethyl alcohol

Toluene

Chlorobenzene

Bromobenzene

Acetone

1.5014 (at 20°C)

1.4486 (at 20°C)

1.4631

1.3312

1.3620

1.500

1.5248 (at 20°C)

1.5598

1.3588 (at 20°C)

0.8794 (at 20°C)

1.4985

1.595

0.7952

0.7881

0.8715

1.066 (at 20°C)

1.4991

0.792 (at 20°C)

Appendix - III : Transition temperature :

Substance Transition temperature

Na2SO4, 10H2O

Na2CO3, 10H2O

NaBr, 2H2O

32.38°C

32.22°C

50.7°C

Appendix - IV : Dissociation constants of acids at 25°°°°C :

Substance Ka at 25°°°°C

Acetic acid

Benzoic acid

Monochloroacetic acid

Oxalic acid

Propionic acid

Formic acid

1.75 × 10−5

6.3 × 10−6

1.4 × 10−3

6.5 × 10−2

1.4 × 10−5

1.7 × 10−4

T.Y.B.Sc. Practical Chemistry (Physical Practicals) Appendices

140

Salicylic acid 1.06 × 10−3

Appendix - V : Liquids and boiling points :

Substance Boiling point

(normal)

Density (20°°°°C)

Chlorobenzene

Bromobenzene

Nitrobenzene

132°C

155 − 156°C

210.9°C

1.107

1.500

1.980

Appendix - VI : Vapour pressure of water and temperature in.

Temp./

°°°°C

0.0 0.2 0.4 0.6 0.8 Temp./

°°°°C

0.0 0.2

85

86

87

88

89

90

91

92

93

433.6

450.9

468.7

487.1

506.0

525.8

546.1

567.00

588.60

437.0

454.4

472.4

491.0

510.0

529.77

550.20

571.26

593

440.0

458.0

476.0

494.7

513.9

533.8

554.4

575.6

597.4

440.0

461.6

479.8

498.5

517.8

537.9

558.5

579.9

601.9

447.5

456.2

483.4

502.2

521.8

542.0

562.8

584.2

606.4

94

95

96

97

98

99

100

610.9

633.9

657.62

682.10

707.30

733.24

760.00

615.42

638.60

662.5

687.0

712.4

738.5

765.5

Appendix - VII : Specific conductance of 0.01 M KCl :

Temperature 20 21 22 23 24

Specific

conductivity

0.001278 0.001305 0.001335 0.001359 0.001380

Structure of Practical Examination Marks

1. One Experiment from Group – A 35

2. One Experiment from Group – B 35

3. Oral 10

✍ ✍ ✍

141

Course - I

CH-348

INORGANIC CHEMISTRY

PRACTICALS

T.Y.B.Sc. Practical Chemistry (Inorganic Practicals) Gravimetric Estimations

142


COURSE – I CH – 348

Inorganic Chemistry Practicals

Experiments Marks Q.1. Qualitative Analysis (Compulsory Experiment) 35 OR Gravimetric Experiment* 30 Q.2. Volumetric Experiment (20 marks) 35 Preparation (10 marks) OR Flame Photometry (20 marks) Preparation (10 marks) OR Column Chromatography (20 marks) Preparation (10 marks) OR Colorimetric Estimation (25 marks) Preparation (10 marks) Q.3. Oral 10


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(A) Gravimetric Estimations (Any Three)

Experiment No. 1 Fe as Fe2O3

Aim : Gravimetric estimation of iron (Fe) as ferric oxide (Fe2O3) from ferrous ammonium sulphate. Principle : Ferrous is oxidized to ferric state by boiling with concentrated HNO3, then it is precipitated as ferric hydroxide (red precipitate) with ammonium hydroxide. This precipitate can be filtered through Whatman filter paper No. 41. Ferric hydroxide on ignition gives ferric oxide. Reactions : 6FeSO4 + 3H2SO4 + 2HNO3 ⎯→ 3Fe2(SO4)3 + 2NO + 4H2O Fe2(SO4)3 + 6NH4OH ⎯→ 2Fe(OH)3 ↓ + 3(NH4)2SO4

2Fe(OH)3 Δ⎯→ Fe2O3 + 3H2O

Requirements : (i) Ferrous ammonium sulphate solution → Given (ii) Dil. H2SO4 (2N) → 10 ml (iii) Conc. HNO3 → 3 – 4 ml (iv) Ammonium nitrate solution (2%) → for washing (v) Ammonium chloride → 1 gm (vi) 1 : 1 ammonia (vii) Silica crucible (viii) Whatman filter paper No. 41. Procedure : 1. Dilute the given solution of ferrous ammonium sulphate (FAS) with distilled water

upto the mark in a 250 ml volumetric flask and shake well. 2. Pipette out 50 ml of this diluted solution in a 250 ml beaker. To this add two test

tubes distilled water. 3. Add 10 ml diluted H2SO4 and heat upto boiling. 4. Then add 3-4 ml conc. HNO3, boil for 5 minutes. At this state ferrous is oxidized to

ferric state. 5. To this boiling solution add 1 gm NH4Cl. Stir with glass rod.


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6. Then add 1 : 1 ammonia solution. The precipitation is of ferric hydroxide (red precipitate). Add slight excess ammonia solution. Then boil the solution for 5 minutes, allow the precipitate to settle.

7. Filter the solution through Whatman filter paper No. 41. Transfer all the precipitate over the filter paper.

8. Wash the precipitate with hot 2% ammonium nitrate for 3 - 4 times. Then wash with hot water till the filtrate is free from chloride and sulphate radicals.

9. Dry the precipitate of Fe(OH)3 in oven or on metal cone. Ignite the precipitate along with the filter paper in a previously weighed silica crucible.

10. Heat the crucible along with the precipitate at 700 - 800°C in an electric furnace for about 45 minutes.

11. Cool the crucible and weigh it. Repeat the process of heating till a constant weight is obtained.

12. Difference in two weights give the weight of residue (Fe2O3). Let it be ‘X’ gms. 13. Calculate the amount of FeSO4(NH4)2SO4 ⋅ 6H2O (Mohr’s salt) in the given solution. Observation Table :

Sr. No. Description Symbol Weight

1. 2. 3.

Weight of empty crucible Weight of crucible + Residue Weight of residue (Fe2O3) (W2 − W1)

W1 W2

‘X’

= ………. gms = ………. gms = ………. gms

Calculations : 50 ml diluted solution contains = ‘X’ gms of Fe2O3 ∴ 250 ml diluted solution contains = ‘X’ × 5 gms of Fe2O3 = ‘Y’ gms of Fe2O3 From equation, Fe2O3 = 2[FeSO4(NH4)2SO4 ⋅ 6H2O] ≡ 2Fe 159.7 784 11.70 1. To find out quantity of ferrous ammonium sulphate in the given solution : 159.7 gms of Fe2O3 obtained from = 784 gms of FAS.

∴ ‘Y’ gms of Fe2O3 will be obtained from = 784

159.7 × ‘Y’ gms of FAS.

= 4.909 × ‘Y’ gms of FAS = …… Z gms of FAS 2. To find out amount of Fe in the given solution : 159.70 g Fe2O3 ≡ 111.70 gm Fe ∴ Y g Fe2O3 ≡ ?


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= 111.70 × Y

159.70 gms Fe

= 0.6994 × Y gms Fe = ‘B’ gms Fe Result Table :

Sr. No.

Description Value

1. 50 ml of diluted solution gave ‘X’ ….. gms of Fe2O3

2. 100 ml of diluted solution gave ‘Y’ ….. gms of Fe2O3

3. Quantity of FeSO4(NH4)2SO4 ⋅ 6H2O (Mohr’s salt) in the given solution

‘Z’ ….. gms

4. Quantity of Fe in the given solution ‘B’ ….. gms

VIVA VOCE

1. What is gravimetric analysis ?

2. What type of crucible is used in gravimetric estimation of Fe as Fe2O3 ? Why ?

3. Which filter paper is used for filteration ?

4. What is the use of dessicator ?

5. What is present in dessicator ? What is its role ?

6. Why the precipitation of Fe(OH)3 is washed with ammonium nitrate solution ?

7. What do you mean by filteration by decantation ?

8. What is used for precipitation of ferric hydroxide ? Why ?

9. List various operations involved in gravimetric analysis.

10. How Fe2+ is converted into Fe3+ ?

11. Why HNO3 is added to the diluted FAS solution ?

Experiment No. 2 Nickel as Ni-DMG

Aim :

Gravimetric estimation of nickel as nickel dimethyl glyoxamate.

Principle :


To a slight acidic solution of nickel, if we add alcoholic solution of dimethyl glyoxamate and slight excess of ammonia then pinkish red precipitate of nickel dimethyl glyoxamate is formed. It can be filtered and weighed directly.

Dimethyl glyoxime reagent is a bidentate ligand, nickel has coordination numbers 4 and 6. Here two molecules of reagent combine with one molecule of nickel, forming a square planar geometry. Reaction :

Ni2+ + 2H2 DMG Heat⎯⎯⎯→50°C

Ni(HDMG)2 + 2H+

Ni2+

CH3

CH3 C N OH

C N OH

Heat

at 60ºC

CH3 C N N C CH3

O O

H

Ni

CH3 C N N C CH3

OO

H

DMG Nickel dimethyl glyoxamate(Ni-DMG)

+

Requirements : (i) NiSO4 ⋅ 7H2O solution → given (ii) Tartaric acid → 5 gms (iii) 1 : 1 HCl → 5 ml (iv) 2% dimethyl glyoxamine (alcoholic) → 20 ml (v) 1 : 1 ammonia (vi) Gooch crucible (G4) (vii) Oven. Procedure : 1. Dilute the given solution of nickel sulphate with distilled water upto the mark in a

250 ml volumetric flask and shake well. 2. Pipette out 50 ml of this diluted solution in a 250 ml beaker. To this add two test

tubes distilled water. 3. Then add 5 gms of tartaric acid and 5 ml of 1 : 1 HCl till the solution is alkaline and

pinkish red precipitate of nickel dimethyl glyoxamate is obtained. 4. Heat the solution on water bath for about 30 minutes. Allow the precipitate to settle

and cool to room temperature.

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5. Filter the precipitate through previously weighed Gooch crucible (G4). 6. Wash the precipitate with cold distilled water till the filtrate is free from chloride

radical (tested with AgNO3 solution). 7. Dry the precipitate to 150°C in a oven for about one hour. 8. Allow it to cool and weigh the residue as nickel dimethyl glyoxamate. 9. Calculate the amount of NiSO4 ⋅ 7H2O and the quantity of nickel in the given solution. Observation Table :

Obs. No. Description Symbol Weight

1. 2. 3.

Weight of empty crucible Weight of Gooch crucible + Residue Weight of residue (W2 − W1)

W1 W2

‘X’

= ………. gms = ………. gms = ………. gms

Calculations : 50 ml diluted solution gives = ‘X’ gms of Ni-DMG ∴ 250 ml diluted solution gives = ‘X’ × 5 gms of Ni-DMG = ‘Y’ gms of Ni-DMG From equation, NiSO4 ⋅ 7H2O ≡ Ni − (C4H7O2N2)2 ≡ Ni 280.87 289.19 58.71 1. Estimation of nickel : 289.19 gms of Ni-DMG solution = 58.71 gms of Ni

∴ Y gms of Ni-DMG contains = 58.71289.19 × Y gms of Ni

= 0.2330 × ‘Y’ gms of Ni. 2. Estimation of NiSO4 ⋅ 7H2O in the given solution : 289.19 gms of Ni-DMG contains = 280.87 gms of NiSO4 ⋅ 7H2O

∴ ‘Y’ gms of Ni-DMG contains = 280.87289.19 × ‘Y’ gms

= 0.9716 × Y gms = ‘Z’ gms of NiSO4 ⋅ 7H2O 3. To find out quantity of Ni in the solution : 288.71 gm of Ni-DMG ≡ 58.71 gms of Ni

∴ ‘Y’ gm of Ni-DMG ≡ 58.71 × Y

288.71 gms of Ni

= 0.2034 × Y gms of Ni = ‘B’ gms of Ni Result Table :


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Sr. No. Description Value

1. 50 ml of diluted solution gave X = ….. gms of Ni-DMG

2. Quantity of nickel in the given solution Y = ….. gms

3. Quantity of NiSO4 ⋅ 7H2O in the given solution

Z = ….. gms

4. Quantity of Ni in the given solution B = ….. gms

VIVA VOCE 1. Which crucible is used for filteration ? 2. What is Gooch crucible ? 3. What is DMG ? Draw its structure. 4. What type of ligand the DMG is ? 5. What do you mean by ligand ? 6. What are polydentate ligands ? 7. Write the reaction of Ni2+ with DMG. 8. Why DMG is bidentate ligand ? 9. What is the coordination number of Ni2+ in Ni-DMG complex ? 10. What is the geometry of Ni-DMG complex ? 11. Why the precipitate of Ni(H DMG)2 kept in oven and not ignited ?

Experiment No. 3 Al as Aluminium Oxide

Aim : To determine the amount of aluminium gravimetrically as Al2O3. Principle : Aluminium is precipitated as aluminium hydroxide by ammonia solution in the presence of ammonium chloride. The gelatinous precipitate of aluminium hydroxide is washed with a solution of strong electrolyte subjected to strong ignition and finally weighed as Al2O3. Reactions : (1) Al3+ + NH4Cl + NH4OH ⎯→ Al(OH)3 ↓

(2) 2Al(OH)3 Δ

⎯→ Al2O3 + 3H2O ↑. Theory : The gravimetric estimation of Al as Al2O3 must be done following several precautions. Aluminium hydroxide is amphoteric in nature. It is a bulky, gelatinous precipitate, appreciably in ammonia but practically insoluble in the presence of ammonium salts. The precipitation of Al(OH)3 commences in approximately pH 4 and is complete when the pH lies between


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6.5 and 7.5. The required pH can be controlled by adding ammonium chloride before adding ammonium hydroxide which creates buffering effect and also help the coagulation of the initially colloidal precipitate. The addition of ammonium chloride controls the hydroxide concentration. It also reduces the co-precipitation of divalent cations like Ca2+ and Mg2+ and some other cations. Excess of ammonia should not be added because prolong boiling is required to remove excess of ammonia which also makes the precipitate obtained by precipitation in hot solution. The precipitate cannot be washed with hot water because aluminium hydroxide is readily peptised and will pass through the filter paper. It is washed with 2% solution of ammonium nitrate which avoids sol formation. This precipitate is hygroscopic and hence high temperature (1200°C) is necessary for the rapid dehydration of aluminium hydroxide to aluminium oxide. For this reason, the precipitate ignited in a rapid weighing without undue exposure to atmosphere is essential as freshly ignited alumina is extremely hygroscopic.

Requirements :

(i) Ammonium aluminium sulphate (alum) solution (Give 10 to 50 ml) given in a 100 ml volumetric flask.

(ii) Ammonium chloride

(iii) 1 : 1 ammonia

(iv) Methyl red indicator (0.2% alcoholic solution)

(v) Silica crucible

(vi) Ammonium nitrate

(vii) Whatman filter paper No. 41.

Procedure :

1. Dilute the given solution of ammonium aluminium sulphate [(NH4)2SO4 ⋅ Al2(SO4)3 ⋅ 24H2O] with distilled water upto the mark in a 100 ml volumetric flask. Shake the flask and take out all the solution in a clean 250 ml beaker.

2. Take by a burette 25 ml of this diluted solution in a 250 ml beaker. Add 50 ml distilled water to it by a measuring cylinder.

3. Add one gram ammonium chloride to it.

4. Add 2 to 3 drops of methyl red indicator. The solution turns red.

5. Heat the solution just to boiling.

6. Add 1 : 1 ammonia solution drop by drop to precipitate Al as Al(OH)3 till the colour of the solution becomes distinct yellow. Avoid excess addition of ammonia solution. A white gelatinous precipitate of Al(OH)3 is obtained.

7. Boil the solution for two minutes.


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8. Filter the hot solution containing the precipitate of Al(OH)3 through Whatman filter paper No. 41 by decantation method. Then transfer the precipitate carefully to the filter paper.

9. Wash the beaker and the precipitate with hot 2% ammonium nitrate solution; use about 10 ml wash liquid at a time. Wash the precipitate till it is free from sulphate ion. The fresh filtrate, during washing finally should not give a precipitate with barium nitrate. This confirms the removal of sulphate ion on the precipitate.

10. Dry the precipitate with the filter paper on a metallic cone or in an oven.

11. Ignite the precipitate along with ashless filter paper in previously weighed silica crucible. Heat the crucible with the filter paper at about 1200°C on a pipe clay triangle for half an hour without the lid and for 15 minutes with lid partially open. The residue of Al(OH)2 is converted to Al2O3.

12. Transfer the crucible and lid to the desiccator when it is warm. Allow it to cool in the dessicator when it is warm. Allow it to cool.

13. When cool, quickly weigh the crucible and lid along with the residue of Al2O3. Find the weight of Al2O3 residue.

Reactions :

Al2(SO4)3 + 6NH4OH ⎯→ 2Al(OH)3 ↓ + 3(NH4)2SO4

2Al(OH)3 Δ

⎯→ Al2O3 + 3H2O ↓

Observation Table :

Sr. No. Description Symbol Weight

1.

2.

3.

Weight of empty silica crucible + lid

Weight of crucible + lid + Al2O3 residue

Weight of Al2O3 residue (W2 − W1)

W1

W2

‘X’

= ………. gms

= ………. gms

= ………. gms

Calculations :

25 ml diluted solution of ammonium

aluminium sulphate ≡ X g of Al2O3

∴

100 ml diluted solution of ammonium

aluminium sulphate ≡ 4 × g of Al2O3

≡ Y g of Al2O3

1. To find out the quantity of ammonium aluminium sulphate in the given solution :

From the equation,


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(NH4)2SO4 ⋅ Al2(SO4)3 ⋅ 24H2O ≡ Al2O3 ≡ 2Al

906 ≡ 102 ≡ 54

102 g of Al2O3 ≡ 906 g of (NH4)2SO4 ⋅ Al2(SO4)3 ⋅ 24H2O

∴ Y g of Al2O3 ≡ 906102 × Y g of (NH4)2SO4 ⋅ Al2(SO4)3 ⋅ 24H2O

≡ 8.88 × Y g

≡ Z g of (NH4)2SO4 ⋅ Al2(SO4)3 ⋅ 24H2O 2. To find out the quantity of aluminium in the given solution : 102 g of Al2O3 ≡ 54 g of Al

∴ Y g of Al2O3 ≡ 54102 × Y g of Al

≡ 0.5294 × Y g of Al ≡ B g of Al Result Table :


1. 25 ml of diluted solution of alum gave X ….. g of Al2O3

2. Quantity of alum in the given solution Z ….. gms

3. Quantity of aluminium in the given solution

B ….. gms

VIVA VOCE 1. Give the reaction involved in gravimetric estimation of Al as Al2O3. 2. Which filter paper is used in this gravimetric estimation ? Why ? 3. What is the nature of precipitate in this experiment ? 4. What is the role of NH4Cl in gravimetric estimation of Al as Al2O3 ? 5. Why hot solution is used for precipitation ? 6. Washing with hot water is never done in this experiment. Why ?

Experiment No. 4 Ba as BaSO4

Aim : Gravimetric estimation of barium as barium sulphate by using sulphamic acid. (Homogeneous precipitation technique). Principle :


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Sulphamic acid is hydrolysed to ammonium sulphate above 60°C. Barium (Ba+2)

combines with sulphate ion (SO2−4 ) and barium sulphate is precipitated slowly. This type of

precipitation minimizes coprecipitation and thus impurities are reduced. Similarly precipitate is compact and granular as precipitation is slow and through the solution. This precipitate can be filtered through Whatman filter paper No. 42 and weighed as BaSO4. Reactions :

NH2SO3H + H2O Boil⎯⎯⎯→ NH+4 + SO

2−4 + H+

Ba2+ + SO2−4 ⎯⎯⎯→ BaSO4 ↓ (white precipitate)

Requirements : (i) BaCl2 ⋅ 2H2O solution → given (ii) Sulphamic acid → 5 gm (iii) Whatman filter paper No. 42 (iv) Silica crucible, watch glass (v) Oven furnace etc. Procedure : 1. Dilute the given solution of BaCl2 ⋅ 2H2O with distilled water upto the mark in a

250 ml volumetric flask and shake well. 2. Pipette out 50 ml of this diluted solution in 250 ml beaker. To this add two test tubes

distilled water. 3. Then add 5 gm of sulphamic acid, stir with glass rod. 4. Cover the beaker with glass. Heat slowly on an sand bath for about 30 minutes. 5. The white precipitate of BaSO4 settles at the bottom. 6. Filter the precipitate through Whatman filter paper No. 42. Then wash the precipitate

with hot distilled water, till the filtrate is free from chloride and sulphate radicals. 7. Dry the precipitate in a oven and finally ignite it in a previously weighed silica

crucible. 8. Heat the crucible along with the precipitate at 700 - 800°C in an electric furnace for

about 30 minutes. 9. Cool the crucible and weigh it. Repeat the process of heating till a constant weight is

obtained. 10. Difference in two weights is the weight of BaSO4. Let it be ‘X’ gms. 11. Calculate the amount of BaCl2 ⋅ 2H2O in the given solution. Observation Table :

Obs. No. Description Symbol Weight

1. Weight of empty silica crucible W1 = ………. gms


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2. 3.

Weight of crucible + residue (BaSO4) Weight of residue (W2 − W1)

W2

‘X’ = ………. gms = ………. gms

Calculations : Since 50 ml diluted solution gives = ‘X’ gms of BaSO4 ∴ 250 ml diluted solution gives = ‘X’ × 5 gms of BaSO4 = ‘Y’ gms of BaSO4 From equation, BaSO4 ≡ BaCl2 ⋅ 2H2O ≡ Ba 233.36 244.36 137.36 1. Quantity of BaCl2 ⋅ 2H2O in the given solution : ∴ 233.36 gms of BaSO4 obtained from = 244.36 gms of BaCl2 ⋅ 2H2O

∴ ‘Y’ gms of BaSO4 = 244.36233.36 × ‘Y’ gms of BaCl2 ⋅ 2H2O

= 1.047 × ‘Y’ gms of BaCl2 ⋅ 2H2O = ………. Z gms of BaCl2 ⋅ 2H2O 2. To find out the amount of Ba in the given solution : 233.36 gm of BaSO4 ≡ 137.36 gms of Ba

∴ Y gm of BaSO4 ≡ 137.36 × Y

233.36 gms of Ba

≡ 0.5886 gm × Y gms of Ba ≡ B gms of Ba Result Table :


1. 50 ml of diluted solution gave X = ….. gms of BaSO4

2. Quantity of BaCl2 ⋅ 2H2O in the given solution

Z = ….. gms

3. Quantity of Ba in the given solution B = ….. gms

VIVA VOCE 1. Which filter paper is used in this experiment ? 2. Which crucible is used ? 3. How Ba is precipitated ? 4. What do you mean by precipitation from homogeneous solution ? 5. How sulphamic acid is used in this experiment ?


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6. What are the advantages of precipitation from homogeneous solutions in comparison to conventional method of precipitation ?

7. What is gravimetric analysis ? 8. What is quantitative analysis ? 9. How the quantity of Ba2+ present in the solution is calculated ?

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(B) Volumetric Estimations (Any Four)

Introduction : The term volumetric analysis (or called as titrimetric analysis) refers to quantitative analysis carried out by determining the volume of a solution of accurately known concentration which is required to react quantitatively with a measured volume of solution of the substance to be determined. The solution of accurately known strength is called the standard solution. The weight of the substance to be determined is calculated from the volume of standard solution used and the chemical equation and relative molecular masses of the reacting compounds. It is one of the used analytical techniques. It is simple, rapid and good accuracy can be obtained from it. The process of adding the standard solution until the reaction is just complete is termed as a titration, and the substance to be determined is titrant. The most important point in the process of titration is the determination of the end point, that is, the point at which the reaction is just completed. To determine the end point of a reaction, we use certain chemicals by which the change of colour indicates that the reaction is completed. Such chemicals are called indicators.

Experiment No. 1 Manganese (Mn) by Volhard’s Method

Aim : To determine the amount of manganese (Mn) volumetrically by Volhard's method. Principle : The details of standardization of KMnO4 solution is already studied in S.Y.B.Sc. Volhard's method for determination of manganese is based on the principle that when warm solution of manganese (here manganese sulphate) is titrated with standard solution of KMnO4 the

dark brown precipitate of manganese dioxide is obtained in the flask. Thus, MnO−4 ions are

reduced to MnO2 and Mn2+ (manganous) ions are oxidized to MnO2. This can be observed in the following reaction : 2KMnO4 + 3MnSO4 + 2H2O → K2SO4 + 5MnO2 ↓ + 2H2SO4

i.e. 2MnO−4 + 3Mn2+ + 2H2O → K2SO4 + 5MnO2 ↓ + 4H+.

The precipitate that is hydrated manganese oxide is acidic in nature which absorbs manganese hydroxide and prevent complete oxidation. To avoid this difficulty, Volhard introduced the use of ZnO paste. Following are the functions of zinc oxide paste :

1. It settles the MnO2 precipitate quickly.

T.Y.B.Sc. Practical Chemistry (Inorganic Practicals) Volumetric Estimations

155

2. It neutralises excess of H2SO4. 3. It avoids the formation of KMnO4. 4. If Fe impurities are present in the solution it forms zinc ferrites. Thus the interference

of Fe is masked. 5. It avoids the formation of manganous acid. Thus precipitation of manganous

manganite is prevented. Hint : Potassium permanganate attacks organic matter. So burette with rubber tubing and pinch cock cannot be used. The strength of potassium permanganate is altered if it is kept in contact with rubber for long time. So the burette used in permanganate titration must be glass stoppered. Chemicals : (i) 0.05 N (approximate) KMnO4 (ii) Given unknown solution of manganese (MnSO4) (iii) A.R. grade oxalic acid (iv) Zinc oxide paste (v) 2 N HNO3. Procedure : This experiment is performed in the 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. 2. Transfer it in a clean beaker and rinse the watch glass with distilled water. 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. 5. Take out all the solution in a clean beaker to make homogeneous solution. This is

0.05 N oxalic acid solution.

6. Use this solution for standardization of given KMnO−4 solution.

Part B : Standardization of KMnO4 solution : 1. Fill the burette No.1 with 0.05 N (approximate) 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.


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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 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 : Estimation of Manganese : 1. Dilute the given solution of manganese sulphate to 100 ml with distilled water in a

volumetric flask. Take out whole solution in a beaker and stir it well. 2. Fill the burette No.1 with standardized KMnO4 solution. 3. Clean the burette No. 2 and fill it with diluted MnSO4 solution. 4. Take 9 ml of diluted MnSO4 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 of ZnO paste. Also add one test tube of water. 6. Heat the solution on wire gauze to about 60°C and add 2 - 3 drops of 2 N HNO3

solution (which promotes the settling of the precipitate of MnO2 at the bottom). 7. Titrate this hot solution with standardized KMnO4 solution from burette No. 1 with

constant shaking. 8. Continue the addition of KMnO4 till the supernatant solution in the conical flask

becomes faint pink which persists even after vigorous shaking. Record the burette reading and call this as Y1 ml.

9. Add to the same conical flask 1 ml solution of MnSO4 from burette No. 2 with constant shaking. The pink colour disappears. Heat the solution on water bath to 60°C.

10. Add KMnO4 from burette No. 1 till a faint pink colour appears to the supernatant 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 KMnO4 from burette No. 2. The pink colour disappears. Heat the solution on water bath to 60°C.


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12. Add KMnO4 solution from burette No. 1 till faint pink colour appears to the supernatant solution. Record the burette reading and call this as Y3 ml.

Note : During the titration, allow the precipitate to settle down and observe the colour of the supernatant solution against white background at each addition.

Observation Table :

Part B : Standardization of KMnO4 solution.

To find : Exact normality of KMnO4 solution

Burette 1 : KMnO4 solution (approximate 0.05 N)

Burette 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 ↑

Obs. No. Burette 1 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

∴ Exact normality of KMnO4 = N4 = …… N

Part C : Estimation of Manganese.

Given : KMnO4

To find : Amount of Mn in the given solution.

Burette - 1 : KMnO4 solution.

Burette - 2 : MnSO4 solution


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Indicator : KMnO4 self indicator.

End Point : Faint pink colour to supernant solution. Reaction : 2KMnO4 + 3MnSO4 + 2H2O ⎯→ K2SO4 + 5MnO2↓ + 2H2SO4

Burette - 1 (KMnO4) Y1 ml Y2 ml Y3 ml

Burette - 2 (MnSO4) 9 ml 10 ml 11 ml

To calculate the amount of Mn :

We have reaction,

2KMnO4 + 3MnSO4 + 2H2O ⎯→ K2SO4 + 5MnO2↓ + 2H2SO4

2 moles of KMnO4 ≡ 3 moles of Mn

∴ 1 mole of KMnO4 ≡ 32 moles of Mn

1N KMnO4 = 32 ×

15 moles of Mn

(Since 1 g equivalent of KMnO4 = 15 mol. wt. of KMnO4)

= 310 × 54.94 gm Mn

= 16.48 g Mn

1000 ml 1 N KMnO4 = 16.48 g Mn

∴ 1 ml 0.05 N KMnO4 = 0.000824 g Mn

∴ Y1 ml of N4 KMnO4 = 0.000824 × N4 × Y1

0.05 = A1 g Mn

Similarly, Y2 ml of N4 KMnO4 = 0.000824 × N4 × Y2

0.05 = A2 g Mn

and Y3 ml of N4 KMnO4 = 0.000824 × N5 × Y3

0.05 = A3 g Mn

Now, 9 ml diluted solution contains A1 g Mn

∴ 100 ml diluted solution contain B1 g Mn

Similarly, 10 ml diluted solution contain A2 g Mn


and 11 ml diluted solution contain A3 g Mn



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Mean = B1 + B2 + B3

3 = B g Mn

Thus, amount of Mn in 100 ml diluted solution = B g Mn Result Table :

1. Exact normality of KMnO4 solution = N4 ………. N

2. The amount of Mn in the given solution = B ………. g

VIVA VOCE 1. Explain the term 'Volumetric analysis'. 2. Define the terms : Standardization, titration, standard solution. 3. What do you mean by normal solution ? 4. How 1 g equivalent of Mn is calculated ? 5. How 100 ml 0.05 N oxalic acid solution is prepared ? 6. What are the functions of ZnO paste in the estimation of manganese ? 7. Why ZnO paste is added in the estimation of manganese ? 8. What is normal solution ? How will you prepare 0.01 N solution of oxalic acid ? 9. Why this method is called Volhard's method ?

Experiment No. 2

Estimation of NO−2 using KMnO4

Aim : To determine the amount of nitrite from a given salt by titrating it against a standard solution of 0.05 N KMnO4. Theory: An acidified solution of sodium nitrite is oxidized by potassium permanganate to nitrate:

2MnO−4(aq) + 6H

+(aq) + 5NO

−2(aq) → 2Mn

2+(aq) + 5NO

−3(aq) + 3H2O(l)

The nitrite solution cannot be titrated with permanganate solution from a burette in the usual way, because, as soon as it is acidified, the nitrous acid formed begins to decompose.

2NaNO2(aq) + H2SO4(aq) → Na2SO4(aq) + 2HNO2(aq) 3HNO2(aq) → HNO3 + H2O(l) + 2NO(g)

Thus, the nitrite solution is placed in the burette and is added, SLOWLY and with constant swirling to an acidified solution of potassium permanganate. Chemicals : (i) Na2C2O4 (0.025 N exact).


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(ii) KMnO4 solution (approximate 0.025 N). (iii) KNO2. (iv) H2SO4 (2 N). Procedure : This experiment is performed in the following three parts : Part A : Preparation of standard solution of sodium oxalate (MW = 134) (0.025 N) : 1000 ml 1 N oxalic acid solution contains 67 g Na2C2O4. 1000 ml 0.1 N oxalic acid solution contains 6.7 g Na2C2O4. ∴ 100 ml 0.025 N oxalic acid solution contains 0.1675 g Na2C2O4.

7. Weigh accurately 0.1675 g of A.R. grade sodium oxalate crystals on watch glass.

8. Transfer it in a clean beaker and rinse the watch glass with distilled water.

9. Stir the solution with glass rod and dissolve the crystals. Pour the solution into 100 ml volumetric flask.

10. Dilute the solution with distilled water upto the mark.

11. Take out all the solution in a clean beaker to make homogeneous solution. This is 0.025 N oxalic acid solution.

12. Use this solution for standardization of given KMnO4 solution. Part B : Standardization of KMnO4 solution :

12. Fill the burette No. 1 with 0.025 N (approximate) KMnO4 solution. 13. Fill the burette No. 2 with standard 0.025 N sodium oxalate solution. 14. Take 9 ml of sodium oxalate solution from burette No. 2 in a 100 ml conical flask.

Call this volume as V2 ml. 15. Add to it 1/2 test tube (10 ml) of 2 N (dilute) H2SO4. 16. Heat the solution on wire gauze to about 70-80°C. 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.

17. Continue the addition of KMnO4 till the solution in the conical flask changes from colourless to permanently faint pink. Record the burette reading and call this as ‘X1’ ml.

18. Add to the same conical flask 1 ml solution of 0.025 N sodium oxalate from burette No. 2. Heat the flask again so that solution in the flask acquires a temperature of about 70-80°C and solution becomes colourless.

19. 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.


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20. To the same solution in the conical flask add 1 ml solution of 0.025 N sodium oxalate from burette No. 2 (See that the solution in the flask is hot, if not, heat it again to 70-80°C). The pink colour disappears.

21. 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 : Estimation of Nitrite from the given solution :

1. Weigh accurately 0.200 g (W) of sodium nitrite and dissolve it in a small quantity of distilled water.

2. Dilute the above solution to prepare 100 ml of sodium nitrite solution using volumetric flask .

3. Fill burette No. 1 with diluted sodium nitrite solution and burette No. 2 with potassium permanganate solution.

4. Pipette out 10 ml potassium permanganate solution from burette No. 2 into a conical flask. Add 10 ml (1/2 test tube) dilute sulphuric acid (2 N) to acidify the nitrate solution.

5. Titrate this acidified potassium permanganate solution by the addition of the sodium

nitrite solution from burette No. 1 slowly and with continual stirring till the pink colour is just disappeared. Record this reading as Y1 ml.

6. Similarly perform two more titrations to obtain at least two concordant results. (difference < 0.2 ml). Record these as Y2 and Y3 respectively.

7. Find out the quantity of nitrite present in the sample of given sodium nitrite.

Remarks:

1. The end-point of this reaction is not very sharp. Be careful not to use too much titrant.

2. The normality of KMnO4 stated in this manual is only an approximate value. Do not use this for calculations.

Observation Table : Part B : Standardization of KMnO4 solution.

To find : Exact normality of KMnO4 solution

Burette 1 : KMnO4 solution (~ 0.025 N) Burette 2 : Sodium oxalate (0.025 N)

Indicator : KMnO4 (self-indicator) End point : Colourless to pink

Reaction : 2 KMnO4 + 5 Na2C2O4 + 8 H2SO4 → K2SO4 + 2 MnSO4 + 5 Na2SO4 + 10 CO2 + 8H2O

Obs. No. Burette 1 ‘X’ ml

Burette 2 V2 ml

Normality N = 0.025 × V2

X 1. X1 = ….. 9.0 N1 = 0.025 × 9


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X1 2. X2 = ….. 9 + 1 = 10.0 N2 = 0.025 × 10

X2 3. X3 = ….. 10 + 1 = 11.0 N3 = 0.025 × 11

X3

Calculations: 1. To calculate exact normality of KMnO4:

Mean Normality N4 = N1 + N2 + N3

3 = .......... N

∴ Hence, the exact normality of KMnO4 is ………..(N4) N. Observation Table: Part C: Estimation of Nitrite from Sodium nitrite. To find : Amount of nitrite from sodium nitrite. Burette 1 : Sodium nitrite Burette 2 : KMnO4 solution (accurate N4) Indicator : KMnO4 (self-indicator) End point: Pink to colourless Reaction:

2MnO−4 + 5NO

−2 + 6H+ → 2Mn2+ + 5NO

−3 + 3H2O

Obs. No. Burette 1 ‘Y’ ml

Burette 2 V2 ml

Constant Burette Reading (CBR)

1. Y1 = ….. 10 ‘Y’ ml 2. Y2 = ….. 10

3. Y3 = ….. 10

Calculations:

2. To calculate amount of nitrite:

2 MnO−4 + 5 NO

−2 + 6 H+ → 2 Mn2+ + 5 NO

−3 + 3 H2O

2 moles of KMnO4 ≡ 5 moles of NO−2

1 mole of KMnO4 = 5/2 moles of NO−2

1 N KMnO4 ≡ 5/2 × 1/5 moles of NO−2

(As 1 g equiv. of KMnO4 = 1/5 Mol. wt. of KMnO4)

= 5/10 × 46.00 g NO−2 = 23 g NO

−2

As 1000 ml 1 N KMnO4 ≡ 23 g NO−2


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∴ Y ml of N4 KMnO4 = Y1 × N4 × 23

1000 × 1 = ‘A’ g NO−2

As 10 ml diluted solution contains ‘A’ g of NO−2

∴ 100 ml diluted solution contains = 100 × A

10 = 10 × ‘A’ = ‘B’ g of NO−2

∴ Hence, the amount of nitrite present in the given sample = ……… ‘B’ g of NO−2

3. To calculate percentage of nitrite:

As W grams of sodium nitrite sample contains ‘B’ grams of nitrite

∴ 100 grams of sodium nitrite sample contains = 100 × 'B'

W = …… (Z) % of NO−2

Result Table :

1. Exact normality of KMnO4 N4 = ………. N

2.

10 ml KMnO4 solution required Y = ………. ml of sodium nitrite solution

3. Amount of nitrite in the given solution B = …….. g

4. Amount of nitrite in the given solution ‘Z’ = …….. %

VIVA VOCE

1. Explain the term 'Volumetric analysis'.

2. Define the terms : Standardization, Titration, Standard solution.

3. What do you mean by normal solution ?

4. Calculate the equivalent weight of KMnO4 in this reaction.

5. How 100 ml 0.025 N sodium oxalate solution is prepared ?

6. What is the type of this titration ?

7. Why KMnO4 is not titrated directly?

Experiment No. 3 Percentage Purity of NaCl

Aim : Estimation of percentage purity of given sample of sodium chloride. Principle : Sodium chloride reacts quantitatively with silver nitrate as per the following equation : AgNO3 + NaCl → AgCl + NaNO3


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169.89 58.5 So the purity of sodium chloride is determined by titrating it with standard silver nitrate solution. The titration is carried out in neutral cold solution using precipitation or adsorption indicator. This method is known as Mohr’s method of estimation. 169.89 g AgNO3 ≡ 58.5 g NaCl So, 1000 ml 1 N solution of AgNO3 ≡ 58.5 g NaCl ∴ 1 ml 1 N solution of AgNO3 ≡ 0.085 g NaCl

Chemicals :

(i) Sodium chloride sample

(ii) 0.05 N silver nitrate solution

(iii) 5% potassium chromate solution

Procedure :

1. Weigh accurately 0.5 g of sodium chloride on a watch glass. Record the weight as 'W' g.

2. Transfer this weighed sample in 100 ml beaker. Rinse the watch glass. Dissolve the crystals in about 50 ml distilled water.

3. Transfer the resulting solution carefully in 100 ml volumetric flask. Rinse the beaker with distilled water. Dilute the solution upto the mark with distilled water. Take out all the solution in a beaker.

4. Fill the burette No. 1 with standard 0.05 N solution of AgNO3.

5. Fill the burette No. 2 with sodium chloride solution.

6. Take 9 ml sodium chloride solution in a 100 ml conical flask.

7. To this add about 3 - 4 drops of 5% potassium chromate solution.

8. Titrate this solution with standard silver nitrate solution from burette No. 1 with swirling the conical flask until the brick red colour formed by each drop begins to disappear more slowly indicating the most of the chloride has been precipitated.

9. Make the addition dropwise towards the end and finish the titration when faint brick red colour appears in the precipitate which persists even after brisk shaking. The precipitate will also coagulate suddenly at this point. Record the burette reading and call this as X1 ml.

10. To the same solution in the conical flask add 1 ml solution of NaCl from burette No. 2. Shake the flask, brick red colour disappears.

11. Add AgNO3 solution from burette No. 1 with swirling the conical flask. Make the addition dropwise and finish the titration when faint brick red colour appears in the


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precipitate which persist even after brisk shaking. Record the burette reading and call this as X2 ml.

12. To the same solution in the conical flask add 1 ml solution of NaCl from burette No. 2. Shake the flask, brick red colour disappears.

13. Add AgNO3 solution from burette No. 1 with swirling the conical flask. Make the addition dropwise and finish the titration when faint brick red colour appears in the precipitate which persist even after brisk shaking. Record the burette reading and call this as X3 ml.

Note :

1. The above method can be employed for the determination of KCl, NH4Cl, KBr in neutral solutions.

2. Burette with rubber tube should not be used as the rubber may react with silver nitrate solution.

3. Silver nitrate addition should be very slow. It should not be added in large amounts at a time.

4. All solutions required for this practical should be prepared in distilled water only. Rinse all the apparatus required with distilled water.

Observation Table :

Burette 1 : Standard 0.05 N AgNO3 solution

Burette 2 : Diluted solution of NaCl

Indicator : 5% potassium chromate solution

End point : Colourless to faint brick red

Obs. No. Burette 1 Y ml

Burette 2 V2 ml

1.

2.

3.

X1 =

X2 =

X3 =

9.0

9 + 1 = 10.0

10 + 1 = 11.0

Mean burette reading 1

= X1 + X2 + X3

3 = X ……. ml

Mean burette reading 2

= 9 + 10 + 11

3 = 10 ml

Calculations :

1. To calculate amount of NaCl in the solution :

1 ml 1 N solution of AgNO3 ≡ 0.0585 g NaCl


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∴ X ml 0.05 N solution AgNO3 ≡ 0.0585 × 0.05 × X

1

0.00292 × X = Y g of NaCl

∴ 10 ml diluted NaCl solution = Y g of NaCl

∴ 100 ml diluted NaCl solution = Y × 10 g NaCl

= Z g NaCl

The amount of NaCl in the solution = Z = …….. g. 2. To calculate purity of NaCl sample : W g sample = Z g NaCl

∴ 100 g sample = Z × 100

W = ………. % NaCl

= Percentage purity of NaCl Result Table :

1. Weight of NaCl sample taken W = ………. g 2. Amount of AgNO3 required for 10 ml

diluted NaCl solution X = ………. ml

3. Percentage purity of NaCl = ………. %

VIVA VOCE 1. What will happen if some one use tap water for preparation of solutions ? 2. Why the burette with rubber tube is not advisable ? 3. In which class will you classify the Mohr’s method of titration ? 4. What are adsorption indicators ? 5. Is it possible to have a situation like Z > W ? Justify your answer.

6. What impurities do you expect in the NaCl sample collected from sea water ?

Experiment No. 4 Analysis of Brass

Aim : To determine the amount of copper from brass iodometrically. Principle : Brass is an alloy which consists of mostly copper (50 to 90%) and zinc (20 to 40%) along with small amounts of lead (0 to 2 %), tin (0 to 6 %) and iron (0 to 1%). The experiment consists of dissolving a known quantity of brass in nitric acid, removing the nitrate by fuming with sulphuric acid.


167

Copper from the brass may be iodometrically determined while zinc is determined gravimetrically as Zn2P2O7. In iodometric determination, Cu ions liberate iodine from KI and the liberated I2 is titrated against standard thiosulphate solution. When the brass sample is dissolved in nitric acid, the copper is brought into solution in the form of cupric ions. The Cu2+ ions are reducible by iodide. Then the Cu2+ can be titrimetrically determined by the iodometric method. 2Cu2+ + 4I− → 2CuI + I2

2S2O−3 + I2 → S4O

−6 + 2I−

Cu ≡ I ≡ S2O−3

Chemicals : (i) K2Cr2O7 crystals (ii) Na2S2O3 solution (approximate 0.05 N) (iii) 5% KI solution (iv) Starch indicator (v) 1 : 1 HNO3 (vi) 2 N NaOH (vii) 2 N CH3COOH (viii) Brass sample. Procedure : Part A : Preparation of standard solution of 0.05 N K2Cr2O7 (Eq. wt. 49) : 1000 ml 1 N K2Cr2O7 solution ≡ 49 g of K2Cr2O7 ∴ 100 ml 0.05 N K2Cr2O7 solution ≡ 0.245 g of K2Cr2O7

1. Weigh accurately 0.245 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. Stir the solution with glass rod and dissolve the crystals.

3. Pour this solution in a clean 100 ml volumetric flask. Add carefully little water and dilute upto the mark.

4. Take out the solution in a clean beaker.

5. This becomes 0.05 N K2Cr2O7 solution. This solution is used for standardization of Na2S2O3 solution.

Part B : Standardization of Na2S2O3 solution :

1. Fill the burette No. 1 with 0.05 N (approximate) Na2S2O3 solution.

2. Fill the burette No. 2 with 0.05 N K2Cr2O7 solution.

3. Take 9 ml K2Cr2O7 solution from burette No. 2 in a 100 ml conical flask or stoppered bottle. Call this volume as V2 ml.


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4. Add 3 ml concentrated HCl and add 10 ml 5% KI solution. Shake the flask/bottle well and keep it for five minutes.

5. Titrate the liberated iodine with Na2S2O3 solution till a pale yellow colour appears. Then add 1 - 2 ml starch indicator. The solution will become blue in colour. Continue the titration.

6. The end point of the 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 flask/bottle add 1 ml of 0.05 N K2Cr2O7 from burette No. 2 (need not add any indicator now). The solution turns blue.

8. Titrate this against Na2S2O3 solution from burette No. 1 till the colour of the solution changes from blue to green. Record this reading as X2 ml.

9. To the same flask/bottle add 1 ml of 0.05 N K2Cr2O7 by burette No. 2 (do not add any indicator). The solution turns blue.

10. Titrate this against Na2S2O3 solution from burette No. 1 till the colour of the solution changes from blue to green. Record this reading as ‘X3' ml.

11. From the burette readings X1, X2 and X3 find out the normality of Na2S2O3 solution. Part C : Preparation of brass solution :

1. Weigh accurately about 0.2 g of given brass sample on a watch glass. Call this as W g. 2. Transfer it carefully to 100 ml conical flask. Add about 15 ml of 1:1 HNO3 by a

measuring cylinder. Cover the flask with stem cut glass funnel and heat it on wire gauze at low flame.

3. Dissolve the alloy (brass sample) completely and then cool the solution. To this cold solution add 10 ml water and 5 ml conc. H2SO4 to remove the excess of HNO3. Boil the solution to eliminate brown fumes. (Heat the solution till white fumes are evolved).

4. If the solution is clear, transfer it carefully to 100 ml volumetric flask and dilute upto the mark with distilled water. (If the solution is not clear, then filter it through Whatmann filter paper No. 42. Wash the residue with water. Collect the filtrate and washings and dilute to 100 ml).

5. Take out whole solution in a beaker and stir well. Part D : Iodometric determination of copper :

1. Fill the burette No. 1 with standardized thiosulphate solution. 2. Fill the burette No. 2 with diluted solution of brass. Take 10 ml of this solution in a

stoppered bottle. 3. 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 bottle well. Add 2 to 3 drops of acetic acid in excess.


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4. Add 10 ml of 5% KI solution by means of measuring cylinder. 5. Titrate the liberated iodine with Na2S2O3 solution from burette No. 1 till the colour of

the solution becomes pale yellow. 6. Then add 1 to 2 ml freshly prepared starch indicator. The solution becomes blue. 7. Continue the titration. The end point of the titration is recorded when colour changes

from dark blue to colourless. Record this reading as Y1 ml. 8. Take two more readings by repeating step No. 2 to 7 and record the constant burette

reading. 9. From this reading determine the amount of copper in the given sample of brass.

Observation Table : Part B : Standardization of Na2S2O3 solution To find : Exact normality of Na2S2O3 solution Burette 1 : 0.05 N (approximate) Na2S2O3 solution Burette 2 : 0.05 N (exact) K2Cr2O7 Indicator : Starch solution (freshly prepared) End point : Blue to green. Equations : 1. K2Cr2O7 + 6KI + 14HCl → 8KCl + 2CrCl3 + 7H2O + 3I2 ↑ 2. 2Na2S2O3 + I2 → Na2S4O6 + 2NaI


Burette 2 V2 ml

Normality

N = 0.05 × V2

X

1. X1 = ….. 9.0 N1 =

0.05 × 9X1

2. X2 = ….. 9 + 1 = 10.0 N2 =

0.05 × 10X2

3. X3 = ….. 10 + 1 = 11.0 N3 =

0.05 × 11X3

Mean normality N4 = N1 + N2 + N3

3 = .......... N

Exact normality of Na2S2O3 = N4 = ………. N

Part D : Estimation of Copper To find : Amount of copper in the brass sample


Burette 1 : Standardized Na2S2O3 solution Burette 2 : Diluted brass solution Indicator : Starch solution (freshly prepared) End point : Blue to colourless. Reactions : 2CuSO4 + 4KI → 2CuI2 + 2K2SO4 2CuI2 → Cu2I2 + I2 2CuSO4 + 4KI → Cu2I2 ↓ + 2K2SO4 + I2 ↑ Liberated I2 is titrated against Na2S2O3. I2 + Na2S2O3 → 2NaI + Na2S4O6

Readings Pilot Reading ml

Burette Reading, ml Constant B.R.

ml I II III

Final Initial Difference

0.00

0.00

0.00

0.00

‘Y’ ml

Calculations : 1. To calculate amount of copper in the solution : 2CuSO4 ⋅ 5H2O ≡ I2 ≡ 2Na2S2O3 ⋅ 5H2O = 2Cu ∴ Na2S2O3 ⋅ 5H2O ≡ Cu. 248 ≡ 63.5 g 100 ml 1 N Na2S2O3 ≡ 63.5 g Cu

∴ ‘Y’ ml N4 Na2S2O3 ≡ Y × N4 × 63.5

1000 × 1 = A g of Cu

10 ml diluted solution of brass contains A g Cu ∴ 100 ml diluted solution of brass contains A × 10 = B g Cu 2. To calculate percentage of copper in the brass sample : W g of brass sample = B g of copper ∴ 100 g of brass sample = ?

= 100 × B

W = ………. % of copper

Result Table :

1. Weight of the brass sample taken W = ………. g 2. Exact normality of Na2S2O3 solution N4 = ………. N 3. Volume of 0.05 N Na2S2O3 solution

required to titrate 10 ml diluted solution of brass

Y = ………. ml

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171

4. Amount of Cu in the 100 ml solution B = ………. B g 5. Percent of Cu in brass alloy = ………. %

VIVA VOCE 1. Why always excess of KI is used for iodometric titrations ? 2. Why should fresh solution of starch be used ? 3. Write the chemical equations involved in this analysis. 4. What are the major and minor constituents of brass alloy ? 5. What is the principle of estimation of Cu in brass ? Explain with chemical equations. 6. What is iodometry and iodimetry ? 7. What is the role of K2Cr2O7 solution in this titration ?

Experiment No. 5 Aim : To determine the amount of Phosphate (PO4)3− from fertilizer sample by volumetric method. Principle : When solution of phosphate is treated with excess ammonium molybdate in the presence of concentrated nitric acid then formation of yellow precipitate of ammonium molybdo-phosphate [(NH4)3 ⋅ PO4 ⋅ 12 MoO3] take place. Precipitate is filtered using Whatman filter paper - 42 and then washed till free from acid. The washed precipitate of ammonium molybdophosphate gives rise to the formation of Na2MoO4, Na2HPO4, H2O and NH3. This solution is boiled to evolve excess of ammonia present in the solution and then excess NaOH is back titrated with standard HCl solution using phenolphthalein indicator. From amount of NaOH required for dissolution of ammonium molybdophosphate (phosphate as P2O5) in the precipitate was calculated using the following equation. Equation : 1000 ml 1 N NaOH ≡ 3.088 g P2O5 Chemicals : (i) Conc. HNO3

(ii) 1 N NaOH (iii) 0.1 N HCl (iv) 3% ammonium molybdate (v) 0.1 N potassium hydrogen phthalate Apparatus : (i) Beakers (ii) Conical flask


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(iii) Volumetric flask (iv) Burette (v) Funnel (vi) Whatman paper - 42 Procedure : Part A: Preparation of standard solution of potassium hydrogen phthalate, KHP (0.1 N): 1. Weigh accurately 2.045 g of A.R. grade potassium hydrogen phthalate crystals on

watch glass. 2. Transfer it in a clean beaker and rinse the watch glass with distilled water. 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.

5. This is 0.1 N KHP solution. Use this solution for standardization of given NaOH solution.

Part B : Standardization of NaOH solution :

1. Fill the burette No. 1 with 0.1 N KHP solution.

2. Fill the burette No. 2 with diluted NaOH (10 ml 1 N NaOH is diluted to 100 ml) solution (0.1 N approximate).

3. Take 9 ml NaOH solution from burette No. 2 in a 150 ml conical flask.

4. Add 2 - 3 drops of phenolphthalein indicator. The solution becomes pink in colour.

5. Titrate this solution against 0.1 N KHP solution from burette No. 1, till pink colour just disappears. Call this burette reading as X1’ ml.

6. To the same flask add 1 ml of 0.1 N NaOH from burette No. 2. The solution turns pink.

7. Titrate this against 0.1 N KHP solution from burette No. 1 till the colour just disappears. Record this reading as ‘X2’ ml.

8. To the same solution in conical flask add 1 ml of NaOH solution from burette No. 2. The solution turns pink.

9. Titrate this solution against 0.1 N KHP from burette No. 1 till the colour of solution disappears. Record this reading as ‘X3’ ml.

10. From the burette readings X1, X2 and X3, find out the normality of NaOH solution as N1, N2 and N3.

Part C : Precipitation of phosphate as ammonium molybdophosphate from fertilizer sample :


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1. Weigh accurately about 0.2 - 0.3 g of fertilizer (DAP) sample on a watch glass. Call this as ‘W’g.

2. Transfer it carefully in 100 ml conical flask. Add about 50 ml distilled water and 2 ml conc. HNO3.

3. If necessary, filter and wash the residue with 10 ml distilled water.

4. Heat the filtrate upto 40°C, to it add 3 ml conc. HNO3 and then add 3% ammonium molybdate solution, till complete precipitation of phosphate take place.

5. Then add 5 ml excess 3% ammonium molybdate.

6. Keep this precipitate in hot condition for 20 - 30 minutes. Filter this precipitate through Whatman filter paper 42 and wash with 1% ammonium nitrate, till free from acid.

Part D : Back titration : Estimation of phosphate, PO3−4 :

1. Dissolve the above precipitate by dropwise addition of 20 ml 1 N NaOH solution on the filter paper. After complete dissolution of the precipitate, wash the Whatman filter paper with 20 ml hot distilled water.

2. Boil this solution for 10 minutes, so that complete evolution of NH3 takes place from the solution.

3. Cool the solution to room temperature and then dilute this to 100 ml in volumetric flask.

4. Fill the burette No. 1 with 0.1 N HCl solution. 5. Fill the burette No. 2 with above diluted solution. 6. Take 10 ml of diluted solution of fertilizer sample from burette No. 2 in a 100 ml

conical flask. 7. Add 2 - 3 drops of phenolphthalein indicator. The solution becomes pink in colour. 8. Titrate this solution against 0.1 N HCl from burette No. 1, till pink colour disappears. 9. This is the end point of titration. Record this reading as first burette reading. 10. Take two more readings by repeating step 6 to 8 and record the constant burette

reading as ‘Y’ ml. Part E : Blank titration : 1. Take 0.1 N HCl solution in burette No. 1. 2. Take diluted solution of NaOH (Dilute 20 ml 1 N NaOH to 100 ml in volumetric flask)

in burette No. 2. 3. Take 10 ml of this NaOH solution in a 100 ml conical flask from burette No. 2. 4. Add 2 - 3 drops of phenolphthalein indicator. The solution becomes pink in colour.


174

5. Titrate this solution against 0.1 N HCl from burette No. 1, till pink colour disappears. 6. This is the end point of titration. Record this reading. 7. Take two more readings by repeating step 3 to 5 and record the constant burette

reading as ‘Z’ ml.

Observation Table :

Part B : Standardization of NaOH solution

Burette 1 : 0.1 N KHP (Potassium hydrogen phthalate)

Burette 2 : Diluted NaOH solution

To find : Exact normality of NaOH solution

Indicator : Phenolphthalein

End point : Pink to colourless


Burette 2 ‘V’ ml Normality, N =

0.1 × VX

1. X1 = ….. 9.0 N1 =

0.1 × 9X1

2. X2 = ….. 9 + 1 = 10.0 N2 =

0.1 × 10X2

3. X3 = ….. 10 + 1 = 11.0 N3 =

0.1 × 11X3

Mean Normality, N4 = N1 + N2 + N4

3 = …… N

Part E : Blank titration

Burette 1 : 0.1 N HCl

Burette 2 : Diluted NaOH solution




Burette Readings, ml Constant Burette Reading

(C.B.R.) ml

I II III


0.00

0.00

0.00

0.00

Y = …… ml


175

Part D : Back titration : Estimation of phosphate, PO3−4

Burette 1 : 0.1 N HCl

Burette 2 : Diluted molybdate solution




Burette Readings, ml Constant Burette Reading

(C.B.R.) ml

I II III


0.00

0.00

0.00

0.00

Z = …… ml

Calculations :

(1) To find out the amount of phosphate (PO3−4 ) from given fertilizer sample:

1000 ml 1 N NaOH = 3.088 g P2O5 ∴ (Blank - Back) i.e. (Y − Z) ml ‘N4’ N NaOH = ? g P2O5

= (Y − Z) × N4 × 3.088

1 × 1000

= A = …… g of P2O5 10 ml diluted solution = ‘A’ = g of P2O5 ∴ 100 ml diluted solution = A × 10 = ‘B’ = g of P2O5 = B = …… g of P2O5 (2) To calculate the percentage of phosphate in the fertilizer sample : “W’ g of fertilizer sample = ‘B’ g of P2O5 ∴ 100 g of fertilizer sample = ?

= 100 × B

W

= …… % of P2O5 Result Table :

1. Weight of fertilizer sample taken W = ………. g

2. Exact normality of NaOH solution N4 = ………. N

3. Amount of phosphate in the 100 ml solution

B = ………. g

4. Percent of phosphate in the given fertilizer = ………. %


176

sample in the form of P2O5

− − −

176

(C) Inorganic Preparations (Any Four)

Introduction :

Inorganic preparation includes preparation of a double salt and three coordination complexes. Coordination complexes differ from the double salts. The combination of two different single salts which crystallizes together as a single substance is known as double salt. Double salt does not loose identity of individual ions in solid and in aqueous solution (i.e. they give characteristic tests for their individual ions).

Coordination complexes consist of a complex ion formed by combination of metal ion and ligands. Here ligands are bounded to central metal ion by a coordinate linkage. These complex ions may have positive or negative charge and are soluble in water. In complexes, metal ions lose their identity in solution and solid state. These complexes do not undergo dissociation in aqueous solution.

Experiment No. 1 Preparation of Hexamminenickel(II) Chloride [Ni(NH3)6]Cl2

Aim :

To prepare hexamminenickel(II) chloride.

Principle :

The nature as well as strength of the ligand is one of the important factors that determines the formation and stability of a metal complex. A stronger ligand can replace a weaker ligand from a metal centre. Therefore, ammonia can easily replace OH− or H2O from a metal centre (as NH3 is a stronger ligand than OH− or H2O). Similarly, ethylenediamine can replace ammonia because it is more stronger than ammonia.

[Ni(NH3)6]Cl2 is an octahedral complex having purple colour. When nickel chloride hexahydrate, NiCl2 ⋅ 6H2O is dissolved in water, hexaquo complex of nickel is formed. Upon addition of ammonia initially there may be formation of nickel hydroxide (indicated by formation of green ppt.) which further reacts with ammonia to give ammine complex. This reaction depends on the stoichiometry of the reactants and if excess ammonia is not used following mixtures of products with varying composition may be obtained. For example,

[Ni(NH3)5(H2O)]Cl2,

[Ni(NH3)3(H2O)3]Cl2,

T.Y.B.Sc. Practical Chemistry (Inorganic Practicals) Inorganic Preparations

[Ni(NH3)5Cl]Cl etc. NH3

NH3

Ni

H N3 NH3

H N3NH3

Cl2

Structure of [Ni(NH3)6]2+Cl2

The complex can be prepared by taking nickel chloride hexahydrate as a water solution to which excess of ammonia is added. The colour of the solution changes from pale dark green to purple. Upon cooling the reaction mixture in an ice bath [Ni(NH3)6]2+, complex will be precipitated out as purple coloured crystals. The structure of the complex is as shown in the structure above.

Chemicals :

(i) Nickel chloride hexahydrate NiCl2 ⋅ 6H2O.

(ii) Liquor ammonia (sp. gr. 0.9).

Apparatus :

100 ml beakers, Buchner funnel, glass rod, ice bath.

Procedure :

1. Weigh about 1.5 g of nickel chloride hexahydrate. Transfer it in a 100 ml beaker.

2. Dissolve the solid in minimum quantity of water (about 4 ml).

3. Take 4 ml of concentrated aqueous ammonia in another beaker and fill it in burette.

4. Add the ammonia slowly to a rapidly stirred solution of the nickel chloride using burette.

5. Colour of the solution is found to change from pale green to intense violet (if not, add more ammonia).

6. Allow the solution to stand at room temperature for five minutes, and then cool it in an ice bath without disturbance for about 20 minutes.

7. If violet crystals of [Ni(NH3)6]Cl2 are not formed, try to crystallize it by gently scratching the sides of the beaker with a glass rod.

8. Filter the crystals and wash the product with concentrated ammonia solution (5 ml).

9. Dry the crystals by pressing it between filter papers. (Do not expose the product to IR lamp).

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178

10. Record the yield of the dried product and calculate % yield.

Chemical Reactions :

NiCl2 ⋅ 6H2O + H2O → [Ni(H2O)6]Cl2

[Ni(H2O)6]Cl2 + 6NH3 → [Ni(NH3)6]Cl2 + 6H2O

Observations :

Weight of the dried product = X = ………. g

Calculations :

(i) Theoretical yield of the complex :

NiCl2 ⋅ 6H2O = [Ni(NH3)6]Cl2

237.6 g of salt = 231.6 g of complex

∴ 1.5 = ?

= 1.5 × 231.6

237.6 = 1.46 g

∴ Theoretical yield of the complex = 1.46 g

(ii) Percent practical yield of the complex :

Theoretical yield of the complex is 1.46 g.

∴ 1.46 g product = 100% yield

∴ X g product = ?

= X × 100

1.46 = ………… % yield of the complex

Result Table :

1. Practical yield of the complex = X = ……….. g

2. Theoretical yield of the complex = 1.46 gm = ………. 1.46 gm

3. % Practical yield of the complex = ……….. %

4. Colour of the complex Purple

VIVA VOCE 1. What are the factors that decide the strength of ligand ? 2. Arrange the following ligands in the order of increasing strength :

I−, OH−, C2O2−4 , en, CN−, NCS−, H2O, Py, NH3, CO.

3. How will you determine the formation of correct complex ? 4. What is your opinion about the stability of the complex ?


5. Explain the change in colours observed during this preparation.

Experiment No. 2 Preparation of Potassium trioxalatoferrate(III) trihydrate,

K3[Fe(C2O4)3] ⋅ 3H2O Aim : To prepare potassium trioxalatoferrate(III) trihydrate and its purity determination by estimation of oxalate using permanganometry. Principle : Iron forms octahedral complexes with monodentate and bidentate ligands. The oxalate ion is a bidentate ligand which acts as a chelating agent and forms a stable ring around the central metal ion in a metal chelate. This ring structure is called a chelate and formation of such a ring structure is called chelation. Chelated complexes are more stable than simple complexes formed by monodentate ligands, because dissociation of complex involves breaking of two bonds rather than one bond per ligand molecule. The preparation of K3[Fe(C2O4)3] involves two steps. Step one is a preparation of ferrous oxalate from ferrous ammonium sulphate and step two is the conversion of ferrous oxalate to potassium trioxalato ferrate (III) trihydrate. During this conversion hydrogen peroxide works as oxidizing agent i.e. to oxidize Fe(II) to Fe(III). It is water soluble complex. Thus, on addition of alcohol to the solution containing complex, formation of green crystals of trioxalato ferrate (III) complex takes place in solution.

C

C

O:O

O O: Structure of C2O

2−4 ligand

Oxalate ion (C2O2−4 ) is a bidentate ligand i.e. one ion of it forms two coordinate bonds

with central metal ion as shown in the figure above. In the present complex three such oxalate ions form a stable complex (chelate) with Fe3+ ion as shown in structure below.

179


O

CO

C

O

O

Fe

O

CO

C

O

O

O

O

C

C

O

O

K3 3H O2

Structure of K3[Fe(C2O4)3] ⋅ 3H2O

Chemicals : (i) Ferrous ammonium sulphate. (ii) Oxalic acid. (iii) Potassium oxalate monohydrate. (iv) Hydrogen peroxide. (v) Ethanol. (vi) Acetone. Apparatus : 100 ml beaker, Buchner funnel, conical flask, burettes. Procedure : Step I : Preparation of iron (II) oxalate : 1. Dissolve 3.75 g of ferrous ammonium sulphate in 15 ml warm distilled water. Acidify

the resulting solution by adding 3-4 drops of concentrated sulphuric acid so that solution becomes clear.

2. Add 20 ml 10% oxalic acid solution with constant rapid stirring to the above clear solution.

3. Heat the reaction mixture gently and cautiously to boiling and then allow the yellow precipitate of ferrous oxalate to settle.

4. Remove the precipitate by filtration on Buchner funnel. Wash it thoroughly with hot distilled water and finally with acetone.

5. Allow the product to dry and then use it for the preparation of potassium trioxalatoferrate(III) trihydrate.

Step II : Preparation of potassium trioxalatoferrate(III) trihydrate : 1. Add the above precipitate of ferrous oxalate in a warm solution of potassium oxalate

(2.5 g in 15 ml of distilled water).

180


181

2. Add slowly 30% hydrogen peroxide (about 6.0 ml). Stir the solution and maintain its temperature upto 40°C. At this stage ferric hydroxide is formed. Ferrous ion is oxidized to ferric ion.

3. Dissolve the precipitate by heating the solution to boiling and add 3 ml 10% oxalic acid solution. Further add small amount of 10% oxalic acid (1-2 ml) dropwise until precipitate just dissolves. (During addition of oxalic acid solution should be maintained near to boiling.)

4. Filter the hot solution, collect the filtrate in a china dish and boil the solution to reduce the volume upto 10-15 ml. Stop heating, cool the solution and add 15 ml ethyl alcohol to the filtrate. Dissolve any crystal formed by gentle heating on water bath (Caution : Alcohol is inflammable) and keep the solution in dark for crystallization. Now keep the dish in powdered ice for 5 minutes to complete crystallization.

5. Remove the bright green crystals formed by filtration on Buchner funnel. Wash with 1 : 1 alcohol and finally with acetone.

6. Dry the product in air (complex is photosensitive and should not be exposed to light) and record the yield. Call this as X g.

Chemical Reactions : Fe2+ + H2C2O4 ⎯→ FeC2O4 + 2H+ 6FeC2O4 + 6K2C2O4 + 3H2O2 ⎯→ 4K3[Fe(C2O4)3] + 2Fe(OH)3 2Fe(OH)3 + 3H2C2O4 + 3K2C2O4 ⎯→ 2K3[Fe(C2O4)3] ⋅ 3H2O + 3H2O Observations :

Weight of the dried complex = X = .......... g

Calculations : 1. Theoretical yield of the complex : (NH4)2Fe(SO4)2 ⋅ 6H2O ≡ K3[Fe(C2O4)3] ⋅ 3H2O 391.85 g of salt ≡ 490.85 g of complex ∴ 3.75 = ?

3.75 × 490.85

391.85 = 4.7 g

∴ Theoretical yield of the complex = 4.7 g 2. Percent practical yield of the complex : Since, 4.7 g of the complex = 100% yield ∴ X g of the complex = ?

X × 100

4.7 = ……….. %

Result Table :

1. Colour of the complex Bright green


2. Practical yield of the complex ‘X’ = ………. g

3. Percent practical yield of the complex = ………. %

VIVA VOCE 1. What are the different types of ligands ? 2. How oxalate ion interact with metal ion ? 3. What do you mean by chelate ? Comment on stability of chelate. 4. What are polydentate ligands ? 5. What are other reagents that can be used for the standardization of KMnO4 ? 6. Explain the mechanism of self-indicator action of KMnO4. 7. What is the role of ethyl alcohol in this preparation ? 8. What is the type of titration in this experiment ? 9. What are different types of titrations ?

Experiment No. 3 Preparation of Tetramminecopper(II) sulphate monohydrate,

[Cu(NH3)4]SO4 ⋅ H2O Aim : Preparation of tetramminecopper(II) sulphate monohydrate [Cu(NH3)4]SO4⋅H2O and estimation of copper iodometrically. Principle : When the aqueous ammonia is added in a solution of copper sulphate, ammonia molecules coordinate with copper ion (Cu2+ metal ion) and forms a square planar complex [Cu(NH3)4]SO4 ⋅ H2O as shown in structure below. The complex is soluble in water but insoluble in alcohol. Therefore, the complex is precipitated by addition of ethanol to the aqueous solution of the complex and cooling in ice bath. After cooling formation of dark blue coloured crystals takes place in solution which are removed by filteration. As complex is soluble in water, its solution can be prepared easily. Hence, Cu2+ ions in the solution can be estimated by iodometry.

H N :3 Cu : NH3

NH3:

NH3

:

SO H O4 2

Structure of [Cu(NH3)4]SO4 ⋅ H2O

Chemicals : 182


183

(i) Copper sulphate = 1.5 g. (ii) Liquor ammonia = 5 ml. (iii) Rectified spirit = 5 ml. Procedure : 1. Add 1.5 g finely powdered copper sulphate in 10 ml distilled water in a beaker and

dissolve it completely (warm if necessary). 2. Add about 5 ml liquor ammonia (1 ml at a time) to the cooled solution of copper

sulphate with constant stirring. First a bluish white precipitate is formed which dissolves further on addition of ammonia and finally the solution becomes dark blue. Add little excess of liquor ammonia, if necessary.

3. Cool this solution in a water bath and add about 5 ml rectified spirit slowly with the help of burette with constant stirring.

4. Keep the beaker in ice cold water and continue the stirring. Since the crystals (dark blue) of tetramminecopper(II) sulphate are insoluble in alcohol, they get crystallize out and settle at the bottom.

5. If the supernatant solution is dark blue in colour then add more rectified spirit with stirring and cool it well.

6. Filter the complex on Buchner funnel. Wash the crystals with little amount of ethyl alcohol. (Never wash the crystals with water as they are soluble in water).

7. Dry the crystals in a desiccator, weigh and record the yield. Chemical Reaction :

CuSO4 ⋅ 5H2O + 4NH3 ⎯→ [Cu(NH3)4]SO4 ⋅ H2O + 4H2O Observation :

Weight of the dried product = X = ……….. g

Calculations : 1. Theoretical yield of the complex : CuSO4 ⋅ 5H2O ≡ [Cu(NH3)4]SO4 ⋅ H2O 249.6 g of salt ≡ 245.6 g of complex ∴ 1.5 g of salt ≡ 1.476 g of complex ∴ Theoretical yield of complex = 1.476 g 2. Percent practical yield of the complex : Theoretical yield of the complex is 1.476 g ∴ 1.476 g of the complex ≡ 100% yield

∴ ‘X’ g of product ≡ ? = X × 1001.476 = ……….. % yield

∴ % practical yield of complex = ……….. g Result Table :


184

1. Colour of the complex Dark blue

2. Practical yield of the complex X = ………. g

3. % yield of the complex = ………. %

VIVA VOCE 1. What do you mean by coordinate bond and coordinate complex ? 2. Why coordinate complexes are coloured in nature ? 3. Draw the structure of [Cu(NH3)4]SO4 ⋅ H2O. 4. What is the difference between double salt and coordinate complex ? 5. Explain the hybridization in this complex and hence explain the geometry. 6. What are the types of ligand ?

Experiment No. 4 Aim : Preparation of Manganese(III) acetylacetone, [Mn(acac)3]. Principle : Manganese shows variable oxidation states in compounds (+II to +VII). Manganese(III) complexes are comparatively stable. Manganese forms octahedral complexes with acetylacetone ligand. The acetylacetone is a bidentate ligand which acts as a chelating agent and forms a stable ring around the central metal ion in a metal chelate. Chelates are more stable than simple complexes. Complex can be prepared by taking manganese(II) chloride (MnCl2 ⋅ 4H2O) as a water soluble salt. This Mn(II) chloride solution is oxidised with KMnO4 in the presence of acetyl acetone producing the brown coloured [Mn(acac)3] complex.

Chemicals :

(i) Manganese(II) chloride (MnCl2 ⋅ 4H2O) = 1.05 g

(ii) Sodium acetate (CH3COONa) = 2.80 g

(iii) Acetylacetone (CH3COCH2COCH3) = 4 ml

(iv) Potassium permanganate (KMnO4) = 0.22 g

Apparatus :

100 and 50 ml beakers, glass rod, micro-pipette etc.

Procedure :

1. Weigh about 1.05 g of manganese(II) chloride tetrahydrate on watch glass. Transfer it in a 100 ml beaker. Dissolve the solid in 60 ml water.


2. Weigh 0.22 g KMnO4 on a watch glass. Dissolve it in 25 ml distilled water in 50 ml beaker.

3. Weigh 2.8 g sodium acetate trihydrate on a watch glass. Dissolve it in 15 ml distilled water in a separate 50 ml beaker.

4. Add the acetylacetone (4 ml) slowly by constant stirring in manganese chloride solution using micro-pipette or dropper.

5. Add dropwise solution of KMnO4 to the above solution with constant stirring.

6. Stirring is continued for another 5 minutes. Here KMnO4 acts as an oxidising agent.

7. Add solution of sodium acetate dropwise with constant stirring. (Sodium acetate maintains the pH of the solution upto 5).

8. Heat the above solution near to boiling (do not boil) for 10 minutes.

9. Cool the reaction mixture to room temperature. 10. Filter the brown crystals and the product distilled water. 11. Dry the crystals by pressing between filter papers. Then transfer the solid into a

porcelain dish. Dry it in under IR lamp for at least 30 minutes. 12. Record the yield of the complex and calculate % yield of [Mn(acac)3] complex. Reaction :

4MnCl2 + KMnO4 + 15 CH3.CO.CH2COCH3 pH = 5⎯⎯⎯→ = 7 HCl + KCl + 5Mn(CH3COCHCOCH3)3 + 4H2O

Structure of the Complex :

OO

OO

Mn

C CH3

CH

CH3C

O

CC

CH3

CH C

O

CH3C C

H

C CH3CH3

OO

OO

Mn

O

O

acac

acac

acac

OR

Acetyl acetone = acac Fig. 1

Calculations : (i) Theoretical yield : Reactant Product (complex) 4 MnCl2 ⋅ 4 H2O ≡ 5[Mn(CH3COCHCOCH3)3] 792 g ≡ 1760 g 792 g MnCl2 ⋅ 4H2O = 1760 g of [Mn(acac)3] complex

185


186

∴ 1.05 g MnCl2 ⋅ 4H2O = 2.33 g of complex (ii) Practical yield of the complex = ‘W’ = …… g complex (iii) % Practical yield of the complex : 2.33 g of complex = 100% yield

∴ ‘W’ g of complex = W × 100

2.33 % yield

= …… % Practical yield Result Table :

1. Colour of the complex Dark brown

2. Practical yield of the complex W = ………. g

3. Theoretical yield 2.33 g

4. % Practical yield of the complex …… %


Preparation of Tris (Thiourea) Copper(I) chloride, [Cu(Thiourea)3]Cl.

Principle :

Cu(I) forms stable octahedral complex with thiourea. The complex has a Cu(I) as a central metal ion is surrounded by three coordinating molecules of thiourea forming octahedral complex. This is a chelate complex. Thiourea forms coordinate bonds to Cu(I) ion through nitrogen atom. Copper(I) compounds and complexes are usually colourless. The soluble compounds of Cu(I) are unstable in aqueous solution, since they disproportionate to Cu2+ and Cu ions as

2Cu+ → Cu2+ + Cu

Chemicals :

(i) Thiourea, [(NH4)2CS] = 3.2 g

(ii) Copper powder or turnings = 0.7 g

(iii) conc. HCl = 6 ml

(iv) Acetone.

Apparatus : 100 ml beakers, glass rod, water bath, etc.

Procedure :

1. Weigh 3.2 g of thiourea. Transfer it in 100 ml beaker.

2. Dissolve the solid in about 15 ml of distilled water.


3. Add to the above solution 0.7 g copper turnings or powder, then add 3 ml of conc. HCl.

4. Heat the above solution on water bath. Copper turnings dissolve with liberation of hydrogen gas. Add water to maintain constant level in the flask. (Do not boil the solution.)

5. Then filter the above solution through a funnel using ordinary filter paper while hot and allow it to cool slowly. White opaque crystals will form.

6. Filter the crystals using Buchner funnel and wash with little acetone.

7. Allow the crystals to air dry and record the yield. Call it as ‘W’ g.

8. Calculate % practical yield of the complex.

Chemical Reaction :

Cu + 3(NH2)2CS conc. HCl⎯⎯⎯⎯→

H2O [Cu((NH2)2CS)3]Cl + H+

Structure of the complex :

Cu

NH2 C S

NH2H2N

NH2H2N

S C

H N2 C S

Cl

Fig. 1

Calculations : (i) Theoretical yield of the complex : Starting material Finished product Cu(powder) ≡ [Cu((NH2)2CS)3] ⋅ Cl 63.54 g 263.5 g 63.54 g of Cu ≡ 263.5 g complex ∴ 0.7 g of Cu = 2.9 g complex (ii) Practical yield of the complex = ‘W’ g = …… g (iii) % Practical yield of the complex : 2.9 g complex = 100% practical yield of the complex

∴ ‘W’ g complex = W × 100

2.9 %

= …… % practical yield of the complex Result Table :

1. Colour of the complex White opaque

187


188

2. Practical yield of the complex W = ………. g

3. Theoretical yield of the complex 2.9 g

4. % practical yield of the complex …… %

- - -

188

(D) Colorimetric Estimations (Any Two)

Introduction : Colorimetric estimation deals with the measurement of colour intensity. It is instrumental method of quantitative analysis. The colour of a compound is due to absorption of light waves of certain wavelengths. If a solution does not absorb light, it is transparent and colourless. Colorimetry is connected with the determination of the concentration of coloured substances by measurement of the relative absorption of light with respect to a known concentration of the substance. When visible radiations of definite wavelength are passed through coloured solution, then some of the radiations are absorbed by solution and remaining are transmitted through solution. The extent to which absorption of light takes place depends on the concentration of solution. This is the basic principle of colorimetry. Colorimeter is used for measuring absorption in visible region i.e. colorimetric analysis is concerned with visible region with wavelength between 400 nm to 800 nm. The concentration of metals which gives rise to formation of coloured, soluble compound or complex with reagent (ligand) can be determined by colorimetric methods. In these estimations, a particular wavelength of light is passed through a solution of known and unknown concentration under similar conditions. This particular wavelength is called λmax. Thus, λmax is the wavelength at which the solution shows maximum absorbance. The advantage of colorimetric analysis is that it requires much less time, it is more accurate and highly sensitive method than usual chemical analysis. Colorimeter is useful for the determination of constituents which are present in quantities less than 1 or 2 %. For the laws of colorimetry, refer the colorimetric experiments of physical chemistry.

Experiment No. 1 Determination of Iron

Aim : To determine the concentration of given unknown solution of Fe(III) using ammonium thiocyanate at 540 mμ by colorimetric method. Principle : Fe(II) ion does not react with thiocyanate. But Fe(III) ion reacts with thiocyanate to give a series of intensely blood-red coloured complexes. The formation of series of complexes depends upon the concentration of thiocyanate ion. At very low concentration of thiocyanate the predominant coloured species is [Fe(SCN)]2+. Thus, Fe3+ + 6 SCN− → [Fe(SCN)6]3−

T.Y.B.Sc. Practical Chemistry (Inorganic Practicals) Colorimetric Estimations

189

blood-red colour This blood-red coloured solution gives maximum absorbance at 540 nm. The absorbance is directly proportional to concentration, hence absorbance is measured at 540 nm and by graphical method concentration of unknown is determined. At 0.1 M thiocyanate concentration, the coloured species is mainly [Fe(SCN)2]+. At very high concentration of thiocyanate the species is [Fe(SCN)6]3−. In colorimetric determination, a large excess of thiocyanate should be used, since this increases the intensity and also the stability of the colour. Strong acids (HCl or HNO3 with concentration of 0.05 M to 0.5 M) should be present to suppress the hydrolysis of Fe(III) ions as,

Fe3+ + 3H2O ≡ Fe(OH)3 + 3H+ which gives rise to formation of small amount of Fe(OH)3. Generally, sulphuric acid is not used because sulphate ions have a tendency to form complexes with Fe(III) ions. The elements like zinc, cadmium, mercury, bismuth, molybdenum, uranium, titanium, cobalt, nickel, copper and silver, interfere during the analysis. Stannous (Sn2+) and mercurous

(Hg2+2 ) salts if present, should be converted into stannic (Sn4+) and mercuric (Hg2+) salts,

otherwise the colour is destroyed. Fluorides, oxalates, phosphates, arsenates and tartarates interfere, since they form fairly stable complexes with Fe(III) ions. The influence of phosphates and arsenates is reduced by the presence of comparatively high concentration of acid. When large quantities of interfering substances are present then generally one of the following procedure is used. (i) Iron is separated by precipitation with a slight excess of ammonia solution and

dissolving the precipitate in dil. HCl. (ii) The ferric thiocyanate is extracted three times either with pure ether or better, with a

mixture of amyl alcohol and pure ether in the ratio 5 : 2. The organic layer is used for the colour comparison.

Chemicals : Fe(III) unknown solution, conc. HCl and HNO3, 20% ammonium thiocyanate, H2O2, distilled water. Apparatus : Colorimeter, volumetric flask (25 ml), 50 ml beaker (11), water bath, measuring cylinder, burette. Preparation : (A) Preparation of standard solution (stock solution) of ferric ion (conc. 1 mg/ml) : This known solution is prepared by using either of the following methods. Method - 1 : Dissolve 7.022 g of A.R. ferric ammonium sulphate in 100 ml of distilled water and 50 ml of 1 : 5 H2SO4. Then add dropwisely a dilute solution of KMnO4 (1%) using


190

burette until a slight pink colouration remains after stirring well. Dilute to 1 litre with distilled water and shake well. Concentration of solution is 1 mg/ml Fe. Method - 2 : Dissolve 8.64 g of A.R. ferric ammonium sulphate in little distilled water, add 100 ml of A.R. conc. HCl and dilute to 1 litre with distilled water. Concentration of solution is 1 mg/ml Fe. A solution of Fe(III) with known concentration (0.1 mg/ml of Fe) is prepared by diluting ten times the stock solution of 1 mg/ml concentration i.e. 10 ml is diluted to 100 ml. (Distribute 8 to 12 ml of stock solution (conc. 1 mg/ml) in 100 ml volumetric flask as an unknown solution.) Procedure : (A) Preparation of a set of known iron (III) solution (0.1 mg/ml) : 1. Take 1 to 5 numbered clean and dry 50 ml beakers. 2. Fill Fe(III) solution of known concentration (0.1 mg/ml) in burette. (Do not forget to rinse the burette before filling of solution.) 3. Take 25 ml volumetric flask and add 1 ml solution of known concentration of iron (III)

(0.1 mg/ml) from the burette. 4. Add 1.5 ml 1 : 1 HCl solution to the above flask and then add 0.5 ml 3% H2O2 using

common burette. 5. Add 10 ml distilled water using measuring cylinder to this flask. 6. Add 2.5 ml 20% NH4SCN solution to the above flask using common burette. Dilute

the solution upto the mark using distilled water, shake well. Transfer the solution to 50 ml beaker numbered 1.

7. Similarly prepare the solution for set of 2 ml, 3 ml, 4 ml and 5 ml by repeating the procedure from 3 to 6 and using same volumetric flask of 25 ml capacity and transfer the solutions to 50 ml beakers numbered 2 to 5.

Important Note : Record absorbance within 30 minutes after addition of thiocyanate solution. (B) Preparation of a set of unknown iron (III) solution (1 mg/ml) : 1. Dilute the given unknown iron (III) solution to 100 ml with distilled water in a 100 ml

volumetric flask and shake well. 2. Take 6 to 10 numbered clean and dry 50 ml beakers. 3. Fill the diluted Fe(III) solution of unknown concentration in a burette. 4. Take 25 ml volumetric flask and add 1 ml diluted unknown Fe(III) solution from the

burette. 5. Add 1.5 ml 1 : 1 HCl solution to the above flask and then add 0.5 ml 3% H2O2 by

means of common burette. 6. Add 10 ml distilled water using measuring cylinder to this flask.


7. Add 2.5 ml 20% NH4SCN solution with the help of common burette and dilute the solution upto the mark using distilled water, shake well. Transfer the solution to 50 ml beaker numbered 6.

8. Similarly, prepare the solutions for set of 2 ml, 3 ml, 4 ml and 5 ml by repeating the procedure from 4 to 6 and using same volumetric flask. Transfer the solutions to 50 ml beakers numbered 7 to 10.

(C) Preparation of a blank solution : Take 25 ml clean and dry volumetric flask, add to it 1.5 ml of 1 : 1 A.R. HCl followed by 0.5 ml 3% H2O2 solution. Then add to it 2.5 ml 20% NH4SCN solution and dilute upto the mark with distilled water, shake well. Transfer this solution to beaker number 11. Use this as a blank solution. (D) Measurement of absorbance : 1. To measure the O.D. use filter of wavelength 540 nm. 2. Set 100% transmittance or zero optical density for blank solution. 3. Then measure the optical density of each known and unknown solution. Observation Table : (i) Filter used = 540 nm. (ii) Concentration of known solution = 0.1 mg/ml of Fe(III). Beaker

No. Known Fe(III) solution (ml)

O.D. for known

Beaker No.

Unknown Fe(III) solution (ml)

O.D. for unknown

1 2 3 4 5

6 7 8 9

10

Graph :

O.D

.

ml of Fe( ) solutionIII

b a

Unknown

Fig. 1

Plot a graph of optical density (absorbance) against ml of known and unknown Fe(III) solution on the same graph paper. From the graph find out the concentration of the given unknown solution.

Calculation : Calculate the concentration of unknown solution by graphical and by calculation method. (i) By graphical method :

191


Amount of Fe(III) = ba × Concentration of known × 25

= ml of known

ml of unknown × 0.1 × 25 = ………. mg

(ii) By calculation method :

Concentration of unknown

Concentration of known = O.D. of unknown

O.D. of known

∴ Concentration of unknown = O.D. of unknown

O.D. of known × Concentration of known × 25 = …. mg

Using above formula calculate the concentration of unknown for five flasks and take their mean. Result Table :

By graph By calculations

Amount of Fe(III) in the given solution

.......... mg ………. mg

Experiment No. 2 Determination of Cobalt

Aim : To determine the concentration of a given unknown cobalt solution using R-nitroso salt at 520 mμ by colorimetric method. Principle : The colorimetric determination of cobalt using R-nitroso salt is based on the principle that when the hot solution of cobalt is treated with aqueous solution of R-nitroso salt in presence of sodium acetate it produces red coloured solution due to formation of complex [Co (R-nitroso salt)3]. R-nitroso salt (1-nitroso-2-naphthol 4, 6-disulphonate) is a bidentate ligand. Cobalt has co-ordination number six. Therefore, three molecules of reagent combine with ion forming red coloured cobalt complex having octahedral structure.

192

ONO

OH

SO Na3

SO Na3

3

ONO

OH

SO Na3

SO Na3

Co

3

60 Co

Co3+

Along with cobalt other metals like Cu, Fe, Cr, Ni also form a complex with R-nitroso salt but complexes are decomposed by addition of 1 : 2 HNO3 except with cobalt complex.


193

Dilution is done with distilled water. The intensity of colour of solution depends upon concentration of cobalt. The absorbance (O.D.) of known and unknown solutions are measured by colorimeter. By comparing the absorbance (O.D.) of known and unknown graphically the concentration of unknown cobalt solution is determined. Apparatus : Colorimeter, 1 to 11 numbered 50 ml beakers, 25 ml volumetric flask, burettes etc. Chemicals : (i) Known cobalt solution = 0.1 mg/ml of cobalt. (ii) Unknown cobalt solution = Given in 100 ml volumetric flask (1 mg/ml. Give 8 to

12 ml). (iii) 1% aqueous R-nitroso salt solution (freshly prepared). (iv) 50% sodium acetate solution. (v) 1 : 2 HNO3 solution. Procedure : (A) Preparation of a set for known cobalt solution : 1. Take 1 to 5 numbered clean 50 ml beakers. 2. Fill the solution of known cobalt concentration (0.1 mg/ml) in burette. 3. Add 1, 2, 3, 4 and 5 ml solution of known concentration of cobalt (0.1 mg/ml) from

burette to beakers numbered 1 to 5. 4. Add 2.5 ml of 50% sodium acetate solution and 2.5 ml 1% R-nitroso salt solution. 5. Heat these beakers on a boiling water bath for 5 minutes. Cool and add 2.5 ml

1 : 2 HNO3 using burette to each of these beakers. 6. Again heat these beakers on boiling water bath for 5 minutes. Red coloured complex

is formed. Cool the beakers. 7. Transfer the solution from beaker to 25 ml volumetric flask. Wash the beaker with few

ml distilled water and transfer to volumetric flask and finally dilute it with distilled water upto the mark.

8. Transfer the solutions again to 50 ml beakers marked 1 to 5. (B) Preparation of a set for unknown cobalt solution : 1. Dilute the given unknown solution of cobalt with distilled water in 100 ml volumetric

flask (supplied) and shake well and fill it in burette. 2. Take 6 to 10 numbered beakers. 3. Add with the help of burette 1 ml, 2 ml, 3 ml, 4 ml and 5 ml of diluted unknown

cobalt solution in the respective beakers. 4. Add 2.5 ml of 50% sodium acetate solution and 2.5 ml 1% R-nitroso salt solution. 5. Heat these beakers on a boiling water bath for 5 minutes. Cool and add 2.5 ml 1 : 2

HNO3 using burette to each of these beakers.


6. Again heat these beakers on boiling water bath for 5 minutes. Red coloured complex is formed. Cool the beakers.

7. Transfer the solution from beaker to 25 ml volumetric flask. Wash the beaker with few ml distilled water and transfer to volumetric flask and finally dilute it with distilled water upto the mark.

8. Transfer the solutions again to the 50 ml beakers numbered 6 to 10. (C) Preparation of blank solution : In a 50 ml beaker (No. 11) take 2.5 ml of 50% sodium acetate solution + 2.5 ml of 1% R-nitroso salt solution. Heat the beakers on water bath for 5 minutes. Cool and add 2.5 ml 1 : 2 HNO3 solution, again heat for 5 minutes. Cool and transfer the solution from beaker to 25 ml volumetric flask. Dilute to 25 ml with distilled water. Transfer this solution again to 50 ml beaker numbered 11. Use this solution as a blank solution. (D) Measurement of absorbance : 1. To measure O.D. of [Co(R-nitroso salt)3] complex solution, use 520 nm filter. 2. Set 100% transmittance or zero optical density for blank solution. 3. Then measure the optical density of each known and each unknown solution. Observation Table : (i) Concentration of known cobalt solution = 0.1 mg/ml. (ii) Filter used = 520 nm.

Beaker No.

Known Co solution (ml)

O.D. for known

Beaker No.

Unknown Co solution (ml)

O.D. for unknown

1

2

3

4

5

6

7

8

9

10

Graph :

O.D

.

ml of cobalt solution

b a

Unknown

Fig. 1

Plot a graph of optical density (absorbance) against ml of known and unknown cobalt solution on the same graph paper. From the graph find out the concentration of the given unknown solution.

194


195

Calculations : (i) By graphical method :

Amount of Co = ba × Concentration of known × 25

= ml of known

ml of unknown × 0.1 × 25 = ………. mg




O.D. of known



Using above formula calculate the concentration of unknown for five flasks and take the mean. Result Table :


Amount of cobalt in the given solution

.......... mg ………. mg

Experiment No. 3 Determination of Titanium

Aim : To determine the concentration of a given unknown titanium solution using H2O2 by colorimetric method. Principle : The determination of titanium is based on the principle that when an acidic solution of Ti(IV) is treated with H2O2 it produces yellow coloured complex. The intensity of yellow colour is proportional to the amount of titanium (IV) present in the solution. The reaction occurs in strongly acid solution with excess H2O2 as,

Ti(SO4)2 + H2O2 ≡ H2[TiO2(SO4)2]

Colourless Yellow This yellow coloured solution gives maximum absorbance at 440 nm. Dilution of solution is done with 2N H2SO4 only. The absorbance on colorimeter is measured for known and unknown solutions. The concentration of unknown solution is determined by graphical method. Chemicals : (i) Known titanium solution = 1 mg/ml of TiO2.


196

(ii) Unknown titanium solution − (Given in 100 ml volumetric flask) 5 mg/ml TiO2 (Give 13 to 18 ml).

(iii) 3% H2O2. (iv) 2N H2SO4. Apparatus : Colorimeter, 25 ml volumetric flask, 50 ml beakers numbered from 1 to 11.

Procedure :

(A) Preparation of set for known Ti solution :

1. Take 1 to 5 numbered clean and dry beakers.

2. Fill Ti solution of known concentration in burette.

3. Take 25 ml volumetric flask and add 1 ml solution of known concentration of Ti (0.1 mg/ml) from the burette.

4. Add 1.5 ml of 3% H2O2 solution by a burette to the above volumetric flask.

5. Dilute the solution in flask with 2N H2SO4 upto the mark, shake well and transfer the solution into 50 ml beaker.

6. Similarly, prepare the solutions for set of 2 ml, 3 ml, 4 ml and 5 ml by repeating the procedure from 3 to 5 and using same volumetric flask.

7. Transfer the solutions to 50 ml beakers numbered 2 to 5.

(B) Preparation of set for unknown Ti solution :

1. Dilute the given unknown solution of titanium with 2 N H2SO4 in a 100 ml volumetric flask (supplied) and shake well.

2. Take 6 to 10 numbered 50 ml beakers.

3. Fill the diluted titanium solution of unknown concentration in burette.

4. Take 25 ml volumetric flask and add 1 ml diluted unknown Ti solution from the burette.

5. Add 1.5 ml of 3% H2O2 solution by a burette to the above volumetric flask.

6. Dilute the solution in flask with 2 N H2SO4 upto the mark, shake well and transfer the solution into 50 ml beaker numbered 6.

7. Similarly, prepare the solutions for set of 2 ml, 3 ml, 4 ml and 5 ml by repeating the procedure from 4 to 6 and using same volumetric flask.

8. Transfer the solutions to 50 ml beakers numbered 7 to 10.

(C) Preparation of a blank solution :


Take 25 ml volumetric flask and add to it 1.5 ml 3% H2O2. Dilute it upto the mark with 2N H2SO4. Shake well and transfer the solution to beaker numbered 11. Use this solution as a blank solution.

(D) Measurement of optical density :

1. To measure O.D. of Ti(IV) solution use blue filter (440 nm).

2. Set 100% transmittance or zero optical density for blank solution.

3. Measure the O.D. of each known and unknown solution.

Observation Table :

(i) Concentration of known titanium solution : 1 mg/ml of TiO2.

(ii) Filter used = 440 nm.

Beaker No.

Known Ti solution (ml)

O.D. for known

Beaker No.

Unknown Ti solution (ml)

O.D. for unknown

1

2

3

4

5

6

7

8

9

10

Graph :

O.D

.

ml of Ti solution

b a

Unknown

Fig. 1

Plot a graph of optical density (absorbance) against ml of known and unknown titanium solution on the same graph paper. From the graph find out the concentration of the given unknown solution.

Calculations :

Calculate the concentration of unknown solution by graphical and by calculation method.

(i) By graphical method :

Amount of Ti = ba × Concentration of known × 25

= ml of known

ml of unknown × 0.1 × 25 = ………. mg

197


198




O.D. of known



Using above formula calculate the concentration of unknown for five flasks and take the mean. Result Table :


Amount of titanium in the given solution

.......... mg ………. mg

VIVA VOCE

1. What are the regions of electromagnetic spectrum ?

2. Explain Radiant energy, Radiant power, Transmittance, Absorbance, Absorptivity and Molar absorptivity.

3. Define : Lambert’s law, Beer’s law, Lambert-Beer’s law.

4. How Lambert-Beer’s law is varied ?

5. Give the construction and working of a simple colorimeter.

6. Describe essential parts of a colorimeter.

7. How Lambert-Beer’s law is used in quantitative analysis ?

8. What are the various detectors used in visible region ? Describe them.

9. Explain the differences between :

(a) Colorimeter and Spectrophotometer. (b) Absorbance and Absorptivity.

10. Describe the various applications of colorimetry.

11. What are light filters ? Explain the absorption filters. Comment on the choice of a suitable filter.

12. What are the disadvantages of filter photometers ? How are they removed ?

13. What are photo-electric cells ?

14. Give reasons for deviation from Lambert-Beer’s law.

15. What are the advantages of colorimetric analysis ?

16. What is the relationship between absorbance and percentage transmission ?


199

17. What do you understand by λmax ?

18. What are the units used to describe the wavelength ?

19. What are the units of absorptivity and molar absorptivity ?

20. Explain the concept of additivity of absorbance.

21. What do you mean by complementary colour ?

22. What is monochromator ?

23. What are the series of complexes formed by Fe+3 ion with SCN− ion ?

24. Why a strong acid should be present in the colorimetric estimation of Fe+3 ?

25. What is the filter used in the colorimetric estimation of Fe+3 with SCN− ion ?

26. What is the colour produced by Fe+3 ion with SCN− ion ? Why does the colour fade away after sometime ?

27. How will you determine the concentration of unknown solution of Fe+3 by graphical method in colorimetric estimation ?

28. Which colour is produced by cobalt solution with aqueous solution of R-nitroso salt.

29. Draw the octahedral structure of cobalt complex with R-nitroso salt.

30. What is the use of 1 : 2 HNO3 in this estimation ?

31. What is the concentration of unknown cobalt solution ? Which filter we have used for estimation of cobalt ?

32. How is the amount of unknown cobalt calculated by calculation method ?

33. Which colour is produced when titanium reacts with H2O2 ?

34. What is the concentration of the standard (known) titanium solution ? Which filter you have used for estimation of titanium ?

35. How is amount of unknown titanium calculated by calculation method ?

36. Write the structure of complex formed by R-nitroso salt with cobalt.

- - -

(E) Column Chromatograpy (Three Mixtures)

[One Mixture should be Colourless] Separation of Binary Mixture of Cations by Column Chromatography

Introduction : In all chromatographic methods there are two phases, one stationary and other mobile. Separation depends upon the relative movements of these two phases. Chromatography methods may be classified according to the nature of the stationary phase, which may be solid or liquid. If stationary phase is solid, the method is known as adsorption chromatography. If the stationary phase is liquid, the method is known as partition chromatography. Since in each case the mobile phase may be either liquid or a gas. There are in all four types of chromatographic systems.

Chromatography

PartitionLiquid Stationary Phase

1.Liquid mobile phase

Liquid-liquidChromatography

e.g. Paper chromatography

2.Gaseous mobile phase

Gas-liquidChromatography

AdsorptionSolid Stationary Phase

3.Liquid mobile phase

Liquid-solidChromatography

e.g. TLC, on exchangeI

4.Gaseous mobile phase

Gas-solidChromatography

Column Chromatography is one of the best methods used for separation of components of mixtures based on the partition. This method uses suitable substance as a stationary phase or support. Although this method is time consuming, we can get the constituents in highly pure form. The separation of components from a given mixture (inorganic cations) by using cellulose column is mostly partition chromatography. In this technique, we use two solvents (liquid phase) such as acetone and HCl − H2O. The separation of mixture is done between the two phases. (i) Aqueous phase (Stationary phase) (ii) Organic solvents - acetone phase (Mobile phase) In partition chromatography one of the two solvents is held on a column as a stationary phase. Generally silica gel is used for separation of organic mixtures, while powder is used for separation of inorganic mixtures. The method depends on the separation of components of a mixture between water held on the column and non-polar solvents (or less polar than water) percolating down the column. The more hydrophilic components having more affinity for water are held at the top

200

T.Y.B.Sc. Practical Chemistry (Inorganic Practicals) Column Chromatography

of the column, while more hydrophobic components being soluble in less polar medium move down the column and are eluted (removed) first.

In the present syllabus we have to separate following binary mixtures by column chromatography. Out of these at least one colourless mixture is to be separated.

(a) Fe+3 + Al3+

(b) Co2+ + Ni+2

(c) Cu+2 + Ni+2

(d) Cu+2 + Al+3

(e) Zn2+ + Al3+

(f) Zn2+ + Mg2+

Cotton orGlass wool

Glass columnor

Burette

Adsorbent

Cotton orGlass wool

Sample

Eluate

Fig. 1: Typical chromatographic column

Metal ion reacts with HCl and forms chloride complexes and such complex moves along with the solvent acetone first, The metal ion which does not move with acetone is collected afterwards by using 1 M HCl.

It is found by experiment that Fe+3, Co+2, Cu2+ and Zn2+ move with acetone and will be collected first. While Al+3, Ni+2 and Mg+2 are eluted by using 1 M HCl as a second fraction.

201


202

We know that Rf values of Fe+3, Co+2, Cu+2, Zn2+ are higher than those of Al+3, Ni+2, Mg+2. Therefore, from the binary mixture we shall first get Fe+3, Co+2, Cu+2 and Zn+2 components.


From the given mixture separate the cations by column chromatography using cellulose.

Apparatus :

Glass column (30 cm long and 3 cm in diameter), 25 ml glass burette with glass stopper, 100 ml conical flasks, glass wool, cotton plug, filter paper etc.

Chemicals :

(i) A mixture of two cations (given in test tube).

(ii) Organic solvent, 2% HCl - Acetone (2 ml pure conc. HCl and 98 ml distilled acetone)

(iii) 1 M HCl (second solvent) - 20 to 30 ml

(iv) Cellulose powder - 4 g

Principle :

In column chromatography, the stationary phase is a finely divided solids like cellulose, silica or alumina. A column is prepared by carefully packing this solid material in a long glass column or a burette. A typical column is shown in Fig. 1 of previous experiment.

A sample solution is placed in a small volume at the top of the column. The components of the sample are adsorbed over the column material. The adsorbed components are eluted with a suitable solvent (mobile phase) which is called as eluant. The less strongly adsorbed components are eluated first and more strongly adsorbed components are eluated later. Thus, separation is effected. The process is called elution and the liquid collected after eluation is called eluate. The eluate is further analysed qualitatively.

Procedure :

[A] Preparation of Cellulose Column :

1. Take 25 ml glass burette and clean with little amount of acetone and dry it before use.

2. Take 4 g of cellulose powder in 100 ml beaker and add to it 25 ml acetone to form a slurry. Stir with glass rod.

3. Place a small glass wool or cotton pad at the bottom of the burette with the help of long glass rod.

4. Close the tap and clamp the burette in a vertical position, transfer the slurry to the burette.


203

5. Cellulose will settle on the glass wool (cotton).

6. Prepare a uniform column of 15 - 20 cm in length, without any air gap.

7. Cover the column with little cotton plug.

8. Open the tap of the burette and allow a little solvent to flow and close the tap again.

9. Observe that there will be some solvent 2 cm above the top level of cellulose. Precaution : The column should not become dry.

10. Now use this column for separation of cations. [B] Loading of the Mixture : 1. Open the tap of the burette and allow the solvent to flow down. 2. Immediately pour the given sample (3 ml) solution to the column, and get a uniform

layer. Rinse the test tube with 2% HCl acetone mixture. 3. Add 5 - 10 ml solvent and keep a clean conical flask below the burette to collect the

first component of the mixture. 4. The rate of flow of the solvent should be very slow. 5. As the solvent passes through, various components will pass through with different

rates and after some time separate out into two coloured bands or zones, if both cations are coloured.

6. This process is known as development of the column. [C] Separation of Bands : 1. The first component is collected in a conical flask using 2% HCl. Acetone solvent, till

the first band disappears completely. It will require 20 ml of solvent. 2. The second component is collected in another dry conical flask (100 ml) using 1M

HCl as second solvent, till the second component is completely removed from the column, it will require about 20 ml 1 M HCl.

[Hint : (i) For Fe+3/Al3+ mixture only one band is observed (ii) For Co+2/ Ni2+ and Cu+2 mixture two bands are seen.]

3. For Zn2+ and Mg2+ mixture, no bands are observed. However, sometimes one cation gives two different coloured bands due to formation of two different complexes e.g. CoCl3 ⋅ 2H2O is pink and CoCl3 ⋅ H2O is blue. So formation of colour band cannot be taken as a sure test for separation. [D] Concentration of the Components : After separation of two components, concentrate the first fraction by heating on steam bath, so the acetone will be evaporated off and we get concentrated solution of corresponding metal chlorides. Concentrate the second fraction by heating on a wire gauze till the final volume becomes 10 to 15 ml.


204

(Note : Student should show the two separated and concentrated components to the examiner.) [E] Identification of Cations : After separation and concentration of two cations, confirm them by performing following spot tests. [I] First fraction : Fe+3, Co+3, Cu2+, Zn2+ (a) Tests for Fe+3 : 1. Take 2 drops of the solution in a test tube and 1 - 2 drops of distilled HCl + 1 ml 3%

H2O2 + 1 - 2 drops of KCNS or NH4CNS (1%) ⎯→ Red colour, therefore Fe+3 is present and confirmed.

2. Take 2 drops of solution in a test tube + 2 drops of K4[Fe(CN)6] ⎯→ Blue colour, therefore Fe+3 is present and confirmed.

3. Place a drop of solution on a filter paper, add a drop of phenanthroline reagent (0.1 % aq.) ⎯→ Red colour, therefore Fe+3 is present and confirmed.

(b) Tests for Co+2 : 1. Take 2 drops of the solution in a test tube + 2 drops of α nitroso [β-naphthol (1% in

50 % acetic acid)] ⎯→ Red brown colour, therefore Co+2 is present and confirmed. 2. 1 drop of the solution on a test paper + 1 drop of sodium acetate (50% aq.)

+ 1 drop of R-nitroso salt (1% aq.) ⎯→ Red spot, therefore Co+2 is present and confirmed.

3. Take 1 drop of the solution in a test tube. Add few drops of acetone + 2 drops of acetone + 2 drops of NH4CNS ⎯→ Blue or bluish green colour, therefore Co+2 is present and confirmed.

4. Place a drop of the solution on a test paper. Add two drops of rubeanic acid and one drop of benzidine. Expose to ammonia vapours ⎯→ Orange yellow colour, therefore Co+2 present and confirmed.

5. Place a drop of the solution on a filter paper. Add five drops of saturated solution of NH4CNS ⎯→ A green to blue colour which changes to pink on dilution, therefore Co+2 present and confirmed.

(c) Tests for Cu+2 : 1. 2 drops of solution in a test tube + 5 drops of dilute acetic acid + 4 drops of

K4[Fe(CN)6] ⎯→ A chocolate red precipitate, therefore Cu+2 present and confirmed. 2. Place a drop of the solution on a test paper. Add few drops of alcoholic solution of

rubeanic acid and expose to ammonia vapours ⎯→ Olive green colour, therefore Cu+2 present and confirmed.

3. Place a drop of the solution on a test paper, add one drop of rubeanic acid and one drop of benzidine. Expose to ammonia vapours ⎯→ Grey green colour, therefore Cu+2 present and confirmed.


205

4. Take a drop of the solution in a test tube, add 2 drops of cupron and a drop of NH4OH ⎯→ Green colour, therefore Cu+2 present and confirmed.

(d) Tests for Zn :

1. Take two drops of solution. Add few drops of acetic acid and pass H2S ⎯→ White precipitate. Then Zn is present.

2. Two drops of solution + Few drops of acetic acid + 4 drops of K3[Fe(CN)6] + Few drops of diphenylamine in acetic acid ⎯→ Green precipitate. Zn is present and confirmed.

3. Few drops of solution + Few drops of dil. HNO3 + Few drops of Co(NO3)2 solution. Heat to dryness ⎯→ A green mass is obtained. Then zinc is present and confirmed.

[II] Second Fraction : Al+3, Ni2+, Mg2+

(a) Tests for Al+3 :

1. 1 drop of the solution on a test paper. Dry the spot under I.R. lamp + 1 drop of alizarin reagent (1% aq.) and expose to ammonia ⎯→ Violet colour around the spot, therefore Al3+ present and confirmed.

2. 2 ml of the solution in a test tube + 1 ml dil. HCl + 4 ml 1:1 Ammonia (solution should become alkaline). Heat to boiling ⎯→ White gelatinous precipitate of Al(OH)3 is formed, therefore Al3+ present and confirmed.

(b) Tests for Ni2+ :

1. 1 drop of the solution on a test paper + 1 drop of dimethyl glyoxime and exposed to ammonia ⎯→ Red colour, therefore Ni+2 present and confirmed.

2. 1 drop of the solution on a test paper + 1 drop of saturated solution of rubeanic acid in ethanol and expose to ammonia ⎯→ Violet colour, therefore Ni+2 present and confirmed.

3. Take 2-3 drops of the solution in a test tube. Add 2 drops of α-nitroso β-naphthol solution ⎯→ Brown precipitate soluble in dil. HCl, therefore Ni+2 present and confirmed.

4. Place a drop of the solution on a test paper. Add a drop of rubeanic acid and a drop of benzidine. Expose to ammonia vapours ⎯→ Blue coloured spot, therefore Ni+2 present and confirmed.

(c) Tests for Mg2+ :

1. Two drops of solution on watch glass + 2 drops of titan yellow solution + 2 drops of dil. NaOH solution ⎯→ a rose red colour or precipitate ⎯→ Then Mg2+ is present.

2. Two drops of solution + Hypoiodite reagent ⎯→ A reddish brown precipitate confirms the presence of Mg2+.


206

[Hypoiodite reagent : NaOH solution + Equal amount of KI solution + Iodine solution till the solution becomes just yellow].

3. 2 drops of solution + 3 drops of NH4OH solution + 2 drops of Na2HPO4 solution → white ppt. confirms the presence of Mg2+.

Result Table :

Fractions collected Amount of fraction concentrated

Cation present

(i) First fraction (ii) Second fraction

........ ml

........ ml

VIVA VOCE 1. What do you mean by chromatography ? 2. What are different types of chromatographic methods ? 3. What is column chromatography ? What is its principle ? 4. What is the principle behind separation of a mixture by column chromatography ? 5. What is stationary phase and mobile phase in column chromatography ? 6. What is the role of cotton plug in column chromatography ? 7. Give one confirmatory test for Fe+3, Co+2, Cu+2, Zn2+ and Al3+, Ni2+, Mg2+. 8. What is meant by elution ? What are the different solvents used for elution ? 9. What is meant by eluate ? 10. What is meant by eluant ? 11. What do you mean by column chromatography ? 12. What is Rf value ?

- - -

(E) Flame Photometry Analysis (Any Three Experiments)

Introduction : Flame photometry is the instrumental method of analysis and used for analysis of limited elements mainly Li+, Na+, K+ and Ca2+. Ca2+ can be analysed by gravimetric or volumetric methods of analysis, but there is no any other suitable method of analysis for Li+, Na+ and K+. Na+, K+ and Ca2+ from soil, drinking water, from food samples, from blood, from fertilizers, etc. is analysed by this method. This method of estimation is highly sensitive and even can be applied at 1 ppm level accurately. The typical flame photometer consists of sample inlet system (nebulizer), burner (and fuel gas), detector and read out device.

FlameFilter

Lens

Photodetector

Oxidant

Fuel gasBurn

er

Nebuliser

To drain

Analytesolution

Mixing chamber

Amplifierand

Readout

Fig. 1: Flame photometer

Sample of Li+, Na+, K+ and Ca2+ ions in solution form is used for analysis. The solid sample is disintegrated by suitable method such as acid digestion, or microwave digestion diluted to suitable volume and used for analysis. The concentration of these ions in analyte solution should be less than 100 ppm and typically in the range from 1 to 25 ppm. All reagents used for flame photometric analysis should be free off these ions or alternatively ‘atomic spectroscopic grade’ reagents must be utilized. By flame photometry one element can be detected in presence of another without interference. In a typical flame photometric analysis, a solution containing analyte is aspirated into the burner in the form of fine mist. This process is called nebulisation. In the flame, the solvent evaporates due to heating, leaving behind finely divided solid particles of analyte salt. In the

207

T.Y.B.Sc. Practical Chemistry (Inorganic Practicals) Flame Photometry Analysis

208

flame, due to high temperature, salt particles are converted to gaseous state molecules. Then, gaseous state molecules auto decompose to atoms by redox reaction.

e.g. NaCl(g) Δ

⎯→ Na(g) + 12 Cl2(g)

Na+ + 1e− → Na(g) and Cl− → 12 Cl2 (g) + e−

The atoms in flame absorb heat energy and are excited from ground state to excited state (3s orbital to 3p orbital). Excited state of atom is highly unstable. Thus, the excited atoms return to a ground state with emission of energy. This energy is emitted in the form of electromagnetic radiations of characteristic wavelength. Emitted wavelength is a characteristic of the element. The intensity of the emitted radiation from the flame is then determined. Intensity of emitted radiations can be related to the concentration of the analyte ions in the solution. At sufficiently low concentration, intensity of emitted radiation is directly proportional to the concentration of Na+ ions in the analyte solution.

Experiment No. 1 Estimation of Sodium (Na)

Aim: To determine the concentration of given unknown samples of Sodium (Na) by flame photometry using calibration curve method. Principle: In a typical flame photometric analysis of Na+, a solution containing Na+ is aspirated into the burner and dispersed into the flame as a fine spray. This process is called nebulisation. In the flame, the solvent evaporates first, leaving finely divided solid particles of NaCl which move to hottest region of the flame where gaseous Na atoms and Na+ ions are produced. The atoms are excited from ground state to excited state (3s orbital to 3p orbital) by absorbing energy from the flame. Excited state of atom is highly unstable. Thus, the excited atoms return to a ground state with the emission of energy. This energy is emitted in the form of electromagnetic radiations of characteristic wavelength i.e. 586 nm. This wavelength is a characteristic of the Na. The intensity of the emitted radiation from the flame is then determined. Intensity of emitted radiations can be related to the concentration of the Na+ ions in the solution. At sufficiently low concentration, intensity of emitted radiation is directly proportional to the concentration of Na+ ions in the analyte solution. Chemicals:

A. R. grade Sodium chloride (NaCl), Distilled water. Apparatus: Burette, Graduated pipette, Beakers (250 ml, 50 ml – 6 beakers), volumetric flask (100 ml, 25 ml).


209

(Note: All solutions must be prepared in glass distilled water. Before use all glass apparatus must be rinsed with distilled water.)

Procedure:

Part A: Preparation of standard solution (stock solution) of Na+ (100 ppm):

Prepare Na+ stock standard solution of 100 ppm 250 ml from A.R. grade NaCl. 100 ppm Na+ means 100 mg Na+ ions in 1000 ml solution. Thus, 250 ml solution will require 25 mg Na+ i.e. 0.025 g Na+ (1 ppm = 1 mg Na+/ltr.).

58.5 g NaCl = 23 g Na+.

Therefore, ‘X’ g NaCl = 0.025 g Na+

Therefore, X = 0.025 × 58.5

23 = 0.06358 g NaCl

Thus, weigh 0.064 g i.e. 64 mg NaCl and prepare 250 ml solution in volumetric flask with distilled water.

Part B: Preparation of series working standard solutions and unknown samples of Na+:

Prepare series working standard solutions of 4, 8, 12, 16 and 20 ppm by using following procedure.

1. Take 1 to 5 numbered clean and dry 25 ml volumetric flasks.

2. Fill Na+ stock solution of known concentration in burette (100 ppm).

3. Take out 1, 2, 3, 4 and 5 ml standard of 100 ppm by burette in 1 to 5 numbered 25 ml volumetric flasks. Dilute these solutions upto the mark with distilled water. Transfer the solutions to 50 ml beakers numbered 1 to 5.

4. Prepare two unknown samples of Na+, the concentration of these samples should be less than 20 ppm. These samples are used for flame photometric analysis to find out the amount of Na+.

Part C: Measurement of intensity of flame by using Flame Photometer:

1. Switch ON the power of apparatus. Ignite the burner. Set sodium filter. Properly adjust the flame of the flame photometer. For this purpose follow the instructions given in the manual.

2. Aspirate distilled water in flame and set zero using respective ‘setting knob’.

3. Now aspirate 20 ppm standard solution of Na+ for calibration process. For this standard solution adjust intensity such as 20 or 40 or 60 or 80. Use standardization knob for this purpose.


4. Record the intensity of flame for all Na+ standards and unknown sample solutions one by one. Aspirate distilled water after every reading. Tabulate the recorded observations.

Observation Table:

Solution Conc. of Na+

in ppm Intensity of flame for Na+ solutions

Standard-1 4

Standard-2 8

Standard-3 12

Standard-4 16

Standard-5 20

Sample-1 Unknown

Sample-2 Unknown

Graph:

Plot a graph of concentration of Na+ against flame intensity.

Inte

nsity

of

fla

me

R = 0.9872

80

ppm ofsample

Concentration of Na( ), ppmI

0 4 8 12 16 20 24 20

10

20

30

40

50

60

70

8

Fig. 1

Calculations: By using calibration curve method: Use calibration curve method for calculation of concentration of sample. Plot a graph of concentration of Na+ (x-axis) against flame intensity (y-axis). Place perpendicular from the intensity axis to the plot and then extend to concentration axis. From the point of intercept obtain the concentration of Na+ in the given solution.

210


211

Result Table :

Amount of Na+ in the given sample of water Sample-1 Sample-2

----------- ppm --------------- ppm

Experiment No. 2 Aim: To determine the concentration of given unknown samples of Sodium (Na) by flame photometry using least square (regression analysis) method. Principle, Chemicals, Procedure is same as that of Experiment No. 1 (Flame Photometer Analysis of sodium), only method of calculation is different. Perform the same procedure as per given in the Experiment No. 1. Observation Table:

Solution Conc. of Na+ in ppm (xi)

Flame intensity (yi)

Standard-1 x1 = 4 y1 =





Sample-1 Unknown

Sample-2 Unknown

Calculation by least square method: There are two assumptions in the least-square method. 1. The linear relationship actually exists between measured response (y-axis value) and

standard analyte concentration. The mathematical relationship describing this assumption is called as regression model which is given by the following equation:

y = ax + b, where ‘a’ is slope and ‘b’ is intercept to y-axis. 2. Any deviation from the individual point from the straight line arises from the error in

the measurement i.e. we assure that there is no error in the x value of the point. Both these assumptions are appropriate for many analytical methods. But whenever there is significant uncertainty in the x data, then basic least-squares method do not give rise straight line.

The calculations are as follows: The slope of line is given by the following equation:


212

a = n ∑ xi yi − ∑ xi ∑ yi

n ∑ x2i − (∑ xi)2

where, n = 5 (number of data points), xi = value of individual concentration in ppm (x1 to x5) yi = value of individual flame intensities (y1 to y5) ∑ xi = (4 + 8 + 12 + 16 + 20), ∑ yi = sum of flame intensities for five readings of standards

∑ x2i = 42 + 82 + 122 + 162 + 202,

∑ y2i = make a square of each flame intensity and then take sum.

(∑ xi)2 = (4 + 8 + 12 + 16 + 20)2, n ∑ xi yi = 5[(x1 y1) + (x2 y2) + (x3 y3) + (x4 y4) + (x5 y5)]

b = −y + a−x

where, −x = (4 + 8 + 12 + 16 + 20)

5

−y = (sum of flame intensities of five standards)

5

From slope. concentration of sample is calculated. y = ax + b, where ‘a’ is slope and ‘b’ is intercept to y-axis. and, y = flame intensity of sample, x = concentration of sample, ppm (to be calculated). Regression of the line can be calculated by the following equation:

r = n ∑ xi yi − ∑ xi ∑ yi

[n ∑ x2i − (∑ xi)2] [n ∑ y

2i − (∑ yi)2]

r is called as regression of line. This value indicates the linearity of line. For 100% linearity of five standard readings, the value of r is ±1. Practically this value is expressed as r2 and is 1 for 100% linearity. Experimentally this value should be greater than 0.95. If this value is less than 0.95 then we can say that there exists large error in the determination of flame intensities. Such readings cannot be accepted for calculations. Result Table :

Amount of Na+ in the given sample of water Sample-1 Sample-2

----------- ppm ---------- ppm

Experiment No. 3 Estimation of Potassium

Aim:


213

To determine the concentration of given unknown samples of Potassium (K) by flame photometry using calibration curve method. Principle : In a typical flame photometric analysis of K+, a solution containing K+ is aspirated into the burner and dispersed into the flame as a fine spray. This process is called nebulisation. In the flame, the solvent evaporates first, leaving finely divided solid particles of KCl which move to hottest region of the flame where gaseous K atoms and K+ ions are produced. The atoms are excited from ground state to excited state (3s orbital to 3p orbital) by absorbing energy from the flame. Excited state of atom is highly unstable. Thus, the excited atoms return to a ground state with the emission of energy. This energy is emitted in the form of electromagnetic radiations of characteristic wavelength i.e. 586 nm. This wavelength is a characteristic of K. The intensity of the emitted radiation from the flame is then determined. Intensity of emitted radiations can be related to the concentration of K+ ions in the solution. At sufficiently low concentration, intensity of emitted radiation is directly proportional to the concentration of K+ ions in the analyte solution. Chemicals: A.R. Grade Potassium chloride (KCl), Distilled water. Apparatus: Burette, Graduated pipette, Beakers (250 ml, 50 ml – 6 beakers), Volumetric flask (100 ml, 25 ml). (Note: All solutions must be prepared in glass distilled water. Before use all glass apparatus must be rinsed with distilled water.) Procedure: Part A: Preparation of standard solution (stock solution) of Na+ (100 ppm): Prepare K+ stock standard solution of 100 ppm 250 ml from A.R. grade KCl. 100 ppm K+ means 100 mg K+ ions in 1000 ml solution. Thus, 250 ml solution will require 25 mg K+ i.e. 0.025 g K+ (1 ppm = 1 mg Na+/ltr.). 73.58 g KCl = 38.08 g K+. Therefore, ‘x’ g KCl = 0.025 g K+

Therefore, x = 0.025 × 73.58

38.08 = 0.0489 g KCl

Thus, weigh 0.0489 g i.e. 49 mg KCl and prepare 250 ml solution in the volumetric flask with distilled water. Part B: Preparation of series of working standard solutions and unknown samples of K+: Prepare series working standard solutions of 4, 8, 12, 16 and 20 ppm using following procedure. 5. Take 1 to 5 numbered clean and dry 25 ml volumetric flasks. 6. Fill K+ stock solution of known concentration in burette (100 ppm).


7. Take out 1, 2, 3, 4 and 5 ml standard of 100 ppm by burette in 1 to 5 numbered 25 ml volumetric flasks. Dilute these solutions upto the mark with distilled water. Transfer the solutions to 50 ml beakers numbered 1 to 5.

8. Prepared two unknown samples of K+, the concentration of these samples should be less than 20 ppm. These samples are used for the flame photometric analysis to find out amount of K+.

Part C: Measurement of intensity of flame by using Flame Photometer: 1. Switch ON the power of apparatus. Ignite the burner. Set sodium filter. Properly

adjust the flame of the flame photometer. For this purpose follow the instructions given in the manual.

2. Aspirate distilled water in the flame and set zero using respective ‘setting knob’. 3. Now aspirate 20 ppm standard solution of K+ for calibration process. For this

standard solution adjust intensity such as 20 or 40 or 60 or 80. Use standardization knob for this purpose.

4. Record the intensity of flame for all K+ standards and unknown sample solutions one by one. Aspirate distilled water after every reading. Tabulate the recorded observations.

Observation Table: Solution Conc. of K+ in

ppm Intensity of flame for K+ solutions

Standard-1 4 Standard-2 8 Standard-3 12 Standard-4 16 Standard-5 20 Sample-1 Unknown Sample-2 Unknown

Graph: Plot a graph of concentration of K+ against flame intensity.

Inte

nsity

of

fla

me

R = 0.9872

80

ppm ofsample

Concentration of K( ), ppmI

0 4 8 12 16 20 24 20

10

20

30

40

50

60

70

8

214


215

Fig. 1 Calculations: By using calibration curve method: Use calibration curve method for calculation of concentration of sample. Plot a graph concentration of K+ (x-axis) against flame intensity (y-axis). Place perpendicular from the intensity axis to the plot and then extend to concentration axis. From point of intercept obtain the concentration of K+ in the given solution. Result Table :

Amount of K+ in the given sample of water Sample - 1 Sample - 2

----------- ppm ---------- ppm Test your Knowledge 1. Give the emitted wavelength by excited Na atoms in flame. What is the colour of

Na flame? 2. Give the emitted wavelength by excited K atoms in flame. What is the colour of

K flame? 3. Why distilled water must be used for analysis of metal by flame photometry? 4. Give the equation for number of atoms in the excited state. Calculate fraction of

excited state Na atoms in the flame of temperature 800°C. 5. What is the effect of drop size on flame temperature and atomization? 6. Why transition metals are not analyzed by flame emission spectroscopy?

Experiment No. 4 Estimation of K

Aim: To determine the concentration of given unknown samples of Potassium (K) by flame photometry using least square (regression analysis) method. Principle, Chemicals, Procedure is same as that of Experiment No. 3 (Flame Photometer Analysis of Potassium), only method of calculation is different. Perform the same procedure as per given in the Experiment No. 3. Observation Table:

Solution Conc. of K+ in ppm (xi)

Flame intensity (yi)

Standard-1 x1 = 4 y1 = Standard-2 x2 = 8 y2 = Standard-3 x3 = 12 y3 = Standard-4 x4 = 16 y4 = Standard-5 x5 = 20 y5 = Sample-1 Unknown Sample-2 Unknown


216

Calculation by least square method: There are two assumptions in the least-squares method. 3. The linear relationship actually exists between measured response (y-axis value) and

standard analyte concentration. The mathematical relationship describing this assumption is called as regression model which is given by following equation:

y = ax + b, where ‘a’ is slope and ‘b’ is intercept to y-axis. 4. Any deviation from the individual point from the straight line arises from the error in

the measurement i.e. we assure that there is no error in the x value of the point. Both these assumptions are appropriate for many analytical methods. But whenever there is significant uncertainty in the x data, then basic least-squares method do not give rise straight line.

The calculations are as follows: The slope of line is given by following equation:

a = n ∑ xi yi − ∑ xi ∑ yi

n ∑ x2i − (∑ xi)2

where, n = 5 (number of data points) xi = value of individual concentration in ppm (x1 to x5) yi = value of individual flame intensities (y1 to y5) ∑ xi = (4 + 8 + 12 + 16 + 20), ∑ yi = sum of flame intensities for five readings of standards

∑ x2i = 42 + 82 + 122 + 162 + 202,

∑ y2i = make a square of each flame intensity and then take sum.

(∑ xi)2 = (4 + 8 + 12 + 16 + 20)2, n ∑ xi yi = 5[(x1 y1) + (x2 y2) + (x3 y3) + (x4 y4) + (x5 y5)]

b = −y + a−x

where, −x = (4 + 8 + 12 + 16 + 20)

5

−y = (sum of flame intensities of five standards)

5

From slope concentration of sample is calculated. y = ax + b, where ‘a’ is slope and ‘b’ is intercept to y-axis. and, y = flame intensity of sample, x = concentration of sample, ppm (to be calculated). Regression of the line can be calculated by following equation:

r = n ∑ xi yi − ∑ xi ∑ yi

[n ∑ x2i − (∑ xi)2] [n ∑ y

2i − (∑ yi)2]

r is called as regression of line. This value indicates the linearity of line. For 100% linearity of five standard readings the value of r is ±1. Practically this value is expressed as r2 and is 1 for 100% linearity. Experimentally this value should be greater than 0.95. If this value is less


217

than 0.95 then we can say that there exists large error in the determination of flame intensities. Such readings cannot be accepted for calculations. Result Table :

Amount of K+ in the given sample of water Sample-1 Sample-2 ----------- ppm --------------- ppm

- - -

217

(F) Inorganic Qualitative Analysis (Four Mixtures)

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 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 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.

T.Y.B.Sc. Practical Chemistry (Inorganic Practicals) Inorganic Qualitative Analysis

218

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 etc.) before using them.

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


219

(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−, SOA

2−4 E A, NO A

−2 E A, NO A

−3 E A.

(b) Amorphous Insoluble salts containing CO A

2−3 E A, S2−, PO A

3−4 E A,

BO A

3−3 E A, 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).

Observation Inference (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.


220

(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) decompo-sition of hydroxides.

(d) Change in colour

(Contd.)


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.


221

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, NHA

+4E

A 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. CO A

2−3 (Carbonate)E A

(b) Evolution of colourless H2S gas having

smell of rotten eggs and turning lead acetate paper black.

S2− (Sulphide)


222

(c) Brownish fumes acidic to litmus, NO2 gas.

NO A

−2 (Nitrite)E A

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 NOA

−3 E

Amay 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, POA

3−4 E

A 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 :


223

Take a little mixture in a porcelain dish. Add to it a 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 thin paste on glass rod and hold it in the flame. If a green edged flame of ethyl borate is

observed, BO A

3−3 E

A, 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 (mix. 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,


224

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+.


225


226

ANALYSIS OF BASIC RADICALS

Separation of Basic radicals into group

Original solution + dil. HCl O.S.* + dil. HCl (warm), pass H2S gas

Precipitate (white) Group I is present Ag+, Hg+, Pb2+. If present See Pg. 226

Precipitate (coloured) Group II is present Hg2+, Pb2+, Bi2+, Cu2+, Cd2+, As3+, Sb3+, Sn2+, Sn4+.

If present See Pg. 226

O.S./Filtrate : If group II present, boil well to expel H2S completely and

then

(a) Take 2 drops of filtrate on the tile + K3[Fe(CN)]6. If blue colour Fe2+

present.

(b) If Fe2+ is present add 2 - 3 ml of conc. HNO3 in the filtrate and boil

well so that Fe2+ is oxidised to Fe3+. (c) This solution + NH4Cl (solid) + 1 : 1 Ammonia (NH4OH) till alkaline to

litmus.

Precipitate Group III A is present

Fe2+, Fe3+, Al3+, Cr3+.

If present See Pg. 230

O.S./Filtrate* + NH4Cl (Solid) + 1:1 Ammonia (NH4OH) till

alkaline + H2S

Precipitate

Group III B is present Zn2+, Mn2+, Co2+, Ni2+. If present See Pg. 231

Filtrate boil well to expel H2S completely

O.S./Filtrate** + NH4Cl (solid) + 1 : 1 Ammonia + (NH4)2CO3

Precipitate

(white)

Group IV is

present

Ca2+, Ba2+, Sr2+

If present

See Pg. 231

Filtrate + NH4Cl (solid)

+ 1 : 1 Ammonia + Na2HPO4

Precipitate

(white) Group V present is Mg2+

If present

See Page 232

If no ppt upto Group V

or only one group is

detected, then Group VI

is present

NHA

+4E

A, K+,

Na+. If present See Page 233

* Use fairly dilute solution for testing of Group II and III B. * Use concentrated solution for testing of Group IV.


227

SEPARATION TABLE FOR GROUP I :

Transfer the ppt. of group I metal chloride 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+ is confirmed.

SEPARATION TABLE FOR GROUP II :

Transfer the ppt. of group II metal sulphide 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. 228).

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.


228

Residue Filtrate

Black ppt. HgS. Dissolve the residue in aquaregia (HCl + HNO3

3 : 1) by boiling, dilute with water.

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.

C.T. Residue Filtrate

(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

It may contain sulphates or nitrates of Bi, Cu, Cd. Add 1 : 1 ammonia in excess. Filter or centrifuge.

(ii) Solution + KI if ammonium acetate to Residue Filtrate

yellow ppt. of HgI2.

Hg2+ present and confirmed.

get solution.

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 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 dil. 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.


229

SEPARATION TABLE FOR GROUP II B : Transfer the ppt. of Group II B metal sulphide into a 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 + Rhodamine B. If red colour is seen and

(iii) Solution + KI + pyridine. If coloured solution, then Sb3+ present and confirmed.

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+ present and confirmed.

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.


230

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. 230

Filtrate contains phosphate of III B, IV, V group

Add neutral FeCl3 solution till phosphate is precipitated and solution acquires brownish red colour. Boil and filter.

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

Add to the filtrate NH4Cl solid + 1 : 1 NH3 till alkaline

3

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 III B, IV and V group according to the chart given on Page No. 225.


231

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 the solution). Then use this solution for detection of group III A onwards according to chart given on Page No. 225.

SEPARATION TABLE FOR GROUP III A : Transfer the ppt. of Group IIIA metal hydroxide 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 x 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+ present and confirmed. 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.


232

SEPARATION TABLE FOR GROUP III B :

Transfer the ppt. of group IIIB sulphide 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

(or in aquaregia). Boil the excess

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 is seen 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 may contain MnCl2 and ZnCl2. Boil the solution to expel H2S. Add excess of NaOH solution. Filter.

Residue Filtrate

It is white residue turning brown due to atmospheric oxi-dation. Dissolve this ppt. in conc. HNO3 + a pinch of PbO2, boil and

dilute with the 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 group IV metal carbonate 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.


233

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. Solution of BaCl2 is

It may contain Sr and Ca acetate. Add excess of ammonium sulphate solution and heat to boiling. Allow to stand for 5 minutes. Filter.

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+ present and confirmed.

ANALYSIS OF GROUP V : Transfer the ppt. of group V in a test tube and dissolve it in a little dil. H2SO4. Use this

solution for C.T. of Mg2+.

C.T. for 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.)


234

SEPARATION TABLE FOR GROUP VI : (i) Use water solution only.

(ii) Test the water solution for NHA

+4 E A, Na+ and K+ as follows :

(1) Test for NH A

+4 E A:

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

red, NHA

+4 E Apresent.

C.T. for NHA

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

then NHA

+4 E Aconfirmed.

(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]

A

If yellow ppt.then K+ presentE A A

If no yellow ppt.then Na+ presentE

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 CO A

2−3 E A, S2−, PO A

3−4 E A, BO A

3−3 E Aare easily detected during the dry tests for

anions. Halides (Cl−, Br−, I−) and nitrate (NOA

−3 E A) may also be detected in dry tests if the tests for

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

2−4 E A. 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.


235

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

S2−, CO A

2−3 E Ato 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, POA

3−4E

A,

BO A

3−3 E Aand SOA

2−4E

A, while use sodium carbonate extract only for the detection of S2−, NOA

−2 E A, NOA

−3 E

Aetc.

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 A

3−4 E Apresent

(e) White ppt. soluble in dil. HNO3

BO A

3−3 E Apresent

(2) Neutral solution or Water solution + Ba(NO3)2 solution.

White ppt. insoluble in dil. HNO3.

SO A

2−4 E Apresent

(3) Sodium carbonate extract + Lead acetate.

Black ppt. soluble in HNO3 S2− present


236

(4) Sodium carbonate extract or W.S. + dil. acetic acid + FeSO4 solution.

Brown coloured solution NO A

−2 E Apresent

(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 A

−3 E Apresent

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

2−3 E Apresent

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 A( )SO2−4 E A:

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

PbSO4, SOA

2−4 E Ais 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, SOA


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.


237

6. Nitrite (NOA

−2E A) :

(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, NOA

−2 E Ais confirmed.

OR

(b) Mixture + dil. H2SO4 and warm. If brown fumes of NO2 gas, then NOA

−2 E Ais

confirmed.

7. Nitrate (NOA

−3E A) :

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 NOA

−3 E Ais confirmed.

8. Carbonate (CO A

2−3 E A) :

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

which turns lime water milky, COA


(Hint : No confirmatory test for POA

3−4 E Aand BO A

3−3 E Ais 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 : (NOA

−3 E Aand 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 NOA

−3 E A.

RESULT: Basic Radicals (positive) Acidic Radicals (negative) 1. 2.

1. 2.

VIVA VOCE 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 ?


238

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 COA

2−3 E A?

18. What is chromyl chloride test ?

19. What is C.T. for NOA

−3 E A?

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 ?

- - -

238

Appendices

STOCK SOLUTIONS (A) Gravimetric Estimations :

1. Ferrous ammonium sulphate solution : Weigh 80.0 g of Ferrous ammonium sulphate. Transfer it to a 250 ml beaker. Add 100 ml distilled water and stir the solution to dissolve the salt. Add 2 ml concentrated H2SO4 to avoid hydrolysis and precipitation of iron. Transfer this clear solution to a 250 ml volumetric flask and dilute it upto the mark with distilled water. (Give only 10 - 15 ml of this solution for estimation in a 100 ml volumetric flask.)

2. Nickel sulphate solution : Weigh 6.25 g NiSO4.7H2O. Transfer it to a 250 ml beaker. Add 100 ml distilled water and stir the solution to dissolve the salt. Add little concentrated H2SO4 (few drops) to make clear solution. Transfer this clear solution to a 250 ml volumetric flask and dilute it upto the mark with distilled water. (Give only 10 - 15 ml of this solution for estimation in a 100 ml volumetric flask.)

3. 1% alcoholic solution of DMG : Dissolve 0.25 g of dimethyl glyoxime in 100 ml ethyl alcohol. Transfer this solution to 250 ml volumetric flask and dilute upto the mark with ethyl alcohol.

4. Ammonium aluminium sulphate solution : Weigh 75.0 g of ammonium aluminium sulphate [(NH4)2SO4.Al2(SO4)3.24H2O]. Transfer it to a 250 ml. beaker. Add 100 ml distilled water and stir the solution to dissolve the salt. Add little concentrated H2SO4 (few drops) to make clear solution. Transfer this clear solution to a 250 ml volumetric flask and dilute it upto the mark with distilled water. (Give only 10 - 15 ml of this solution for estimation in a 100 ml volumetric flask.)

5. Barium chloride solution : Weigh 11.0 g of BaCl2.2H2O. Transfer it to a 250 ml. beaker. Add 100 ml distilled water and stir the solution to dissolve the salt. Add little concentrated HCl (few drops) to make clear solution. Transfer this clear solution to a 250 ml volumetric flask and dilute it upto the mark with distilled water. (Give only 10 - 15 ml of this solution for estimation in a 100 ml volumetric flask.)

(B) Volumetric Estimations :

1. Mn by Volhard's Method :

(a) 0.05 N KMnO4 solution : Weigh 1.58 g of KMnO4. Transfer it to a 250 ml beaker. Add 100 ml distilled water to dissolve the solid. Transfer the resulting solution to a 1000 ml volumetric flask and dilute upto the mark with distilled water.

T.Y.B.Sc. Practical Chemistry (Inorganic Practicals) Appendices

239

(b) Manganese stock solution : Weigh 6.336 g of MnSO4.H2O. Transfer it to a 250 ml. beaker. Add 100 ml distilled water and stir the solution to dissolve the salt. Add little concentrated H2SO4 (few drops) to make clear solution. Transfer this clear solution to a 250 ml volumetric flask and dilute it upto the mark with distilled water. (Give only 10 ml of this solution for estimation in a 100 ml volumetric flask).

2. Estimation of phosphate (PO3−4 ) from fertilizer :

(a) 1 N NaOH : 40 g of NaOH dissolved in 1000 ml water, dilute to 1000 ml with distilled water.

(b) 3% ammonium molybdate solution : Dissolve 15 g ammonium molybdate in 500 ml distilled water.

3. Analysis of Brass :

(a) 0.05 N sodium thiosulphate solution : Weigh 12.5 g of Na2S2O3 ⋅ 5H2O. Transfer it to a 250 ml beaker. Add 100 ml distilled water and stir the solution to dissolve the solid. Transfer this clear solution to a 1000 ml volumetric flask and dilute it upto the mark with distilled water.

(b) 2 N NaOH solution : Weigh 80.0 g of NaOH. Transfer it to a 250 ml. beaker. Add

100 ml distilled water and stir the solution to dissolve the salt. Transfer this clear

solution to a 1000 ml volumetric flask and dilute it upto the mark with distilled water.

(c) 2 N acetic acid : Accurately measure 118.0 ml of glacial acetic acid. Carefully transfer

it to 1000 ml volumetric flask. Add distilled water to make upto the mark that is final

volume of 1000 ml.

(d) 5 % KI solution : Weigh 12.5 g of KI. Transfer it to 250 ml beaker. Add about 100 ml

distilled water. Dissolve the salt and transfer this clear solution to a 250 ml volumetric

flask. Dilute the solution upto the mark with distilled water.

4. % Purity of Sodium Chloride :

(a) 0.05 N silver nitrate solution : Weigh 8.494 g of AgNO3. Transfer it to a 250 ml

beaker. Add 100 ml distilled water and stir the solution to dissolve the solid. Transfer

this clear solution to a 1000 ml volumetric flask and dilute it upto the mark with

distilled water.

(b) 5 % potassium chromate solution : Weigh 5.0 g of K2CrO4. Dissolve it in little

distilled water and dilute it to 100 ml with distilled water.


240

(C) Inorganic Preparations :

1. Potassium Trioxalato Ferrate :

(a) 10% oxalic acid : Weigh 25.0 g of H2C2O4 ⋅ 2H2O. Transfer it to a 250 ml. beaker. Add 100 ml distilled water and stir the solution to dissolve the solid. Transfer this clear solution to a 250 ml volumetric flask and dilute it upto the mark with distilled water.

(b) 0.05 N KMnO4 solution : As given under the experiment Manganese by Volhard's method.

(c) 2 N H2SO4 : Accurately, measure 56.0 ml of concentrated sulphuric acid. Take about 500 ml of distilled water in a 1000 ml volumetric flask. Add above measured sulphuric acid slowly into the volumetric flask with constant stirring. Finally dilute the contents of the flask with distilled water upto the mark.

2. Tetramine Copper (II) Sulphate :

As given under the experiment Analysis of Brass.

3. Potash Alum :

(a) 0.01 M zinc sulphate solution : Weigh 2.87 g of ZnSO4 ⋅ 7H2O. Transfer it to a 250 ml beaker. Add 100 ml distilled water and stir the solution to dissolve the solid. Transfer this clear solution to a 1000 ml volumetric flask and dilute it upto the mark with distilled water.

(b) 0.01 M EDTA : Weigh 3.722 g of EDTA. Transfer it to a 250 ml beaker. Add 100 ml distilled water and stir the solution to dissolve the solid. Transfer this clear solution to a 1000 ml volumetric flask and dilute it upto the mark with distilled water.

(c) Eriochrome black T-indicator : Weigh 0.2 g of indicator powder. Add to it 15 ml triethanolamine and dilute it by adding 5 ml ethyl alcohol. Stir well to dissolve the powder. Store in an indicator bottle.

(D) Colorimetric Estimations :

1. Estimation of Iron :

(a) Unknown iron solution : 8.64 g ferric ammonium sulphate + 100 ml conc. HCl dilute to 1 litre with distilled water (concentration 1 mg/ml).

(b) Known iron solution : Take 100 ml unknown iron solution, dilute to 1 litre with distilled water (concentration 0.1 mg/ml).

2. Estimation of Cobalt :

(a) Unknown cobalt solution : 2.018 g cobaltous chloride in one litre distilled water (concentration 1 mg/ml). [Give 6 to 9 ml stock solution in a 100 ml volumetric flask].

(b) Known cobalt solution : Take 100 ml unknown cobalt solution and dilute it to 1 litre with distilled water (concentration 0.1 mg/ml).


241

3. Estimation of Titanium :

(a) Unknown titanium solution : 11.045 g of potassium titanyl oxalate + 23 g (NH4)2SO4 + 250 ml concentrated H2SO4 boil, cool and dilute to one litre with distilled water (concentration 5 mg/ml). [Give 15 to 20 ml stock solution in a 100 ml volumetric flask].

(b) Known titanium solution : Take 200 ml of unknown titanium solution and dilute to 1 litre with 2 N H2SO4 (concentration 1 mg/ml).

4. Estimation of Iron by Solvent Extraction :

(a) Known Fe3+ solution : Weigh 0.86 g of ferric ammonium on a watch glass. Transfer it to 250 ml beaker, add 50 ml and stir the solution to dissolve the salt. And 1 ml concentrated H2SO4 to it to avoid hydrolysis and precipitation of Fe3+ as Fe(OH)3. If solution is not clear add few drops of H2SO4, to make it clear. Transfer this clear solution to a 1000 ml volumetric flask and dilute upto the mark with distilled water. This solution contains 0.1 mg of Fe3+ per ml of the solution.

(b) Unknown Fe3+ solution : Weigh 1.08 g of ferric ammonium sulphate on a watch glass. Transfer it to 250 ml beaker add 50 ml distilled water and stir the solution to dissolve the salt. And 1 ml concentrated H2SO4 to it to avoid hydrolysis and precipitation of Fe3+ as Fe(OH)3. If solution is not clear add few drops of H2SO4 to make it clear. Transfer this clear solution to a 1000 ml volumetric flask and dilute upto the mark with distilled water. This solution contains 0.125 mg of Fe+3, per ml of the solution.

(c) 6% 8-hydroxyquinoline solution : Weigh 60 g of 8-hydroxyquinoline on a watch glass. Transfer it to a 250 ml beaker. Add 100 ml chloroform and stir the solution well to dissolve 8-hydroxyquinoline. Transfer the clear solution to a 1000 ml volumetric flask and dilute it upto the mark with chloroform. This is 6% solution of 8-hydroxyquinoline.

(E) Column Chromatography :

(a) Ferric ion solution : 13.5 g FeCl3 ⋅ 2H2O + 20 ml conc. HCl. Dissolve and dilute it to 100 ml with distilled water.

(b) Cobalt solution : 8 g CoCl2 ⋅ 6H2O + 20 ml conc. HCl. Dissolve and dilute it to 100 ml with distilled water.

(c) Copper solution : 11.2 g CuSO4 ⋅ 5H2O + 20 ml conc. HCl. Dissolve and dilute it to 100 ml with distilled water.


242

(d) Zinc solution : 7.9 g ZnSO4 ⋅ 7H2O + 10 ml conc. HCl. Dissolve and dilute it to 100 ml with distilled water.

(e) Aluminium solution : 27.9 g Al2(SO4)3 ⋅ 18H2O + 20 ml conc. HCl. Dissolve and dilute it to 100 ml with distilled water.

(f) Nickel solution : 4.42 g NiSO4 ⋅ 7H2O + 10 ml conc. HCl. Dissolve and dilute it to 100 ml with distilled water.

SOLUTIONS

(C) Inorganic Preparations – Solutions for Volumetric Analysis :

1. 0.05 N KMnO4 : Dissolve 1.58 g A.R. grade KMnO4 in 500 ml distilled water and dilute to 1 litre.

2. 0.05 N HCl : Dilute 5 ml conc. HCl to 1 litre with distilled water.

3. 0.025 N Na2S2O3 : Dissolve 6.25 g Na2S2O3 in 500 ml distilled water and dilute to 1 litre.

4. 0.01 M EDTA : Dissolve 3.72 g of disodium salt of EDTA in 500 ml distilled water and finally dilute to 1 litre.

(D) Colorimetry : (1) Colorimetric Estimation of Iron : (i) Iron (III) solution of unknown concentration (1 mg/ml) : Dissolve 7.022 g of A.R.

ferric ammonium sulphate Fe2(SO4)3 ⋅ (NH4)2SO4 ⋅ 24H2O in 100 ml distilled water. Add 100 ml of A.R. grade HCl and dilute it to 1 litre using distilled water. This gives solution of concentration 1 mg/ml of Fe. (Give 8–12 ml of this solution in 100 ml volumetric flask to student.)

(ii) Iron (II) solution of known concentration (0.1 mg/ml) : Take 10 ml of Fe(III) solution of 1 mg/ml concentration in 100 ml volumetric flask and dilute it to 100 ml. This gives solution of 0.1 mg/ml concentration.

(iii) 1 : 1 HCl : Take 50 ml A.R. HCl and dilute it to 100 ml in volumetric flask. (iv) 3% H2O2 solution : Take 10 ml of 100 volume H2O2 and dilute it to 100 ml in

volumetric flask. (v) 20% NH4SCN solution : Dissolve 20 gm of ammonium thiocyanate in distilled water

and dilute it to 100 ml in volumetric flask. (2) Colorimetric Estimation of Cobalt : (i) Cobalt solution of unknown concentration (1 mg/ml) : Dissolve 2.018 gm of

cobaltous chloride CoCl2 ⋅ 6H2O in 50 ml distilled water and dilute it to 1 litre. This gives concentration 1 mg/ml of cobalt. (Give 8 - 12 ml solution to student as an unknown.)


243

(ii) Cobalt solution of known concentration (0.1 mg/ml) : Take 10 ml of cobalt

solution of 1 mg/ml and dilute it to 100 ml in volumetric flask. This gives

concentration 0.1 mg/ml of cobalt.

(iii) 50% sodium acetate solution : Dissolve 50 g sodium acetate in 50 ml distilled water

and dilute it to 100 ml in volumetric flask.

(iv) 1% R-nitroso salt solution : Dissolve 1 g R-nitroso salt in 50 ml distilled water and

dilute it to 100 ml in volumetric flask.

(v) 1 : 2 HNO3 solution : Dissolve 33 ml conc. HNO3 in 50 ml distilled water and dilute

to 100 ml in volumetric flask.

(3) Colorimetric Estimation of Titanium :

(i) Titanium solution of unknown concentration (5 mg/ml) : Take 2.76 g of

potassium titanyl oxalate, K2TiO(C2O4)2 ⋅ 2H2O and 5.75 g of ammonium sulphate,

(NH4)2SO4. Add carefully 62 ml of conc. H2SO4 to it. Boil, cool and dilute to 250 ml

with distilled water. This gives concentration 5 mg/ml. (Give 13 to 18 ml of this

solution to students as an unknown.)

(ii) Titanium solution of known concentration (1 mg/ml) : Take 50 ml of titanium

5 mg/ml solution and dilute it to 250 ml using volumetric flask. This gives

concentration of 1 mg/ml of titanium.

(iii) 3% H2O2 : (As above).

(iv) 2 N H2SO4 : Dilute 57 ml conc. H2SO4 to 1 litre.

(4) Solvent Extraction of Iron and Colorimetric Estimation :

(i) Iron (III) solution of unknown concentration (1.5 mg/ml) : Dissolve 12.96 g of

A.R. grade ferric ammonium sulphate, Fe2(SO4)3 ⋅ Al2(SO4)3 ⋅ 24H2O in 100 ml distilled

water and add 25 ml conc. HCl. Dilute the contents to 1 litre in volumetric flask using

distilled water. This gives concentration 1.5 mg/ml. (Give 13 – 18 ml of this solution to

students as an unknown.)

(ii) Iron (III) solution of known concentration (0.1 mg/ml) : Dilute 6.7 ml 1.5 mg/ml

iron solution to 100 ml in volumetric flask. This gives 0.1 mg/ml concentration of iron.

(iii) 6% 8-hydroxyquinoline in chloroform : Dissolve 6 g of 8-hydroxyquinoline in 40 ml

chloroform and dilute it to 100 ml using chloroform.


244

Column Chromatography :

(a) Ferric ion solution : 13.5 g FeCl3 ⋅ 2H2O + 20 ml concentrated HCl. Dissolve and dilute it to 100 ml with distilled water.

(b) Cobalt solution : 8 g CoCl2 ⋅ 6H2O + 20 ml concentrated HCl. Dissolve and dilute it to 100 ml with distilled water.

(c) Copper solution : 11.2 g CuSO4 ⋅ 5H2O + 20 ml concentrated HCl. Dissolve and dilute it to 100 ml with distilled water.

(d) Zinc solution : 7.9 g ZnSO4 ⋅ 7H2O + 10 ml concentrated HCl. Dissolve and dilute it to 100 ml with distilled water.

(e) Aluminium solution : 27.9 g Al2(SO4)3⋅18H2O + 20 ml concentrated HCl. Dissolve and dilute it to 100 ml distilled water.

(f) Nickel solution : 4.42 g NiSO4 ⋅ 7H2O + 10 ml concentrated HCl. Dissolve and dilute it to 100 ml distilled water.

- - -

245

Course - I

CH-349

ORGANIC CHEMISTRY

PRACTICALS

T.Y.B.Sc. Practical Chemistry (Organic Practicals) Separation of Binary Mixtures ….

246


COURSE – I

CH – 347

Organic Chemistry Practicals

Marks

Q.1. Binary Mixture Separation and Qualitative Analysis 40

Q.2. Organic Estimations/Preparations 30

Q.3. Oral 10


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(A) Separation of Binary Mixtures and Qualitative Analysis

(Minimum 8 Mixtures)

In S.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 [4 Mixtures] :

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 [2 Mixtures] or

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

If the nature of the binary mixture is solid-liquid, the liquid component (usually low boiling) is first separated by simple distillation. In case of liquid-liquid 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), shake

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 or 1 : 1 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

248


249

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 to 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 recrystallize 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 recrystallize 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 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. (1) Type determination of volatile liquid : Before determining the type of volatile liquid, check its miscibility in water, by mixing few drops of liquid with few drops of water in a test tube. If it is immiscible in water, then follow the procedure for type determination as given in (1) for solid-solid mixture. Note that the individual component will not precipitate out during regeneration from the filtrate (as it is a liquid) but it will separate as an oil which in turn can be separated out using a Pasteur pipette. (2) Type determination of water miscible liquids : If the liquid to be analysed is miscible in water, then for the determination of type of this liquid, we cannot use the procedure given in (1), because the reagents used are aqueous reagents in which the water miscible liquid will not separate as an oil. In such cases, following procedure is adopted.

Sr. No. Test Observation Inference 1. Litmus test* Blue litmus turns red. Liquid is acidic or

phenolic. 250


251

2. Liquid + 2-3 drops of neutral FeCl3.

Red litmus turns red. Blue/Green/Violet colour.

Liquid is basic. Liquid is phenolic.

*To test with litmus paper, dip the end of a small glass rod into the liquid and then gently touch the litmus paper with the rod. Do not dip the litmus paper into the solution. If both of the above tests are negative then the given liquid is Neutral in nature. Many volatile liquids are neutral in nature. 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’s 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 little amount of given substance in CCl4 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 color disappears with no white fumes on the glass rod.

Compound is saturated Compound is saturated Compound is unsaturated


252

C C

R

R'

R

R'

Alkene

+ Br2 C C

R

R'

R

R'

CCl4

Br

Br

colour Dibromo derivativeBrown colour disappears

Brown


2. Baeyer’s test : Take a little amount 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 Compound is saturated

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 braces. 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


253

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 :

Test Observation Inference 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

2Na + S → Na2S

Na2S + 2CH3COOH → 2CH3COσOρNa + 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

2Na + S → Na2S Na2S + Na2[Fe(CN)5NO] → Na4[Fe(CN)5NOS] Sodium nitroprusside Violet colour

Test for Nitrogen : Test Observation Inference

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 colored) and mix it with glass rod. Then

Prussian blue or bluish green color

Nitrogen present


add 1 drop of dil. H2SO4 or HCl.

Na + C + N → NaCN FeSO4 + 6 NaCN → Na4[Fe(CN)6] + Na2SO4 Sodium ferrocyanide 3Na4[Fe(CN)6] + 2Fe2(SO4)3 → Fe4[Fe(CN)6]3 + 6Na2SO4 Ferric ferrocyanide Prussian blue colour

Test for Nitrogen and Sulphur together :


Sodium fusion extract and few drops of FeCl3 solution

Blood red colour Nitrogen and sulphur both present.

Na + C + N + S → NaCNS FeCl3 + 3NaCNS → Fe(CNS)3 + 3NaCl Blood red colour

Test 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 2 NaCl + Br2 (Brown colour)CCl4

NaI + Cl2 2 NaCl + I2 (Violet colour)CCl4

Conclusion : 254


255

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


2. For phenolic (–OH) group :

Take a pinch of the given compound + 1 ml of alcohol + 2-3 drops of aq. FeCl3.

Blue / green / violet colour or red brown colour

Phenol is confirmed.

3 ArOH + aq. FeCl3 ⎯⎯→ Fe(OAr)3 + 3HCl Blue/Green/Violet or Red brown colour


3. Carbonyl (–CO–) group : Take a pinch of the given

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'

H

2, 4-DNP Yellow or Red ppt.

Tests for Aldehydes : (a) Tollen’s test :

256


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 )]3 23OH Ar-COO + 2H O2 + 2Ag + 4NH3

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+ H O2

(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

Aldehyde group present

H N2

NH2

SO H3

C

O

R H

Aldehyde

+ 2

Schiff's reagent (colourless)

C

H N2

CH-HNR

HO S3

Pink

N-CHR

SO H3

NH2

�2H O2

+ 2H SO2 3

257


If aldehyde group is absent assume 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

Methyl ketone Iodoform

Na

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

�

(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) RCOOR’ + NH2OH.HCl NaOH

⎯⎯⎯→ R−CO−NH−OH + R’−OH + NaCl + H2O

258


(ii)

3R−CO−NH−OH + FeCl3Hydroxamic acid ⎯⎯→

O

HN

O

Fe

3

Red-violet colour

R

+ 3HCl


5. Carbohydrate group : (a) Molish test : Compound + 3-4 drops of α-naphthol in alcohol. Shake and then add 5 drops of conc. H2SO4 from the walls of the tube.

Violet red colouration at the junction of the two layers

Carbohydrate present

OH

CHO

(CHOH)4

CH OH2

+H SO2 4

Glucose

OHC

(CHOH)4

CH OH2

�-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 + 1ml of alcohol + 6 drops CaCl2 + pinch of Zn dust. Heat to boiling for at least 4 minutes and then filter into Tollen’s reagent.

Black or grey precipitate

Nitro group present.

259


NO2

CaCl /Zn2

NHOH

Tollen's

NO

reagent+ Ag + NH3

Blackppt.

Hydroxylamine

�H O2

+ H O2

(b) Ferrous hydroxide test : Compound + freshly prepared 5% ferrous ammonium sulphate solution + 1 drop of dil. H2SO4 + excess of KOH. Shake the tube quickly

Red-brown precipitate

Nitro group is confirmed

Ar-NO2 + 6 Fe(OH)2 + 4 H2O Ar-NH2 + 6 Fe(OH)3

BrownBluecolour


7. Test for Primary/ Secondary/Tertiary amines :

Take a pinch of the given 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

(ii) C H N6 5

R

R

(i) HONO

(ii) NaOHN

R

R

C H6 4 N O

Tertiaryamine

Green ppt. (p-Nitroso compound)

(iii) C H NH6 5 2

HONO

Primary amine

C H -N6 5 N�-Naphthol

in NaOH

N

OH

N C H6 5

Orange dye

(i) C H NHR6 5

HONO

Secondaryamine

C H6 5(NaNO + H SO )2 2 4

Yellow oil (N-Nitroso compound)or solid

N N O

R

(NaNO + H SO )2 2 4

260



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 + HCl�

Ar-NH2

Primaryamine

+ R-COO Na

(ii)HONO

Primary amine

Ar N� NCl�-Naphthol

in NaOH

N

OH

N Ar

Orange dye

Ar NH� 2 (NaNO + HCl)2

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

C

X

NH2R

+

C

X

R O Na

X = O, AmideX = S, Thioamide

NaOH�

+ NH3


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

261


262

R-X + NaOH Ri)

ii) NaX + AgNO3

-OH + NaXHNO3

X

AgX + NaNO3

= 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 holds 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.


Thermometerattached with capillary

Thiele's tube

Paraffin oil

Burner

Sealed capillary(ii)

Sealed capillarywith compound

(iii)

Sealed capillaryattached to thethermometer

(iv)

Capillary(i)

(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 :

263


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)

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. 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 264


265

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

Lessaigne’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.

- - -

265

(B) Organic Estimations (Any Four)

Experiment No. 1 Estimation of Acetamide

Aim :

To determine the amount of amide (acetamide) in the given solution.

Principle :

Amides are carboxylic acid derivatives when it is heated with alkali it undergoes hydrolysis to form the salts of carboxylic acids with evolution of ammonia gas.

Acetamide also on hydrolysis by an alkali forms the salts of carboxylic acids with evolution of ammonia gas. For estimation (i.e. to find out the amount of acetamide) a known amount of standard alkali (NaOH or KOH) is added to acetamide solution. After hydrolysis the unused alkali is determined by titrating it against a standard 0.05 N HCl solution.

The amount of acetamide present in the given sample solution is estimated by measuring the volume of alkali required for hydrolysis in terms of 0.05 N HCl solution.

Reactions :

General reaction

RCONH2 + KOH Hydrolysis

⎯⎯⎯⎯⎯→ RCOOK + NH3 ↑

Reactions with acetamide

1. Back titration :

CH3CONH2 + KOH Hydrolysis

⎯⎯⎯⎯⎯→ CH3COOK + NH3 ↑

Acetamide (Excess) Potassium acetate

2. Blank titration :

KOH + HCl ⎯⎯→ KCl + H2O

(Unreacted)

Chemicals :

(i) Acetamide solution (10-15 ml) (unknown concentration)

(ii) 0.5 N KOH

(iii) 0.05 N HCl (exact)

(iv) Phenolphthalein indicator.

T.Y.B.Sc. Practical Chemistry (Organic Practicals) Organic Estimations

266

Procedure : Part I : Back Titration : 1. Transfer the given acetamide solution in the 150 ml conical flask. Add 10 ml 0.5 N

KOH solution by means of common burette to the given solution of acetamide and one test tube of water.

2. Add one porcelain piece in the flask (used for even heating). Place funnel in the mouth of the flask and heat it on a sand bath (one hour) till the evolution of ammonia is complete (as tested by turmeric paper, it should not turn red).

3. Then cool the flask and dilute the contents to 100 ml in volumetric flask with distilled water. Mix it well and fill this diluted solution in the burette No. 1.

4. Fill the burette No. 2 with 0.05 N HCl (exact). 5. Take exactly 9 ml of diluted acetamide solution in a conical flask from burette No. 1.

Add one test tube of distilled water and two drops of phenolphthalein indicator. Pink colour is developed.

6. Titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as Y1 ml.

7. To the same conical flask add 1 ml of diluted acetamide solution from burette No. 1 and titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as Y2 ml.

8. Again to the same conical flask add 1 ml of diluted acetamide solution from burette No. 1 and titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as Y3 ml.

Procedure : Part II : Blank Titration : 1. Take 10 ml of 0.5 N KOH (approx.) solution from the common burette and dilute to

100 ml in a 100 ml volumetric flask. And fill in burette No. 1 again. This will form the 0.05 N KOH solution.

2. Fill the burette No. 2 by 0.05 N HCl (exact). 3. Take exactly 9 ml of 0.05 N KOH solution in a conical flask from burette No. 1. Add

one test tube of distilled water and two drops of phenolphthalein indicator. Pink colour is developed.

4. Titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as X1 ml.

5. To the same conical flask add 1 ml of 0.05 N KOH solution from burette No. 1 and titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as X2 ml.

6. Again to the same conical flask add 1 ml of 0.05 N KOH solution from burette No. 1 and titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as X3 ml.


267

Observation Table : Part I : Back Titration : Burette No. 1 : Diluted acetamide solution Burette No. 2 : 0.05 N HCl Indicator : Phenolphthalein End point : Pink to colourless

Reading Reading (B.R.) ml

Mean B.R. (ml) I II III

Burette No. 1 (acetamide solution)

9 ml 10 ml 11 ml 10 ml

Burette No. 2 (HCl) Y1 = …… ml

Y2 = …… ml

Y3 = …… ml

Y = Y1 + Y2 + Y3

3

Y = …… ml.

Part II : Blank Titration : Burette No. 1 : 0.05 N KOH solution Burette No. 2 : 0.05 N HCl Indicator : Phenolphthalein End point : Pink to colourless



Burette No. 1 (KOH) 9 ml 10 ml 11 ml 10 ml

Burette No. 2 (HCl) X1 = …… ml X2 = …… ml X3 = …… ml X =

X1 + X2 + X3

3

X = …… ml.

Calculations : Volume of KOH required for 10 ml of dil. acetamide solution in terms of 0.05 N HCl = x − y = …… ml

RCONH2 + KOH Hydrolysis⎯⎯⎯⎯→ RCOOK + NH3↑ and

KOH + HCl ⎯⎯⎯⎯→ KCl + H2O According to reactions : 1 mole of KOH = 1 mole of acetamide 1000 ml 1 N KOH = 59 gms of acetamide i.e. 1000 ml 1 N HCl = 59 gms of acetamide

So, 1 ml 0.05 N HCl = 0.05 × 59

1000 = 0.00295 gms of acetamide.

(X − Y) ml 0.05 N HCl = (X − Y) × 0.00295 g of acetamide = ‘A’ gms of acetamide i.e. 10 ml diluted solution contains = ‘A’ gms of acetamide 100 ml diluted solution contains = (A × 10) = Z gms of acetamide


Result Table :

1. Volume of 0.05 N HCl (exact) required for blank titration. X = …….. ml

2. Volume of 0.05 N HCl (exact) required for back titration. Y = …….. ml

3. Volume of KOH used by 10 ml of diluted solution in terms of 0.05 N HCl (exact)

(X − Y) = …… ml

4. Amount of acetamide in the given solution ‘Z’ = …….. gms.

Experiment No. 2

Estimation of Glucose Aim : To determine the amount of glucose in the given solution by hypoiodite method. Reactions :

I2 + Na CO2 3 NaI + NaOI + CO2

Sodium hypoiodide

CHO

(CHOH)4

CH OH2

Glucose

+ NaOI

COOH

(CHOH)4

CH OH2

Gluconic acid

+ NaI

NaOI I+ Na + 2HCl 2NaCl I+ H O +2 2

(Unreacted)

I2 + 2Na S O2 2 3 2NaI + Na S O2 4 6 Principle : Glucose has an aldehydic functional group and it is a reducing sugar. Iodine in Na2CO3 solution produces excess NaOI. This oxidizes quantitatively glucose into gluconic acid. On acidifying the reaction mixture with HCl, the unreacted NaOI liberates free iodine which can be titrated against standard Na2S2O3 solution using starch indicator. Requirements :

1. Glucose solution given in a 100 ml measuring flask. 2. 0.05 N Na2S2O3 solution. 3. 0.05 N I2 solution. 4. 15% Na2S2O3 solution. 5. 1 N HCl. 6. Starch solution.

268


269

Procedure :

Part I : Back Titration

1. Fill the burette-1 with 0.05 N Na2S2O3 solution.

2. Fill the burette-2 with 0.05 N I2 solution.

3. Dilute the given glucose solution with distilled water upto the mark in a 100 ml measuring flask and fill this diluted solution in burette-3.

4. Take three 250 ml stoppered bottles and number them as 1, 2 and 3.

5. Take 10 ml of diluted glucose solution by using burette - 3 into a bottle No. 1. Add 5 ml of 15% Na2CO3 solution by measuring cylinder. Add to it 10 ml of 0.05 N I2 solution from burette-2. Similarly prepare other two bottles and shake well.

6. Keep these three bottles for 30 minutes in dark for the completion of reaction.

7. After 30 minutes, take out bottle-1 and add 10 ml of 1 N HCl by a cylinder and shake well. The solution becomes orange red due to liberation of I2.

8. Titrate the liberated iodine slowly against 0.05 N Na2S2O3 from burette-1 till the solution becomes faint yellow.

9. Then add 1 ml of starch indicator, blue color will be developed, add more 0.05 N Na2S2O3 dropwise from the burette-1 till blue color disappears. Record the burette reading (Y1).

10. Similarly, take reading for other two bottles and record the readings (Y2 and Y3).

Part II : Blank Titration

1. Take 10 ml of 0.05 N I2 solution from burette - 2 in a stoppered bottle. Add 5 ml of 15% Na2CO3 solution by measuring cylinder and one test tube of distilled water. (Do not wait for 30 minutes).

2. Then add 10 ml of 1 N HCl by a cylinder and shake well. The solution becomes orange red due to liberation of I2.

3. Titrate the liberated iodine slowly against 0.05 N Na2S2O3 from burette-1 till the solution becomes faint yellow.

4. Then add 1 ml of starch indicator, blue colour will be developed, add more 0.05 N Na2S2O3 dropwise from the burette-1 till blue colour disappears. Record the burette reading (X1).

5. Similarly, repeat the procedure twice and record the readings (X2 and X3).


270

Observation Table : Part I : Back Titration : Burette No. 1 : 0.05 N Na2S2O3 solution Burette No. 2 : 0.05 N I2 Indicator : Starch solution End point : Blue to colourless

Readings I II III C.B.R.

Burette-2 0.05 N I2 10 ml 10 ml 10 ml

Burette-1 0.05 N Na2S2O3 Y1 = …… ml Y2 = …… ml Y3 = …… ml Y = …… ml

Part II : Blank Titration : Burette No. 1 : 0.05 N Na2S2O3 solution Burette No. 2 : 0.05 N I2 Indicator : Starch solution End point : Disappearance of blue colour

Readings I II III C.B.R.

Burette-2 0.05 N I2 10 ml 10 ml 10 ml

Burette-1 0.05 N Na2S2O3 X1 = …… ml X2 = …… ml X3 = …… ml X = …… ml

Volume of iodine consumed in terms of 0.05 N Na2S2O3 solution for 10 ml of diluted glucose solution. 2Na2S2O3 ≡ I2 ≡ Glucose Calculations : From the reactions, 2 moles of Na2S2O3 = 1 mole of glucose = 180 gms of glucose So, 1000 ml 1 N Na2S2O3 = 90 gms of glucose So, 1 ml 0.05 N Na2S2O3 = 0.0045 gms of glucose So, (X − Y) ml 0.05 N Na2S2O3 = (X − Y) 0.0045 gms of glucose = ………. ‘A’ gms of glucose Thus 10 ml diluted solution of glucose contains = ‘A’ gms of glucose So, 100 ml diluted solution of glucose contains = A × 10 = Z gms of glucose Result Table :

1. Volume of 0.05 N Na2S2O3 for blank titration X = ………. ml

2. Volume of 0.05 N Na2S2O3 for back titration Y = ………. ml

3. Volume of iodine consumed in terms of 0.05 N Na2S2O3 solution for 10 ml of diluted glucose solution

(X − Y) = ………. ml

4. Amount of glucose in the given solution Z = ………. g


271

Experiment No. 3

Estimation of Ethyl Benzoate Aim :

To determine the amount of ester (ethyl benzoate) in the given solution.

Principle :

Esters are also carboxylic acid derivatives, when it is heated with alkali it undergoes hydrolysis to form the salts of carboxylic acids and alcohol. The ease of reaction depends on the solubility of esters. Water-soluble esters are generally hydrolyzed by aqueous alkali and water insoluble esters by alcoholic alkali. Ethyl benzoate is insoluble in water.

Ethyl benzoate on hydrolysis by an alcoholic alkali (NaOH or KOH) gives benzoate salts and alcohol. For estimation (i.e. to find out the amount of ethyl benzoate), a known amount of standard alkali (NaOH or KOH) is added to ethyl benzoate solution. After hydrolysis the unused alkali is determined by titrating it against a standard 0.05 N HCl solution.

The amount of ethyl benzoate present in the given sample solution is estimated by measuring the volume of alkali required for hydrolysis in terms of 0.05 N HCl solution.

Reactions :

General reaction

RCOOR' + KOH Hydrolysis

⎯⎯⎯⎯⎯→ RCOOK + R'OH

Reactions with ethyl benzoate

1. Back titration :

C6H5COOC2H5 + Alc. KOH Hydrolysis

⎯⎯⎯⎯⎯→ C6H5COOK + C2H5OH

Ethyl benzoate (Excess) Potassium benzoate

2. Blank titration :

Alc. KOH + HCl ⎯⎯→ KCl + H2O

(Unreacted)

Chemicals :

(i) Unknown ethyl benzoate solution (10-15 ml)

(ii) 0.5 N alcoholic KOH (approx.)

(iii) 0.05 N HCl (exact)

(iv) Phenolphthalein indicator.


272

Procedure :

Part I : Back Titration :

1. Transfer the given ethyl benzoate solution in 100 ml R.B. flask. Add to it 10 ml 0.5 N alcoholic KOH solution by means of common burette.

2. Add one porcelain piece in the R.B. flask (used for even heating). Attach water condenser to the flask and heat it on a water bath for one hour.

3. Then cool the flask and dilute the contents to 100 ml in volumetric flask with distilled water. Mix it well and fill this diluted solution in the burette No. 1.

4. Fill the burette No. 2 by 0.05 N HCl (exact).

5. Take exactly 9 ml of diluted benzoate solution in a conical flask from burette No. 1. Add one test tube of distilled water and two drops of phenolphthalein indicator. Pink colour is developed.

6. Titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as Y1 ml.

7. To the same conical flask add 1 ml of diluted benzoate solution from burette No. 1 and titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as Y2 ml.

8. Again to the same conical flask add 1 ml of diluted benzoate solution from burette No. 1 and titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as Y3 ml.

Part II : Blank Titration : 1. Take 10 ml of 0.5 N KOH (approx) solution from the common burette and dilute to

100 ml in a 100 ml volumetric flask. And fill in burette No. 1 again. This will form the 0.05 N KOH solution.

2. Fill the burette No. 2 with 0.05 N HCl (exact). 3. Take exactly 9 ml of 0.05 N KOH solution in a conical flask from burette No. 1. Add

one test tube of distilled water and two drops of phenolphthalein indicator. Pink colour is developed.

4. Titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as X1 ml.

5. To the same conical flask add 1 ml of 0.05 N KOH solution from burette No. 1 and titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as X2 ml.

6. Again to the same conical flask add 1 ml of 0.05 N KOH solution from burette No. 1 and titrate it against 0.05 N HCl (exact) solution from burette No. 2 till the pink colour disappears. Note down the burette reading as X3 ml.


273

Observation Table : Part I : Back Titration : Burette No. 1 : Diluted ethyl benzoate solution Burette No. 2 : 0.05 N HCl Indicator : Phenolphthalein End point : Pink to colourless



Burette No. 1 (benzoate solution)

9 ml 10 ml 11 ml 10 ml

Burette No. 2 (HCl) Y1 = …… ml

Y2 = …… ml

Y3 = …… ml

Y = Y1 + Y2 + Y3

3

Y = …… ml.

Part II : Blank Titration : Burette No. 1 : 0.05 N KOH solution Burette No. 2 : 0.05 N HCl Indicator : Phenolphthalein End point : Pink to colourless



Burette No. 1 (KOH) 9 ml 10 ml 11 ml 10 ml

Burette No. 2 (HCl) X1 = …… ml

X2 = …… ml

X3 = …… ml

X = X1 + X2 + X3

3

X = …… ml.

Calculations : Amount of KOH used by 10 ml of diluted ethyl benzoate solution in terms of 0.05 N HCl = (x − y) = …… ml

RCOOR' + alc. KOH Hydrolysis⎯⎯⎯⎯→ RCOOK + H2O

KOH + HCl ⎯⎯⎯⎯→ KCl + H2O According to reactions : 1 mole of KOH = 1 mole of ethyl benzoate 1000 ml 1 N KOH = 150 gms of ethyl benzoate i.e. 1000 ml 1 N HCl = 150 gms of ethyl benzoate

So, 1 ml 0.05 N HCl = 0.05 × 150

1000 = 0.0075 gms of ethyl benzoate.

(X − Y) ml 0.05 N HCl = (X − Y) × 0.0075 g of ethyl benzoate = ‘A’ gms of ethyl benzoate i.e. 10 ml diluted solution contains = ‘A’ gms of ethyl benzoate 100 ml diluted solution contains = (A × 10) = Z gms of ethyl benzoate


274

Result Table :

1. Amount of 0.05 N HCl (exact) required for blank titration X = …….. ml

2. Amount of 0.05 N HCl (exact) required for back titration Y = …….. ml

3. Amount of KOH used by 10 ml of diluted solution in terms of 0.05 N HCl (exact)

(X − Y) = …… ml

4. Amount of acetamide in the given solution ‘Z’ = …….. gms.

Experiment No. 4 Molecular Weight of Monobasic/Dibasic Organic Acid

Aim : Determination of molecular weight of given monobasic/dibasic acid by volumetric method. Principle : The molecular weight of a compound is the weight of one molecule of it, as compared to the weight of hydrogen as one. The methods commonly employed for the determination of molecular weight of organic substances are physical and chemical methods. The determination of molecular weight of an organic acid is usually carried out by two methods : (i) The volumetric method. (ii) The silver salt method. The volumetric method is based upon the neutralization of a weighed amount of the acid by a standard alkali. The equivalent weight as well as molecular weight of organic acid can be determined by this method provided that basicity is known. Reactions : For monobasic acid R⎯COOH + NaOH ⎯⎯→ R⎯COONa + H2O For dibasic acid

COOH⏐COOH

+ 2NaOH ⎯⎯→

COONa⏐COONa

+ 2H2O

When basicity of acid is given, its molecular weight can be calculated from its equivalent weight, e.g. For monobasic acid, Molecular weight = Basicity × Equivalent weight = 1 × Equivalent weight. For dibasic acid, Molecular weight = Basicity × Equivalent weight = 2 × Equivalent weight.


275

Chemicals : (i) Monobasic/dibasic acid sample solution in 100 ml volumetric flask (ii) 0.05 N NaOH (approx.) (iii) Potassium hydrogen phthalate (iv) Phenolphthalein indicator. Procedure : Part I : Preparation of 0.05 N potassium hydrogen phthalate : Weigh accurately 1.02 g of A.R. grade potassium hydrogen phthalate. Dissolve it in little amount of water. Transfer it carefully to 100 ml volumetric flask. Dilute it upto the mark with distilled water. Pour the whole solution in a beaker. Part II : Standardization of 0.05 N NaOH : 1. Fill the burette No. 1 with 0.05 N NaOH. 2. Fill the burette No. 2 with 0.05 N K-H-Phthalate. 3. Take exactly 9 ml 0.05 N K-H-Phthalate solution in a conical flask from burette No. 2.

Add one test tube of distilled water and two drops of phenolphthalein indicator. 4. Titrate it against 0.05 N NaOH solution from burette No. 1 till the faint pink colour

appears. Note down the burette reading as X1 ml. 5. To the same conical flask add 1 ml of 0.05 N K-H-Phthalate solution from burette

No. 2 and titrate it against 0.05 N NaOH solution from burette No. 1 till the pink colour appears. Note down the burette reading as X2 ml.

6. Again to the same conical flask add 1 ml of 0.05 N K-H-Phthalate solution from burette No. 2 and titrate it against 0.05 N NaOH solution from burette No. 1 till the pink color appears. Note down the burette reading as X3 ml.

Part III : Determination of molecular weight of monobasic/dibasic acid : 1. Dilute the given acid solution upto the mark with distilled water. 2. Fill the burette No. 1 with 0.05 N NaOH. 3. Fill the burette No. 2 with diluted acid solution. 4. Take exactly 9 ml diluted acid solution in a conical flask from burette No. 2. Add one

test tube of distilled water and two drops of phenolphthalein indicator. 5. Titrate it against 0.05 N NaOH solution from burette No. 1 till the faint pink colour

appears. Note down the burette reading as Y1 ml. 6. To the same conical flask add 1 ml of diluted acid solution from burette No. 2 and

titrate it against 0.05 N NaOH solution from burette No. 1 till the pink colour appears. Note down the burette reading as Y2 ml.

7. Again to the same conical flask add 1 ml of diluted acid solution from burette No. 2 and titrate it against 0.05 N NaOH solution from burette No. 1 till the pink colour appears. Note down the burette reading as Y3 ml.

Observation Table : Part II : Standardization of 0.05 N NaOH Burette No. 1 : 0.05 N NaOH solution Burette No. 2 : 0.05 N K-H-Phthalate Indicator : Phenolphthalein End point : Colourless to pink


276



Burette No. 2 (K-H-phthalate)

9 ml 10 ml 11 ml 10 ml

Burette No. 1 (NaOH) X1 = …… ml

X2 = …… ml

X3 = …… ml X =

X1 + X2 + X3

3

X = …… ml. Let the constant burette reading be ‘X’ ml. Part III : Determination of Molecular Weight of Monobasic/Dibasic Acid : Burette No. 1 : 0.05 N NaOH solution Burette No. 2 : Diluted acid solution Indicator : Phenolphthalein End point : Colourless to pink



Burette No. 2 (Diluted acid solution)

9 ml 10 ml 11 ml 10 ml

Burette No. 1 (NaOH) Y1 = …… ml

Y2 = …… ml

Y3 = …… ml Y =

Y1 + Y2 + Y3

3

Y = …… ml. Let the constant burette reading be ‘Y’ ml. Calculations : 1. To find the exact normality of NaOH : NaOH = C8H5O4K (K-H-Phthalate) N1V1 = N2V2 N1 × (C.B.R. = X) = 0.05 × 10

N1 = 0.05 × 10

X = ………. N

Exact normality of NaOH = N1 = ………. N 2. The strength of given acid is calculated as follows : W = Weight of acid present in the given solution [Ask for the weight (W) of acid to the examiner] Volume of the alkali used for complete neutralization of the acid solution = Y ml. Normality of alkali solution = N1 Then we can write,

Y ml of alkali (NaOH) solution having normality N1 = W10 gms acid.

[Since W gms. of acid is present in 100 ml of diluted solution] 1000 ml of alkali solution having normality

N = W × 1000

Y × N1 × 10 gms of acid.


277

We know that, 1000 ml of 1 N alkali solution = 1 gm equivalent of alkali. Since acids and alkalies (bases) neutralize each other in equivalent proportions,

hence, 1 gm Equivalent of acid = W × 1000


So, Equivalent weight of an acid (Z) = W × 1000


3. The molecular weight of an acid : [Ask for the basicity of acid to the examiner] Now molecular weight = Equivalent weight × Basicity = Z × Basicity For monobasic acid : Molecular weight = Equivalent weight = M1 gms For dibasic acid : Molecular weight = 2 × Equivalent weight = 2 × Z = M2. Result Table :

1. Amount of NaOH required for 10 ml of 0.05 N potassium hydrogen phthalate

X = …….. ml

2. Amount of NaOH required for 10 ml of given acid solution Y = …….. N 3. Exact normality of NaOH N1 = …… N 4. Molecular weight of given acid

VIVA VOCE 1. How will you prepare (a) 0.05 N KOH, (b) 0.05 N NaOH, (c) 0.05 N HCl ? 2. What is the difference between normal solution and molar solution ? 3. How to check the completion of hydrolysis of acetamide by NaOH solution ? 4. Write the reaction of (a) hydrolysis of acetamide by NaOH, (b) hydrolysis of ethyl

benzoate with aqueous KOH. 5. What is the principle of estimation of ethyl benzoate ? 6. What is standard solution ? Why standardization of NaOH is required ? Which

compound is used to standardize NaOH solution ? 7. Explain the terms : (a) Monobasic acid, (b) Dibasic acid, (c) Tribasic acid. Give one

example of each type of acid. 8. Write the relation between molecular weight and equivalent weight. 9. How will you prepare (a) 0.05 N and 0.02 N potassium hydrogen phthalate, (b) 0.02 N

and 0.05 N NaOH solution ? 10. What is the structure of glucose ? Why glucose is a reducing sugar? 11. Write the reaction of titration of iodine against Na2S2O3. 12. What is the reaction of glucose with sodium hypoiodide. 13. How will you prepare 0.05 N, 0.1 N and 1 N I2 solution ? 14. In iodometric titration after addition of Na2CO3 and I2 solution in glucose solution,

the reaction bottles are kept in dark for 30 minutes, why ? - - -

(C) Organic Preparations (Any Eight)

Experiment No. 1 Organic Preparations

Aim: To study the oxidation reaction. Experiment: Preparation of Adipic acid from Cyclohexanone. Principle: Cyclohexanone can be readily oxidized to adipic acid by using concentrated nitric acid as oxidizing agent. Reaction:

conc. HNO3

Adipic acid

O

Cyclohexanone

COOH

COOH

Chemicals:

(i) Cyclohexanone = 0.5 ml

(ii) Conc. HNO3 = 2.5 ml

(iii) Distilled water = 1.5 ml

Procedure:

Place 2.5 ml of conc. HNO3 in 25 ml round bottom flask. To this add 1.5 ml distilled water and 0.5 ml cyclohexanone dropwise so that the temperature does not rise above 85°C. After the addition is over reflux on a wire gauze for half an hour. Cool the flask in ice bath, filter the product on Buchner funnel. Record the practical yield and re-crystallize from hot water.

Re-crystallization:

Take the crude product and water (5 ml) in a beaker and dissolve by heating on a water bath. Filter the hot solution and cool the filtrate. Filter, dry and record the melting point and TLC (using toluene as a solvent).

Calculations: (i) Theoretical yield:

Cyclohexanone (C6H10O) ≡ Adipic acid (C6H10O4)

98g ≡ 146g

278

T.Y.B.Sc. Practical Chemistry (Organic Practicals) Organic Preparations

Therefore, 0.5g ≡ 0.74g

(ii) Practical yield = ‘A’ g

(iii) Practical % yield:

Since, 0.74 g of product = 100% yield

Therefore, ‘A’ g = A × 100

0.74 = …… %

Result Table:

Practical yield

M.P. of adipic acid

Practical % yield

Rf value of adipic acid

Solvent used for TLC

…… g

…… °C

…… %

……

……

Note:

Cyclohexanone do not show any spot on TLC

Experiment No. 2 Aim:

To study the oxidation reaction.

Experiment:

Preparation of Quinone (p-Benzoquinone) from Hydroquinone.

Principle:

Hydroquinone can be readily oxidized to quinone by using oxidizing agent like potassium dichromate and conc. H2SO4 or KBrO3.

Reaction:

279

OH O

OH O

Hydroquinone Quinone

K Cr O /conc. H SO2 2 7 2 4

or KBrO3

Chemicals: (i) Hydroquinone = 0.5 g (ii) Potassium dichromate = 1.0 g (iii) conc. H2SO4 = 1 ml


280

(iv) Distilled water = 15 ml Procedure: Take 0.5 g hydroquinone and 5 ml distilled water in a beaker (25 ml). Heat on a wire gauze to obtain a clear solution. Take 1 g potassium dichromate (K2Cr2O7) in a conical flask and dissolve in 10 ml water and add 1 ml conc. H2SO4. Shake and cool the conical flask in ice water. To this ice cold solution add hydroquinone solution (prepared above) dropwise over a period of 30 minutes with constant shaking. Do not allow the temperature to rise above 20oC. After complete addition, continue shaking for further 10 minutes. Yellow crystals of quinone separate out. Filter on a Buchner funnel and dry it well. (Note: Do not wash with water as the product is water soluble). Record the practical yield and re-crystallize from ethyl alcohol.

OR Chemicals: (i) Hydroquinone = 0.5 g

(ii) Potassium bromate = 0.4 g

(iii) conc. H2SO4 = 1 ml

(iv) Distilled water = 10 ml

Procedure:

Take 0.5 g hydroquinone, 5 ml distilled water and 0.4 g potassium bromate (KBrO3) in a beaker (25 ml). Heat on a wire gauze to obtain a clear solution. Cool the beaker and add 1 ml conc. H2SO4 dropwise with constant shaking. Cool the beaker in ice water. Yellow crystals of quinone separate out. Filter on a Buchner funnel and dry it well. (Note: Do not wash with water as the product is water soluble). Record the practical yield and re-crystallize from ethyl alcohol.

Re-crystallization:

Dissolve the crude product in minimum amount of ethyl alcohol in a beaker by heating on a water bath. Filter the hot solution and cool the filtrate. The yellow needles of quinone separate out. Filter, dry and record the melting point and TLC (using toluene as a solvent).

Calculations: (i) Theoretical yield:

Hydroquinone (C6H6O2) ≡ Quinone (C6H4O2)

110 g ≡ 108 g

Therefore, 0.5 g ≡ 0.49 g



Since 0.49 g of product = 100% yield



0.49 = …… %

Result Table:

Practical yield

M.P. of quinone

Practical % yield

Rf value of hydroquinone

Rf value of quinone


…… g

…… °C

…… %

……

……

……


To study the aromatic electrophilic substitution reaction.

Experiment:

Preparation of p-nitro acetanilide from acetanilide.

Principle: Aromatic compounds can be conveniently nitrated by the use of the nitrating mixture, which is normally a mixture of concentrated nitric acid and concentrated sulphuric acid. The function of sulphuric acid is to convert nitric acid into a highly reactive, electrophilic,

nitronium ion, NO+2, which is the effective nitrating agent. Nitration of activated aromatic

compounds is carried out under milder condition whereas deactivated rings require drastic conditions. Activating groups are o, p-directing, while deactivating groups are m-directing for electrophilic substitution.

Reaction:

HNO3 + 2H2SO4 ⎯→ NO+2 + H3O+ + 2HSO

−4

NHCOCH3

conc. HNO3

Acetanilide p-Nitro acetanilide

NHCOCH3

NO2

conc. H SO2 4

Chemicals:

(i) Acetanilide = 0.5 g

(ii) conc. HNO3 = 0.25 ml

281


282

(iii) conc. H2SO4 = 1.2 ml

(iv) Glacial acetic acid = 0.6 ml Procedure: Take 0.5 g acetanilide in 0.6 ml glacial acetic acid and add 1 ml conc. H2SO4 in a beaker. Cool the mixture in ice salt bath (5-10°C). To this add a mixture of 0.2 ml conc. H2SO4 and 0.25 ml conc. HNO3 dropwise. Maintain the temperature below 10°C during addition. After addition is over allow the mixture to attain room temperature and leave it for 30 minutes. Pour the mixture on to crushed ice (20 g) with stirring. Filter the separated product and wash with cold water. Dry the product, record the practical yield and re-crystallize it. Re-crystallization: Dissolve the crude product in minimum amount of ethyl alcohol in a beaker by heating on a water bath. Filter the hot solution and cool the filtrate. The crystals of the product separate out. Filter, dry and record the melting point and TLC (using toluene as solvent). Calculations: (i) Theoretical yield: Acetanilide (C8H9ON) ≡ p-Nitro acetanilide (C8H8O3N2) 135 g ≡ 180 g Therefore, 0.5 g ≡ 0.66 g (ii) Practical yield = ‘A’ g (iii) Practical % yield: Since 0.66 g of product = 100% yield


0.66 = …… %

Result Table:

1. Practical Yield 2. M.P. of p-nitro acetanilide 3. Practical % yield 4. Rf value of acetanilide 5. Rf value of p-nitro acetanilide 6. Solvent used for TLC

…… g …… °C …… % …… …… ……

Experiment No. 4 Aim: To study the etherification reaction. Experiment: Preparation of 2-naphthyl methyl ether from 2-naphthol.


Principle: Phenols can be methylated to give methyl ethers. Methylation can be done either by using diazomethane or dimethyl sulphate in alkaline medium. Reaction:

Dimethyl sulphate

2-Naphthol 2-Naphthyl methyl ether

NaOH

OH OCH3

Chemicals: (i) 2-Naphthol = 0.5 g (ii) Dimethyl sulphate = 0.35 ml (iii) NaOH = 0.2 g Procedure: Take 0.5 g 2-naphthol and 0.2 g NaOH in 5 ml distilled water in a beaker (25 ml). Heat on a wire gauze to obtain a clear solution. Cool the solution (10-15°C) and then add 0.35 ml dimethyl sulphate drop wise. After the addition is over, warm the mixture for one hour at 70-80oC and then cool. Filter the product and wash it with 10% sodium hydroxide solution and then with water. Dry the product, record the practical yield and re-crystallize it. Re-crystallization: Dissolve the crude product in minimum amount of ethyl alcohol in a beaker by heating on a water bath. Filter the hot solution and cool the filtrate. Filter the white crystals of the product. Dry and record the melting point and TLC (using toluene as solvent). Calculations: (i) Theoretical yield: 2-Naphthol (C10H8O) ≡ 2-Naphthyl methyl ether (C11H10O) 144 g ≡ 158 g Therefore, 0.5 g ≡ 0.54 g (ii) Practical yield = ‘A’ g (iii) Practical % yield: Since 0.54 g of product = 100% yield


0.54 = …… %

Result Table:

Practical yield

M.P. of 2-naphthyl methyl ether

Practical % yield

…… g

…… °C

…… %

283


Rf value of 2-naphthol

Rf value of 2-naphthyl methyl ether


……

……

……


To study the benzoylation reaction.

Experiment:

Preparation of Benzoyl glycine (Hippuric acid) from Glycine.

Principle:

The hydroxy and amino functions can be easily benzoylated using Benzoyl chloride in aqueous alkaline conditions. This reaction is known as Schotten-Baumann reaction. Reaction:

284

PhCOCl

Glycine Benzoyl glycine(Hippuric acid)

NaOH or KOH

O

NH

H N2COOH COOH

Chemicals:

(i) Glycine = 0.5 g

(ii) Benzoyl chloride = 0.9 ml

(iii) NaOH = 1.34 g in 14 ml water

(iv) conc. HCl

(v) CCl4 = 30 ml

Procedure:

Dissolve 0.5 g of glycine in 5 ml of 10% NaOH solution in a 25 ml conical flask. Add 0.9 ml of benzoyl chloride in five portions to the solution. Stopper the flask and shake vigorously after each addition until all the benzoyl chloride is reacted. Transfer the solution to a beaker and add few grams of ice and add conc. HCl slowly with stirring until the solution is acidic to litmus paper. Filter the crystalline precipitate of the product on a Buchner funnel, wash with cold water and drain well. Place the solid in a conical flask with 10 ml CCl4 and boil gently for 10 min. Allow the mixture to cool slightly, filter under gentle suction and wash with 10-20 ml CCl4. Record the practical yield and re-crystallize it.

Re-crystallization:


285

Dissolve the crude product from boiling water (5 ml), with addition of decolorizing charcoal if necessary, filter the hot solution and cool the filtrate. The crystals of the product separate out. Filter, dry and record the melting point and TLC [using Butanal : acetic acid: water (4 : 1 : 1) as solvent or CHCl3:methanol (9 : 1) or toluene]. Calculations: (i) Theoretical yield: Glycine (C2H5O2N) ≡ Benzoyl glycine (C9H9O3N) 75 g ≡ 179 g Therefore, 0.5 g ≡ 1.19 g (ii) Practical yield = ‘A’ g (iii) Practical % yield: Since, 1.19 g of product = 100% yield


1.19 = …… %

Result Table:

1. Practical yield

2. M.P. of benzoyl glycine

3. Practical % yield

4. Rf value of glycine

5. Rf value of benzoyl glycine

6. Solvent used for TLC

…… g

…… °C

…… %

……

……

……

Experiment No. 6 Aim: To study the aromatic substitution reaction (replacement of diazo group, Sandmeyer reaction). Experiment: Preparation of p-Iodo nitrobenzene from p-Nitro aniline. Principle: Diazonium salts undergo a large number of reactions in which the diazo group is lost as molecular nitrogen and is replaced by variety of other groups (e.g. OH, I, Br, Cl, F, CN, NO2, H, SO2H, Ar) which become attached to the aromatic ring. Reaction:


NH2

NaNO2

Cl N

KI

I

HCl

NO2 NO2

N

NO2

p-Nitro aniline Diazonium salt p- odo nitrobenzeneI Chemicals:

(i) p-Nitro aniline = 0.5 g

(ii) NaNO2 = 0.25 g

(iii) conc. H2SO4 = 0.41 ml

(iv) KI = 1 g

Procedure:

Take 0.5 g p-nitro aniline, 0.41 ml conc. H2SO4 and 3 ml water in a conical flask. Stir it well and cool to 0-5°C. Add a cold solution of 0.25 g NaNO2 in 0.75 ml distilled water to the above solution. Filter the resultant cold solution, and add the filtrate with stirring to a solution of 1 g KI in 3 ml water taken in a beaker. Filter the product separated out and wash it with water. Dry the product, record the practical yield and re-crystallize it.

Re-crystallization:

Dissolve the crude product in minimum amount of ethyl alcohol in a beaker by heating on a water bath. Filter the hot solution and cool the filtrate. The shiny brown crystals of p-iodo nitro benzene separate out. Filter, dry and record the melting point and TLC (using toluene as a solvent).

Calculations:

(i) Theoretical yield:

p-Nitro aniline (C6H6O2N) ≡ p-iodo nitro benzene (C6H4O2NI)

124 g ≡ 202 g

Therefore, 0.5 g ≡ 202 × 0.5

124 g = 0.81 g


(iii) Practical % yield: Since 0.81 g of product = 100% yield

Therefore, ‘A’ g = A × 1000.81 = …… %

Result Table:

286


1. Practical yield 2. M.P. of p-iodo nitrobenzene 3. Practical % yield 4. Rf value of p-nitro aniline 5. Rf value of p-iodo nitrobenzene 6. Solvent used for TLC

…… g …… °C …… %

…… …… ……

Experiment No. 7 Aim: To study the ester hydrolysis. Experiment: Preparation of Benzoic acid from Ethyl benzoate. Principle: Esters can be hydrolyzed either under basic or acidic conditions. Aqueous KOH or NaOH readily hydrolyze the esters into salts of acids from which the acids are liberated by acidification. The acid hydrolysis of esters using HCl or H2SO4 directly yields the corresponding acids. Reaction:

(i) aq. NaOH

Ethyl benzoate Benzoic acid

(ii) HCl

O

OH

O

OC H2 5

+ EtOH

Chemicals: (i) Ethyl benzoate = 0.5 ml (ii) 10% NaOH = 7 ml (iii) dil. HCl Procedure: Take 0.5 ml ethyl benzoate, 7 ml 10% NaOH in a round 25 ml bottom flask fitted with water condenser. Boil the mixture for at least 1 h. Disappearance of the ester layer confirms the completion of the reaction. Cool the mixture by adding small amount of ice and then add dil. HCl with stirring until the solution is acidic to litmus. Filter the product separated out and wash it with water. Dry the product, record the practical yield and re-crystallize it. Re-crystallization:

287


288

Take the crude product and water (10 ml) in a beaker and dissolve by heating on a wire gauze. Filter the hot solution and cool the filtrate. The shiny white crystals of benzoic acid separate out. Filter, dry and record the melting point and TLC (using toluene as solvent). Calculations: (i) Theoretical yield: Ethyl benzoate (C6H6O2N) ≡ Benzoic acid (C6H4O2NI) 150 g ≡ 122 g

Therefore, 0.5 g ≡ 122 × 0.5

150 g = 0.4 g



Since, 0.4 g of product = 100% yield

Therefore, ‘A’ g = A × 1000.4 = …… %

Result Table:

1. Practical yield

2. M.P. of benzoic acid

3. Practical % yield

4. Rf value of ethyl benzoate

5. Rf value of benzoic acid

6. Solvent used for TLC

…… g

…… °C

…… %

……

……

……

Experiment No. 8

Aim:

To study the aromatic electrophilic substitution reaction.

Experiment:

Preparation of p-Bromo acetanilide from Acetanilide.

Principle:

Aromatic compounds can be conveniently brominated by the use of brominating agent, which is normally a solution of bromine in acetic acid. Bromination of activated aromatic compounds usually give 2, 4, 6-tribromo derivatives while moderately activating group like anilide give preferably the para bromo product.

Reaction:

T.Y.B.Sc. Practical Chem Organic Preparations istry (Organic Practicals)

NHCOCH3

Bromine

Acetanilide p-Bromo acetanilide

NHCOCH3

Br

AcOH

Chemicals:

(i) Acetanilide = 0.5 g

(ii) 25% (w/v) Bromine in AcOH = 2.5 ml Procedure: Dissolve 0.5 g acetanilide in 0.6 ml glacial acetic acid and add 2.5 ml bromine solution in acetic acid (25% w/v). Shake the mixture for 1 h. After 1 h, pour the mixture on to crushed ice (20 g) with stirring. Filter the separated product and wash with cold water. Dry the product, record the practical yield and re-crystallize it. Re-crystallization: Dissolve the crude product in minimum amount of ethyl alcohol in a beaker by heating on a water bath. Filter the hot solution and cool the filtrate. The crystals of the product separate out. Filter, dry and record the melting point and TLC (using toluene as a solvent). Calculations: (i) Theoretical yield: Acetanilide (C8H9ON) ≡ p-Bromo acetanilide (C8H8ONBr) 135 g ≡ 214 g Therefore, 0.5 g ≡ 0.79 g (ii) Practical yield = ‘A’ g (iii) Practical % yield: Since, 0.79 g of product = 100% yield


79 = …… %

Result Table:

1. Practical yield 2. M.P. of p-bromo acetanilide 3. Practical % yield 4. Rf value of acetanilide 5. Rf value of p-bromo acetanilide 6. Solvent used for TLC

…… g …… °C …… %

…… …… ……

289


Experiment No. 9 Aim: To study the acetylation reaction. Experiment: Preparation of Paracetamol from p-Hydroxy aniline. Principle: The hydroxy and amino functions both can be easily acetylated using acetic anhydride or acetyl chloride. However, amino function being more nucleophilic than hydroxy function, can be acetylated selectively using acetic anhydride under neutral conditions. Reaction:

NH2

(CH CO) O3 2

p-Hydroxyaniline Paracetamol

NHCOCH3

OH

Water

OH

Chemicals: (i) p-Hydroxy aniline = 0.5 g (ii) Acetic anhydride = 0.6 ml Procedure: Suspend 0.5 g p-hydroxy aniline in 3.0 ml water and add 1.2 ml acetic anhydride. Shake the mixture vigorously and warm on water bath till the solid dissolves. After 10 min, cool, filter the solid acetyl derivative and wash with cold water. Dry the product, record the practical yield and re-crystallize it. Re-crystallization: Dissolve the crude product in water (10 ml) in a beaker by heating on a water bath. Filter the hot solution and cool the filtrate. The crystals of the product separate out. Filter, dry and record the melting point and TLC (using toluene as a solvent). Calculations: (i) Theoretical yield: p-Hydroxy aniline (C6H7ON) ≡ Paracetamol (C8H9O2N) 109 g ≡ 151 g Therefore, 0.5 g ≡ 0.69 g (ii) Practical yield = ‘A’ g (iii) Practical % yield: Since 0.69 g of product = 100% yield

290



0.69 = …… %

Result Table:

1. Practical yield 2. M.P. of paracetamol 3. Practical % yield 4. Rf value of p-hydroxy aniline 5. Rf value of paracetamol 6. Solvent used for TLC

…… g …… °C …… %

…… …… ……

Experiment No. 10

Aim:

To study the carbonyl reduction reaction.

Experiment:

Preparation of Ethyl benzene from Acetophenone.

Principle: The carbonyl function can be completely reduced to methylene under acidic, basic or neutral conditions. Amongst these the most common methods are Clemmensen’s reduction, Wolf Kishner reduction and Mozingo reduction. The reduction of carbonyl group using hydrazine in basic medium using KOH is called as Wolf-Kishner reduction.

Reaction:

291

NH NH H O2 2 2.

Acetophenone Ethyl benzene

KOH

O CH3

Chemicals:

(i) Acetophenone = 1.0 ml

(ii) 90% Hydrazine hydrate = 2 ml (iii) KOH = 1.1 g (iv) Ethylene glycol = 10 ml (v) Diethyl ether = 20 ml Procedure:


292

Place 0.5 ml acetophenone, 5.0 ml ethylene glycol and 1.0 ml of 90% hydrazine hydrate solution and 1.1 g KOH pellets in a round bottom flask fitted with water condenser. Warm the reaction mixture of boiling water bath till KOH dissolves and then reflux for 1 h on wire gauze. Distill using Hickmann head till the temperature is 175oC. Keep the distillate and again reflux for 2 h. Cool this and extract twice with 10 ml ether. Combine the ether extracts with the distillate removed earlier and dry over sodium sulphate, decant and evaporate the ether. Re-distillation: Distill the residual compound and record the boiling point and TLC (using toluene as a solvent). Note: Ethyl benzene will not show spot on TLC. Formation of the product can be confirmed by negative 2, 4-DNP test. Calculations: (i) Theoretical yield: Acetophenone (C8H8O) ≡ Ethyl benzene (C8H10) 120 g ≡ 106 g Therefore, 0.5 g ≡ 0.44 g (ii) Practical yield = ‘A’ g (iii) Practical % yield: Since, 0.44 g of product = 100% yield

Therefore, ‘A’ g = A × 1000.44 = …… %

Result Table: 1. Practical yield 2. B.P. of ethyl benzene 3. Practical % yield 4. Rf value of acetophenone 5. Solvent used for TLC

…… g …… °C …… % …… ……

Note: Ethyl benzene will not show spot on TLC.

VIVA VOCE 1. What is oxidation? Mention four oxidizing agents. 2. Give one oxidation reaction (a) in which hydrogens are removed, (b) in which oxygen

is added, (c) in which electrons are removed. 3. How will you distinguish hydroquinone from quinone (a) by chemical method,

(b) by spectroscopy ? 4. What is electrophilic aromatic substitution reaction? 5. Diazotization is carried out at low temperature. Why? 6. Sulphuric acid is preferred over hydrochloric acid in the preparation of diazonium

salt. Why?


293

7. Acidic pH is maintained in the reaction involving replacement of diazonium group. Why?

8. What is Sandmeyer reaction? 9. What is the role of sulphuric acid in the nitrating mixture? 10. p-Nitro aniline is not prepared by nitration of aniline. Why? 11. Electron withdrawing groups are meta directing in electrophilic substitution reactions.

Why? 12. What are the various reagents used for methylation reaction? 13. Why NH2 undergoes acetylation in preference to hydroxyl group? 14. What are the different methods for reducing carbonyl group to methylene group? 15. Ester hydrolysis is carried out preferably under basic conditions. Explain.

- - -

( 293 )

Appendices

STOCK SOLUTIONS 1. Acetamide Solution : Dissolve 25 g acetamide in 200 ml water and dilute to one litre. (Give 10-15 ml in conical flask for estimation) 2. Ethyl Benzoate Solution : Dissolve 60 ml ethyl benzoate in 200 ml alcohol and dilute to one litre by ethyl alcohol. (Give 10-15 ml in conical or R.B. flask for estimation) 3. Oil Solution : Dissolve 100 g of castor oil in ethyl alcohol (200 ml) and dilute to one litre by ethyl alcohol. (CHCl3 or CCl4 can be used instead of ethyl alcohol). Give 10 to 15 ml in measuring flask for estimation) 4. Monobasic or Dibasic Acid Solution : Dissolve 47.5 ml acetic acid or 50 g of oxalic acid in 200 ml water and dilute to one litre by water. (Give 10-15 ml for estimation in measuring flask) 5. Wij’s Solution (Iodine-monochloride Solution) : Two methods : (a) Dissolve 13 g of iodine powder in 750 ml acetic acid in R.B. flask. (Warm water bath if

necessary using air condenser). Pass slow stream of dry Cl2 gas till colour changes to clear orange. This is Wij’s solution. It should be stored in well stoppered amber colour bottle and keep in dark cool place. (Do not keep the solution for more than 3 days.)

(b) Dissolve 25 g I2 in 500 ml ethyl alcohol and 30 g HgCl2 in 500 ml ethyl alcohol. Mix these solutions and keep for 12 hours in an amber colour bottle.

6. Glucose Solution : Dissolve 25 g. glucose in 200 ml water and dilute it to one litre by water. (Give 10-15 ml for estimation in measuring flask) Other Solutions : 1. 0.5 N KOH : 28 g in one litre water. 2. 0.5 N alcoholic KOH : 28 g in one litre alcohol. 3. 0.05 N NaOH : 2 g NaOH in one litre water. 4. 0.05 N Na2S2O3 : 12.4 g Na2S2O3 in one litre water. 5. 10% KI : 10 g KI in 100 ml water.

6. 1 N HCl : 100 ml conc. HCl diluted to one litre by water.

T.Y.B.Sc. Practical Chemistry (Organic Practicals) Appendices

294

7. 0.05 N HCl : 5 ml conc. HCl diluted to one litre by water.

8. 0.05 N I2 : 6.35 g I2 and 15 g KI in 10 ml water. Stirr well, go on adding water with stirring till iodine dissolves completely. Dilute the

solution to one litre by water. 9. Schiff’s reagent : Dissolve 0.2 g of rosaniline in 100 ml water. Pass SO2 gas till the

solution becomes colourless (pink colour vanishes). Indicators : 1. Phenolphthalein : Dissolve 1 g phenolphthalein in 500 ml ethyl alcohol and dilute to

one litre by water. 2. Starch : Take 1 g starch and make a paste with water. Add this slowly to 200 ml

boiling water. Boil further for few minutes and allow to cool.

List of Organic Mixtures :

(a) Solid - Solid :

1. Benzoic acid + Acetanilide 2. Benzoic acid + Naphthalene

3. Benzoic acid + β-naphthol 4. Benzoic acid + m-dinitrobenzene

5. Benzoic acid + p-nitroaniline 6. Cinnamic acid + α-naphthol

7. Cinnamic acid + Anthracene 8. Cinnamic acid + β-naphthol

9. Cinnamic acid + m-nitroaniline 10. Salicylic acid + Acetanilide

11. Salicylic acid + α-naphthol 12. Salicylic acid + m-dinitrobenzene

13. Aspirin + Naphthalene 14. Aspirin + β-naphthol

15. Aspirin + Acetanilide 16. β-naphthol + Naphthalene

17. β-naphthol + m-dinitrobenzene 18. β-naphthol + p-toluidine

19. β-naphthol + Acetanilide 20. phthalic acid + p-toluidine

21. o-nitrophenol + Anthracene 22. p-nitrophenol + m-dinitroaniline

23. α-naphthol + m-dinitrobenzene 24. α-naphthol + m-nitroaniline

25. o-nitrophenol + p-nitroaniline 26. p-toluidine + Naphthalene

27. p-toluidine + Acetanilide 28. p-nitroaniline + m-dinitrobenzene

29. m-nitroaniline + Naphthalene 30. m-nitroaniline + Anthracene

(b) Solid - Liquid :

31. Acetone + Oxalic acid 32. Acetone + Acetanilide

33. Chloroform + Naphthalene 34. Chloroform + m-dinitrobenzene

35. Ethylmethyl ketone + urea 36. Chloroform + m-nitroaniline


295

37. Acetone + Acetamide 38. Chloroform + Acetanilide (c) Liquid - Liquid : 39. Chloroform + Chlorobenzene 40. Ethyl acetate + Nitrobenzene 41. Acetone + Aniline 42. Chloroform + Dimethylaniline 43. Methylacetate + Acetophenone 44. Ethylmethyl ketone + Chlorobenzene

USEFUL DATA Atomic Numbers and Atomic Weights of Elements

Element Symbol At. No.

At. Wt. Element Symbol At. No.

At. Wt.

Actinium Ac 89 227 Einsteinium Es 99 254

Aluminium Al 13 26.9815 Erbium Er 68 167.36

Americium Am 95 243 Europium Eu 63 151.96

Antimony Sb 51 121.75 Fermium Fm 100 257

Argon Ar 18 39.948 Fluorine F 9 18.9984

Arsenic As 33 74.9216 Francium F 87 223

Astatine At 85 210 Gadolinium Gd 64 157.25

Barium Ba 56 137.34 Gallium Ga 31 69.72

Berkelium Bk 97 247 Germanium Ge 32 72.59

Berylium Be 4 9.0122 Gold Au 79 196.9665

Bismuth Bi 83 208.9804 Hafnium Hf 72 178.49

Boron B 5 10.81 Helium He 2 4.0026

Bromine Br 35 79.904 Holmium Ho 67 164.9304

Cadmium Cd 48 112.40 Hydrogen H 1 1.0079

Calcium Ca 20 40.08 Indium In 49 114.82

Californium Cf 98 251 Iodine I 53 126.9045

Carbon C 6 12.011 Iridium Ir 77 192.22

Cerium Ce 58 140.12 Iron Fe 26 55.847

Cesium Cs 55 132.9054 Krypton Kr 36 83.80

Chromium Cr 24 51.996 Lawrencium Lr 103 260


296

Cobalt Co 27 58.9332 Lead Pb 82 207.2 (Contd.)

Element Symbol At. No.

At. Wt. Element Symbol At. No.

At. Wt.

Copper Cu 29 63.546 Lithium Li 3 6.941

Curium Cm 96 247 Lutetium Lu 71 174.97

Dysprosium Dy 66 162.50 Magnesium Mg 12 24.305

Manganese Mn 25 54.9380 Ruthenium Ru 44 101.07

Mendelevium Md 101 256 Samarium Sm 62 150.4

Mercury Hg 80 200.59 Terbium Tb 65 158.9254

Molybdenum Mo 42 95.94 Scandium Sc 21 44.9559

Neodymium Nd 60 144.24 Selenium Se 34 78.96

Neon Ne 10 20.183 Silicon Si 14 28.086

Neptunium Np 93 237.0482 Silver Ag 47 107.868

Nickel Ni 28 58.71 Sodium Na 11 22.9898

Niobium Nb 41 92.9064 Strontium Sr 38 87.62

Nitrogen N 7 14.0017 Sulphur S 16 32.06

Nobelium No 102 255 Tantalum Ta 73 180.9479

Osmium Os 76 190.2 Technetium Tc 43 99

Oxygen O 8 15.9994 Tellurium Te 52 127.60

Palladium Pd 15 30.9738 Thallium Tl 81 204.37

Platinum Pt 78 195.09 Thorium Th 90 232.0381

Plutonium Pu 94 242 Thulium Tm 69 168.9342

Polonium Po 84 210 Tin Sn 50 118.69

Potassium K 19 39.08 Titanium Ti 22 47.90

Praseodymium Pr 59 140.9077 Tungsten W 74 183.85

Promethium Pm 61 147 Uranium U 92 238.029

Protactinium Pa 91 231.0359 Vanadium V 23 50.9414

Radium Ra 88 226.0254 Xenon Xe 54 131.30

Radon Rn 86 222 Ytterbium Yb 70 173.04

Rhenium Re 75 186.207 Yttrium Y 39 88.9059

Rhodium Rh 45 102.9055 Zinc Zn 30 65.38

Rubidium Rb 37 85.4678 Zirconium Zr 40 91.22


297

Chlorine Cl 17 35.453 Lanthanum La 57 138.9055

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Documents

PRACTICAL CHEMISTRY B. Sc G. S. Gugale A. V. Nagawade R. A. Pawar S. S. Jadhav V. D. Bobade A. D. Natu D. R. Thube P. C. Mhaske L. K. Nikam Nirali Prakashan