INTRODUCTION TO THE TECHNIQUES OFEXPERIMENTAL CHEMISTRY

Authored by

Jacob VincentAssistant Professor

Department of ChemistryDr. Sivanthi Aditanar College of Engineering

Tiruchendur

Gyandhara International Academic PublicationPublishing Research Worldwide>>>Thane, Maharashtra, India

‘INTRODUCTION TO THE TECHNIQUES OF EXPERIMENTAL CHEMISTRY’by Jacob Vincent

© Copyright lies with GIAP Journals, IndiaAll rights reserved. No part of this publication may be reproduced or transmitted, in any form or byany means, without permission. Any person who does any unauthorized act in relation to thispublication shall be liable to criminal prosecution and civil claims for damages.First Edition, July 2016

Price: 100 INR

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ISBN: 978-93-83006-15-1

Publisher: Gyandhara International Academic Publications (GIAP)www.giapjournals.orgEmail: publications.giap@gmail.comContact: +968 97362302

DisclaimerThe views and contents of this book are solely of authors only. This book is being sold on the condition and understandingthat its contents are merely for information and reference.The author and publisher specifically disclaim all and any liability for any loss, damage, risk, injury, distress etc. sufferedby any person, whether or not a purchaser of this book, as a consequence whether direct or indirect, of any action takenor not taken on the basis of the contents of this book.The publisher believes that the contents of this book do not violate any existing copyright / intellectual property of othersin any manner what so ever. However, in the event the author has been unable to track any source and if any copyrighthas been inadvertently infringed, please notify the publisher in writing for corrective action.

Dedicated to YHWH

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AcknowledgementWe record our deep sense of indebtness to our Founder President Padma Shri.Dr.B.Sivanthi Adityan andour honourable chairman Sri.Balasubramanian S Adityan for having blessed this venture. We are thankfulto our secretary Dr.Gopala Krishnan and the management of Dr.Sivanthi Aditanar College of Engineering,Tiruchendur for their continued guidance and constant encouragement.

We express our sincere gratitude to Dr.G.Wiselin Jiji, Principal, Dr.Sivanthi Aditanar College ofEngineering, Tiruchendur for the constant inspiration and encouragement. We are deeply indebted to anumber of colleagues and students who have read this material and given us many valuable suggestionsfor improving the presentation.

We owe a special debt of gratitude to the faculties from department of chemistry. Dr.Sivanthi AditanarCollege of Engineering, Tiruchendur, Tuticorin district, Tamilnadu who have read the original manuscriptin detail and gave us many perceptive comments. We have received, and gratefully acknowledge, reportsof course testing of the notes for this book by our previous batch students.

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Book prefaceThis course has been specially designed as an intensive introduction to the techniques of experimentalchemistry. Our goals in this class are twofold. First, since freshmen cannot enroll in any of the regularchemistry lab courses, this book has been created to give interested first-year students an opportunity toget hands-on experience with chemistry. A second aim is to prepare freshmen for laboratory in theChemistry Department. Freshmen often have a difficult time finding a comfortable position in ourdepartment because they don't yet have the experimental skills and experience developed in our regularchemistry lab course sequence. Unlike other laboratory classes, the goal is not just to successfullyperform an experiment and write a report; instead, the focus will be on mastering the techniques andskills necessary to carry out experiments. During the next month, you will mix, stir, and measure until youreach a "professional level" of skill in various techniques fundamental to chemical experiments.

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About AuthorAuthor has an experience of Assistant Professor in chemistry for 7 years in Engineering College and 3years in Polytechnic College. He has completed M.Sc. M.Phil (Chemistry), M.B.A., Ph.D. (Nano polymers-pursuing). He has published 3 papers in international journals and 2 papers in national journals. He haspresented one paper in international conference and one paper in national conference.

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Salient features of the bookLogically explained and student friendly approach.

Simple, systematic and well-organized text.

Well-illustrated with nicely drawn graphs.

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Table of Contents

Chapter 1 Determination of Dissolved Oxygen Content of Water Sample by Winkler’s Method......................................................................................................................................................... 1

Chapter 2 Determination of Chloride Content of Water Sample by Argentometric Method ........ 4

Chapter 3 . Determination 0f Strength of Given Hydro Chloric Acid Using Ph Meter ................. 7

Chapter 4 Determination of Strengh of Acids In A Mixture Using Conductivity Meter ............. 11

Chapter 5 Estimation 0f Iron Content 0f Water Sample Using Spectrophotometer(1,10-Phenanthroline Method)................................................................................................................ 15

Chapter 6 Determination of Molecular Weight of Polyvinyl Alcohol Using Ostwald Viscometer....................................................................................................................................................... 18

Chapter 7 Conductometric Titration of Strong Acid vs Strong Base .......................................... 21

1

Chapter 1Determination of Dissolved Oxygen Content of Water Sample byWinkler’s Method

Aim

To find out the quantity of dissolved oxygen present in the given water sample being supplied with 0.0094N standard Potassium dichromate solution and solution of Sodium thiosulphate by Winkler’s method.

Principle

In this part study was performed to see the effect of variation of Cerium(V) concentration on the reactionvelocity in the oxidation of 2-phenyl ethanol in aqueous sulphuric acid midium Tables from 3.12 to 3.21show the individual sets for the variation of oxidant concentrations. First column of each table containstime while the second column shows absorbance at various time intervals. In the third column slopevalue obtained from the individual graphs plotted between absorbance versus time, are given. Theseslope values were converted into -dc/dt values as discussed in section 2.2, which are given at the base ofeach individual table. Table 3.22 shows the summarized variation in which first column showsconcentration of cerium(IV) and second column shows corresponding -dc/dt values. In the third columnvalues calculated for molar concentration of the oxidant are given.

2 KOH + MnCl2 Mn(OH)2 + 2KCl

(White ppt)

Mn(OH)2 + O2 2 MnO(OH)2

(Brown ppt)

This brown precipitate manganic oxide reacts with Sulphuric acid to give nascent oxygen which isresponsible to liberate free iodine.

MnO(OH)2 - + H2SO4 MnSO4 + 2 H2O + [O]

2KI + H2SO4 + [O] K2SO4 + H2O + I2

2Na2S2O3 + I2 2Na2S4O6 + 2 NaI

Starch is added as indicator which forms an intense blue coloured loose adsorption complex with iodine.This complex gets broken when thiosulphate is added and hence a sharp colour change occurs at theend point.

2

Procedure

Titration 1: Standarisation Of Sodium Thiosulphate

The burette is washed and filled with sodium thiosulphate solution. 20 ml of potassium dichromatesolution is pipetted out into a clean conical flask. About 20 ml of dilute sulphuric acid is added followed bythe addition of 10 ml of 10% potassium iodide solution. Shake the flask well. The liberated iodine istitrated against thio solution taken in the burette. When the solution becomes pale yellow in colour, 2 mlof freshly prepared 1% starch solution is added and the titration is slowly continued. The end point is thecolour change from blue to colourless. Titrations are repeated to get concordant titre values. Thereadings are tabulated (Table-1). From the titre value the normality of sodium thiosulphate solution iscalculated.

Titration 2: Estimation of Dissolved Oxygen

Water samples are collected in BOD bottle. To this 1 ml of manganous chloride solution and 1 ml ofalkaline potassium iodide solution were added to fix the dissolved oxygen in it. The manganous hydroxideis oxidised to brown colour manganic oxide and precipitated. Shake the solution well and allow 20minutes time for all oxygen to react and then to settle as precipitate. When half of the precipitate settlesdown, the stopper is removed and 2 ml of Sulphuric acid solution is added. The bottle is restoppered andmixed by inverting many number of times until the precipitate is completely dissolved.

60 ml of this solution was pipetted out into a clean conical flask and titrated against 0.01 N sodiumthiosulphate solution using starch indicator. The end point is the disappearance of blue colour. Titrationsare repeated till we get concordant values.

Result

The amount of dissolved oxygen present in the given water sample = __________ mg / lit

Normality of K2Cr2O7 = 0.01 N

Table 1: Standardisation of Sodium thiosulphate

Sl.No Volume of

K2Cr2O7 (ml)

Burette reading (ml)

Initial Final

Volume ofSodium

thiosulphate(ml)

Indicator

1

2

Starch

Concordant titre value = __________ ml

3

Calculations

Volume of potassium dichromate solution (V1) = 20 ml

Normality of potassium dichromate solution (N1) = 0.01N

Volume of sodium thiosulphate solution (V2) = ml

Normality of sodium thiosulphate solution (N2) = ?

N2 = V1 N1 /V2

= N

Table 2: Estimation of Dissolved Oxygen

Sl.No Volume of

Water sample(ml)

Burette reading (ml)

Initial Final

Volume ofNa2S2O3 (ml)

Indicator

1

2

Starch

Concordant titre value = __________ ml

Calculation

V N x 8 x 1000

DO = ___________ x ______________ x v

(V-2) 60

270 N x 8 x 1000

DO = ___________ x __________________ x v

(270-2) 60

=

= _______________ mg / lit

Volume of BOD bottle (270ml)

Strength of Sodium thiosulphate

Volume of thio consumed for 60 ml of acidified solution

4

Chapter 2Determination of Chloride Content of Water Sample by ArgentometricMethod

Aim

To determine the amount of chloride present in the water sample being supplied with silver nitratesolution and standard sodium chloride solution of 0.0198N.

Principle

It is an example of precipitation reaction. The reaction between chloride and silver nitrate is direct andsimple. It proceeds as follows.

Ag+NO3- + Na+Cl- AgCl + NaNO3

(in water) (White precipitate)

Since one equivalent of AgNO3 reacts with one equivalent of NaCl

Equivalent weight of AgNO3 = 170

Equivalent weight of chloride = 35.46

The completion of the reaction in this case is observed by employing K2CrO4 solution as the indicator. Atthe end point the yellow colour of chromate changes into reddish brown due to the reaction

2 AgNO3 + K2CrO4 Ag2CrO4 + 2KNO3

(Yellow) (Reddish brown)

K2CrO4 indicator will not be precipitated as Ag2CrO4 until all the chlorides in the solution have beenprecipitated as AgCl. This observation is in keeping with the difference in the solubility product Ag2CrO4

and AgCl [ Ks (Ag2CrO4) < Ks (AgCl))], Ks - solubility product.

Procedure

Titration 1: Standardisation Of Silver Nitrate Solution

Exactly 20 ml of the standard NaCl solution is pipetted out into a clean conical flask. 1 ml of 2% K2CrO4indicator solution is added to it. The solution turns yellow in colour. It is titrated against the silver nitrate(AgNO3) solution taken in the burette. During each addition of AgNO3 solution, the conical flask isshaken well and stirred with the glass rod. Nearness of the end point is noted by the formation of

5

coagulation of silver chloride precipitate at the bottom of the flask. Now, the addition of AgNO3 is donedrop by drop until the supernatant solution gets a faint reddish brown tinge. The titration is repeated toget concordant values. From the volume of AgNO3 solution, the strength of AgNO3 is calculated.

Titration II: Estimation of chloride

Exactly 20 ml of water sample is pipetted out into a clean conical flask. To this solution, one ml of 2% ofK2CrO4 indicator solution is added. It is titrated against the standardised AgNO3 solution from theburette. The addition of AgNO3 solution is continued, until the solution produced a reddish brown tinge.The titration is repeated for concordancy. From the volume of AgNO3 consumed, the strength of chlorideand hence its amount is calculated.

Result

The amount of chloride present in the 100ml of the given water sample = ____________ g

Normality of sodium chloride = 0.0198N

Table –1: Standardisation of Silver Nitrate

Sl. No Volume ofNaCl (ml)

V1

Burette reading (ml) Volume ofAgNO3 (ml)

V2

Indicator

Initial Final

1 20 0 Potassiumchromate

2 20 0

3 20 0

Concordant titre value = ----------- ml

Calculation Of The Normality Of Silver Nitrate

Volume of sodium chloride (V1) = 20 ml

Normality of sodium chloride (N1) = ___________ N

Volume of silver nitrate (V2) = ____________ ml

Normality of silver nitrate (N2) =____?_______

According to the law of volumetric analysis

V1 N1 = V2 N2

N2 = V1 N1 /V2= 20 x 0.0198

V2 Normality of silver nitrate = __________ N

6

Estimation of Chloride

Table –2: Estimation of Chloride

Sl. No Volume ofWater

sample(ml)

Burette reading (ml) Volume of AgNO3 (ml) Indicator

Initial Final

1 20 PotassiumChromate

2 20

Concordant titre value = ----------- ml

Calculation of the Normality of water sample (Chloride)

Volume of water sample (V1) = 20 ml

Normality of water sample (chloride) (N1) = ?

Volume of silver nitrate (V2) = ____________ml

Normality of silver nitrate (N2) = ____________N

According to the law of volumetric analysis (V1 N1) = V2 N2

N1 = V2 N2 / V1

Normality of chloride in water sample = ____________N

Estimation of Chloride

Amount of chloride present in 100ml of water sample =

Equivalent weight of chloride x Normality of chloride / 10 = 35.46 x Normality of chloride /10 = ________ g

7

Chapter 3Determination 0f Strength of Given Hydrochloric Acid Using pH Meter

Aim

To determine the amount of HCl by pH metric titration using a standard solution of NaOH of 0.0099 N isprovided

Principle

pH is mathematically defined as the negative logarithm to the base ten of H+ ion concentration.

pH = - log10 [H+]

In this experiment, initially HCl solution is only present. As HCl solution contains only H+ ions, pH may bebelow 3.

When NaOH is added slowly from the burette, the pH will increase gradually. At the end point, all the H+ions from HCl are replaced by OH-from NaOH which leads to complete neutralization of H+ ions.

H+ Cl- + Na+OH- NaCl + H2O

After the end point, further addition of NaOH increases the pH sharply as there is an excess of fastmoving OH- ions.

Materials Required

1) pH meter

2) Glass Electrode

Cell representation of Glass Electrode:

Pt, 0.1 M HCl / Glass

Procedure

The burette is filled with Standard NaOH solution. Exactly 20 ml of the given HCl solution is pipetted outinto a clean beaker. It is then diluted to 20 ml by adding distilled water. The glass electrode is dipped in itand connected with a pH meter.

8

Titration-I

First a preliminary titration is carried out by adding standard NaOH solution in portions of 1ml from theburette to the HCl solution taken in the beaker and pH of the solution is noted for each addition. By thegradual addition of NaOH, sudden jump occurs in the pH value. After the sudden jump take 7 morereadings and the range at which the end point lies is found out by plotting volume of NaOH added againstpH(Graph I). This preliminary titration is used to find out the range of the end point.

Titration-II

Another titration is carried out by adding standard NaOH solution in portions of 0.2ml near the end point(ie).just before 1ml from the end point and pH of the solution is noted after each addition. The addition ofNaOH is continued even after the end point for further 1ml. The accurate end point is determined byplotting ΔpH / ΔV against Volume of NaOH added (Graph II). From the end point the strength of HClsolution and hence its amount can be calculated.

Equivalent weight of NaOH = 40

Equivalent weight of HCl = 36.5

Result

Amount of HCl present in 100ml of the given solution = ________________ g

Normality of sodium hydroxide = 0.1N

Table –1: HCl Vs NaOH

Sl.No Volume of NaOH (ml) pH

1 0

2 1

3 2

4 3

5 4

6 5

7 6

8 7

9 8

10 9

9

11 10

12 11

13 12

14 13

15 14

Model Graph I

Volume of NaOH Vs pH

Table II: HCl Vs NaOH

Sl.No Volume ofNaOH (ml)

pH Δ pH Δ V (ml) Δ pH / Δ V Mean volumeof NaOH

1

2

3

4

5

6

7

8

9

10

10

11

12

13

14

15

Model Graph II

Mean Volume of NaOH Vs Δ pH / Δ V

Calculation

Calculation of strength of HCl

Volume of HCl (V1) = 20 ml

Normality of HCl (N1) = ___________?

Volume of NaOH (V2) = ___________ ml

Normality of NaOH(N2) = 0.00995 N

According to the law of volumetric analysis V1 N1 = V2 N2

N1 = V2 x N2 / V1

= V2 x 0.00995 / 20

Strength of HCl = __________ N

The amount of HCl present in

100ml of the given solution = ____________ N x Eq.wt of HCl /10

= _____________ N x 36.5/10

11

Chapter 4Determination of Strengh of Acids In A Mixture Using Conductivity Meter

Aim

To determine the amount of a strong acid and a weak acid (HCl & CH3COOH) present in 100ml of thegiven mixture of acid solution by conductometric titration using standard NaOH of 0.0295 N

Principle

The conductance of a electrolytic solution is due to the presence of ions. Since specific conductance ofa solution is proportional to the concentration of the ions in it, conductance of the solution is measuredduring titration.

When the sodium hydroxide is added slowly from the burette to the solution, HCl (strong acid) getsneutralised first. Since the fast moving H+ ions are replaced by slow moving Na+ion decrease inconductance takes place until the end point is reached.

HCl + NaOH NaCl + H2O (1st neutralization)

After the complete neutralization of all HCl, the neutralization of CH3COOH starts,

CH3COOH + NaOH CH3COONa + H2O (IInd neutralization)

Since CH3COONa is stronger than CH3COOH conductivity slowly increases until all CH3COOH iscompletely neutralised. When the end point is reached, the addition of NaOH will cause sudden increasein the conductance. This is due to the presence of fast moving OH- ions.

Materials Required

1) Conductivity bridge

2) Conductivity cell

Procedure

The burette is filled with NaOH solution upto the zero level. 40 ml of the given mixture of acid (HCl &CH3COOH) is pipetted out into a clean 100 ml beaker. The conductivity cell is then dipped in it.Theconductivity of the solution is measured using conductivity meter. The two terminals of the cell areconnected with a conductivity bridge.

Now 1 ml of NaOH from the burette is added to the solution, taken in the beaker stirred for sometime andthen conductivity is measured. The conductivity goes on decreasing upto the end point (A).

12

After the end point, again NaOH is gradually added, which causes increase in conductance. Theincrease in conductance is observed until the end point (B) is reached.

After the second end point, sudden increase in conductance is observed on further addition of NaOH.The reading (conductivity) is continuously measured for each addition of NaOH and are tabulated. Nowthe graph is plotted between the volume of NaOH Vs Conductance of the given solution. From the graphthe first end point (A) and the second end point (B) is noted. From the end points the amount of HCl andCH3COOH present in 100ml of the solution is calculated.

Advantages of conductometric titrations:

i) It gives more accurate end point.

ii) It is also used for the analysis of dilute solutios and weak acids.

iii) It is used in the case of coloured solutions where colour change of the indicator is not clear.

iv) Since the end point is detected graphically, no keen observation is necessary near the end point.

Equivalent weight of NaOH = 40

Equivalent weight of HCl = 36.5

Equivalent weight of CH3COOH = 60

Result

1. The amount of HCl present in 100ml of the given solution =____________ g

2. The amount of CH3COOH present in 100ml of the given solution =____________ g

Normality of NaOH = 0.0295N

13

Table: Volume of mixture (HCl + CH3COOH) Vs NaOH

Sl.No Volume of NaOH (ml) Conductance (mho)1 02 13 24 35 46 57 68 7

Model Graph

Volume of NaOH Vs Conductance

Calculation-I

Volume of mixture (HCl) (V1) = 40 mlNormality of mixture (HCl) (N1) = ?Volume of NaOH (V2) =______________ ml (1st titre value)Normality of NaOH (N2) = 0.02950 NAccording to the law of volumetric analysis (V1 N1) = V2 N2

N1 = V2 x N2 / V1= V2 x 0.02950 / 40

Strength of HCl =_______________NThe amount of HCl present in100ml of the given solution = N x Eq.Wt of HCl/10

= N x 36.5/10= ____________ g

14

Calculation-II

Volume of mixture (CH3COOH) (V1) = 40 ml

Normality of mixture (CH3COOH ) (N1) = ?

Volume of NaOH (V2) =___________ml (IInd titre value)

Normality of NaOH (N2) = 0.02950N

According to the law of volumetric analysis V1 N1 = V2 N2

N1 = V2 x N2 / V1

= V2 x 0.2950 / 40

Normality of CH3COOH =_________________N

The amount of CH3COOH present in

100 ml of the given solution =

Normality of CH3COOH x Eq.wt of CH3COOH /10 = N X 60/10

15

Chapter 5Estimation 0f Iron Content 0f Water Sample Using Spectrophotometer(1,10-Phenanthroline Method)

Aim

To estimate the amount of iron present in the given water sample using spectrophotometer(1,10-phenanthroline method).

Principle

The Beer – Lambert’s law is

log I /I = εCl

Where log I / I0 = A

A = εCl

Given

I = Intensity of incident light

I = Intensity of transmitted light

C = Concentration of solution (mole / litre)

L = Thickness of cell ( in cm)

= molar absorption coefficient ( L /cm/mole)

A = Absorbance

Wave length of light to be used is 510 nm.

Iron (II ) forms a red orange complex with 1, 10 – phenanthroline. The complexing reagent is weak base,that reacts to form phenanthrolinium ion.

Fe2+ + 3 phenH [ Fe(phen)32+] +3 H+

16

Procedure

About 25 ml of standard iron solution is taken in a 100 ml standard flask and 25 ml of distilled water istaken in another 100 ml standard flask (blank). 1 ml of hydroxylamine,10 ml of sodium acetate, and 10 mlof 1, 10 Phenanthroline solution is added to each of these flasks and the mixtures are allowed to standfor 5 min. Then the mixtures are diluted to the mark. The cells which the standard and the blank are goingto be taken are cleared well, with distilled water, and also with the samples. Now the standard ironsolution and the blank solution are placed in the cell and the absorbance of the standard is obtained withrespect to blank. The above procedure is repeated with (3, 6,9,12,15ppm,) of standard iron solutions .Simillarly absorbance for unknown concentration is measured.

A calibration graph is drawn between absorbance and concentration which passes through theorigin.From this unknown concentration of the solution is determined.

Result

The amount of iron present in the given sample of water = ______________ ppm.

Table Measurement of absorbance of given solution

S.NO Concentration of Fe2+ (ppm) Absorbance1 Blank2 33 64 95 126 157 unknown

Model Graph

Absorbance Vs Concentration (ppm)

17

18

Chapter 6Determination of Molecular Weight of Polyvinyl Alcohol Using OstwaldViscometer

Aim

To determine the molecular weight Mv of a given sample of polyvinyl alcohol by Viscometry method usingOstwald Viscometer. (Solutions of polyvinyl alcohol in water of concentrations 0.15, 0.30, 0.45, 0.60, and0.75 g/dl are given)

Principle

Molecular weight of the polymer means average molecular weight of the polymer. This can bedetermined from intrinsic viscosity of a dilute polymer solution. This is related to the molecular weight bythe relation.

[] = k (Mv) a

Where,

k and a are constants for a given polymer solution at a definite temperature. Mv represents averagemolecular weight by viscometric method.

For polyvinyl alcohol and water combination, ‘k’and ‘a’ values are 4.53 x 10-4 and 0.64 respectively.Intrinsic viscosity [] is related to the specific viscosity of the polymer solution by the relation.

] = [sp/ C] C0

Where,

sp = Specific viscosity

C = Concentration of the solution (g / dl)

(sp) is related to r by,

[sp] = r –1

where,

r = t / t 0

19

where ,

t and t 0 are flow time for solution and solvent respectively. [] can be known by drawing a graphbetween sp / C Vs C. the intercept of sp / C at zero concentration is known as intrinsic viscosity.

Molecular weight of repeating unit of polyvinyl alcohol is = 44

Structure of Polyvinyl alcohol is Cl Cl

[–CH2-CH-CH2-CH-]n

Procedure

Ostwald viscometer is filled with distilled water in the capillary limb upto the mark without air bubbles andthe flow time is noted by the following procedure. Using a rubber bulb, the water is sucked in the capillaryportion and above, upto the higher mark. The bubble is removed and the liquid is allowed to flow out.When the liquid reaches the top mark, stop clock is started. When the liquid reaches a lower mark, thestop clock is stopped. During the experiment, the viscometer should be in vertical position. Flow timemeasurements are repeated to get concordant values.

The water is drained out completely and the solutions of polyvinyl alcohol of different concentrations aretaken in the viscometer one after another by following the

above procedure. The flow time for various concentrations are determined. The solutions are handled inthe order 0.15, 0.30, 0.45, 0.60 and 0.75 g /dl . The flow times are tabulated and sp / C is calculatedfor each concentrations. A graph is drawn between

sp / C and C, from which [] can be known.

Result

The molecular weight of the given sample of polyvinyl alcohol =____________

Flow time for solvent water t0 =___________seconds

Table 1:

Sl.No Concentration ‘C’(g /dl)

Flow time (sec) t - t 0

sp = _____t 0

sp/ C

(dl / g)t 1 t 2 t

1. 0.15

2. 0.30

3. 0.45

4. 0.60

5. 0.75

20

Molecular Weight

Intrinsic viscosity [] = k (Mv) a

(from graph) = 4.53 x 10-4+ (Mv) 0.64

=

(Mv) 0.6 4 = [] /( 4.53 x 10-4)

0.64 log (Mv) = log [] / ( 4.53 x 10-4)

log (Mv) = log []/( 4.53 x 10-4)

___________________ 0.64

Model Graph

Concentration (g/dl) Vs sp/ C (dl / g)

21

Chapter 7Conductometric Titration of Strong Acid vs Strong Base

Aim

To determine conductometrically the amount of strong acid (hydrochloric acid) present in the givensolution by titrating with standard sodium hydroxide of 0.06875N.

Principle

The electrolytic solution conducts electricity due to the presence of ions. The specific conductance ofsolution is proportional to the concentration of the ions in it. When a solution of hydrochloric acid istitrated with NaOH , the fast moving hydrogen ions (H+) are progressively replaced by slow movingsodium ions(Na+). As a result, conductance of the solution decreases. This decrease in conductancewill take place until the complete neutralization of hydrochloric acid. Further addition of alkali, raise theconductance sharply because of an excess of free hydroxide ions (OH-)

NaOH + HCl NaCl + H2O

Materials Required

1. Conductivity bridge.

2. Conductivity cell.

Procedure

The burette is filled with standard sodium hydroxide solution. The given hydrochloric acid is transferredinto a 100 ml standard measuring flask and made upto the mark. Exactly 40 ml of the made uphydrochloric acid solution is pipette out into a clean 100 ml beaker. The conductivity cell is then dipped init. The conductivity of the solution is measured using a conductivity meter. Now 1ml of standard sodiumhydroxide solution is added from the burette and the conductivity is noted after each addition. Theconductance decreases gradually till the end point and then increases. These values are tabulated(Table – 1). Now the graph is plotted between volume of sodium hydroxide Vs conductance . From theend point the amount of hydrochloric acid present in 100ml of the given solution is calculated.

Advantages of conductometric titrations:

i) It gives more accurate end point.

ii) It is also used for the analysis of dilute solutios and weak acids.

iii) It is used in the case of coloured solutions where colour changes of the indicator is not clear.

22

iv) Since the end point is detected graphically, no keen observation is necessary near the end point.

Equivalent weight of NaOH = 40Equivalent weight of HCl = 36.5

Result

The amount of hydrochloric acid present in the 100ml of the given solution = ___________ g

Normality of NaOH = 0.06875N

Table -1: HCl Vs NaOH

Sl. No Volume of NaOH (ml) Conductance (mho)

1 02 13 24 3--- ------ ------- ---

15 1416 15

Model Graph

Volume of NaOH (ml) Vs conductance (mho)

23

Calculation-I

Volume of NaOH solution (V1) =_____________ml

Normality of NaOH solution (N1) = 0.06875 N

Volume of HCl solution (V2) = 40 ml

Normality of HCl solution (N2) =_______________?

According to the law of volumetric analysis

V1 N1 = V2 N2

N2 = V1 x N1 / V2

= V1 x 0.06875 / 40

Strength of HCl = ________________ N

The amount of HCl present in

100ml of the given solution = Normality of HCl x Eq.Wt of HCl/10

= N x 36.5/10

=________________ g.