INTRODUCTION TO THE TECHNIQUES OF EXPERIMENTAL CHEMISTRY
Authored by Jacob Vincent
Assistant Professor Department of Chemistry Dr. Sivanthi Aditanar College of Engineering Tiruchendur
Gyandhara International Academic Publication Publishing Research Worldwide>>> Thane, Maharashtra, India
‘INTRODUCTION TO THE TECHNIQUES OF EXPERIMENTAL CHEMISTRY’ by Jacob Vincent
© Copyright lies with GIAP Journals, India All rights reserved. No part of this publication may be reproduced or transmitted, in any form or by any means, without permission. Any person who does any unauthorized act in relation to this publication shall be liable to criminal prosecution and civil claims for damages. First Edition, July 2016
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ISBN: 978-93-83006-15-1 Publisher: Gyandhara International Academic Publications (GIAP) www.giapjournals.org
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Dedicated to YHWH
Acknowledgement We record our deep sense of indebtness to our Founder President Padma Shri.Dr.B.Sivanthi Adityan and our honourable chairman Sri.Balasubramanian S Adityan for having blessed this venture. We are thankful to 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 of Engineering, Tiruchendur for the constant inspiration and encouragement. We are deeply indebted to a number of colleagues and students who have read this material and given us many valuable suggestions for improving the presentation. We owe a special debt of gratitude to the faculties from department of chemistry. Dr.Sivanthi Aditanar College of Engineering, Tiruchendur, Tuticorin district, Tamilnadu who have read the original manuscript in detail and gave us many perceptive comments. We have received, and gratefully acknowledge, reports of course testing of the notes for this book by our previous batch students.
ii
Book preface This course has been specially designed as an intensive introduction to the techniques of experimental chemistry. Our goals in this class are twofold. First, since freshmen cannot enroll in any of the regular chemistry lab courses, this book has been created to give interested first-year students an opportunity to get hands-on experience with chemistry. A second aim is to prepare freshmen for laboratory in the Chemistry Department. Freshmen often have a difficult time finding a comfortable position in our department because they don't yet have the experimental skills and experience developed in our regular chemistry lab course sequence. Unlike other laboratory classes, the goal is not just to successfully perform an experiment and write a report; instead, the focus will be on mastering the techniques and skills necessary to carry out experiments. During the next month, you will mix, stir, and measure until you reach a "professional level" of skill in various techniques fundamental to chemical experiments.
iii
About Author Author has an experience of Assistant Professor in chemistry for 7 years in Engineering College and 3 years in Polytechnic College. He has completed M.Sc. M.Phil (Chemistry), M.B.A., Ph.D. (Nano polymerspursuing). He has published 3 papers in international journals and 2 papers in national journals. He has presented one paper in international conference and one paper in national conference.
iv
Salient features of the book Logically explained and student friendly approach. Simple, systematic and well-organized text. Well-illustrated with nicely drawn graphs.
v
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(1,10Phenanthroline 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
vi
Chapter 1
Determination of Dissolved Oxygen Content of Water Sample by Winkler’s Method Aim To find out the quantity of dissolved oxygen present in the given water sample being supplied with 0.0094 N 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 reaction velocity in the oxidation of 2-phenyl ethanol in aqueous sulphuric acid midium Tables from 3.12 to 3.21 show the individual sets for the variation of oxidant concentrations. First column of each table contains time while the second column shows absorbance at various time intervals. In the third column slope value obtained from the individual graphs plotted between absorbance versus time, are given. These slope values were converted into -dc/dt values as discussed in section 2.2, which are given at the base of each individual table. Table 3.22 shows the summarized variation in which first column shows concentration of cerium(IV) and second column shows corresponding -dc/dt values. In the third column values 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 is responsible to liberate free iodine. MnO(OH)2 -
+ H2SO4
2KI + H2SO4 + [O] 2Na2S2O3
MnSO4
K2SO4
+ 2 H2O + [O] + H 2O + I 2
+ 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 the end point.
1
Procedure Titration 1: Standarisation Of Sodium Thiosulphate The burette is washed and filled with sodium thiosulphate solution. 20 ml of potassium dichromate solution is pipetted out into a clean conical flask. About 20 ml of dilute sulphuric acid is added followed by the addition of 10 ml of 10% potassium iodide solution. Shake the flask well. The liberated iodine is titrated against thio solution taken in the burette. When the solution becomes pale yellow in colour, 2 ml of freshly prepared 1% starch solution is added and the titration is slowly continued. The end point is the colour change from blue to colourless. Titrations are repeated to get concordant titre values. The readings are tabulated (Table-1). From the titre value the normality of sodium thiosulphate solution is calculated.
Titration 2: Estimation of Dissolved Oxygen Water samples are collected in BOD bottle. To this 1 ml of manganous chloride solution and 1 ml of alkaline potassium iodide solution were added to fix the dissolved oxygen in it. The manganous hydroxide is oxidised to brown colour manganic oxide and precipitated. Shake the solution well and allow 20 minutes time for all oxygen to react and then to settle as precipitate. When half of the precipitate settles down, the stopper is removed and 2 ml of Sulphuric acid solution is added. The bottle is restoppered and mixed 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 sodium thiosulphate solution using starch indicator. The end point is the disappearance of blue colour. Titrations are 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
Burette reading (ml)
K2Cr2O7 (ml) Initial
Final
Volume of Sodium thiosulphate (ml)
1
Indicator
Starch
2
Concordant titre value = __________ ml
2
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)
=
?
=
V1 N1 /V2
=
N
N2
Table 2: Estimation of Dissolved Oxygen Sl.No
Volume of
Burette reading (ml)
Water sample(ml) Initial
Volume of Na2S2O3 (ml)
Indicator
Final
1
Starch
2 Concordant titre value = __________ ml Calculation V DO
DO
N x 8 x 1000
= ___________ x ______________ x v (V-2)
60
270
N x 8 x 1000
= ___________ 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
3
Chapter 2
Determination of Chloride Content of Water Sample by Argentometric Method Aim To determine the amount of chloride present in the water sample being supplied with silver nitrate solution 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 and simple. It proceeds as follows. Ag+NO3- + Na+Cl(in water)
AgCl
+ NaNO3
(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. At the end point the yellow colour of chromate changes into reddish brown due to the reaction 2 AgNO3 + K2CrO4 Ag2CrO4 + (Yellow)
2KNO3
(Reddish brown)
K2CrO4 indicator will not be precipitated as Ag2CrO4 until all the chlorides in the solution have been precipitated 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% K2CrO4 indicator 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 is shaken well and stirred with the glass rod. Nearness of the end point is noted by the formation of
4
coagulation of silver chloride precipitate at the bottom of the flask. Now, the addition of AgNO3 is done drop by drop until the supernatant solution gets a faint reddish brown tinge. The titration is repeated to get 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% of K2CrO4 indicator solution is added. It is titrated against the standardised AgNO3 solution from the burette. 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 chloride and 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 â&#x20AC;&#x201C;1: Standardisation of Silver Nitrate
Sl. No
Volume of NaCl (ml) V1
Initial
1
20
0
2
20
0
3
20
0
Concordant titre value
Burette reading (ml) Final
Volume of AgNO3 (ml) V2
Indicator
Potassium chromate
= ----------- 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 = __________ N
V2 Normality of silver nitrate
5
Estimation of Chloride Table â&#x20AC;&#x201C;2: Estimation of Chloride Sl. No
Volume of Water sample(ml)
1
20
2
20
Burette reading (ml) Initial
Volume of AgNO3 (ml)
Indicator
Final Potassium Chromate
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
6
Chapter 3
Determination 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 is provided
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 be below 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-
ď&#x201A;Ž
NaCl
+ H 2O
After the end point, further addition of NaOH increases the pH sharply as there is an excess of fast moving 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 out into a clean beaker. It is then diluted to 20 ml by adding distilled water. The glass electrode is dipped in it and connected with a pH meter.
7
Titration-I First a preliminary titration is carried out by adding standard NaOH solution in portions of 1ml from the burette to the HCl solution taken in the beaker and pH of the solution is noted for each addition. By the gradual addition of NaOH, sudden jump occurs in the pH value. After the sudden jump take 7 more readings and the range at which the end point lies is found out by plotting volume of NaOH added against pH(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 of NaOH is continued even after the end point for further 1ml. The accurate end point is determined by plotting ΔpH / ΔV against Volume of NaOH added (Graph II). From the end point the strength of HCl solution 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)
1
0
2
1
3
2
4
3
5
4
6
5
7
6
8
7
9
8
10
9
pH
8
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 of NaOH (ml)
pH
Δ pH
Δ V (ml)
Δ pH / Δ V
Mean volume of NaOH
1 2 3 4 5 6 7 8 9 10
9
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
= V 2 N2
N1
= V 2 x N 2 / 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
10
Chapter 4
Determination 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 the given 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 of a solution is proportional to the concentration of the ions in it, conductance of the solution is measured during titration. When the sodium hydroxide is added slowly from the burette to the solution, HCl (strong acid) gets neutralised first. Since the fast moving H+ ions are replaced by slow moving Na+ion decrease in conductance takes place until the end point is reached. HCl
+ NaOH ď&#x201A;Ž
NaCl
+ H2O (1st neutralization)
After the complete neutralization of all HCl, the neutralization of CH3COOH starts, CH3COOH
+ NaOH
ď&#x201A;Ž
CH3COONa
+ H2O (IInd neutralization)
Since CH3COONa is stronger than CH3COOH conductivity slowly increases until all CH3COOH is completely neutralised. When the end point is reached, the addition of NaOH will cause sudden increase in 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.The conductivity of the solution is measured using conductivity meter. The two terminals of the cell are connected with a conductivity bridge. Now 1 ml of NaOH from the burette is added to the solution, taken in the beaker stirred for sometime and then conductivity is measured. The conductivity goes on decreasing upto the end point (A).
11
After the end point, again NaOH is gradually added, which causes increase in conductance. increase in conductance is observed until the end point (B) is reached.
The
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. Now the graph is plotted between the volume of NaOH Vs Conductance of the given solution. From the graph the first end point (A) and the second end point (B) is noted. From the end points the amount of HCl and CH3COOH 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
12
Table: Volume of mixture (HCl + CH3COOH) Vs NaOH Sl.No 1 2 3 4 5 6 7 8
Volume of NaOH (ml) 0 1 2 3 4 5 6 7
Conductance (mho)
Model Graph Volume of NaOH Vs Conductance
Calculation-I Volume of mixture (HCl) (V1) Normality of mixture (HCl) (N1) Volume of NaOH (V2) Normality of NaOH (N2) According to the law of volumetric analysis (V1 N1) N1 Strength of HCl The amount of HCl present in 100ml of the given solution
= 40 ml =? =______________ ml (1st titre value) = 0.02950 N = V2 N2 = V2 x N2 / V1 = V2 x 0.02950 / 40 =_______________N =N x Eq.Wt of HCl/10 =N x 36.5/10 = ____________ g
13
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
14
Chapter 5
Estimation 0f Iron Content 0f Water Sample Using Spectrophotometer(1,10Phenanthroline Method) Aim To estimate the amount of iron present in the given water sample using spectrophotometer(1,10phenanthroline 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+
15
Procedure About 25 ml of standard iron solution is taken in a 100 ml standard flask and 25 ml of distilled water is taken in another 100 ml standard flask (blank). 1 ml of hydroxylamine,10 ml of sodium acetate, and 10 ml of 1, 10 Phenanthroline solution is added to each of these flasks and the mixtures are allowed to stand for 5 min. Then the mixtures are diluted to the mark. The cells which the standard and the blank are going to be taken are cleared well, with distilled water, and also with the samples. Now the standard iron solution and the blank solution are placed in the cell and the absorbance of the standard is obtained with respect 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 the origin.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 1 2 3 4 5 6 7
Concentration of Fe2+ (ppm) Blank 3 6 9 12 15 unknown Model Graph
Absorbance
Absorbance Vs Concentration (ppm)
16
17
Chapter 6
Determination of Molecular Weight of Polyvinyl Alcohol Using Ostwald Viscometer Aim To determine the molecular weight Mv of a given sample of polyvinyl alcohol by Viscometry method using Ostwald Viscometer. (Solutions of polyvinyl alcohol in water of concentrations 0.15, 0.30, 0.45, 0.60, and 0.75 g/dl are given)
Principle Molecular weight of the polymer means average molecular weight of the polymer. This can be determined from intrinsic viscosity of a dilute polymer solution. This is related to the molecular weight by the relation. [] = k (Mv) a Where, k and a are constants for a given polymer solution at a definite temperature. Mv represents average molecular 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
18
where , t and t 0 are flow time for solution and solvent respectively. [] can be known by drawing a graph between 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 Structure of Polyvinyl alcohol is
Cl
= 44 Cl
[–CH2-CH-CH2-CH-]n
Procedure Ostwald viscometer is filled with distilled water in the capillary limb upto the mark without air bubbles and the flow time is noted by the following procedure. Using a rubber bulb, the water is sucked in the capillary portion 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, the stop clock is stopped. During the experiment, the viscometer should be in vertical position. Flow time measurements are repeated to get concordant values. The water is drained out completely and the solutions of polyvinyl alcohol of different concentrations are taken in the viscometer one after another by following the above procedure. The flow time for various concentrations are determined. The solutions are handled in the order 0.15, 0.30, 0.45, 0.60 and 0.75 g /dl . The flow times are tabulated and sp / C is calculated for 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’
Flow time (sec)
(g /dl) t1 1.
0.15
2.
0.30
3.
0.45
4.
0.60
5.
0.75
t2
t
t- t0
sp/ C
sp = _____
(dl / g)
t0
19
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)
20
Chapter 7
Conductometric Titration of Strong Acid vs Strong Base Aim To determine conductometrically the amount of strong acid (hydrochloric acid) present in the given solution by titrating with standard sodium hydroxide of 0.06875N.
Principle The electrolytic solution conducts electricity due to the presence of ions. The specific conductance of solution is proportional to the concentration of the ions in it. When a solution of hydrochloric acid is titrated with NaOH , the fast moving hydrogen ions (H+) are progressively replaced by slow moving sodium ions(Na+). As a result, conductance of the solution decreases. This decrease in conductance will take place until the complete neutralization of hydrochloric acid. Further addition of alkali, raise the conductance sharply because of an excess of free hydroxide ions (OH-) NaOH + HCl
ď&#x201A;Ž 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 transferred into a 100 ml standard measuring flask and made upto the mark. Exactly 40 ml of the made up hydrochloric acid solution is pipette out into a clean 100 ml beaker. The conductivity cell is then dipped in it. The conductivity of the solution is measured using a conductivity meter. Now 1ml of standard sodium hydroxide solution is added from the burette and the conductivity is noted after each addition. The conductance decreases gradually till the end point and then increases. These values are tabulated (Table â&#x20AC;&#x201C; 1). Now the graph is plotted between volume of sodium hydroxide Vs conductance . From the end 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.
21
iv) Since the end point is detected graphically, no keen observation is necessary near the end point.
Result
Equivalent weight of NaOH Equivalent weight of HCl
= 40 = 36.5
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)
1
0
2
1
3 4
2 3
---
---
------
-----
15 16
14 15
Conductance (mho)
Model Graph Volume of NaOH (ml) Vs conductance (mho)
22
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 =N
x
x
Eq.Wt of HCl/10
36.5/10
=________________ g.
23