practical phys chemistry
TRANSCRIPT
According to revised CBCS
PRACTICAL PHYSICALCHEMISTRY
Dr. ASHOK SANTUKRAO KULKARNI
Member of Faculty
Bahirji Smarak Mahavidyalaya, Basmath
SIDDHI PUBLISHING HOUSE
Nanded, Maharashtra, India.
CBCS syllabus of S. R. T. M. University, Nanded
A TEXT-BOOK OF
PRACTICAL PHYSICALCHEMISTRY
For B.Sc. degree course
By
Dr. ASHOK SANTUKRAO KULKARNI M.Sc., B.Ed., Ph.D.
Member – BOS in Chemistry & Member of Faculty of Science and Technology
(SRTMU, Nanded) Associate Professor
Department of Chemistry, Bahirji Smarak Mahavidyalaya, Basmathnagar Dist. Hingoli (India)
FIRST EDITION
SIDDHI PUBLISHING HOUSE
Nanded, Maharashtra, India.
syllabus of S. R. T. M. University, Nanded
PRACTICAL PHYSICAL
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ISBN No.978-81-940206-8-4
Authors
Dr. Ashok Santukrao Kulkarni Mob. No.: 9420886667 / 8766884753
All Rights Reserved
No part of this publication may be reproduced, in retrieved system or transmitted in any form by any means without prior written permission.
Published By
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Typesetting
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Price : 100 /-
First Edition : 17 December 2020
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PREFACE TO FIRST EDITION
It is a great pleasure to place the A text book of practical physical chemistry, at the hands of the student’s. The laboratory experimental B.Sc. course book has been written according to the prescribed latest syllabus of UGC & S.R.T.M.U, Nanded. It is also useful to other universities. I have tried to make this book more useful to meet the needs of chemist’s and B.Sc. Chemistry degree course students. It serves as a text on the experimental physical chemistry and provides both fundamental concepts, theoretical and practical approach in a simple, lucid and comprehensive manner even an average learner chemist can perform the experiments introduced in the separate chapter.
Its object is to take the chemistry learners with very easy way and to motivate them in order to get practical knowledge of physical chemistry. So, this book includes basic definitions, terminology, systematic observation tables, plots, numerical data and Appendix. It is an essential practical text book for the degree course students and teachers. I hope the present book will serve better the students performing experiment at degree course in all Indian Universities.
I’m grateful to Hon. Jayprakashaji Dandegaonkar [President], Hon. Adv. Munjajirao Jadhav [Vice-president], Hon. Panditraoji Deshmukh [Secretary], Bahirji Education Society, Vapati. Principal, Dr. R. N. Ingle and Vice Principal Dr. N. N. Lokhande, B.S. Mahavidyalaya Basmathnagar, District Hingoli, for their constant inspiration and encouragement.
I wish to express special thanks to Prof. Bhaskar S. Dawane, Chairman B.O.S. Chemistry & all Members of B.O.S. Chemistry S.R.T.M.U., Nanded. Mr. Omkar, Miss. Manasi and members of the family for their co-operation in this pursuit and all dear friends for their helpful co-operation. I am also thankful to book publisher Siddhi Prakashan,Nanded. Dr. Rajesh G. Umbarkar & Shri Jai Rathod, in bringing this book in its present. Any suggestions from the reader for the improvement of the text book are always welcome. Finally it is our duty to thank who have co-operated & encourage to this endeavor.
Thanking you
Dr. Ashok Santukrao Kulkarni
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C O N T E N T S
CHAPTER TITLE PAGE
NO.
1 FUNDAMENTAL CONCEPT 8-10
1.1 Introduction 8
1.2 Concept of Acids and Bases 9
2 CONDUCTOMETRY 11-26
Introduction 11
Expt. No. 1
To determine the normality and strength of strong acid (HCl/ H2SO4/ HNO3) conductometrically by using standard Solution of strong base (NaOH/ KOH)
15
Expt. No. 2
To determine the normality and strength of weak acid (CH3COOH/ HCOOH) conductometricaly by using standard Solution of strong base (NaOH/ KOH).
17
Expt. No. 3
Determination of solubility of sparingly soluble salt likePbSO4, BaSO4, etc at room temperature 19
Expt. No. 4
Determine the concentration of KCl solution by titrating it with standard solution of AgNO3 conductometricaly 21
Expt. No. 5
Determine the amount of oxalic acid conductometrically by using standard NaOH Solution 23
Expt. No. 6
Determine the equivalent conductance of a strong electrolyte at several concentrations and verify the Onsager’s equation, conducto-metrically
25
3 POTENTIOMETRY 27-43
Introduction 27
Expt. No. 7
To determine the normality and strength of strong acid (HCl/ H2SO4/ HNO3) potentiometrically by using standard Solution of strong base (NaOH/ KOH)
35
Expt. No. 8
To determine redox potential of Fe+ + + / Fe + + system by titrating with standard K2Cr2O7ptentiometrically 37
Expt. No. 9
To determine the dissociation constant of weak acid (CH3COOH) potentiometric measurement 39
Expt. No. 10
To determine the normality and strength of acids in mixture of strong acid (HCl) and weak acid (CH3COOH) potentiometrically using standard solution of strong base (NaOH)
42
4 pH-METRY 44-47
Expt. No. 11
To determine pKa value of the given organic acid by pH- measurement. 44
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Expt. No. 12
Determine dissociation constant organic acid [acetic acid] using various buffer solutions of sodium acetate and acetic acid in aqueous solution pH –metrically.
46
5 COLORIMETRY 48-53
Expt. No. 13
Verify Lambert-Beer’s law using KMnO4 Colorimetrically and determine concentration of unknown KMnO4 solution. 50
Expt. No. 14
To study the complex formation between Fe3+ and 5-SSA by Job’s method using colorimeter 52
6 POLARIMETRY 54-56
Introduction 54 Expt. No. 15
To study the inversion of cane sugar by polarimeter 54
7 SURFACE TENSION 57-58
Expt. No. 16
Determine the interfacial tension between two miscible liquids say benzene and water at room temperature 57
8 MEASUREMENT OF DENSITY 59-60
Expt. No. 17
To measure the density of a given liquid by Pyknometer 59
Expt. No. 18
To measure the density by using density bottle 60
9 VISCOSITY 61-63
Introduction 61 Expt. No. 19
Determine the viscosity of a liquid by Ostwald's viscometer 62
Expt. No. 20
Determine the composition of the given mixture consisting of two miscible liquids, A & B by viscosity measurement 63
10 CHEMICAL KINETICS 64-72
Introduction 64
Expt. No. 21
Determine energy of activation of a reaction between KI and K2S2O8. (Potassium iodide and Potassium persulphate) 65
Expt. No. 22
To study the kinetics of dissolution of magnesium metal in dilute HCl 67
Expt. No. 23
To study the kinetics of decomposition of sodium thiosulphate by a mineral acid 69
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Expt. No. 24
Determine the rate constant of the reaction between K2S2O8 and KI with equal concentration of reacting species (a=b). 70
11 HETEROGENEOUS EQUILIBRIA 73-79
Introduction 73
Expt. No. 25
Determine the solubility of benzoic acid in water at difference temperature and hence it’s heat of solution 74
Expt. No. 26
To study the effect of NaCl (or succinic acid) on critical solution temperature of phenol water system and determine the concentration of that solute in the given system
76
Expt. No. 27
Determine the partition coefficient of iodine between carbon tetrachloride and water 78
12 THERMODYNAMICS 80-83
Expt. No. 28
Determine the enthalpy change of neutralization of strong acid by strong base. 80
Expt. No. 29
Determine the heat of neutralization of acetic acid against strong base and determine enthalpy of ionization of weak acid 82
13 MOLECULAR WEIGHT DETERMINATION 84-86
Expt. No. 30
To determine molecular weight of a nono-volatile solute by the rast’s method 84
Expt. No. 31
To determine the molecular weight of a nonvolatile solute (KCl, urea, glucose etc.) by Beckmann's freezing method (using water as solvent).
85
Appendix 88-102
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SAFETY PRECAUTIONS
to be observed during Laboratory Work
• The important precautions should be taken in the chemical laboratory. • Never work in the chemical laboratory without proper guidance. Use laboratory coat
or apron while performing. • Experiment in the laboratory. • Don’t smell or taste the chemicals in the laboratory. • The experiment is carried out with proper exhaust system. • If you have to dilute an acid always add acid to water with constant stirring. • Don’t weigh a any substance on a piece of paper use proper glassware. • Keep the digital balance and its surroundings clean don't spill the chemicals on the
pan of the balance. • Always handle the chemicals, glassware, equipments carefully. • All waste materials, papers, broken glassware, always place in the dustbin. • While preparing the solutions, always label and cover the reagents or solutions
properly. • Never pour back the reagents taken by you in the reagent bottles. • Measuring cylinders, standard flask should not be heated. • Before leaving the laboratory make sure that you clean workplace, equipment,
glassware and keep safety.
LABORATORY FIRST-AID
A First-Aid Box, clearly labeled, should be kept in a readily accessible place in the laboratory, and should contain the following articles. All bottles and packages should be clearly labeled. Bandages, lint, gauze, cotton-wool, adhesive plaster, and a sling. Delicate forceps, needles, thread, safety-pins, and scissors. Fine glass dropping-tube. Glass eye-bath.
Vaseline, Castor oil, Olive oil, Sal volatile, Zinc oxide ointment, Boric acid (powder). Chloramine-T (fine hydrated crystals, obtainable). Savlon Antiseptic cream and emulsion, Savlon cream is a very effective and harmless antiseptic; it is unaffected by long storage.
Bottles of the following:
- Acriflavine Emulsion (in quantity). - Savlon Antiseptic (aqueous solution).
Lime-water (in quantity):
- 2% Iodine solution, 1% Boric acid, 1% Acetic acid. - 8% Sodium bicarbonate solution (i.e., saturated in the cold). - 1% Sodium bicarbonate solution. - Amyl nitrite capsules.
Burns:
Whatever the cause of the burn or scald, e.g., flames, hot metal, hot liquid, acids, caustic alkalis, etc.:
(1) Remove any smouldering or acid/alkali-soaked clothing.
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(2) Thoroughly and profusely wash the burnt area with cold tap water. (3) Gently dry with a clean (or preferably sterile) towel or cotton wool. (4) Cover burnt or scalded area with sterile dressing. (5) Transfer patient to hospital or doctor for consideration of further treatment (a course
of anti-tetanus toxoid is frequently indicated if the patient is not already immunised, i.e., has had a full course of booster doses of ATT within the last five years). Minor scalds may be treated as above (1 to 3) and then savlon or acriflavine emulsion
applied before application of a sterile dressing.
SPECIAL CASES
Bromine Burns: When experiments involving the use of liquid bromine are being performed, small bottles of petroleum (b.p. 80- 100°) should be immediately available. If bromine is spilt on the hands, immediately wash it off with an ample supply of petroleum, when the bromine will be completely removed from the skin. (If subsequently the skin which has been in contact with the petrol feels tender and ‘smarts’ owing to removal of the normal film of grease, cover gently with savlon emulsion or with olive oil.)
Sodium Burns: Most frequently caused by small molten pellets ejected from heated tubes. If a small pellet of sodium can still be seen, remove carefully with forceps. Then wash the burn thoroughly in cold water, then in 1% aqueous acetic acid and then cover with gauze soaked in savlon emulsion or olive oil.
Phosphorus Burns: Wash well in cold water, and then immerse in ca. 1% aqueous silver nitrate solution. If serious, wash again with water, and apply the savlon treatment.
Methyl sulphate: If spilt on hands: wash immediately with an ample supply of concentrated ammonia solution. Then dab gently with a wad of cotton wool soaked in the ammonia solution.
EYE ACCIDENTS
Whatever liquid (e.g., acid or caustic alkali) or solid (e.g., glass) has entered the eye, the eye should be thoroughly and profusely washed with tap water (e.g., put head under tap) and the eye-lids should be held widely open, especially where caustic alkalis have entered the eye.
No attempt should be made to remove imbedded glass. This is a hazardous procedure and should be left to a doctor-more damage can result if attempted by someone not skilled in this technique. Soreness which may follow very minor accidents to the eye may be relieved by placing 1 drop of castor oil in the corner of the eye.
CUTS
Minor: wash thoroughly with soap or disinfectant and water until absolutely clean. Apply sterile dressing.
For more serious cuts, where bleeding is profuse, apply pressure with a thick sterile (or at least clean) pad, dressing, or towel over the area. If an artery is spurting, try to minimize bleeding by applying pressure immediately above and below the cut.
If glass is still thought to be present in the cut, wash thoroughly before applying pressure. If bleeding is profuse, the application of pressure to prevent bleeding is more
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important than the removal of the glass. In all cases the patient should be transferred to hospital or to a doctor, for further treatment of the wound, and also for a course of anti-tetanus toxoid, which is indicated if the patient is not already fully immunized.
POISONS
Solid and Liquid Poisons
(a) In the mouth but not swallowed. Spit out at once, and wash the mouth out repeatedly with water.
(b) If swallowed. Whatever the substance swallowed (e.g., acids, alkalis, arsenic or mercury compounds, etc.), dilute by drinking approximately one pint of milk, preferably, or water.
No attempt to make the patient vomit should be carried out, especially in the cases of swallowed acids or alkalis.
Hydrogen cyanide (inhaled) or alkali cyanides (taken by mouth); inhale amyl nitrite from freshly opened capsules. Obtain medical attention urgently.
GAS POISONING
Remove the patient to the fresh air, and loosen clothing at the neck. If breathing has stopped or is extremely weak, give artificial respiration and continue until the patient is transferred to hospital or until a doctor arrives.
To counteract chlorine or bromine fumes if inhaled in only small amounts, inhale ammonia vapour. Afterwards suck eucalyptus pastilles, or drink warm dilute peppermint or cinnamon essence, to soothe the throat and lungs.
ELECTRIC SHOCK
Switch off. Treat for burns as necessary.
If the patient is in a state of ‘shock’ (i.e., pale, faint or collapsed, sweating, cold) treat by lying flat, or preferably with the legs raised approximately one foot, loosen clothing around the neck, keep warm but not hot (one to two blankets) and transfer to hospital or obtain medical attention urgently.
TREATMENT OF FIRES
Clothes: Laboratories should be equipped with a sufficient number of fireproof blankets, so that a blanket is available at any point of the laboratory at a few seconds’ notice. Each blanket should be kept in a clearly labeled box, the lid of which is closed by its own weight and not by any mechanical fastening, which might delay removal of the blanket. The box itself should be kept in some open and unencumbered position in the laboratory.
The blanket when required should be at once wrapped firmly around the person whose clothes are on fire, the person then placed in a prone position on the floor with the ignited portion upwards, and water poured freely both over the blanket and in between the blanket and the person’s clothes until the fire is extinguished.
BENCH FIRES
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Most of the available fire-extinguishers are unsuitable for chemical laboratory use. Those which give a stream of water are useless for extinguishing burning ether, benzene, petrol, etc., and exceedingly dangerous if metallic sodium or potassium are present. Those which give a vigorous and fine stream of carbon tetrachloride, frequently serve merely to fling the burning material (and particularly burning solvents) along the surface of the bench without extinguishing the fire, the area of which is thus actually increased. Contact of the tetrachloride with metallic sodium or potassium may cause violent explosions. The following methods should therefore generally be used:
(1) Sand. Buckets of dry sand for fire-extinguishing should be available in the laboratory and should be strictly reserved for this purpose, and not encumbered with sand-baths, waste-paper, etc. Most fires on the bench may be quickly smothered by the ample use of sand. Sand once used for this purpose should always be thrown away afterwards, and not returned to the buckets, as it may contain appreciable quantities of inflammable, non-volatile materials (e.g., nitrobenzene), and be dangerous if used a second time.
(2) Carbon tetrachloride. Although sand is of great value for extinguishing fires, it has the disadvantage that any glass apparatus around which the fire centers is usually smashed under the weight of the sand. Alternatively, therefore, for small fires carbon tetrachloride may be poured in a copious stream from a Winchester Bottle on to the fire, when the ‘blanketing’ effect of the heavy carbon tetrachloride vapor will quickly extinguish the fire. In such cases it should be remembered, however, (a) not to use carbon tetrachloride if metallic sodium or potassium is present, as violent explosions may result, (b) to ventilate the laboratory immediately after extinguishing the fire, in order to disperse the phosgene vapour which is always formed when carbon tetrachloride is used in this way.
(3) Carbon dioxide. The above disadvantages of carbon tetrachloride have caused it to be now superseded by carbon dioxide; these are the most effective extinguishers for use in the laboratory. Carbon dioxide has the advantages that in use (a) adverse chemical reactions are extremely unlikely, (b) there is no electrical hazard, and (c) damage to apparatus is minimal.
(4) If a liquid which is being heated in a beaker or a conical flask catches fire, it is frequently sufficient to turn off the gas (or other source of heating) below and then at once to stretch a clean duster tightly over the mouth of the vessel. The fire quickly dies out from lack of air, and the (probably valuable) solution is recovered unharmed. Students should bear in mind that the majority of bench fires arise from one of three causes, all of which result from careless manipulation by the student himself. These causes are: (i) the cracking of glass vessels which are being heated (usually for distillation purposes) whilst containing inflammable liquids. This cracking may occasionally be due to faulty apparatus, but is almost invariably caused by an unsuitable method of heating, the latter furthermore being often hastily applied. (ii) The addition of unglazed porcelain to a heated liquid which is ‘bumping’ badly-with the result that the previously superheated liquid suddenly froths over and catches fire. Porcelain should never be added to a ‘bumping’ liquid until the latter has been allowed to cool for a few minutes and therefore has fallen in temperature below its boiling-point. (iii) The heating of volatile, inflammable liquids in flasks not fitted with reflux condensers.
The most dangerous solvent in the laboratory is carbon disulphide, the flash-point of which is so low that its vapour is ignited, e.g., by a gas-ring 3-4 minutes after the gas has been turned out. Carbon disulphide should therefore never be used in the laboratory unless an adequate substitute as a solvent cannot be found. Probably the next most dangerous liquid for general manipulation is ether, which, however, has frequently to be employed.
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SPECIAL CAUTION
Safety goggles should always be worn over the eyes when carrying out potentially dangerous operations, e.g. vacuum distillations, distillation of large volumes of inflammable liquids and experiments employing large quantities of metallic sodium.
CHAPTER
1
FUNDAMENTAL CONCEPT 1.1 Introduction: The general knowledge of practical fundamental concept is very essential particularly while performing the experiments. Students should know the basic terminology of chemistry practicals. They should know the different methods of expressing concentrations, pertaining to basic concept of preparation of solutions, standard solution, molecular formula, molecular weight, equivalent weight, normality, molarity, molality & fundamental definitions of acids, bases, electrolytes, measurements of physical properties, physical constant and their units. Students should be well aware of the experimental procedure, observation tables.
Solution: A solution is a single phase homogeneous mixture of one or more components
Standard solution: The solution whose concentration or strength is known is called as standard solution.
Titrate: A known volume of solution is taken in a conical flask by using a pipette is called as titrate or aliquot.
Titrant: The solution is taken in a burette and added to titrate till the reaction is exactly complete is called as titrant.
Methods expressing concentration of solution:
1. Normality (N): It is the no. of gram equivalents of solute in 1 lit. of solution.
No. of gram equivalent of solute
N =
Volume of solution in litre
Weight of solute in grams ×1000
N =
Eq. weight of solute × volume of solution in ml
1 gm of equivalent weight means the equivalent weight of a substance expressed in gms. e.g. 1gm of equivalent weight of oxalic acid means 63 of oxalic acid.
Gm equivalent = weight of solute in gm / Equivalent weight of solute
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2. Molarity (M):It is the no. of moles of solute per litre of solution is called molarity. It is expressed as mol / litre.
M = Moles of solute/ Volume of solutionin litre.
M = Weight of solute in gm × 1000 / M. Wt. of solute × vol. of sol. in ml.
Moles (n) = Wt. of solute in gms / Molecular wt. of solute.
3. Molality (m):It is the moles solute per kg of solvent is called molality.
m = moles of solute / weight in kg of solvent.
m = Wt. of solute in gm × 1000 / M. Wt. of solute × Wt. of solvent in gm.
Normal solution (1N): When 1 gm equivalent of substance is dissolved in water to make the volume of solution 1 liter then, the solution is known as 1 normal solution.
Molar solution (1M): When one gram mole of substance is dissolved in water to make the volume of solution one litre is known as one molar solution
Molal solution (1m):When one gram mole of substance is dissolved in 1000 gms of the solvent then this concentration of the solution is called 1 molal.
Mole Fraction (x1): The no. of moles of that component divided by the total no. of moles of all components in the solution.
mole fraction of solute = moles of solute / moles of solute + moles of solvent.
mole fraction of solvent = moles of solvent / moles of solute + moles of solvent.
The mol fraction of solution = mol fraction of solute + mol fraction of solvent = 1
Volume fraction: If two liquids are miscible to form a solution then the volume in fraction can be determined as,
Volume fraction of one liquid (V1) =V1/ V1+V2
Volume fraction of other liquid (V2)=V2/V1+V2
eg. 100 ml solution of ethyl alcohol & water, If 30 ml ethyl alcohol is dissolved in 70 ml of distilled water then,
Volume fraction of alcohol=Volume of alcohol / Volume of alcohol + Volume of water
Volume fraction of alcohol=30/30+70=0.30
Volume fraction of water=70/30+70=0.70
That means solution contain 30% alcohol &70%water
ppm = mg/kg
1 million grams=1000 kilo grams =1000 lit. of liquid.
1.2 Concept of Acids and Bases:
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Arrhenius Theory Concept (1887): In 1887, Arrhenius, the Swedish chemist, proposed theory of ionization of electrolyte in aqueous solution. According to Arrhenius theory, an acid is the substance that releases H+ ions and base releases OH-ions in the water.
eg: HCl (Acid) + Water H+ + Cl- NaOH (Base) + Water OH- + Na+
J.N. Bronsted – Lowry Theory Concept (1923): This concept is independent of solvent and called protonic concept. According to this theory an acid is a molecule which donatesa proton [H+] (Proton donor) and base which can accept proton (Proton acceptor) [H+].
eg: (acid) HCl + H2O (Base) = H3O + Cl-
Gilbert N. Lewis Theory Concept (1930): This concept is based upon electronic theory of valency. According to this theory an acid is an electron pair acceptor and base is an electron pair donor.
eg: A ( Acid)+:B (Base) A-B (Co-ordinate Complex)
H+ (Acid)+:NH3(Base) NH4 + (Co-ordinate Complex)
Lewis Acid: Any species (molecule, atom or ion) which can accept a pair of electrons to term co-ordinate bond. These are electrophiles. e.g. All cations like Ag+3, Mg++, Cu++, etc. & salts, AlCl3, BF3, BeCl2, SO3 etc.
Lewis Base: Any species (molecules, atom, ion) which can donate a pair of electrons to form coordinate bond. There are nuclephiles. e.g. NH3, H2O, all anions, Cl-, SO4--, Br- etc.
Electrolyte: The substance which produces oppositely charged ions in water& conductivity of ions depends on number of ions present in the solution. eg: All acids, bases &salts are electrolytes. Electrolytes classified into two types:
A) Strong electrolyte: A substance which is almost completely disassociated or ionized in water is called as strong electrolyte. Which is good conductors of electricity and have high value of equivalent conductance e.g.: Strong acids: Hcl, H2SO4, HNO3, H3PO4, HClO4, HBr, HI etc.
Strong bases: NaOH, KOH, Ca(OH)2, Mg(OH)2 etc. Strong salts: KCl, NaCl, NH4Cl etc.
B) Weak electrolyte: ‘A substance which dissociate to a limited extent in aqueous solution’ or A substance which is partially ionized in water is called as weak electrolyte. It has low value of equivalent conductance. eg: Weak acids : Acetic acid, Benzoic acid, Oxalic acid, Pthalic acid, all Organic acids. Weak bases: NH4OH, Na2CO3, NH4CO3, Alkyl Amines. Weak salts: Lead (II) acetate, Mercury (II) chloride etc.
Normality of liquid acids: To prepare standard solution, we must know the normality of liquid acids, these are given below.
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1) Conc. HCl=10N
2) Conc. HNO3= 16N 3) Conc. H2SO4= 36N 4) Conc.CH3COOH= 17N
CHAPTER
2
CONDUCTOMETRY
Concept of Conductivity: The power of electrolytes to conduct electric currents is termed as conductivity or conductance. Like metallic conductors, electrolytes obey Ohm’s law. According to this law, the current (I) flowing through a metallic conductor is given by the relation.
I = E/R
E = Potential difference at two ends on volts R = Resistance in ohms (Ω) The resistance of conductor is directly proportional to its length (L) and inversely proportional to the area of its cross-section (A)
R α L/A
R = ρ × L/A
ρ (Rho) = Proportionality constant is called resistivity orspecific resistance. Its value depends upon the material of the conductor.
• Specific resistance (ρ): Specific resistance of a conductance is the resistance in ohms
which 1 cm cube of it offers to the passage of electricity. R = ρ × L/A
ρ = R × A/L
If L = 1cm and A = 1sq.cm, then, ρ = R
• Specific conductance (Kc): The reciprocal of specific resistance is termed as specific conductance or conductivity or the conductance of 1cm3 of a solution of an electrolyte. It is denoted by K (Kappa).
K =
=
×
Unit of specific conductivity, K=1/R ×L/A=1/Ω⋅Cm/cm2
K=Ohm-1 cm-1= mhos cm-1=S cm-1
Internationally recommended unit of mhos or ohm-1 is Siemen(S)
Equivalent conductance (λ): It is defined as the conductance of one gram of an electrolyte placed between electrodes at dilution of “V” cc solution.
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The equivalent conductance is denoted by ‘λ’
It is the product of specific conductivity & volume (V) of solution containing one gram equivalent of the electrolyte.
λ= Kc⋅V
If an electrolyte contain N gram equivalents in 1000cc of solution. Then volume of solution containing one gram equivalent will be 1000/N; V= 1000/N
λ = Kc × 1000 / N
N = Concentration or Normality
Unit of equivalent conductance = Kc⋅V
λ = 1/ R × L/A ×V
λ = 1/Ohm × cm/ cm2 × cm3 / eqvt-1
λ = Ohm-1 cm2 eqvt-1 or (Siemens)S cm2 eq.-1
Molar conductance (λm):It is the conducting power of all the ions produced by one mole of the electrolyte in the given solution. It is denoted by λm.It is given by –
λm = k V
Where, V=the volume in ccof one mole of the electrolyte.
λm = 1000 k /m
The unit of molar conductance is ohm-1 cm2 mol-1 or S cm2 mol-1 or mhos cm2 mol-1
Variation of equivalent conductance with dilution electrolytes: The equivalent conductance of solution does not give linear plot with concentration. It is studied by plotting equivalent conductance against square root of concentration. It has been found that the equivalent conductance depends on nature of electrolyte.
Variation of equivalent conductance using strong electrolyte: As we know the strong electrolytes are almost completely ionized at all dilutions. It has been found that, when increase the dilution decrease in forces of attractions between the ions of opposite charges, each ion is present in the vicinity of oppositely charged ions. Dilution overcomes forces of attraction; hence ions become free to move independently. So the equivalent conductance increases till it reaches a limitary value. This value is known as equivalent conductance at infinite dilution or zero concentration [λ∞ or λ0].
Variation of equivalent conductance using weak electrolyte: Weak electrolytes have low ionic concentrations, here inter ionic forces are negligible, hence on increasing dilution increase the degree of dissociation or ionization. Thus, increase in equivalent conductance is due to the increase in number of ions. In case of weak electrolyte the ratio of equivalent
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conductance of given concentrations to the equivalent conductance at infinite dilution is equal to the degree of dissociation.
α = λc /λ∞
On dilution specific conductance decreases while equivalent conductance and molar
conductance increases: As we know that specific conductance increases with increase in ionic concentration and speed of the ions. It is the important to know that, the specific conductance is the conductance of one cc (cubic centimeter) of the solution. Upon dilution the concentration of ions per cc decreases. Hence decrease in ionic concentration, decrease the specific conductance.
On the other hand, the equivalent and molar conductivity increases with increasing dilution because the equivalent and molar conductivity are the product of specific conductivity and volume (Kc⋅V). So on dilution specific conductivity decreases and volume of solution increases, but ions per cc decreases. Equivalent and molar conductivity is the power of all ions produced by one gram equivalent of an electrolyte at that concentration. Dilution increases, volume of solution also increases thereby numbers of free ions increases. Hence, equivalent and molar conductivity increases with dilution.
The ionic conductance: The conductivity is due to mobility of cations and anions. Weak electrolytes are partially ionized in the solution and ionization increases with dilution.
λ∞ = The ionic conductance at infinite dilution
λa= The ionic conductance of anions
λc= The ionic conductance of cations
Then we can calculate the values of ionic conductance at infinite dilution by applying Kohlrausch’s law.
Kohlrausch’s Law: It states that the equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductance of the cations and anions.
It is mathematically expressed as,
λ∞ = λa + λc
Where, λa and λc are the equivalent conductance of the anions and cations.
Ionic equivalent conductivity values at 25oc
Cations λc Anions λa Electrolyte λ∞
H+ 350 CH3COO- 40.9 CH3COOH 390.9
K+ 73.5 Cl- 76.4 KCl 149.9
Na+ 50 Cl- 76.4 NaCl 126.4
K+ 73.5 Br- 78.1 KBr 151.6
Ag+ 62 Cl- 76.4 AgCl 138.4
Page | 16
½ Ba++ 63.6 ½ SO4-- 80 Baso4 143.6
Na+ 50 OH- 199 NaOH 249
H+ 350 Cl- 76.4 HCl 426.4
Conductivity Cell: It is used to measure conductivity of the solution. It is a glass vessel with two electrodes at a definite distance apart. The electrodes in conductivity cell constructed by a conductive material such as graphite, stainless steel or platinum. Conductivity is measured in millisiemens (mS) or microsiemens (µS). A simple type of conductivity cell is used in the laboratory is as shown in the figure.
Fig : Conductivity cell
Cell Constant: This is characteristics of conductivity cell. Sometimes there is no accuracy in distance between electrodes and area of cross sections. Therefore, determination of cell constant is essential. The exact value of cell constant can be determined by measuring the distance between electrodes and their area of cross sections.
As we know observed conductivity is directly proportional to area of cross section (a) and inversely proportional to distance between electrodes (l).
The observed conductivity α
Observed conductivity = K⋅
Specific conductivity (K) = Observed Conductivity × cell constant
Cell constant =
Difference between Conductometric and Volumetric titrations:
Sr. Conductometric titration Volumetric titration
1 It measures the conductance to check the end point.
It measure volume of titrant to check the end point.
2 It can be carried out even with colored It cannot carry out in colored solutions.
l
Platinum pates
Electrolytic solution
Glass tube
Page | 17
solutions. 3 It gives accurate results. It does not give accurate results.
4 End point is determined graphically. End point determined by change in color of indicator.
5 It can be used in polybasic acids titrations. Polybasic acids do not give accurate end point.
6 It can be useful in weak acid and Strong base titration
It can’t be useful in weak acid and Strong base titration.
Conductometric titration:
The conductivity of solution depends upon the number of free ions and speed of the ions present in the solution. During the course of titration a definite volume of titrant is added into titrate. Then note down the observed conductivity. Plot the graph conductance against volume of the titrant added. Two branches of straight lines may be obtained. The point of intersection of the two lines which gives the end point of titration.
Advantages of conductometric titration:
• It is useful in titration of coloured or turbid solution. • There is no need of indicator. • This is also applicable in highly dilute solutions. • It can be used for all kinds of titrations. i.e. Acid-base, redox precipitation & complex
tritration.
Aim: To determine the normality and strength of strong acid (HCl/ H2SO4/ HNO3) conductometrically by using standard Solution of strong base (NaOH/ KOH).
Apparatus: Conductometer, conductivity cell, beakers, burette, pipette, Magnetic stirrer, measuring cylinder
Chemicals: 0.1 N NaOH solution andHCl solutions.
Chemical reaction: Hcl + NaOH → NaCl+ H2O
Theory: The molar conductivity of the solution depends on concentration of acid solution and speed or mobility of ions. H+ and OH- ions are fast mobile ions and Na+ and Cl- are less mobile ions. When strong acid is titrated against strong base the fast mobile H+ ions are replaced by slow mobile Na+ ions. Hence conductivity of solutions decreases till the end point of titration reaches. Further addition of excess NaOH solution; increase the conductance of the solution due to the fast mobile free OH- ions.
Procedure:
E X P E R I M E N T No. 01
Page | 18
1. Rinse and fill the clean burette with standard NaOH solution up to the zero mark. 2. Pipette out require volume of Hcl solution in a clean 100 ml beaker and add about 30-
40 ml of distilled water. Place the magnetic stirrer in the beaker. 3. Dip the conductivity cell in the Hcl solution and connect it to Conductivity Bridge. 4. Titrate the Hcl solution with std. NaOH solution. Add NaOH solution from the
burette, stir well and note down the observed conductance at every addition of NaOH solution.
5. At the beginning of titration acid solution has high conductivity due to the fast mobile H+ ions.
6. When NaOH solution is added to Hcl solution, the fast mobile H+ ions are replaced by less mobile Na+ ions this will result in the decrease in the conductance.
7. After end point further addition of NaOH solution, the conductance rises due to fast mobile free OH- ions.
8. Plot a graph of conductance against volume of NaOH solution, two straight lines are obtained. The point of intersection of two lines gives end point of titration. The nature of graph is ‘V’ shape. It will observe that conductance of solution decreases till equivalence point (end point) reaches. At the end point solution contain only Na+ & Cl– [Na+ + OH–]+ [H+ + Cl– ] [ Na+ + Cl– ]+ H2O neutralization reaction.
After end point further addition of NaOH solution, due to fast mobile OH– ions.
Observations:
i) Given volume acid = __________ ml ii) Observation table
Sr.
No. Volume of NaOH added in ml
Conductance
Mhos cm2
eqvt-1
1 2 3 4 5 6 7 8 9
Page | 19
Calculations:
From graph volume NaOH required to complete neutralization of acid = end point=V2=_____________ml.
HCl v/s NaOH
N1 X V1 = N2V2
N1 X Given volume of Acid = 0.1 X V2
N1 = .. 012 3 45
6 3
N1 = ………. N
Strength of HCl = N1 X equivalent weigh of HCl.
=_________× 36.5
= _________gm/ litre.
Result: The normality of HCl acid is ______N and strength of HCl acid is _______gm/ litre.
Aim: To determine the normality and strength of weak acid (CH3COOH/ HCOOH) conductometricaly by using standard Solution of strong base (NaOH/ KOH).
Apparatus: Conductometer, conductivity cell, beakers, burette, pipette etc.
Chemicals: 0.1 N NaOH solution and CH3COOH solution.
Theory: The titration of weak acid with strong base like NaOH shows neutralization reaction.
End point
Conductance
Volume of NaOH
E X P E R I M E N T No. 02
Page | 20
[Na+ + OH- ]+ [H+ + CH3COO- ] [ Na+ +CH3COO - ]+ H2O
Before addition of alkali, the acid solution has conductance due to the hydrogen ions. This conductivity is less, as weak acid is less dissociated. When alkali is added, weak acid CH3COOH is converted into strong electrolyte CH3COONa. Hence conductance of solution increases. After complete neutralization, further addition of alkali will cause the sharp increase conductance, due to the excess of highly mobile OH- ions. A graph between conductance v/s volumes of alkali added will consist of two straight line branches intersecting at the neutral point.
Procedure:
1. Take given solution of acid (CH3COOH) solution in a beaker; add nearly 40ml of distilled water. Dip conductivity cell in it and connect to Conductometer.
2. Rinse and fill the burette with std. NaOH solution. Remove air bubble, adjust it to zero mark.
3. Note the conductivity of acid solution before addition of NaOH. Add 0.5ml of NaOH from burette, stir it and note the conductance. Repeat the procedure to get twenty readings.
4. Plot a graph between volume of NaOH added (X- axis) and conductance (on Y axis). Note volume of NaOH required for complete neutralization of CH3COOH. Hence calculate normality and strength of CH3COOH solution.
Observations:
i) Given volume of acetic acid = ………..ml. ii) Observation table
Sr.
No.
Volume of NaOH added
in ml
Conductance ,[Mhos cm2
eqvt-1
]
1 0.0 2 0.5 3 1.0 4 1.5 5 2.0 6 2.5 7 3.0 8 3.5 9 4.0
10 4.5 11 5.0
Conductance
Volume of NaOH
End point
Page | 21
12 5.5 13 6.0 14 6.5 15 7.0 16 7.5 17 8.0 18 8.5 19 9.0 20 9.5 21 10.0
Calculations:
From graph volume NaOH required to complete neutralization of acid =V2=________ml
CH3COOH v/s NaOH
N1 X V1 = N2V2
N1 X Given volume of Acid = 0.1 X V2
N1 = .. 012 3 45
6 3
N1 = _______ N
Strength of Acetic acid= N1 X equivalent weight of acetic acid
Strength of acetic acid =_______ × 60 = _______gm/ litre.
Result: The normality of Acetic acid is ______N and strength of Acetic acid is ______gm/ litre.
Aim: Determination of solubility of sparingly soluble salt likePbSO4, BaSO4, etc at room temperature.
Apparatus: Conduct meter, Beakers, Conductivity cell, Solid KCl, Solid BaSo4, standard volumetric flask.
Theory: According to Arrhenius theory, strong electrolytes are dissociated at all dilution and weak electrolytes are partly or partially dissociated, therefore degree of dissociation depends on concentration of the weak electrolyte. Strong electrolyte or weak electrolyte, both is considerably soluble in water but there are some salt like AgCl, PbSO4 and BaSO4 which are sparingly soluble in water .Their solubility is very less as compared to the weak and strong electrolytes.
The solubility of sparingly soluble salt at a particular temperature can be determined by conductivity measurements by using formula C = 1000Kc/λ ∞
E X P E R I M E N T No. 03
Page | 22
Where, ‘c’ is the normality of sparingly soluble salt
Kc = Specific conductance of the saturated solution at that temperature and
λ ∞= equivalent conductance at infinite dilution.
PROCEDURE:
Determination of cell constant:
1. Weigh accurately 0.745 gmKCl crystal on watch glass dissolved it in 100 ml distilled water and label as 0.1 N KCl ( 1/10 N KCl )
2. Pipette out 20 ml 0.1N KCl solution in 100 ml standard volumetric flask and dilute up to mark. Label this solution as 0.02N (1/ 50 N KCl).
3. Measure the conductivity value of 0.1 N and 0.02 N KCl solutions. This will be observed conductivity of KCl solution.
4. Calculate cell constant.
Determination of solubility:
1. Take about 60-70 ml distilled water in a beaker add BaSO4 Salt in it. Prepare saturated solution, Stir it well and filter it. Note down its temperature. Note down the conductivity of saturated solution of BaSO4 at room temperature.
2. Hence calculate the solubility of BaSO4 at room temperature.
Result: The solubility of BaSO4is (C) ______mole/ litre.
Observations And Calculation:
i) Temperature of experiment = _______°C ii) Determination of cell constant
Sr. No.
Concentration of KCl
Observed conductance
[Mhos cm2 eqvt-1]
Specific conductance
Kc
Cell Constant = 89
:;<=>?=@ 9AB@C9DEB9=
1 F
. KCl
2 F
G. KCl
Mean Value of cell constant = iii) Determination of Solubility
Sr. No. Temperature
Observed conductance ofsatu.
BaSO4 Solution
Specific conductance [Kc] =
Cell constant X observed conductance
[solubility]C = 1000 X Kc
λ∞
1
λ∞ = λa + λc = 127. 2 + 160.00 = 287. 2,
Page | 23
[ λ∞ =The equivalent conductivity of BaSO4 at infinite dilution.]
Aim: Determine the concentration of KCl solution by titrating it with standard solution of
AgNO3 conductometricaly.
Apparatus: Conductometer, conductivity cell, beakers, burette, pipette etc.
Chemicals: 0.1 N AgNO3 solution and KCl solution.
Theory: Let us consider the titration between silver nitrate and potassium chloride the reaction is
[K+ + Cl- ]+ [Ag+ + NO3- ] KNO3+ AgCl↓
At the beginning of the titration there is little change in conductance because the Cl-
ions are replaced by NO-3ions and have almost same ionic conductance.After equivalent point
the excess AgNO3 added causes sharp increase in conductance .Therefore two branches of straight lines obtained may intersect at the point of equivalence .Draw a perpendicular from the equivalence point.we get volume of AgNO3 required for pptAgcl. There is no sharp end point because of less mobility of ions,as compared with acid- base titration.
PROCEDURE:
1. Take given solution of KCl in a beaker adds nearly 40ml of distilled water. Dip conductivity cell in it and connect to conductivity meter.
2. Rinse and fill the burette with std. AgNO3solution. Adjust it to zero mark. 3. Titrate KCl solution against silver nitrate solution; AgCl will precipitate in this
reaction. 4. Add 0.5ml of AgNO3 solution from burett, stir it and note the conductance. 5. Plot a graph between volume of AgNO3 added (X- axis) and conductance (on Y axis).
Note volume of Agno3 required for the complete precipitation reaction. Hence calculate normality and strength of KCl.
Observations:
i) Given volume KCl = ………..ml. Sr. Volume of AgNO3 added in ml Conductance [Mhos cm
2 eqvt-
1]
E X P E R I M E N T No. 04
Page | 24
1 0.0 2 0.5 3 1.0 4 1.5 5 2.0 6 2.5 7 3.0 8 3.5 9 4.0 10 4.5 11 5.0 12 5.5 13 6.0 14 6.5 15 7.0 16 7.5 17 8.0 18 8.5 19 9.0 20 9.5
CALCULATIONS:
From graph volume AgNO3 required to complete the precipitation of KCl =V 2=……….ml.
KCl v/s AgNO3
N1 X V1 = N2V2
N1 X Given volume of KCl = 1.0 X V2
N1 = .. 012 345
6 3
N1 = _________N
Strength of KCl= N1 × equivalentweightofKCl
Conductance
Volume of AgNO3
End point
Page | 25
= ________ × 74.55
= ________gm/ litre.
Result: The normality of KCl solution is …N and strength of KCl solution is …… gm/ litre.
Aim: Determine the amount of oxalic acid conductometrically by using standard NaOH Solution.
Apparatus: Conductometer, conductivity cell, beakers, burette, pipette etc.
Theory: Oxalic acid is a dibasic acid and there are two hydrogen ions are replaceable hence two dissociation constants are well separated one after another. The first dissociation constant is similar to a strong acid and the second one is similar to that of weak acid. While performing titration, conductance will decrease first up to first dissociation is complete then rises slowly up to second dissociation constant. After end point, conductance rises rapidly due to fast mobile OH- ions because of excess addition of NaOH. This titration curve will have two brakes are obtained due to dibasic nature of oxalic acid.
Chemicals: 0.1 N NaOH solution and oxalic acid solution.
Procedure:
1. Pipette out given amount of oxalic acid in beaker and add 40ml distilled water. Dip conductivity cell in it and connect to Conductivity Bridge.
2. Fill micro burette with standard NaOH solution. 3. Titrate the oxalic acid solution with NaOH conductometrically. 4. Note down conductivity at every addition of NaOH solution. 5. Plot a graph conductance against volume of NaOH solution added.
OBSERVATIONS:
i) Given volume Oxalic acid = ………..ml. ii) Observation table:
Sr.
No. Volume of NaOH added in ml
Conductanc [Mhos cm2
eqvt-1]
1
2
3
4
5
E X P E R I M E N T No. 05
Page | 26
6
7
8
9
10
CALCULATIONS:
From graph volume NaOH required to complete neutralization of oxalic acid =V2= _______ ml.
Oxalic acid v/s NaOH
N1 X V1 = N2V2
N1 X Given volume of Oxalic Acid = 1.0 X V2
N1 = .. 012 3 45
6 3 L
N1 = _________ N
Strength of oxalic acid= N1 X equivalent weight.
= ________ × 63
= _________gm/ litre.
Result: The normality of Oxalic acid is ……N and strength of oxalic acid is…….gm/ litre.
Conductance
i dis.const. ii dis.const.
Volume of NaOH
Page | 27
Aim: Determine the equivalent conductance of a strong electrolyte at several concentrations and verify the Onsager’s equation, conducto-metrically.
Apparatus: Conductometer, conductivity cell, beakers, standard volumetric flask etc.
Chemicals: 0.1 N KCl solution.
Theory: According to Arrhenius theory strong electrolytes dissociate at all dilution and increase in equivalent conductance λc with dilution is not due to the dissociation like a weak electrolyte. Here increases in dilution result in the decrease of inter ionic distance. Therefore migration speed of the ions increases. Hence equivalent conductance increases on dilution.
The variation of equivalent conductance with concentration of strong electrolyte in dilute solution is given by Debye-Huckel-Onsager equation.
λc = λ∞ – ( a + b λ∞) √C
Where ‘c’ is concentration of electrolyte, λcis the equivalent conductance at concentration ‘C’, λ∞ is the equivalent conductance at infinite dilution, a and b are constant.
The equation is verified by a plot of λ c V/S√O. The plot will be a straight line. The intercept to Y axis will beλ∞ and slope will be (a + b λ∞).
Procedure:
1. Determine cell constant by using F
. KCl and F
G. KCl solution.
2. Prepare 50 ml of 0.02, 0.01, 0.005, 0.002, 0.001, 0.0005, 0.0002, and 0.0001 N solution of KCl from stock solution.
3. Note down conductivity of each solution. Calculate specific conductance and equivalent conductance for each solution. Plot a graph between λcvalues (Y axis) and
√O (on X axis).The intercept to Y axis will give equivalent conductance of strong electrolyte at infinite dilution (λ∞).
The plot of equivalent cond. v/s √c The straight line nature of plot verifies the Onsager equation. Hence equivalent conductance at zero concentration can be calculated by graphical method.
E X P E R I M E N T No. 06
Page | 28
Observation:
i) Temperature of the experiment (Room temperature) =……..°C. ii) Measurement of cell constant.
Sr. No.
Concentration of KCl
Observed conductance
[Mhos cm2 eqvt-1]
Specific Conductance
Kc Cell constant= 89
.
1 F
. KCl
2 F
G. KCl
Mean cell constant =
iii) Determination of conductivity.
Sr. No.
Concentration of KCl
C
Observed conductance
[Mhos cm2 eqvt-1]
Specific conductance Kc =Cell. Constant X
obs. conductance
Equivalent Conductance λ c= ... 0P
Q
√O
1 2 3 4 5 6 7 8
CALCULATIONS: Determination of cell constant and mean of the cell constant
Cell constant = Specific conductance / Observed conductance Specific conductance (Kc) = Observed conductance × Cell constant
Equivalent conductance (λc) = 1000 x Kc / C
RESULT: The equivalent conductance at infinite dilution is …….
Conductance
KCl
CH3COOH
NaOH
c
Page | 29
CHAPTER
3 POTENTIOMETRY
INTRODUCTION:
Potentiometric study include in electro chemistry, which deals with inter relationship between chemical energy and electrical energy. Certain chemical reaction takes place spontaneously and can produce electrical energy. In such change chemical energy is converted into electrical energy. At the expense of spontaneous redox reaction, electrical energy is generated. On the other hand, at the expense of non-spontaneous redox reaction, chemical energy is generated.
A cell is a device which is used to convert either electrical energy into chemical energy or chemical energy into electrical energy. There are two types of cell: (1) Electrolytic cell(2)Electrochemical cell
1) Electrolytic cell: An electrolytic cell is a device which is used to convert electrical energy into chemical energy. In this cell non spontaneous redox reaction take place by supplying electrical energy, the process is known as electrolysis.
2) Electrochemical cell: An electrochemical cell is a device which is used to convert chemical energy into an electrical energy. Electrochemical cell is also known as Galvanic cell or Voltaic cell.eg: Daniell cell, dry cell, lead accumulator, Leclanche cell
Electrical units:
1) Coulomb (Q): A unit of electricity, one ampere current passing through the circuit for one second is called one Coulomb.
Q = Current in ampere ˟ time in second Q = c.t
2) Faraday (F): The quantity of electricity required to deposite or liberates one gram equivalent of substance is 96500 Coulombs, this quantity of electricity is known as Faraday.
(1F = 96500 Coulombs = 1 mole of electrons). 3) Ampere (A): A unit of rate of flow of electricity, an ampere is a current of one
coulomb of electricity passed through the electric circuit for one second. 4) Ohm (Ω): A unit of electrical resistance measured in ohm. 5) Volt (V): A unit of electromotive force (emf) or potential difference. It is a difference
in electrical potential required to send a current of one ampere through a resistance of one ohm.
6) Joule (J): The work done or energy measured in joules, one joule is the work done per second by a current of one ampere flowing through a resistance of one ohm.
7) Watt (W): The electrical power is measured in watt; watt is the power in a circuit in which a current of one ampere flows across a potential difference of one volt.
1 Watt = 1 Joule/ 1 second
Page | 30
Concept of electrode potential or half cell potential:
Nernst theory: In 1889, Walter Nernst gave his advanced theory of electrode potential.According to Nernst theory,
a) De-electronation: All metallic elements and hydrogen have a tendency to pass into the solution in the form of positive ions and liberate electrons on electrode this process is known as de-electronation or oxidation. This property is due to the solution pressure of metal (Ps), thus the solutions acquire positive charge and metal acquire negative charge. This give rise to electrical double layer, because of electrical double layer a potential difference is developed. As this potential used due to oxidation, it is called oxidation potential. eg: Zinc half cell in Daniell cell, Zn(s)Zn2+
(aq)
Oxidation reaction, Zn(s) → Zn2+(aq) + 2e-
b) Electronation: The process in which an atom or ion of an element gains one or more electrons is called as electronation. The tendency of metal to pass into its ion solution as cations is opposed by a reverse tendency of the ions to be deposited back on the surface of the metal by taking electrons. This reverse tendency of ions is called as osmotic pressure of ions (PO). Due to this, solution acquires excess negative charge and electrode acquires positive charge. This give rise to electrical double layer between the surface of metal and solution. The potential difference is developed. As this potential is due to reduction reaction called as reduction potential. eg: Copper half cell in Daniell cell, Cu2+
(aq) Cu(s)
Reduction reaction, Cu2+(aq) + 2e- → Cu(s)
There are three possibilities: i) Ps >P0 = de-electronation or oxidation takes place, negative potential develops. ii) Ps < P0 = electronation or reduction takes place, positive potential develops. iii) Ps = P0= no potential develops between the metal and the solution, such
electrode is referred as ‘null’ electrode.
Nernst equation:
The metal has tendency to loose or gain electrons. When metal is in contact with its own ion solution, this give rise to oxidation or reduction reaction. Let ‘M’ be the metal electrode and ‘n’ be the number of the electrons involved in the electrode reaction. Then the reactions are as follows,
M → Mn++ ne- …… oxidation
Cu
Cu++
- + + -
- + + -
- + + -
- + + -
- + + -
- + + -
- + + -
-ve
Fig.: Copper half cell
Zn
Zn++
+ - - +
+ - - +
+ - - +
+ - - +
+ - - +
+ - - +
+ - - +
-ve
Fig.: Zinc half cell
Page | 31
Mn++ ne- → M …… reduction
Nernst equation for oxidation reaction:
EM/Mn+ → E0
M/Mn+ - 2.303RT/nF log [Mn+](aq)
Substituting the values of R,T andF, the equation becomes,
EM/Mn+ = E0
M/Mn+ - 0.0591/ n log [Mn+](aq)
Nernst equation for reduction reaction:
EMn+
/M = E0M
n+/M+ 0.0591/ n log [Mn+](aq)
Single electrode or half cell of an electrochemical cell:
A metal rod is dipped in its own salt solution is called half cell or single electrode. eg: Daniell half cell, Zinc half cell, Zn(s)Zn2+
(aq),
Copper half cell, Cu2+(aq)Cu(s)
Single electrode potential:
When a metal rod is dipped in its own salt solution and it has tendency to lose or gain electrons when it is in contact with its own ions solution.This gives rise to the electrical double layer which is setup at the electrode surface. This electrical double layer develops potential difference (emf). If there is oxidation reaction oxidation potential is developed or if there is reduction reaction, called reduction potential.
Electromotive force (Emf) or cell potential:
The flow of current of electrons through the circuit, it determined by the push of electrons at the anode and attraction of electrons at the cathode, these two forces sends electrons through the circuit is called as emf or cell potential. It is measured in volts (V) referred as cell voltage.
The emf of a cell can be calculated from the half cell potentials of the two cells using following formula,
E(cell) = Ecathode– Eanode or
E(cell) = ER – EL
Where, ER and EL are the potential of right hand and left hand electrodes respectively.
Reversible Cells:
A reversible cell must satisfy the following conditions,
1) If an emf exactly equal to that of the cell is applied from an external source, the chemical reaction taking place in the cell will stop.
Page | 32
2) If the external emf infinitesimally less than the actual emf of the cell is applied to it, the current will flow from the cell which is proportional to the chemical change occurring in it.
3) If the external emf is infinitesimally greater than the actual emf of the cell is applied to it, the current will start flowing in opposite direction and cell reaction gets reversed. eg: Daniell cell, Zn(s) / ZnSO4(aq) / CuSO4(aq) / Cu(s).
If the cell is connected to an external force of emf just small than that of the cell, the chemical reaction occurring in the cell proceed from left to right direction.
Zn + cu++ zn++ + cu
It on the other hand, the external emf is just greater than that of the cell itself, the reaction in the cell occurs in the reverse direction. Hence Daniel cell works as a reversible cell.
A combination of any two reversible electrodes gives a reversible cell.
1) Electrodes reversible with anions: Calomel electrode is reversible anion electrode with respect to chloride ions, silver- silver chloride electrode, silver – silver bromide electrode. Reversible electrode consist of ametal in contact with one of its sparingly soluble salts and a solution of soluble salt having a common anion with the sparingly soluble salt called reversible electrode with respect to anion.
2) Reversible electrode with respect to cations: Electrode metal in contact with its own solution ions known as reversible electrode eg. Zn in ZnSO4,Ag in AgNo3.
3) Oxidation-reduction electrode: This type of reversible electrode consists if an inert electrode (Pt or gold) immersed in a mixed solution containing both the oxidized and the reduced forms of a molecule or ion. eg. Pt/ Fe2+, Fe3+, Pt/ Sn2+, Sn4+
Fe2+ Fe3+ + e- Sn2+ Sn4+ + 2e-
Types of electrodes:
1) Indicator electrode: The electrode whose potential depends upon the concentration of ions in solution and is used to find out the concentration of ions in the solution is known as indicator electrode.
eg. Ag/ Ag+(aq), Zn/ Zn2+
(aq)
The potential of indicator electrode depends upon the activity of a particular species, whose concentration is to be determined. There are some important indicator electrodes widely used in potentiometry.
i) Hydrogen electrode ii) Quinhydrone electrode iii) Glass electrode
Page | 33
Hydrogen electrode: It is reference electrode, in addition to its function as a standard electrode; the hydrogen electrode can be used as indicator electrode to measure the hydrogen ion concentration or pH of solution.
Quinhydrone electrode: It is indicator electrode which is reversible with respect to hydrogen ions. Quinhydrone is compound which is in aqueous solution forms equimolar quantities of quinine (Q) and hydroquinone (QH2)
Q + 2H+ + 2e-→ [QH2]
The presence of platinum electrode in a solution containing these two species forms a redox system. The emf develops when Pt electrode dipped in the solution, is given by Nernst equation.
E = E0 + 2.303RT / nFlog [QH2]/ [Q][H+]2
E = E0 + 2.303RT /2F log [QH2]/[Q] -2.303RT/2F log [H+]2
The saturated solution of quinhydrone produces equimolecular quantities of quinone [Q] and hydroquinone [QH2].
∴[QH2]/[Q] = 1 and log 1 = 0.
∴ E = E0 – 2.303RT/2F log [H+]2
E = E0 – 0.0591 log [H+]
E = E0 + 0.0591 pH
The potential of quinhydrone electrode is determined by connecting it with calomel electrode. Oxidation occurs at the calomel and the reduction at the hydroquinone electrode. Therefore emf of the cell,
E(cell) = - 0.246 + 0.699 – 0.0591 pH
0.0591 pH= 0.699 – 0.246 – E(cell)
pH= 0.457- E(cell) / 0.0591
The standard oxidation potential of quinhydrone electrode at 250C is equal to −0.699V.
Glass electrode: It is used for the determination of pH of solution. It is not affected by oxidizing or reducing reagents. The electrode consists of a special soft glass tube of high resistance and low melting point at the terminal of the tube has thin walled glass bulb. The bulb is filled with 0.1 M HCl solutions in which a silver – silver chloride electrode is dipped.
Page | 34
It is represented as, Ag⋅AgCl(s)HCl [0.1M] Glass, the potential of electrode is given by the equation,
EG= E0G – 2.303RT/ F pH
EG = E0G - 0.0591 pH
To measure the pHof the solution, the glass electrode coupled with saturated calomel electrode. The cell is represented as,
Ag.AgCl(s) HCl [0.1M] Glass Unknown pHsoln.KCl(sat) Hg2Cl2Hg.
The potential of calomel electrode is known then, the potential of glass electrode can be easily calculated from the observed emf of the complete cell and the value of pH can be determined using above equation.
The emf of cell is given as, Ecell= Ecalomel- Eglass
Ecell= Ecalomel – E0G –0.0591 pH
Ecell= Ecalomel + 0.0591 pH– E0G
E0G = Ecalomel– Ecell+ 0.0591 pH
Hence, the value of E0G can be determined by measuring emf of the cell.
2) Reference electrode: The electrode whose potential is exactly known or is arbitrarily fixed at given constant temperature is known as reference electrode.
Using reference electrode unknown potential or emf of any other single electrode can be determined.
eg. Standard hydrogen electrode and Calomel electrode.
a) Standard Hydrogen Electrode (SHE):
It is defined as the electrode in which pure and dry hydrogen gas is bubbled at 1 atmospheric pressure and 250 C or 298K temperature on a platinised platinum plate through a solution containing unit activity of H+ ions. It is primary reference electrode; it can be used as indicator electrode to measure the hydrogen ion concentration or pH of solution.
Resin filled tube
Ag-Agcl electrode
pH Sensitive glass0.1N HCl
Page | 35
Construction: It consists of a glass jacket which has small inlet at the top and many outlets at bottom. Inside the glass jacket there is a glass tube closed at both the ends. At the lower end there is platinised platinum plate. In the glass tube platinum wire is sealed in it. At the bottom of the glass tube there is little mercury which is used for good electrical contact. The glass jacket along with glass tube is dipped in a vessel containing 1N HCl solution. Working: Pure and dry hydrogen gas is passed at 1atm pressure and 250C temperature through HCl solution from inlet. Hydrogen gas is adsorbed on the platinum plate. Equilibrium between H+ ions and H2 gas is established across the metal. The half cell reaction can be written as,
H2 → 2H+ + 2e-
As the process is oxidation, a positive potential is developed but it is very small therefore, its value is consider to be zero.
Representation of electrode: Pt, H2(g) H+(aq)1 atm
Uses of SHE:
i) It is used as reference electrode. ii) It is coupled with any other electrode whose emf is to be determined iii) Potential of SHE is zero hence the obtained potential is equalto the potential of
unknown electrode.
Difficulties in setting up of SHE: Preparation of this electrode is tedious process,
i) It is very difficult to maintain 1 atm pressure on hydrogen gas uniformly. ii) It is difficult to maintain unit activity concentration of HCl solution. iii) It is difficult to obtain 100% pure and dry hydrogen gas iv) Platinum used is very expensive. v) It is made up of glass, so it is
delicate to handy.
b) Calomel Electrode:
It is a secondary reversible electrode. It is easily set up and is perfectly reproducible it can acts as oxidation electrode as well as reduction electrode.
It is the most widely used reference electrode due to its easy set up. There are three types of calomel electrodes used according to the concentration of KCl solution which may be 0.1M, 1M or saturated KCl.
Page | 36
Construction: It consists of a glass vessel (tube) having a side arm at the top. There is inner tube where mercury is placed at bottom. Pt wire is sealed at the bottom for good electrical contact. At the bottom of inner tube consist of a paste of mercury, mercurous chloride (Hg2Cl2) or calomel and it’s above mercury is placed. The rest of the glass tube is filled with saturated KCl solution. It is represented as,
HgHg2Cl2(s), KCl(saturated) The potential can be measured very accurately by connecting it to standard hydrogen electrode. The oxidation potential of electrode for different concentrations of KCl solution at 250C or 298K are given below,
Concentration of KCl Emf at 298K
0.1M E = - 0.337 V
1.0M E = - 0.284 V
Saturated E = -0.242 V
Working: Since, it is reversible electrode, two types of reactions are possible depending upon other electrode with which is coupled.
When calomel electrode acts as negative electrode:
Thus oxidation takes place when it is coupled with other electrode at the positive pole.
2Hg(l) → 2Hg+ + 2e-
2Hg+ + 2Cl- → Hg2Cl2(s)
Net reaction,2Hg(l) + 2Cl- → Hg2Cl2(s) + 2e-
E = E0(calomel) – 2.303RT/ 2F log [Cl–]-2
E = E0(calomel)
_2.303RT/F log [Cl–]–1
E = E0(calomel) + 0.0591 log [Cl–]
Where, E0(calomel) = the oxidation potential
If the calomel electrode is connected to positive pole, the reaction is reduction. The net cell reaction is as follows,
Hg2Cl2(s) + 2e- → 2Hg(l) + 2Cl–
E = E0(calomel) – 2.303RT/F log [Cl–]
E = E0(calomel) – 0.0591 log [Cl–]
Page | 37
Aim: To determine the normality and strength of strong acid (HCl/ H2SO4/ HNO3) potentiometrically by using standard Solution of strong base (NaOH/ KOH).
Apparatus: Potentiometer, saturated calomel electrode (SCE), platinum electrode, beakers, micro burette, pipette etc.
Chemicals: 0.1 NNaOH solution and HCl (unknown conc.) solution, quinhydron powder
Theory: The HCl and NaOH being strong electrolyte dissociate completely .The
neutralization reaction is represented as
HCl + NaOH → NaCl + H2O
The indicator electrode used in this titration is the electrode reversible with H+ ions such as
Hydrogen electrode or quinhydron electrode or glass electrode. The reference electrode is
saturated calomel electrode (SCE).The representation of cell is,
anodeHg | Hg2Cl2(s), KCl(sat) | HCl (sat.withquinhydron) | Pt cathode
The emf of the cell is measured after every addition of NaOH solution from burette. When
end point is reached there is sharp change in emf, at this end point all H+ ions get consumed
by OH- ions and the excess of OH-ions gives sharp change in emf.
Procedure:
• Standardisation of potentiometer
1. Connect the standard cell in circuit and adjust the adjustment screw such that the
emf on the screen will be 1.0183V.
2. Do not disturb the adjustment screw throughout the experiment
• Acid - Base Titration
1. Pipette out given amount of acid solution in100ml beaker add to it about 40
ml distilled water to dip the electrodes properly then add a pinch of
quinhydron place the beaker on magnetic stirrer and stir it well
2. Dip calomel electrode (SCE) and platinum electrode in the above solution.
3. Connect these electrodes to potentiometer note down emf of the cell.
4. Add NaOH solution from the burette stir it and note down the emf.
5. Continue the addition of NaOH solution from the burette, repeat the same
procedure till sharp change in emf is obtained.
E X P E R I M E N T No. 07
Page | 38
OBSERVATIONS:
i) Given volume of acid = ________ml. ii) Observation table
Sr.
No.
Volume of NaOH
addedV ml
emf
E(V) ∆E ∆V
STSU
1 0.0 2 0.2 3 0.4 4 0.6 5 0.8 6 1.0 7 1.2 8 1.4 9 1.6 10 1.8
CALCULATIONS:
From graph volume NaOH required to complete neutralization of acid = End point=V2=_________ml.
HCl v/s NaOH
N1 X V1 = N2V2
N1 X Given volume of Acid = 1.0 X V2
N1 = .. 012 345
6 3
N1 = ________ N
Strength of HCl acid = N1 X equivalent weight.
=_________ × 36.5
= _________ gm/ litre.
Result: The normality of acid is _______N and strength of acid is_______gm/ litre
Equivalent point
Equivalent point
YY
Volume of NaOH Volume of NaOH
E E/ V
Page | 39
Aim: To determine redox potential of Fe+ + + / Fe + + system by titrating with standard K2Cr2O7ptentiometrically.
Apparatus: Potentiometer, calomel electrode, platinum electrode, beakers, burette, pipette etc.
Chemicals: 0.1 N K2Cr2O7 solution and FAS solution;2NH2SO4.
Theory: Ferrous ion [Fe++ ] acts as a reducing agent because it can lose one electron to be converted in to ferric ion[Fe+++] and acts as oxidizing agent. If platinum electrode dipped into the solution containing Fe+ + + and Fe + + ions. Then oxidation potential developed on electrode due to the reaction.
Fe + + → Fe+ + + + e-
When platinum electrode is coupled with calomel electrode, then a cell is constructed. It is
represented as.
Pt , Hg | Hg2Cl2(S) , KClSalt | Fe+ + + / Fe + + , Pt
The emf of cell E = E (Fe2 + / Fe 3 + )– E calomel
But E (Fe2 + / Fe 3 + )= E° (Fe2 + / Fe 3 + ) + 2.V.V W
Xlog
Z=V [
Z= 2 [
Hence E = E° (Fe2 + / Fe 3 +) + 2.V.V W
Xlog
Z=V [
Z= 2 [ – E calomel
= E° (Fe2 + / Fe 3 +) + 0.0591 logZ=V [
Z= 2 [ – E calomel
When K2Cr2O7 is added Fe 2+ ion decreases and Fe 3+ ion increases, at the end point
there is sudden increase in emf. At half equivalence point Fe + + ion is equal to Fe+ + + ions.
[Fe + + ]= [Fe+ + + ] and hence log[Z=V []
[Z= 2 [] = log 1 = 0
∴E(Half equivalence) = E° (Fe2 + / Fe 3 +) – E calomel
= E° (Fe2 + / Fe 3 +) – 0.246
E° (Fe2 + / Fe 3 +) =E (Half equivalence) + 0.246
Procedure:
1. Take given amount of ferrous ammonium sulphate solution in a 100 ml beaker; add nearly 40ml of 2N H2SO4 keep it on stirrer and Stir it vigorously.
2. Dip electrodes in solution. Connect platinum electrode to + ve terminal and calomel electrode to –ve terminal ofthe potentiometer.
3. Rinse and fill the burette with std. K2Cr2O7 solution. Remove air bubble, adjust it to zero mark.
E X P E R I M E N T No. 08
Page | 40
4. Note the emf of FAS solution before addition of K2Cr2O7. Add 0.5ml of K2Cr2O7 from burette, stir it and note the emf. Repeat the procedure to get twenty readings.
5. Plot a graph between volume of K2Cr2O7 added (X- axis) and emf (on Y axis), also
Plot a graph between volume of K2Cr2O7 added (X- axis) and ^_
^` (on Y axis). Note the
half equivalence point of titration
Hence calculate formal redox potential.
OBSERVATIONS:
i) Given volume FAS = ………..ml. ii) Observation table
Sr. No.
Volume of K2Cr2O7 added V ml
emf E
∆E ∆V ΔE
ΔV
1 0.0 2 0.5 3 1.0 4 1.5 5 2.0 6 2.5 7 3.0 8 3.5 9 4.0
10 Up to 10.0
Calculations:
I. The emf at half equivalence = …….. II. E° (Fe2 + / Fe 3 +) = E (Half equivalence) + 0.246
Result: The formal redox potential is………..
Equivalent point
Equivalent point
YY
Volume of K2Cr2O
E E/ V
Volume of K2Cr2O
Page | 41
Aim: To determine the dissociation constant of weak acid (CH3COOH) potentiometric measurement.
Apparatus: Potentiometer, calomel electrode(SCE), platinum electrode, beakers, burette, pipette etc.
Chemicals: 0.1 N NaOH solution and CH3COOH solution, Quinhydrone powder. Theory: Acid solution cannot produce potential difference between two electrodes and there
is no change of potential. If acid solution is treated with standard alkali .To convert
this system into reversible redox system, quinhydrone powder is added in to acid
solution.
Quinhydrone is a compound which in aqueous medium forms equimolar
quantities of quinon( Q) and hydroquinone ( QH2 ) by fallowing reaction.
Q + 2H+ + 2e- ←→ QH2
Platinum electrode is used for this half cell and connected to calomel electrode
to make complete cell. The complete cell can be represented as.
Hg,
Hg2Cl2(S )
KCl(salt) Quinhydrone H+ Pt
electrode
The emf of cell is given by
E cell = E Right - E Left
= E°- E Cal - 2.V.VW
XpH
= E°- E Cal - 0.05916pH
pH =..cGde_ 9=ff
...Gg ……. ……………………..( 1)
But according to Henderson’s equation
pH =pKa + log []
[]
At half equivalence [ salt ] = [ Acid ] the equation becomes
pH = pKa …………………………………………(2)
Hence from equation (1) and( 2 )at half neutralization .
Pka = ..cGdeh( 5 )
...Gg
As pKa = - log Ka Where ka is dissociation constant of acid.
Ka = antilog (- Pka )
E X P E R I M E N T No. 09
Page | 42
Thus by potentiometric measurement dissociation constant of an acid can be
calculated.
PROCEDURE:
1. Take the given amount of acid (HCl) solution in a beaker; add nearly 40ml of distilled water. Add about 100 mg of quinhydrone power. Keep it on stirrer and Stir it vigorously.
2. Dip electrodes in solution. Connect platinum electrode to + ve terminal and calomel to –ve terminal of potentiometer.
3. Rinse and fill the burette with std. NaOH solution. Remove air bubble, adjust it to zero mark.
4. Note the emf of acid solution before addition of NaOH. Add 0.5ml of NaOH from burette, stir it and note the emf. Repeat the procedure to get twenty readings. Plot a graph between volume of NaOH added (X- axis) and emf (on Y axis), also Plot
a graph between volume of NaOH added (X- axis) and ^_
^` (on Y axis). Note the equivalent
point of titration and emf at half neutralization point. Hence calculate pka and Ka.
Observations:
i) Given volume acid = ………..ml. ii) Observation table:
VV/AVolume of Alkali
E.M
.F.
E(o
bs.)
Page | 43
Sr.
No.
Volume of NaOH added
V ml
emf
E(V) ∆E ∆V
STSU
1 0.0 2 0.5 3 1.0 4 1.5 5 2.0 6 2.5 7 3.0 8 3.5 9 4.0 10 4.5 11 5.0 12 5.5 13 6.0 14 6.5 15 7.0 16 7.5 17 8.0 18 8.5 19 9.0 20 9.5 21 10.0
Calculations:
1. The equivalent point of titration is ……
2. emf at half neutralization point is …..
Pka = ..cGdeh( 5 )
...Gg
Pka= ………..
pKa = - log Ka
Ka = antilog (- Pka)
Result: The dissociation constant of acid is …
Page | 44
Aim: To determine the normality and strength of acids in mixture of strong acid (HCl) and weak acid (CH3COOH) potentiometrically using standard solution of strong base (NaOH).
Apparatus: Potentiometer, calomel electrode, platinum electrode, beakers, burette, pipette etc.
Chemicals: 0.1 N NaOH solution, HCl solution and CH3COOH solution, Quinhydrone powder.
Theory: When a mixture of strong acid and weak acid is titrated against base. The strong acid neutralizes first and the neutralization of weak acid will commence only when the neutralization of the first is completed.
Thus the titration curve will show two inflection points, first corresponding to the neutralization of strong acid and second to that of weak acid.
Procedure:
1. Take the given mixture of acid (HCl + CH3COOH) solution in a beaker; add nearly 40ml of distilled water. Add about 100 mg of quinhydrone power. Keep it on stirrer and Stir it vigorously.
2. Dip electrodes in solution. Connect platinum electrode to + ve terminal and calomel to –ve terminal of potentiometer.
3. Rinse and fill the burette with std. NaOH solution. Remove air bubble, adjust it to zero mark.
4. Note the emf of acid solution before addition of NaOH. Add 0.5 ml of NaOH from burette stir it and note the emf, repeat the procedure to get twenty readings.
5. Plot a graph between volume of NaOH added (X- axis) and emf (on Y axis), also Plot
a graph between volume of NaOH added (X- axis) and ^_
^` (on Y axis). Note the
equivalent points of titration and calculate normality and strength of HCl solution and CH3COOH solution.
Volume of NaOH
em
f
Volume of NaOH
E/ V
CH3COOH
HCl
HCl
CH3COOH
E X P E R I M E N T No. 10
Page | 45
Observations:
1. Given volume HCl & CH3COOH = _________ + ________ml
Observation table: Sr.
No. Volume of NaOH addedV ml Emf E ∆E ∆V
ST
SU
1 0.0 2 0.5 3 1.0 4 1.5 5 2.0 6 2.5 7 3.0 8 3.5 9 4.0
10 4.5 up to 21
Calculations: From graph the first equivalent point is V1 ……ml second equivalent point is V2______ml. Volume NaOH required to complete neutralization of HCl acid =V 1=_______ml.
HCl v/s NaOH N X V = N1V1 NX Given volume of HCl = 1.0 X V1
N = ..×1 3 45
6 3 lQ
N = ………. N Strength of HCl = NX equivalent weight.
=……X 36.5 = ……..gm/ litre.
Volume NaOH required to complete neutralization of CH3COOH = (V 2 -V 1) =……….ml. CH3COOH v/s NaOH N X V = N2(V 2 -V 1) NX Given volume of CH3COOH = 1.0 X (V 2 -V 1)
N = ..× (` 2 e` )3 45
6 3 mnom::n
N = ………. N Strength of CH3COOH = NX equivalent weight.
=……X 60 = ……..gm/ litre.
Result: The normality of HCl solution is _______N The strength of HCl solution is _________gm/ litre The normality of CH3COOH solution is ________N The strength of CH3COOH solution is __________gm/litre
Page | 46
CHAPTER 4
pH-METRY
pH measurement and control are of great importance in chemical science. Scientist Sorenson in 1909 introduced a convenient scale to express hydrogen ion concentration to decide acidic, alkaline or neutral nature of solution is known as pH scale.
• pH is negative logarithm to the base 10 of hydrogen ion concentration. pH = -log10 [H
+]
• pOH is negative logarithm to the base 10 of OH- ion concentration.
pOH = -log10[OH-]
The sum of pH and pOH is equal to14, pH + pOH = 14.The pH of the solution is equal to 7 is neutral, less than 7 is acidic and more than 7 is basic in nature.
The pH-meter used may be either potentiometric or direct reading pH-meter. It is actually electronic voltmeter. The pH-meter is facilated for the connection of glass electrode. It is necessary to standardize the pH-meter by using buffer solution of known pH, once standardization of pH-meter is done, the test solution can be obtained directly without any calculation.
Aim: To determine pKa value of the given organic acid by pH- measurement.
Apparatus: pH meter, glass electrode, calomel electrode, micro burette, pipette, beaker etc.
Chemicals: 0.1N acetic acid solution, standard NaOH solution.
Theory: The powers of dissociation of acids are classified into strong and weak acids, depending upon the capacity of releasing H+ ions in the solution. Strong acid is almost completely dissociated or ionized and weak acid is partially dissociated in the solution.
It is generally represented as HA.
HA H+ + A-
K = [H+] [A-] / [HA]
K =Dissociation constant
According to Henderson’s equation
E X P E R I M E N T No. 11
Page | 47
pH = pKa +log []
[] ………………(1)
If acid solution is titrated against strong alkali solution at the end point of titration, acid is converted in to salt. At half neutralization point,
[Salt] = [ Acid]
Therefore, log =0, at half equivalence point, equation (1) becomes as,
At half equivalence point, pH =pKa.
Procedure:
1. Keep the knob of pH meter at room temperature and calibrate it. 2. Pipette out as per given volume of acid solution in 100ml beaker. Add to it 40 ml of
distilled water. 3. Dip the glass electrode and calomel electrode into it.Place beaker on magnetic stirrer.
Connect the electrodes to pH-meter. 4. Fill the micro burette with std. NaOH solution. Record the initial pH reading of the
solution at zero ml std. NaOH solution. 5. Titrate the acid solution with std. NaOH solution and record the pH values for every
addition of std. NaOH solution. While measuring the pH do not stir the solution. 6. Plot a graph, pH v/s Volume of NaOH. Calculate pH value at half equivalence.
Sr.
No. Std.NaOH(ml)
(V) pH ∆pH ∆V ∆pH/∆V Mean(V)Ml
1 2 3 4 5 6 7
Volume of NaOH Volume of NaOH
pH pH/ V
Page | 48
AIM: Determine dissociation constant organic acid [acetic acid] using various buffer solutions of sodium acetate and acetic acid in aqueous solution pH –metrically.
Apparatus: pH –meter with glass electrode, calomel electrode, standard volumetric flask 50ml, beakers, measuring cylinder, etc.
Chemicals: 0.2N acetic acid solution, 0.2N sodium acetate solution.
Theory; Acetic acid is weak acid therefore it is partially dissociated in aqueous solution. Sodium acetate is a salt of strong base and weak acid. Various buffer solutions can be prepared by mixing acetic acid solution with sodium acetate solutions and pH of the solution depends upon concentrations of the solutions.
According to Henderson’s equation,
pH = Pka +log[Salt] / [Acid]
Procedure:
1. Keep the knob of pH meter at room temperature and calibrate it. 2. To prepare buffer solutions by mixing 0.2N acetic acid solutionand 0.2N sodium
acetate solution. like, 3. [5ml acetic acid + 45ml sod. acetate],[10ml acetic acid +40ml sod. acetate], [15ml
acetic acid +35ml sodium acetate]……………………… [45ml acetic acid +5ml sodium acetate].Label as1,2,3,…9 buffer solutions
4. Connect calomel and glass electrodes to PH- meter, measure and record the pH of each buffer solution.
Calculations: Determine concentration of each buffer solution,
i) For a mixture of [ 5ml acetic acid +45ml sodium acetate]
[Salt] V/S [Acid] N1V1 = N2 V2
N1 × 50 = 0.2×5 N1 = 0.02, [Salt] =.02N
ii) For [Acid] V/S [Salt] [ 45ml sodium acetate + 5ml acetic acid]
N1 V1 = N2 V2
N1 × 50 = 45×0.2 N1 = 0.18 [Acid] = 0.18 N ⇒ similarly, calculate the concentrations of 2 to 9 buffer solutions. As per the Henderson’s equation,
pH = pKa + log [<EfD]
[y9z@] or this may be written as,
E X P E R I M E N T No. 12
Page | 49
pH =log []
[] + pKa
This is an equation of straight line (Y = mx + c); pKa is the intercept on Y-axis
Plot the graph of pH against log [<EfD]
[y9z@]
The intercept on Y-axis is pKa.
pKa = –log Ka
Ka = Antilog (–pKa), Dissociation constant of acid (Ka) =
OBSERVATION TABLE:
Sr.
No. buff.
solu.
Acetic acid
solution
0.2N[ml]
Sod. Acetate
Solution 0.2N,
[ml]
pH [Salt] [Acid] Log[|~]
[]
1 45 05 0.02 0.18 -0.9542
2 40 10 0.04 0.16 -0.6020
3 35 15 0.06 0.14 -0.3680
4 30 20 0.08 0.12 -0.1761
5 25 25 0.10 0.10 0.0000
6 20 30 0.12 0.08 0.1760
7 15 35 0.14 0.06 0.3679
8 10 40 0.16 0.04 0.6020
9 05 45 0.18 0.02 0.9542
Result:
1. Pka value of the acid = …………. 2. The dissociation constant (Ka) of the acid = ……………….
pH
pKa (intercept)
-ve +velog (Salt)/(Acid)
Page | 50
CHAPTER
5 COLORIMETRY
An instrument widely used for the determination of concentration of colored solutions is called as Colorimetry.
Colorimetric analysis is a method of determining the concentration of a colored substance in a solution. It is applicable to either organic or inorganic compounds. The property of colorimetric analysis is to be determined by the measurement of relative absorption of light with respect to concentration of the colored solution. The colorimetric analysis is concerned with visible region with wave length between 4000Ao and 8000Ao. Moreover, spectro-photometric analysis is concerned with visible and UV region with wavelength 2000A0 to 8000A0.
Colorimetric principle:
When a monochromatic light passes through a colored solution, a part of incident light is reflected, a part is absorbed by the medium and reminder is transmitted.
I0 = Ir + Ia + It
In colorimetric experiment, using glass cells the reflected light is negligible (4%). Therefore,
I0 = Ia+ It
Where, I0 = incident light
Ia = absorbed light
It = transmitted light
The absorbance or optical density (O.D.) is expressed as,
O.D. = log I0 / It
Laws of absorption of light:
Lambert’s law: In 1760 Lambert put forth relationship between absorbed light and thickness of absorbing medium.
Statement: The law state that, “the rate of decrease of intensity of incident light with thickness of the medium is directly proportional to the incident light.”
–dI / dt α It
–dI / dt = k It ……………………………………….(i)
Page | 51
Where, k = constant.
dI / It= –k dt ……………………………………….(ii)
Integrating equation (ii), ∫dI /It = –k ∫dt
∫It = –kt + c …………………………………………(iii)
Where, c = Integration constant
When, t = 0 and I = I0,
∫I0 = c …………………………………………….….(iv)
Put value of ‘c’ in eqn (iii)
∫It = –kt + ∫I0 ………………………………………...(v)
∫I0 – ∫It = kt
∫ I0/ It = kt ……………………………………………(vi)
2.303log10 I0/ It = kt
log10 I0/ It = k/ 2.303 × t, where K = k/ 2.303
log10 I0/ It = Kt ………………………………………(vii)
but log10 I0/ It = O.D …………………………………(viii)
from equation vii & viii
O.D. =Kt …………………………………………….(ix)
K= absorption coefficient
Hence, O.D. is directly proportional to thickness of the absorbing medium.
Beer’s Law: According to Beer’s law, the absorption coefficient is directly proportional to concentration and absorption coefficient is nothing but optical density.
K αc K = εc, Where, ε is molar extinction coefficient. From equation (ix) O.D. = ε c t …………………………………………(x)
This is Lambert –Beer’s law.
The molar extinction coefficient is directly proportional to the O.D. when concentration of the coloured solution is 1 mol/lit & thickness of absorbing layer is1cm.
O. D. = ε ,(when c = 1 & t =1)
Page | 52
Aim: Verify Lambert-Beer’s law using KMnO4 Colorimetrically and determine concentration of unknown KMnO4 solution.
Apparatus: Colorimeter with cell, standard volumetric flask, beakers.
Chemicals: 0.01 N KMnO4 .
Theory: The law states that ‘when a beam of monochromatic light passes through a colored solution. The rate of decrease of intensity of radiation with thickness of absorbing solution is directly proportional to the intensity of incident light and concentration of the solution.
log
= ε × c × t (t = 1 )
OD = ε × c
When a graph is plotted between OD and concentration c, a straight line passing through origin will obtain, with slope is equal to molar extinction coefficient (ε). The Beer’s law is valid only for dilute solution and straight line deviates if concentration is increased.
Procedure:
1. Prepare 0.1 N solution of KMnO4 by dissolving 0.316 gm of crystals in 100ml of distilled water; hence prepare 0.01 N solutions using stock solution.
2. Prepare 0.0001, 0.0002, 0.0004, 0.0006, 0.0008, 0.0010 and 0.0020 N solution of KMnO4 using 0.01N solution.
3. Select λmax by using central solution. 4. By fixing λmax record the OD values of each solution. 5. Find out the optical density of given unknown solution. 6. Plot a graph of OD v/s concentrations. 7. The straight line passing through origin verifies the law, find out slope of line, which
gives value of molar extinction coefficient (ε). 8. To find out concentration of unknown solution, the value of OD is plotted on y-axis
and draw parallel line to X-axis so that it intersect the straight line at a particular point, draw a perpendicular from that point to x- axis which gives the concentration of unknown solution.
Observations:
Determination of λmax:
Sr. Wave length nm O.D.
1 2 3 4 5
E X P E R I M E N T No. 13
Page | 53
Determination of optical density:
Sr.
No. Concentration OD
1 0.0001N
2 0.0002N
3 0.0004N
4 0.0006N
5 0.0008N
6 0.0010N
7 0.0020N
8 Unknown
Result: The straight line passing through origin verifies the law.
The concentration of unknown solution is ………N.
The value of molar extinction coefficient is ……….
Concentration
OD
Unknown Concentration
At High ConcentrationStraight Line Deviation
Page | 54
Aim: To study the complex formation between Fe3+ and 5-SSA by Job’s method using
colorimeter.
Apparatus: Colorimeter, standard flask, test tubes, measuring cylinder etc.
Chemicals: 0.001 N HCl , 0.001 M 5-sulpho salicylic acid in 0.001 N HCl and 0.001 M Fe3+ ion solution in 0.001 N HCl.
Theory: Job’s method is used to detect the molecular composition of complex species in solution which is formed by the reaction of two components as
M + nL → MLn
The quantity of n can be determined if solution of M and L are equimolar.
Fe 3+ ions react with 5 – sulpho salicylic acid and form the complex in acidic medium. The pH of medium is maintained between 2.6 – 2.8.
According to Job’s when equimolar solutions of the reactants are mixed in varying proportions, the maximum amount of complex is formed at equilibrium when proportions of the reactants correspond to the empirical formula of the complex.
The absorbance of each solution is measured at suitable wave length. A graph is plotted between OD and composition. The maximum absorption shows the molar composition on X- axis.
Procedure:
1. Prepare following mixture solutions using 0.001N ferric ion and 0.001M 5- SSA (Sulpho salicylic acid) solution. Fe 3+ ion solution in ml 1 2 3 4 5 6 7 8 9 5- SSA solution in ml 9 8 7 6 5 4 3 2 1
2. Using (5+5) mixture solution determine λ max. 3. Note down O.D. of each solution at λ max. 4. Plot a graph between O.D. (Y axis) V/S compositions (X axis).The maxima of the
curve will show the proportion of the metal ion and 5-SSA in the complex formation.
Observations: Selection of λ max:
Sr.
No. Wave length λ in nm O. D. Selectied of λ max
1 2 3 4 5 6 7
E X P E R I M E N T No. 14
Page | 55
Determination of empirical formula: Sr
No.
Composition
Fe 3+
ion + 5- SSA O.D.
1 1 + 9 2 2 + 8 3 3 + 7 4 4 + 6 5 5 + 5 6 6 + 4 7 7 + 3 8 8 + 2 9 9 + 1
Result: 1:1 complex formation between Fe3+ and 5 – SSA at [5 + 5]ml composition
Composition of Fe3+
+ 5-SSA
(5 + 5)
OD
1:1 Complex formation
Page | 56
CHAPTER
6
POLARIMETRY
Polarimetry: The property of an optically active compound is measured by means of an instrument known as polarimeter.
An ordinary light waves vibrating in many different planes. When it is passed through nicol prism or Polaroid lens. It is found that, light vibrate in only one plane is said to be plane polarised light. Optical activity:
The solutions of some organic compounds have the ability to rotate the plane of
polarised light these compounds are said to be optically active and the property of a compound is called optical activity. Leavo-rotatory
When a solution of a known concentration of an optically active compound is placed in the polarimeter. When the plane polarised light is passed through the certain organic compounds, the plane of polarized light is rotated. A compound that can rotate the plane of polarised light to the left, that is anticlockwise is said to be levorotatory.
Dextro-rotatory
A compound that rotates the plane polarised light to the right that is clockwise is said to be dextrorotatory.
The rotation of plane polarised light is an intrinsic property of optically active molecules.
Aim: To study the inversion of cane sugar by polarimeter.
Theory: The hydrolysis of cane sugar (sucrose) takes place in presence of dilute acid in aqueous phase in excess amount of water; experimentally the order of the reaction is one & molecularity is two sucrose & acid water. The practically concentration of cane sugar changes into glucose & fructose.
The reaction is acid catalyzed; therefore inversion of cane sugar is pseudo-first order unimolecular reaction.
C12H22O11 + H2O C6H12O6 + C6H12O6H+
glucose fructose
E X P E R I M E N T No. 15
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The angles of rotation at the definite time of intervals are recorded. It has been observed that during inversion sign of angle of rotation change from the –ve i.e. angle of rotation decreases. The kinetics of first order reaction is studied by using equation –
K = 2.V.V
log (e)
(– )
When R0 & Rt are angle of rotation at zero initial time & time ‘t’ respectively (R0 – R∞) will be the initial concentration of sucrose i.e. ‘a’ (Rt – R∞) will be concentration of unhdrolyzed sucrose (a – x).
R∞ is angle of rotation after completion of the reaction (constant reading)
Apparatus: Polarimeter, conical flask, beaker, stop watch, std. flask, measuring flask, thermometer, etc.
Chemicals: 20% cane sugar solution, 2N HCl, distilled water etc.
Procedure:
1. Switch on the polarimeter. 2. Dissolve 20gm of sucrose in 100ml distilled water & filter it. 3. Prepare 2NHCl solution, 100ml. 4. Place both the solutions in thermostat to attain constant temperature. 5. Take initial reading for distilled water. 6. Now mix tow solutions 25ml sucrose solution & 25ml 2NHCl solution in Stoppered
bottle & start stop watch immediately, shake the solution well. Fill the polarimeter with this solution immediately & take reading for angle of rotation at zero time i.e. R0
7. Take the similar reading for the angle of rotation time of interval 5, 10, 15, 20, 25, 30 minutes i.e. (Rt). Add or subtract the blank reading. These are the readings at intervals time ‘t’ (Rt).
8. For the final reading the reaction mixture is heated in a stoppered bottle or conical flask at 60-70oC, for half an hour to complete the reaction. It is then cooled & fill in to the polarimeter tube & its angle is determined, this reading is R∞.
Observation Table:
Time
‘t’ (min)
Angle of
rotation Rt (Rt – R∞)/log (Rt – R∞) Kmin
–1
0 R0 = 5 R5 = 10 R10 = 15 R15 = 20 R20 = 25 R25 = 30 R30 =
Page | 58
Plot of the log (Rt – R∞) vs time, it is straight line with intercept on y-axis.
Calculation from graph:
1. The rate constant K = 2.303 × slope
= …………. Min-1
2. Calculate from kinetic first order equation
K = 2.V.V
log (e)
(– )
Find ‘K’ for all time & take average of ‘K’ values
K = …………. Min-1
Result: The rate constant of inversion ‘a’ cane sugar acid catalyzed reaction.
1. From graph ‘K’ = …………. Min-1
2. Form kinetic first order equation K = …………. Min-1
Log (Rt - Rinfinity)
Slope = ...............
Time
SURFACE
The property of liquids arises from the intermolecular forces of interior of a liquid is attracted equally in all directforces are completely balanced and there is no resultant force on the interior molecule. The molecule present in the surface of the liquid will be pulled laterally and downward direction. But the downward pull remains balance it. Therefore the molecules in the surface are always pulled downwards.molecules are under tension .The force or tension acting downwards perpendicular to the plane of the surface is known as surface tension.to the surface per centimeter length.
Aim: Determine the interfacial tension between two miscible liquids say benzene and water at room temperature
Apparatus: Stalgmometer, small beaker, rubber tubing with screw clip, density bottle.
Chemicals: Benzene, distilled water.
Theory: when liquids are in contact, a force exists at the interface. This force or tension at the surface of separation is called as interfacbe determined by dropping the heavier liquid into the lighter one. The size of the drops inside the lighter liquid will be larger than its size in air. Therefore, the number of drops in the liquid will be
Procedure:
1. Determine number of drops of water in air by stalagmometer.2. Fill the stalagmometer with water and dip its lower end into
benzene in a small beaker.3. Allow the drops to fall slowly and count the number.4. Determine the density of benzene by the use of density bottle.5. Hence calculate interfacial tension.
Number of drops, density and surface tension relationship we have the equation,
E
CHAPTER
7
SURFACE TENSION
of liquids arises from the intermolecular forces of attraction; A molecule in the interior of a liquid is attracted equally in all directions by the molecules around it
balanced and there is no resultant force on the interior molecule. The molecule present in the surface of the liquid will be pulled laterally and downward direction.
imbalanced since there is no molecule on the upper balance it. Therefore the molecules in the surface are always pulled downwards.molecules are under tension .The force or tension acting downwards perpendicular to the plane of the surface is known as surface tension. OR The force in dynes acting perpendicular to the surface per centimeter length.
: Determine the interfacial tension between two miscible liquids say benzene and water
Stalgmometer, small beaker, rubber tubing with screw clip, density bottle.
Benzene, distilled water.
when liquids are in contact, a force exists at the interface. This force or tension at the surface of separation is called as interfacial tension. Interfacial tension of two liquids can be determined by dropping the heavier liquid into the lighter one. The size of the drops inside the lighter liquid will be larger than its size in air. Therefore, the number of drops in the liquid will be less than in the air.
Determine number of drops of water in air by stalagmometer. Fill the stalagmometer with water and dip its lower end into benzene in a small beaker. Allow the drops to fall slowly and count the number.
density of benzene by the use of density bottle. Hence calculate interfacial tension.
and surface tension relationship we have the
E X P E R I M E N T No. 16
Page | 59
A molecule in the ions by the molecules around it. These
balanced and there is no resultant force on the interior molecule. The molecule present in the surface of the liquid will be pulled laterally and downward direction.
imbalanced since there is no molecule on the upper side to balance it. Therefore the molecules in the surface are always pulled downwards. Thus surface molecules are under tension .The force or tension acting downwards perpendicular to the
ting perpendicular
: Determine the interfacial tension between two miscible liquids say benzene and water
Stalgmometer, small beaker, rubber tubing with screw clip, density bottle.
when liquids are in contact, a force exists at the interface. This force or tension at ial tension. Interfacial tension of two liquids can
be determined by dropping the heavier liquid into the lighter one. The size of the drops inside
Page | 60
γ 1, 2 = – × B×
×B
Where, γ1,2 is the interfacial tension, na is the number of drops of water in air, nb is the number of drops of water in benzene, γ1 is the surface tension of water. ρ1 and ρ2 are the densities of water and benzene respectively.
Observation table:
Sr.
No Particular I II III
Mean
1 Number of drops of water in air
2 Number of drops of water in liquid (Benzene)nb
Determination of Density of Benzene:
1) weight of empty density bottle =w1gm 2) Weight of density bottle + water =w2 gm 3) Weight of density bottle + benzene = w3 gm 4) Density of water = ……. gm/ ml
Density of benzene /density of water = w3 –w1 /w2-w1
Density of benzene = 45 ×
45
Calculation of interfacial tension:
γ 1, 2 = – × ×
×
Where, γ1,2 = interfacial tension between water and benzene.
na = the number of drops of water in air.
nb = the number of drops of water in benzene.
γ1 =the surface tension of water.
ρ1= the density of water.
ρ2= the density of benzene.
Result: Interfacial tension between water and benzene is ……..dyne/ cm.
MEASUREMENT OF DENSITY
The density of a liquid is defined as its mass per unit volume. It is also specific gravity smce 1gm of water at 4temperature is one. The density of a liquid at any other temperature is expressed relatively to that of density of water at 4°C.
AIM: To measure the density of a given liquid by Pyknome
The density of a liquid can either be measured by Pyknometer or by density bottlePyknometer is a U-shaped apparatus having a bulb there is a mark while the other arm is with caps.
Method: The pyknometer, an apparatus used to determined to density of liquid, itwashed with chromic acid solution and then with distilled water and finally with alcohol and then dried. It is then suspended from the beam oweighs acuurately. Now attaching a rubber tube at one end of pyknometer and placing its other end in a beaker containing distilled water, the water is sucked gently upliquid is filled from mark B to the drawn out tip other arm A.inside the pyknometer. The pyknometer is then again weighed. After weighing, water is removed from the pyknometer, again washed with alcohol and then dried. The pyknometer is then filled with the liquid whose density is to be determined and weighed again.
Observations:
Room temp. = t°C Mass of empty pyknometer = wmass of pyknometer + water = wmass of pyknometer + liquid = W
Calculations:
mass of water = (w2 – w1)gmmass of liquid = (w3– w1)gm density of liquid / density of water = mass of liquid / mass of water
E X
CHAPTER
8
MEASUREMENT OF DENSITY
The density of a liquid is defined as its mass per unit volume. It is also gm of water at 4°C occupies 1 ml volume; its density at this
temperature is one. The density of a liquid at any other temperature is expressed relatively to
density of a given liquid by Pyknometer.
The density of a liquid can either be measured by Pyknometer or by density bottleshaped apparatus having a bulb B and two capillary ends. On one arm,
there is a mark while the other arm is drawn to a point. In some cases, the two ends
, an apparatus used to determined to density of liquid, ittion and then with distilled water and finally with alcohol and
then dried. It is then suspended from the beam of balance with a hook or copper wire and . Now attaching a rubber tube at one end of pyknometer and placing its
other end in a beaker containing distilled water, the water is sucked gently up to mark A.the drawn out tip other arm A. There should be no air bubble
The pyknometer is then again weighed. After weighing, water is removed from the pyknometer, again washed with alcohol and then dried. The pyknometer is then filled with the liquid whose density is to be determined and weighed again.
Mass of empty pyknometer = w1 gm mass of pyknometer + water = w2 gm mass of pyknometer + liquid = W3 gm
)gm )gm
density of liquid / density of water = mass of liquid / mass of water
X P E R I M E N T No. 17
Page | 61
The density of a liquid is defined as its mass per unit volume. It is also known as 1 ml volume; its density at this
temperature is one. The density of a liquid at any other temperature is expressed relatively to
The density of a liquid can either be measured by Pyknometer or by density bottle. ry ends. On one arm,
rawn to a point. In some cases, the two ends are fitted
, an apparatus used to determined to density of liquid, it is first tion and then with distilled water and finally with alcohol and
copper wire and . Now attaching a rubber tube at one end of pyknometer and placing its
to mark A. the There should be no air bubble
The pyknometer is then again weighed. After weighing, water is
density of liquid (d1)/ density of water
Density of liquid (d1) = w
Determination of relative density of liquid with respect to water:
density –
d1/ d2 = w3 – w1 / w2 – w1
Result: The density of a given liquid is
Aim: To measure the density by
The density bottle is slightly round bottomed type of glass vesselglass cork containing a fine capillary.
The density bottle is first washed with chromic acid solution and then with distilled water and finally with alcohol. It is then dried and weighed. The with distilled water and stoppered. There should be no air bubble inside the The density bottle is then again weighed. Water is then poured out and washed with alcohol and dried. The density bottle is then filled with experimental liquid as before and weighed again.
Let: Mass of empty density bottle = w
Mass of density bottle + water =
Mass of density bottle + liquid = w
Then,
¡¢ (¢£)
¤¥¦ (¢§) =
¤¨e ¤
¤§e ¤
d1 = ¤¨e ¤£
¤§e ¤£ × d2
Result: The density of a given liquid is ............ gm ml
E
density of water (d2) = w3 – w1 / w2 – w1
= w3 – w1 / w2 – w1 × d2
elative density of liquid with respect to water: It is the ratio of the
1
The density of a given liquid is ……… gm ml–1
To measure the density by using density bottle.
bottle is slightly round bottomed type of glass vessel. It is fitted with a glass cork containing a fine capillary.
bottle is first washed with chromic acid solution and then with distilled water and finally with alcohol. It is then dried and weighed. The density bottle is then filled with distilled water and stoppered. There should be no air bubble inside the density
bottle is then again weighed. Water is then poured out and washed with alcohol bottle is then filled with experimental liquid as before and weighed
bottle = w1 gm
bottle + water = w2 gm
bottle + liquid = w3 gm
¤£
¤£
of a given liquid is ............ gm ml–1
X P E R I M E N T No. 18
Page | 62
It is the ratio of the
It is fitted with a
bottle is first washed with chromic acid solution and then with distilled bottle is then filled
density bottle. bottle is then again weighed. Water is then poured out and washed with alcohol
bottle is then filled with experimental liquid as before and weighed
Page | 63
CHAPTER
9
VISCOSITY
Viscosity:
It is a general property of liquids to flow under an applied force. A liquid may be considered to be consisting of molecular layers arranged one over the other. The forces of friction between the layers offer resistance to flow.
Viscosity of a liquid is a measure of its frictional resistance which retards the flow of the liquid.
A molecularlayer of the liquid in contact with the wall of the tube remains stationary has zero velocity; whereas the molecular layer in the centre has the highest viscosity in the direction of flow.
Let, these layers are separated by a distance ‘dx’ and have a velocity difference ‘dv’
The force of friction(F) resisting the relative motion of the two layers is directly proportional to the area (A) and the velocity difference ‘dv’, while inversely proportional to the distance between the layers ‘dx’.
F= η A dv/dx
η = F/A × dx/dv
(Eta) η = coefficient of viscosity or viscosity of a liquid.
The fluidity:
The reciprocal of viscosity is called fluidity.
(Φ) Fluidity = 1/η
The common CGS system of units of viscosity for 1gm cm-1 s-1 is called poise (p), centipoises (1×10-2 poise)
SI units are kg per mol per sec. (Kg mol-1 s-1).
Liquid having low viscosity are said to be mobile while having high viscosity value are said to be viscous liquid.
AIM: Determine the viscosity of a liquid by Ostwald's viscometer.
Apparatus: Ostwald's viscometer,
Chemicals: benzene or alcohol or any liquid.
Procedure:
1. Take about 20 ml of benzene or ethyl alcohol and about 20 ml distilled water in two separate beakers and keep in thermostat.
2. Wash and clean the viscometer with chromic 3. Take 10 ml distilled water in a viscometer using pipette.4. Suck up water just above the mark
ready and allowed the water to flownote down the time required to flow the distilled wawithin mark through capillary.
5. Repeat the same procedure 3 to 4 timetime flow of distilled water,
6. Now, wash & clean Viscometer, dry it and introduce 10 ml of given liquid into the viscometer ,suck up liquid just above the mark and allowed to flow the liquid through the capillary note down the time flow of liquid
7. Repeat it 3 to 4 times, find mean of liquid
Calculation of density of liquid &
Weight accurately clean and dry empty with stopper, then take the weight of the bottle with liquid and stopper.
Weight of liquid = wt. of bottle with liquid and stopper
Density = mass of liquid/Volume of liquid.
Similarly, calculate density
The relative density of liquid = d
Calculations:
Viscosity of liquid (η) =
η1= viscosity of water.
η = §
£ ×
¢£
¢§ × η1
Result: The viscosity of a given liquid
E X
Determine the viscosity of a liquid by Ostwald's viscometer.
ld's viscometer, density bottle, measuring cylinders, pipette, stopwatch.
benzene or alcohol or any liquid.
Take about 20 ml of benzene or ethyl alcohol and about 20 ml distilled water in two and keep in thermostat.
Wash and clean the viscometer with chromic acid, rinse with acetone and dry it.Take 10 ml distilled water in a viscometer using pipette.
ck up water just above the mark, keep a stopwatch ready and allowed the water to flow, start the stopwatch note down the time required to flow the distilled water within mark through capillary. Repeat the same procedure 3 to 4 times, find out mean time flow of distilled water, and say it is t1.
wash & clean Viscometer, dry it and introduce 10 ml of given liquid into the viscometer ,suck up liquid just
the mark and allowed to flow the liquid through the capillary note down the time flow of liquid
find mean of liquid, say it is t2.
of liquid & water:
Weight accurately clean and dry empty density bottle hen take the weight of the bottle with liquid and
of bottle with liquid and stopper –wt of empty bottle with stopper.
mass of liquid/Volume of liquid.
density of water.
of liquid = d2/d1
t2/t1 × d1/d2×n1
of a given liquid (η) is ............ poise
X P E R I M E N T No. 19
Page | 64
cylinders, pipette, stopwatch.
Take about 20 ml of benzene or ethyl alcohol and about 20 ml distilled water in two
acid, rinse with acetone and dry it.
wt of empty bottle with stopper.
Page | 65
Aim: Determine the composition of the given mixture consisting of two miscible liquids, A & B by viscosity measurement.
Apparatus: Ostwald’s viscometer, density bottle or pyknometer, stopwatch, pipette.
Chemicals: Miscible liquids A & B, Distilled water.
Theory and procedure:-
Select two liquids, which do not form complex in the mixture compositions.
1. Prepare mixture compositions by mixing two liquids ‘A’ & ‘B’. 2. Prepare 10%, 20%, 30%, 40%, 50%, 60%, 70%, 80%, 90% of ‘A’ liquid in pure ‘B’
liquid by volume. 3. Determine the time flow of 10%, 20%, 30%, 40%, 50%, 60%, 70%, 80%, 90% of ‘A’
liquid with the help of Ostwald viscometer. 4. Note down the time flow of each percentage compositions of ‘A’ liquid and given
unknown % composition of ‘A’ liquid 5. Find the time flow of unknown composition from plot of time of flow against %
compositions of ‘A’ liquid, a straight line is obtained. Mark a line from corresponding time of flow axis to point of location on straight line.
6. A perpendicular line is drawn on the composition axis from point of location. It gives unknown % composition of ‘A’ liquid and % of ‘B’ will be 100 minus % of ‘A’ liquid.
Result:
The unknown% compositions of mixture of two miscible liquids is
1)…….. %A 2)……….%B
Composition %
Vis
co
sity
Unknown % Composition
E X P E R I M E N T No. 20
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CHAPTER
10
CHEMICAL KINETICS
Introduction:
The study of chemical kinetics provides the most useful method, new ideas and scientific innovative educational research. It is important, not only to know, what happens in a chemical reaction but also how it happens? That is to know the rate of reaction which provides very useful information regarding of the reactions. It is interesting to account the relative reactivity’s of varies reagents and what factors influence the of chemical reaction, hence by knowing how the reaction takes place? We can make changes in the experimental conditions and mechanism of the chemical reaction. The chemical reaction takes place, either absorption or evolution of energy. The energy change during chemical reaction is measured in terms as heat evolved or absorbed. The chemical kinetics deals with the study of
measurement of rate of reaction per unit time.
Rate of reaction: The change in concentration of reactants or products per unit time is known as rate of reaction. The rate of reaction is measured by measuring the change in concentration at different time of intervals. The concentration changes are generally measured by volumetric analysis such as conductance or absorbance etc.
Consider a simple reaction,
A → Product
R α [A] OR
–d x/dt α [A]
–dx/dt= K[A]
Where, K is rate constant or velocity constant. The negative sign indicates that concentration of reactants decreases. Hence, the rate of the reaction goes on decreasing with time.
Unit of the rate of reaction, or dx-dt = mol l-1
t-1
Factors affecting the rate of reaction: Experimental conditions like temperature, pressure, concentration of reactants, reagents etc.
Molecularity of the reaction: The number of the reactant molecules taking part in a chemical reaction as represented by stoichiometric equation is known as molecularity of the reaction. In a chemical reaction one, two or three molecules involved, the reactions are known as unimolecular, bimolecular or trimolecular respectively. In some reactions the
Page | 67
moleculariy is more than two but found to be first order, such type of reaction is known as pseudo unimolecular reaction.
Order of reaction: It is defined as the number of reacting molecules whose concentration changes as a result of chemical reaction is known as order of reaction or it is the sum of power of the concentration terms in the rate law equation. It can be determined experimentally only. It can be fractional as well as zero value. It is depending upon number of molecules undergoing change in concentration.
The order of reaction is classified as first order, second order, third order etc. In case of zero order, the concentrations of reactants do not change during the reaction. The rate of the reaction is constant. Its rate is independent of the concentration of reactant molecules is known as zero order reaction.
Effect of temperature on rate of reaction:
Temperature is the most important factor which influences the rate of the reaction. The rate of the chemical reaction depends on temperature that means the rate of reaction increases with increase in temperature.
Activation energy (Ea): Each and every reactant molecule before converted into product must pass through the transition state (TS) or energy barrier. The minimum energy required to reach the reactant molecules at transition state is known as activation energy or it is the amount of energy required to cross the energy barrier or transition state is known as activation energy.
Aim: Determine energy of activation of a reaction between KI and K2S2O8. (Potassium iodide and Potassium persulphate)
Apparatus: Beakers (Test tubes), Pipette, thermostat, stopwatch, measuring cylinders etc.
Theory:
Energy of activation is the minimum thermal energy required by reactant molecules to undergo certain chemical reaction. It is the amount of energy consumed by the reactant to pass over the activated energy barrier called transition state (T.S.).
The reaction between Potassium persulphate and KI is a second order reaction. The liberated iodine is titrated against std. sodium thiosulphate solution. The liberated iodine is the function of time. Hence, the rate constant can be calculated for second order as:
K = £
~×
©
|(|e©)
E X P E R I M E N T No. 21
Page | 68
But, the rate of reaction is sensitive to temp.(T) and reaction between rate constant and temperature is given by Arrenious equation.
K = A.ªeh/W, Ea = energy of activation, A = freq. factor,
logK = − h
2.V.VW + ®¯°……… (1)
But if the reaction is performed under the condition such that the concentration of iodine remain constant.
∴K =
× ±®². and logK = -logt + cont……… (2)
Substitute the eq. (2) in equation (1)
–logt + cont. = − h
2.V.VW + log °
–logt = − h
2.V.VW + ±®².
logt = h
2.V.VW + ±®².
This is equation of straight line the slope = Ea/2.303R
Hence,Ea = slope × 2.303R.
Procedure:
1) Keep the beaker (test tube) A and B in the thermostat with the solution as a given below to attain same temperature. Test Tube (Beaker) A: 20 ml of KI + 10 ml Na2S2O3 Test Tube (Beaker) B : 20 ml of K2S2O8 + 1 ml of starch.
2) Pour one mixture solution into another beaker (Test Tube) and start stopwatch. Shake the solution and keep the test tube in thermostat.
3) Note down the time (t) required to appear blue color in the tube. 4) Now change the temperature of thermostat and repeat the same procedure. 5) Record the time required to appear blue color at different temperature (T). 6) Plot a graph log(t) (Y-axis) v/s 1/T (X-axis) .Find out slope. Hence calculate, Energy
of activation.
Observation Table:
Sr.
No. Temperature(T)
ok Time(t)sec Log(t) 1/T
1
2
3
4
5
Page | 69
Calculation:
1) Slope from graph = ____________ 2) Calculation of Ea
Ea = Slope × 2.303 × R
= _______× 2.303 × 8.314
= ________ J
Result :- The energy of activation (Ea) of reaction between KI and K2S2O8 is _________ J.
Aim: To study the kinetics of dissolution of magnesium metal in dilute HCl.
Apparatus: Six beakers of 250 ml, measuring cylinder of 100 ml, stopwatch.
Chemicals: magnesium ribbon, conc. HCl (5N)
Theory:
Mg reacts with dil. HCl according to the equation.
Mg + 2HCl → mgCl2 + H2↑
The reaction is a first order with respect to metal and is second order with respect to the acid. Let ‘a’ be the initial amount of the metal taken initially with time, t = 0 and out of it, x dissolves in time ‘t’. Let ‘b’ be the concentration of the dilute HCl taken.
The Rate equation will be :
R or L
³ ´(µ − ¶)·2
Slope = y2 - y1/x2 - x1
log t
1/T
Ea = slope x 2.303 x 1.987 cal/mole
Ea = slope x 2.303 x 8.314 J/mole
E X P E R I M E N T No. 22
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This on integrating and simplying gives.
®¯
(eL)³ ´·2²
Both ‘a’ and (a-x) will be kept constant for each working.
It therefore follows:
Kb2t = constant
b2t = New constant
∴ b2 α 1/t And this relation is to be worked out.
Procedure:
1) Take 30, 25, 20, 15 and 10 ml of given HCl and transfer it to five beaker &label them I, II, III, IV and V respectively. Now add 30, 35, 40, 45 and 50 ml of distilled water from I to V beaker respectively to make the volume of each 60ml.
2) Take the magnesium ribbon and cut it into 5 equal pieces of same weight of length 3”. 3) Drop the first piece of ‘mg’ metal to Ist beaker and run the stopwatch. Stop the
stopwatch the moment entire piece dissolves and note the time taken. 4) Repeat the same procedure by dropping ‘mg’ ribbon from II to V beaker one by one
and note time taken. 5) Plot the graph between b2 on Y-axis and reciprocal of time, 1/t on X-axis. A graph of
linear pattern obtained.
Observation Table:
Sr.
No.
Beaker
Number
Amount of acid in
ml ‘b’ b
2
Time required
for dissolution
of mg‘t’ sec
£
~ Sec.
1 I
2 II
3 III
4 IV
5 V
Result: The graph is linear. Hence dissolution mg with respect to HCl acid is second order reaction.
b2
1/ t sec.
Page | 71
Experiment No. 22
Aim: To study the kinetics of decomposition of sodium thiosulphate by a mineral acid.
Apparatus:-Standard volumetric flask, 100 ml beaker, five 250 ml beakers, Stopwatch, thermometer.
Chemical:-Solid sodium thiosulphate (Na2S2O3), 0.5 N HCl.
Theory:- The decomposition of sodium thiosulphate in HCl takes place according to following equation.
Na2S2O3 + 2HCl → 2NaCl + H2O + SO2↑ + S↓
The precipitation of sulphur is marked by turbidity in solution and these indicate the progress of the reaction. The acid is taken in large excess to keep it’s concentration more or less constant and solution of various concentration of sodium thiosulphate are used to study the kinetics.
Procedure:
1) Weigh accuretly 0.10 gm of sodium thiosulphate and transfer it in 100 ml beaker, dissolve it in minimum amount of distilled water, and transfer it in 100 ml standard volumetric flask, and make it to mark by adding distilled water. Transfer the solution in 200 ml beaker and lable as solution I. Similarly weigh 0.30, 0.50, 0.70, 0.90 gm of sodium thiosulphate and prepare solution and lable them as II, III, IV, and V solution.
2) Make a big cross (X) mark on weight paper and put the Ist beaker on it. Add 20 ml of 0.5 N HCl in beaker and start stopwatch. View the cross from above, through solution. The moment cross mark becomes invisible, stop the stopwatch and note the time taken. Repeat the same experiment with remaining four beakers.
3) Plot a graph, taking amount of thiosulphate per 100 ml of solution on Y-axis and 1/t on X-axis.
Observation Table:
Sr.No.
Amount of
Na2S2O3 in gm
per 100ml
Volume of 0.5N
HCl added in ml
Time required
for turbidity
‘t’
1/t
1 0.10 20
2 0.30 20
3 0.50 20
4 0.70 20
5 0.90 20
E X P E R I M E N T No. 23
Page | 72
Result: The straight line graph show that the decomposition of thiosulphate by mineral acid is a first order reaction.
Aim: Determine the rate constant of the reaction between K2S2O8 and KI with equal concentration of reacting species (a=b).
Apparatus: Burette, pipette, conical flask, beakers, stop watch etc.
Chemicals: 0.05NK2S2O8, 0.05N KI, 0.01N Na2S2O3, starch indicator.
Theory: The reaction between potassium per sulphate and potassium iodide is given by,
K2S2O8+2 KI → 2K2SO4 + I2
It is experimentally prove that, the rate of this reaction depends upon two reacting species, therefore this is second order reaction. If the concentration of potassium per sulphate and potassium iodide in reacting mixture solution is equal, then the rate constant of kinetics of second order reaction is represented as –
¸ ³ £
~ ×
©
|(|e©) ……….. (1)
We may write equation (1) for graphical use as follows
£
(|e©) = kt +
£
| ……….. (2)
Am
ou
nt o
f N
a2S
2O
3 in
%1/ t sec.
E X P E R I M E N T No. 24
Page | 73
Procedure:
1. Pipette out 50 ml K2S2O8and 50 ml KI separately in two beakers and keep them in thermostat to maintain constant temperature.
2. Fill the burette with 0.01 M Na2S2O3 solution up to zero mark. 3. Mix one solution into other and start the stop watch. 4. After intervals of 5 minutes, pipette out 10 ml of mixture solution in a conical flask,
add 5-6 drops of starch as an indicator and titrate against sodium thiosulphate. 5. Note down volume of sodium thiosulphate required to decolorize blue colour. 6. Similarly record the reading for 10, 15, 20, 25, 30 minutes.
7. Plot a graph between (y axis)
(eL)and (x axis)× t (time in second). A straight line
graph is obtained; the value of slope = ‘K’ (rate constant) of reaction and intercepts
gives
.
Observation:
Sr.
No.
Time ‘t’
sec
Volume of Na2S2O3
required for
decolorize blue colour in
ml,[BR]
x (a-x) £
(| – ©)
1 05 x 60 = 300
2 10 x 60 = 600
3 15 x 60 = 900
4 20 x 60 = 1200
5 25 x 60 = 1500
6 30 x 60 = 1800
Calculation:
Value of ‘a’ = initial concentration
The reaction mixture contains 50ml 0.05 N K2S2O8and 50ml 0.05 N KI.
Hence, Normality of each reacting species = a = 0.025 N
Calculation of value of x:
Value of ‘x’ = amount converted into product.
Mixture v/ s Na2S2O3
x = N1V1= N2V2
x =N1 x 10 ml = 0.01 x Volume of Na2S2O3 required for decolorize blue color.
N1= ... L¹
.
Page | 74
Result: The rate constant of the reaction between K2S2O8an d KI with equal concentration is _________ sec-1.
Slope (K) = y2 - y1/x2 - x1
1/(a-x)
Time (t) in second
1/a
Page | 75
CHAPTER
11
HETEROGENEOUS EQUILIBRIA
The system which consists of two or more phases is known as heterogeneous system.
In 1874 Williard Gibbs predicted qualitatively the effect of pressure, temperature and concentration on the heterogeneous system in equilibrium, it is given by the phase rule equation.
F = c – p + 2
Where, F = The number of degree of freedom
C = The number of components in the system
P = The number of phases present in the system
Phase (P): Any homogeneous physically distinct and mechanically separable part of the heterogeneous system in equilibrium is known as phase. Ex: –
i) A mixture of gases form a homogeneous mixture and it consist of only one phase. ii) Freezing water consisting three phases ice, liquid water and water vapour all three
phases expressed by smaller chemical component H2O. iii) Solution of phenol and water has two liquid phases.
Component (C):The least independent chemical components or molecule, which is composition of every phase express the system completely. Ex: –
i) Freezing water consisting three phases ice, liquid water and water vapour all three phases expressed by smaller chemical component H2O. Hence water is one component three phases system.
ii) Sulphur system has four phase’s monoclinic sulphur, rhombic sulphur, liquid sulphur and vapour sulphur, all four phases expressed by single component sulphur.
Degree of freedom (F): The minimum number of variables factors such as temperature, pressure and concentration which should be fixed to define the system completely. Ex: –
i) Freezing water consisting three phases ice, liquid water and water vapour all three phases co-exist at freezing water temperature. Since vapour pressure also has fixed value, there is no need of any variables. Hence degree of freedom is zero.
Applying phase rule equation to freezing water system
F = c – p + 2
F = 1 – 3 + 2
Page | 76
F = 0
When, F = 0, the system is nonvariant
F = 1, the system is monovariant
F = 2, the system is bivariant
Pertaining to importance of heterogeneous equilibria different experiments may be studied
a) Distribution of solute between two immiscible liquids b) Partial miscibility of liquids c) Solubility and solubility product
(a) Distribution of solute between two immiscible liquids:-In 1891 Nernst studied several solutes added in two non miscible a mixture of solvents. Solute itself distributes between solvents contact with each other and the solute exists as simple or same molecules in both the solvents.
Statement of Nersnt Distribution Law: “When solute is added in a mixture of two non miscible solvents, solute itself distributes in such a way that solute exist as simple or same molecules in both the solvents. At equilibrium, the ratio of the concentrations of solute in two solvents is constant at given temperature and is independent of amounts of solute and solvents”.
If C1 and C2 are the concentration of the solute in solvent 1 and 2 at equilibrium respectively,
C1/C2 = K (Constant)
Where, K is constant called distribution coefficient or partition coefficient.
Aim: Determine the solubility of benzoic acid in water at difference temperature and hence it’s heat of solution.
Apparatus: Beakers, burette, pipette, Thermometer, Funnel etc.
Chemicals: Solid benzoic acid, 0.05 N NaOH solution, phenolphthalein indicator.
Theory: When solid is dissolved in liquid, heat may be evolved or absorbed when one mole of the solid is dissolved in solution heat energy is evolved or absorbed is known as heat of
solution.
The effect of temperature on solubility is given by Van Hoff’S equation i.e. @fB<
@º = ∆n
ȼ
where s is the solubility and ∆H the heat of solution.
E X P E R I M E N T No. 25
Page | 77
On integrating the above equation, we have
Log s = e∆n
2.V.V ȼ+ constant ,
or integrating between two different temperature ¼ , and ¼2 we get .
log½2- log ½ = ∆n ( º¾º )
2.V.V »ºº
Where, R = 1.987 cal/mole (8.314 J/mole), S1 and S2are then strength of solute at T1 and T2
oK.
Procedure:
1) Prepare a saturated solution of benzoic acid about 100 ml at room temperature. Filter this solution in a clean beaker and cover it.
2) Fill the burette with 0.05 N NaOH solutions. 3) Pipette out 10 ml the benzoic acid solution in a conical flask, add to it 2-3 drops of
phenolphthalein indicator, and titrate this solution against burette solution, till color changed to pink, note down the burette reading. Take two reading at same temperature then determine normality and strength of acid solution.
4) Similarly prepare the saturated solution of benzoic acid at 20o, 35o and 40o and determine the normality and strength of these temperatures.
5) Plot a graph of log‘s’ (y- axis) and 1/T (x- axis). A straight line will obtain, determine slope of line. Calculate the heat of solution ∆H by using equation. ∆H = –2.303 × R × Slope
Observation Table:
Sr.
No.
Temp
(K)
Burette reading Normality
(N)
Solubility/
Strength ‘s’
gm/lit
Log s 1/T(K)
I II Mean
1
2
3
4
Calculation:
Benzoic acid solution v/s NaOH
N1V1 = N2V2
N1=...G×¹
.
Page | 78
Strength of Benzoic acid = N1 × eq. wt of Benzoic acid
= _______ × 122
= _______gm/lit
∆H = 2.303 × R × Slope –
∆H = -2.303 × 1.987 × ________cal
∆H = -2.303 × 8.314 × ________ joules
Result: 1. Solubility of benzoic acid………….
2. Heat of solution of benzoic acid …………
Aim: To study the effect of NaCl (or succinic acid) on critical solution temperature of phenol water system and determine the concentration of that solute in the given system.
Apparatus: Beaker, thermometer, glass tube, measuring cylinder, standard flask, stirrer.
Chemical: Distilled water, NaCl, Phenol.
Theory:
The liquid has limited miscibility with each other is called as partially miscible liquid. When two partially miscible liquids like phenol and water are mixed, two separate- layers are formed. Upper layer of water in phenol and lower layer is phenol in water. As the temperature increases the mutual solubility increases and at a particular temperature the two layers completely merge into each other. That temperature is called critical solution
temperature (or consolute temperature).
The critical solution temperature is sensitive to the presence of impurity in one or both components. If the impurity is soluble only in one of the two components the CST increases. If the impurity is soluble in both components the CST decreases.
The change in CST is found to be linearly proportional to the percentage of impurity.
NaCl is soluble in water only Hence CST increases .where as succinic acid is soluble in both component hence CST decreases.
Procedure:
1) Prepare 20 ml of 1.8%, 1.6%, 1.4%, 1.2%, 1.0%, 0.8%, 0.6%, 0.4% and 0.2% solution of Nacl from 2% Nacl stock solution.
2) Take ten clean and dry glass test tubes & number them as 1 to 10.
E X P E R I M E N T No. 26
Page | 79
3) Take 5ml of each solution in different glass test tubes. 4) Add to each test tube 5ml of phenol. 5) Insert the thermometer and ring stirrer in the 1st test tube and place it in water bath
slowly rise the temperature. Note the temperature at which turbidity disappears. a. Stop heating and allow the water bath to cool slowly with constant stirring. Note
the temperature at which turbidity appears. b. The mean of two temperatures give critical solution temperature for that
composition. c. Similarly determine the critical solution temperature of other composition &
unknown composition. d. Plot a graph between percentage of NaCl (solute) concentration (x-axis) &
miscibility temperature (CST) (on y-axis), a straight line will be obtained. e. Hence find out concentration of solute.
Observation Table:
Sr.
No.
Concentration
of Solute
[NaCl] In %
[NaCl] ml +
ml of Phenol
Temperature T1 + T2/2
Mean of
Miscibility
Temp.
at which
Turbidity
disappears on
Heating (T1)
at which
Turbidity
appears on
cooling (T2)
1 2.0 5ml + 5 ml 2 1.8 5ml + 5 ml 3 1.6 5ml + 5 ml 4 1.4 5ml + 5 ml 5 1.2 5ml + 5 ml 6 1.0 5ml + 5 ml 7 o.8 5ml + 5 ml 8 0.6 5ml + 5 ml 9 0.4 5ml + 5 ml
10 0.2 5ml + 5 ml 11 Unknown 5ml + 5 ml
Result:
1. Concentration of NaCl increases the CST of Phenol water system increase. 2. The concentration of unknown solution is _________%
Unknown Conc. of Solute
Mis
cib
ility T
em
p.
Conc. of Solute (NaCl)
Page | 80
Aim: Determine the partition coefficient of iodine between carbon tetrachloride and water.
Apparatus: Stoppered bottles, burette, pipettes, shaking machine, etc.
Chemical: Iodine crystals, carbon tetrachloride, potassium iodide, sodium thiosulphate, starch, etc.
Theory: The molecular state of iodine in both the solvents is same i.e. it do not show association in any solvents. Hence, the distribution law is strictly obeyed by this system.
In case of distribution of iodine between organic solvent and water at equilibrium, the distribution coefficient or partition coefficient [K] can be given by the expression.
K = Q¿ÀÁ
QÂ
Where, Corg and Caq are the concentrations of iodine in organic solvent (CCl4) and water respectively.
Procedure:
1. Prepare a saturated solution of iodine in about 50ml of pure carbon tetrachloride, 2. Wash the bottles well and label them as 1, 2, 3, 4 etc. prepare following compositions
in those four bottles. 3. Stopper the bottles tightly and shake them vigorously for about an hour. 4. Allow to stand the liquid for some time. The content of the bottle separates into two
layers. The lower layer is of CCl4 and upper layer is of water. 5. Prepare 0.05M solution of sodium thiosulphate [Mol. wt. = 248.18] in 250ml distilled
water. From this solution then prepare 0.01M solutions by quantitative dilution. 6. Pipette out 5ml of CCl4 layer from bottle No. 1 into a conical flask and add to it about
20ml 10% KI solution. 7. Titration the contents of the conical flask against 0.05M Na2S2O3 solution using
freshly prepared starch solution as an indicator. The end point is the disappearance of blue colour. The conical flask should be shaken vigorously during titration to have complete extraction of iodine (I2) from the CCl4 layer. Repeat 2-3 times and take the average.
Similarly, titrate 5ml of CCl4 layer from other bottles using same Na2S2O3
solution. 8. Pipette out 40ml of aqueous layer from bottle no. 1 into a conical flask and 5ml 10%
KI solution. Titrate this solution against 0.01M Na2S2O3 solution, using starch as an indicator. Repeat twice or thrice.
Similarly, titrate 40ml of aqueous layer from other bottles using 0.01M sodium thiosulphate solution.
E X P E R I M E N T No. 27
Page | 81
Observation:
1. Observation Table No. 1
Bottle No.
Volume of
saturated I2
Solution (ml)
Volume of Pure
CCl4 (ml)
Volume of
distilled water
(ml)
1 15ml 10ml 50ml 2 13ml 12ml 50ml 3 10ml 15ml 50ml 4 7ml 18ml 50ml
2. Observation Table No. 2
Bott. No.
Pipette out
organic layer
Burette reading 0.05N Na2S2
O3
Pipette out Aqu. Sol.
Burette reading 0.01N
Na2S2O3
Normality CCl4 (N1)
Normality Aqu. Sol.
(N2)
K= N1/N2
Mean
1 5 20 2 5 20 3 5 20 4 5 20
Calculation: Calculate normality of all bottles for organic and aqueous layer
Organic layer v/s Na2S2O3
N1V1 = N2V2
N1 = N2V2/V1
N1 = ...G×Ã.».
G
Aqueous layer v/s Na2S2O3
N1V1 = N2V2
N1 = N2V2/V1
N1 = ...×Ã.».
2.
Result: Partition coefficient of iodine between carbon tetrachloride and water is ………
Page | 82
CHAPTER
12
THERMODYNAMICS
Aim: Determine the enthalpy change of neutralization of strong acid by strong base.
Apparatus: Thermometer, measuring cylinder, polythene bottle, beaker etc.
Chemicals: 0.5 N HCl and 0.5 N NaOH, distil water.
Theory: The reaction of an acid with a base forming salt and water is called a neutralization reaction. Actually in neutralization H+ from acid combine with OH- ions from the base to form practically unionized water. It is an exothermic reaction. The enthalpy change during neutralization reaction is called enthalpy of neutralization or heat of neutralization. It can define as enthalpy change, when one gram equivalent of an acid is completely neutralized by one gram equivalent of a base in a dilute solution is called as heat of neutralization and the amount of heat energy is evolved during this process. It is exothermic reaction.
The heat of neutralization of a strong acid and strong base is constant and does not change with nature of acid and base. It is -13.7 Kcal or -57.36 kJ mol-1. The neutralization of HCl by NaOH can be represented as fallows.
NaOH Na+ + OH-
HCl H+ + Cl-
Neutralization: NaOH + HCl NaCl + H2O
Na++ OH+ + H+ + Cl NaCl+ H2O
H+ + OH+ H2O
Procedure:
1) Take a 100 ml of 0.5N HCl solution in a Dry and clean polythene bottle. Also take 100 ml of 0.5N NaOH solution in another polythene bottle.
2) Measure the temperature of HCl solution and note as t1°C .Similarly measure the temperature of NaOH solution and note as t2°C.
3) Pour the HCl solution in NaOH and stir it. 4) Insert the thermometer in the solution and note down the maximum temperature to
attained, note as t3°C. 5) Hence calculate the heat of neutralization.
E X P E R I M E N T No. 28
Page | 83
Observation and calculation:
i) Volume of NaOH : 100 ml ii) Volume of HCl : 100 ml iii) Temperature of HCl: t1°C = ________ iv) Temperature of NaOH: t2°C = ________ v) Temperature of Mixture: t3°C = ________
Calculation:
i) Initial temperature of mixture = D°m[ D2°m
2 =
ii) Rise in temperature = ∆t = [t3°C -( D°m[ D2°m
2 ) ]
iii) The heat evolved in neutralization = Heat gained by the solution + Heat gained by the polythene bottle. = (V × d × specific heat × ∆t ) + ( x × ∆t )
Where V = Total volume of reaction mixture =200 ml d = density = 1 gm/ ml Specific heat = 1 x = water equivalent polythene bottle = 1
The heat evolved in neutralization = ( 200 × 1 × 1 × ∆t ) + ( 1 × ∆t ) = 201 ∆t = …………….= Q cal.
iv) Heat liberated in the neutralization of 100 ml of 0.5 N HCl = Q cal.
Heat liberated in the neutralization of 1000 ml of 1 N HCl = ... × × Ä
.. × ..G
∆HN = 20 Q
∆HN = _________ Cal
∆HN = _________ × 4.187 J
Result: The heat neutralization of strong acid and strong base.
Page | 84
Aim: Determine the heat of neutralization of acetic acid against strong base and determine enthalpy of ionization of weak acid.
Apparatus: Calorimeter assembly, thermometers, beakers.
Theory:
• Heat of ionization of an acid may be defined as the amount of heat changed when one gram equivalent of the acid is completely ionized.
• Neutralization of acetic acid takes place in two stages (i) ionization of acid and (ii) formation of unionized water by combination of H+ and OH– ions.
• In any type acid-base neutralization, same reaction takes place. Hence, the heat of neutralization remains constant i.e. 13.7 K cal.
• However, in case of weak acid like acetic acid, the value is found to be less than 13.7 k cal. This is because a part of heat evolved is consumed for further ionization of the weak acid.
Procedure:
• Take 100ml of 0.5N acetic acid solution. A clean thermos flask filled with funnel and thermometer. Note the initial reading, reading temperature of acetic acid. Let it is lies the initial temperature.
• Take 100ml of 0.5N NaOH solution in clean beaker with the help of measuring cylinder & dip thermometer in it. Note the initial temperature of NaOH. Let it is the temperature of acid & alkali changes slowly due to exchange of heat with surrounding.
• Record the temperature of acid & alkali solution after 5 minutes pour NaOH solution into the thermos flask containing acetic acid solution close the mouth of funnel with the cotton & stir it. Note the maximum temperature after mixing.
• Note down the exact time of mixing of solution record the temperature of every minute for 5 minutes till the temperature falls uniformly.
Neutralization of acid can be represented as –
i) CH3COOH ===== CH3COO– + H+ + Q Kcal. ii) CH3COO– + H+ + Na+ + OH– = CH3COO– + Na+ + H2O + 13.7 Kcal.
From these equations, we have
CH3COO H + Na+ + OH– = CH3COO– + Na+ + H2O + (Q + 13.7 K cal)
Observation and calculation:
1. Volume of NaOH : 100 ml 2. Volume of CH3COOH : 100 ml
E X P E R I M E N T No. 29
Page | 85
3. Temperature of CH3COOH: t1°C = ________ 4. Temperature of NaOH: t2°C = ________ 5. Temperature of Mixture: t3°C = ________
Calculation:
1. Initial temperature of mixture = D°m[ D2°m
2 =
2. Rise in temperature = ∆t = [t3°C -( D°m[ D2°m
2 ) ]
3. The heat evolved in neutralization = Heat gained by the solution + Heat gained by the polythene bottle. = (V × d × specific heat × ∆t ) + ( x × ∆t ) Where V = Total volume of reaction mixture =200 ml d = density = 1 gm/ ml Specific heat = 1 x = water equivalent polythene bottle = 1 The heat evolved in neutralization = ( 200 × 1 × 1 × ∆t ) + ( 1 × ∆t ) = 201 ∆t = …………….= Q cal. 4. Heat liberated in the neutralization of 100 ml of 0.5 N CH3COOH = Q cal.
Heat liberated in the neutralization of 1000 ml of 1 N CH3COOH = ... × × Ä
.. × ..G
∆HN = 20 Q
∆HN = _________ Cal
∆HN = _________ × 4.187 J
If ∆HN is the experimentally determined value of heat of neutralization of the acid + the heat of dissociation of acid, then
Heat of ionization of acetic acid = ∆HN – 13.7KCl
Result: Heat if ionization of acetic acid/ weak acid = ……… KCal.
MOLECULAR WEIGHT DETERMINATION
Aim: To determine molecular weight of
Apparatus: Boiling tube thermometer and stop clock, empty testapparatus, thermometer.
Chemicals: Naphthalene or Camphor,
Theory: A simple method developed by Rast’s to determine the molecular weight of nonvolatile solutes, camphor or naphthalene is used melted as non reacting sohomogeneous solution with non-
Procedure: First of all determine the melting point of camphor/ naphthalene using melting point apparatus. Take about 5 gram of heating allows the melting camphor/ find out the melting or freezing point of the naphthalene.
Then add 0.1 gram of given substance containing solvent the mixture shouldmixture solution tube from the bath & allow to cool dip thermometer in it and note down the freezing point of the mixture solution.
Calculate the value of (∆T
∆Tf = T – Ts.
T = freezing point of pure solvent.
Ts = freezing point of the solution.
m = 1000×Kf ×w/W×∆Tf
i) Cryoscopy constant (Kf) for naphthalene = …………deg per kg of Solvent.
ii) Molecular weight of the given solute =…….
E X
CHAPTER
13
MOLECULAR WEIGHT DETERMINATION
To determine molecular weight of a nono-volatile solute by the rast’s method
tube thermometer and stop clock, empty test tube, beaker, Rast’s
or Camphor, parafin liquid.
A simple method developed by Rast’s to determine the molecular weight of nonvolatile solutes, camphor or naphthalene is used melted as non reacting solvent to prepare
-volatile solute.
First of all determine the melting point of camphor/ naphthalene using melting Take about 5 gram of camphor/ naphthalene in a boiling tube
camphor/ naphthalene to cool, note the temperature. find out the melting or freezing point of the naphthalene.
Then add 0.1 gram of given substance non-volatile solute to the boiling tubeshould dipped in the bath until it melts completely, take out
mixture solution tube from the bath & allow to cool dip thermometer in it and note down the freezing point of the mixture solution.
∆Tf) depression in freezing point.
freezing point of pure
freezing point of the
f
deg per mol, per
weight of the given solute
X P E R I M E N T No. 30
Page | 86
MOLECULAR WEIGHT DETERMINATION
method.
tube, beaker, Rast’s
A simple method developed by Rast’s to determine the molecular weight of non-lvent to prepare
First of all determine the melting point of camphor/ naphthalene using melting naphthalene in a boiling tube, melt it by
note the temperature. Using this
to the boiling tube, dipped in the bath until it melts completely, take out
mixture solution tube from the bath & allow to cool dip thermometer in it and note down the
Aim: To determine the molecular weight of a nonvolatile solute Beckmann's freezing method (using water as solvent)
Theory: The temperature at which the solid begins to separate out from liquid is known as freezing point temperature. Or TheStates of substance have the same than that of the pure solvent therefore the freezing point of the solution will bethat of the pure solvent.
When a non volatile solute is dissolved in a pure solvent the decreases hence, the freezing point of the solution decreases than that of the pure solvent this is because of the interaction between solute and solvent particles.depression of freezing point produced by dissolving known amount of the nonin a known amount of solvent, using tmolecular weight is,
m=1000×Kf×w/W×∆Tf
Where, w = the mass of the solute dissolved in pure solvent
W = the mass of the pure solvent
M = the molecular weight of the solute
∆Tf = the depression in freezing point
Kf = the molal depression constant
Beckmann’s thermometer: It is a large size thermometer invented by Otto Beckmann and it does not measure the absolute temperature instead it measure the small difference in temperatures. Apparatus: Beckmann's thermometer, apparatus, two manual stirrers freezing mixture of Ice and NaCl. The procedure:
1. The apparatus consists of a big tube having a side tube for introducing the solute.
2. It is fitted up with stirrer and Beckmann’s thermometer.
3. The inner tube is filled with a weight of the solvent (25 to 40 gm)order to dip the thermometer bulb.
E X
To determine the molecular weight of a nonvolatile solute (KCl, urea, glucose etc.) (using water as solvent).
The temperature at which the solid begins to separate out from liquid is known as The freezing point is the temperature at which solid and liquid
States of substance have the same vapor pressure but the vapor pressure of solution is less than that of the pure solvent therefore the freezing point of the solution will be
When a non volatile solute is dissolved in a pure solvent the vapor pressure of the solution the freezing point of the solution decreases than that of the pure solvent this
interaction between solute and solvent particles. The measurement of depression of freezing point produced by dissolving known amount of the non-volatile solute in a known amount of solvent, using the mathematical expression for the determination of
the mass of the solute dissolved in pure solvent
the mass of the pure solvent
the molecular weight of the solute
the depression in freezing point
the molal depression constant
It is a large size thermometer invented by Otto Beckmann and it does not measure the absolute temperature instead it measure the small difference in
Beckmann's thermometer, ordinary thermometer, Beckmann's freezing point apparatus, two manual stirrers freezing mixture
The apparatus consists of a big tube having a side tube for introducing the
It is fitted up with stirrer and Beckmann’s thermometer. The inner tube is filled with a definite
(25 to 40 gm), in order to dip the thermometer bulb.
X P E R I M E N T No. 31
Page | 87
(KCl, urea, glucose etc.) by
The temperature at which the solid begins to separate out from liquid is known as freezing point is the temperature at which solid and liquid
pressure of solution is less than that of the pure solvent therefore the freezing point of the solution will be lower than
of the solution the freezing point of the solution decreases than that of the pure solvent this
The measurement of volatile solute
he mathematical expression for the determination of
It is a large size thermometer invented by Otto Beckmann and it does not measure the absolute temperature instead it measure the small difference in
eckmann's freezing point
Page | 88
4. Then it is placed in another outer tube to provide sufficient air space between them. 5. This whole system is placed in the vessel containing freezing mixture, whose
temperature is about 5 degree below the freezing point of the pure solvent. 6. The space between the two test tubes serves as an air jacket and prevents direct heat
conduction and consequent super cooling. 7. On placing the whole system in the freezing mixture the temperature of the solvent in
the inner tube gradually falls and due to super cooling, it gives true freezing point of the pure solvent.
8. On rapid stirring, the solid begins to separate, the temperature is noted which is the freezing point of the pure solvent.
9. The inner tube is taken out to remelt the solvent. And a known weight of solute is introduced into it. The inner tube is then placed back in into the freezing mixture. The freezing point temperature of the solution is determined.
10. The difference in two readings gives the depression of the freezing point (∆Tf) Substituting the values of the depression of freezing point in the equation, the molecular weight of the solute can be calculated.
11. Using formula, m = 1000 × Kf × w/ W × ∆Tf Where, w = the mass of the solute dissolved in pure solvent W = the mass of the pure solvent M = the molecular weight of the solute ∆Tf = the depression in freezing point Kf = the molal depression constant
Result: The molecular weight of solute ……………
Page | 89
APPENDIX
Table No.1
Surface tension of water
Temperature oC
Surface tension
(dynes/cm) Temperature
Surface tension
(dynes/cm)
0oC 75.64 25oC 71.97
5oC 74.92 30oC 71.18
10oC 74.22 35oC 70.38
15oC 73.44 40oC 69.56
20oC 72.75 45oC 68.00
Table No.2
Density at 25
Liquid Density
Gm/ml or g/cm3
Liquid Density
Gm/ml or g/cm3
Benzene 0.8734 Toluene 0.8725
Water 0.9971 Ethyl acetate 0.8441
Ethyl alcohol 0.7850 Methyl acetate 0.9277
Carbon tetrachloride 1.5844 Acetone 1.0524
Table No. 3
List of Equivalent weights of substances commonly used in Volumetric Analysis
Titration and Substances used Formula Equivalent Weights
Acid – Alkali Titration:
Sodium carbonate Sodium bicarbonate Potassium carbonate Potassium bicarbonate Sodium hydroxide Barium hydroxide Oxalic acid (hydrated) Hydrochloric acid Sulphuric acid
Na2CO3 NaHCO3 K2CO3 KHCO3 NaOH Ba(OH)2⋅ 8H2O H2C2O4⋅ 2H2O HCl H2SO4
M.W./2 = 53.06 M.W./1 = 84.00 M.W./2 = 69.00 M.W./1 = 100.00 M.W./1 = 40.00 M.W./2 = 157.75 M.W./2 = 63.03 M.W./1 = 35.46 M.W./2 = 49.04
Redox Titration:
Potassium permanganate Potassium dichromate Ferrous sulphate (hydrated) Ferrous sulphate (anhydrous) Ferrous amm. sulphate Oxalic acid (hydrated) Iron (ferrous)
KMnO4 K2Cr2O7 FeSO4⋅ 7H2O FeSO4 FeSO4⋅ (NH4)2SO4⋅ 6H2O H2C2O4⋅ 2H2O Fe
* M.W./5 = 31.606
M.W./6 = 49.03 M.W./1 = 278.00 M.W./1 = 152.00 M.W./1 = 392.12 M.W./2 = 63.04 M.W./1 = 55.84
Page | 90
Iodine Titration:
Iodine Hypo (Sod. Thiosulphate hydrated) Anhydrous Arsenious Oxide Copper sulphate (hydrated)
I2 Na2S2O3⋅ 5H2O Na2S2O3 As2O3 CuSO4⋅ 5H2O
M.W./2 = 126.92 M.W./1 = 248.19 M.W./1 = 158.09 M.W./4 = 49.45 M.W./1 = 249.68
Argentometric Titration:
Silver nitrate Silver Sodium chloride Chloride ion Pot. suphocyanide Amm. sulphocyanid
AgNO3 Ag NaCl Cl– KCNS NH4CNS
M.W./1 = 169.89 M.W./1 = 107.88 M.W./1 = 58.45 M.W./1 = 35.46 M.W./1 = 97.17 M.W./1 = 76.12
Table No. 4
Molecular weights and equivalent weights of some common substances
Name Formula Mol. wt. Eq. wt.
ACIDS:
Hydrochloric acid Nitric acid Sulphuric acid Acetic acid Oxalic acid Benzoic acid Salicylic acid Phosphoric acid Perchloric acid Formic acid Succinic acid Malonic acid Pthalic acid Maleic acid Tartaric acid
HCl HNO3 H2SO4 CH3COOH H2C2O4, 2H2O C6H5COOH C6H4(OH)COOH H3PO4 HClO4 HCOOH HOOC(CH2)2COOH HOOCCH2COOH C6H4(COOH)2 HOOC(CH)2COOH (CHOHCOOH)2
36.5 63.0 98.0 60.0 126.0 122.0 138.0 97.97 105.5 46.0 118.0 104.0 166.0 116.0 150.0
36.5 63.0 49.0 60.0 63.0 122.0 138.0 32.66 100.5 46.0 59.0 52.0 83.0 58.0 75.0
BASES:
Ammo. hydroxide Sodium hydroxide Potassium hydroxide Sodium carbonate Potassium carbonate Sodium bicrabonate Zinc hydroxide Glycine Aniline Pyridine Thiourea P. Toludine
NH4OH NaOH KOH Na2CO3 K2CO3 NaHCO3 Zn(OH)2 H2NCH2COOH C6H5NH2 C5H5N Cs(NH2)2 C6H4NH2CH3
35 40.0 56.0 160.0 138.0 84.0 99.3 75 93 79 76 107
35 40.0 56.0 53.0 69.0 84.0 49.6 75 93 79 76 107
Page | 91
SALTS:
Ammonium ceric sulphate Ammonium ferrous sulphate Barium hydroxide Ceric sulphate Cobalt nitrate Copper sulphate Copper nitrate Iodine Potassium chloride Potassium dichromate Potassium hydrogen phthalate Potassium iodate Potassium iodide Potassium permagnate Potassium thiocyanate Potassium persulphate Potassium ferrocyanide Potassium ferricyanide Silver nitrate Sodium carbonate Sodium chloride Sodium oxalate Sodium thiosulphate Sodium acetate Sodium sulphate Sodium salt of EDTA Zinc suphate Ferric amino sulphate Sodium perchloride
(NH4)4[Ce(SO4)4]2H2O FeSO4(NH4)2SO4⋅ 6H2O Ba(OH)2⋅8H2O Ce(SO4)2 Co(NO3)2⋅6H2O CuSO4⋅5H2O Cu(NO3)2⋅6H2O I2 KCl K2Cr2O7 KC8H4O4 KIO3 KI KMnO4 KSCN K2S2O8 K4Fe(CN)6⋅3H2O K3Fe(CN)6 AgNO3 Na2CO3 NaCl Na2C2O4 Na2S2O3⋅5H2O Na(CH3COO)⋅3H2O Na2SO4⋅10H2O [CH3N(CH2COONa) – CH2COOH]2⋅2H2O ZnSO46H2O Fe2(SO4)3(NH4)2SO4 NaClO4⋅H2O
632.60 392.13 315.48 332.24 291.03 249.68 295.65 253.81 74.55 294.19 204.23 214.00 166.00 158.04 97.18 270.31 422.41 329.26 169.87 105.90 58.44 134.00 248.18 136.08 322.19 372.10 269.43 963.00 140.50
632.60 392.13 157.44 332.24 291.03 249.68 295.65 126.905 74.55 49.032 204.23 214.00 166.00 31.606 97.18 135.15 422.41 329.26 169.87 52.995 58.44 67.00 248.18 136.08 322.19 372.1 269.43 963 140.5
Table No. 5
Dissociation Constants of Some Acids and Bases in Water at 25oC
MONO BASIC ACIDS
Acid Formula Ka pKa
Mono basic acids:
Acetic acid Benzoic acid Chloro acetic acid di-chloro acetic acid tri-chloro acetic acid Formic acid EDTA Lactic acid m-nitrobenzoic acid O-nitro benzoic acid P- nitro benzoic acid Hydrocynic acid
CH3COOH C6H5COOH CH2ClCOOH CHCl2COOH CCl3COOH HCOOH CH3(OH)COOH NO2C6H4COOH NO2C6H4COOH NO2C6H4COOH HCN
1.75 × 10–5
6.3 × 10–5
1.4 × 10–3
5.5 × 10–2
2.2 × 10–1
1.8 × 10–4
1 × 10–2
1.4 × 10–4
3.5 × 10–4
6.8 × 10–3
3.6 × 10–4
7.2 × 10–10
4.75 4.21 2.86 1.26 0.65 3.75 2 3.86 3.45 2.17 3.45 9.14
Page | 92
Nitrous acid Phenol Sulpho salicylic acid Barbituric acid
HNO2 C6H5OH SO3HC6H4(OH)COOH CH2(CONH)2CO
4 × 10–4
1.3 × 10–10
2.0 × 10–1
1.04 × 10–4
3.40 9.89 0.30 3.98
DIBASIC ACIDS
Oxalic acid (COOH)2 5.6 × 10–2 PK1 = 1.25 5.4 × 10–5 PK2 = 4.27
Salicyclic acid C6H6(OH)COOH 1.1 × 10–3 PK1 = 2.97 3.6 × 10–14 PK2 = 13.44
Malonic acid HOOCCH2COOH 1.4 × 10–3 PK1 = 2.84 2.0 × 10–6 PK2 = 5.695
Succinic acid HOOC(CH2)CH 1.6 × 10–5 PK1 = 4.2 2.3 × 10–6 PK2 = 5.6
Glutamic acid HOOC(CH2)CH⋅NH2COOH 4.4 × 10–5 PK1 = 4.35 3.8 × 10–6 PK2 = 5.42
Glycine H2N-CH2COOH 4.57 × 10–3 PK1 = 2.34 1.7 × 10–10 PK2 = 9.78
Adipic acid HOOC(CH2)4COOH 7.5 × 10–5 PK1 = 4.12 7.76 × 10–6 PK2 = 5.11
Pimelic acid HOOC(CH2)5COOH 3.1 × 10–5 PK1 = 4.51 3.8 × 10–6 PK2 = 5.42
Suberic acid HOOC(CH2)6COOH 2.9 × 10–5 PK1 = 4.53 3.98 × 10–6 PK2 = 5.40
Maleic acid HOOC(CH)2COOH 1.2 × 10–2 PK1 = 1.92 5.9 × 10–7 PK2 = 6.27
Pthatic acid C6H4(COOH)2 1.2 × 10–3 PK1 = 2.90 3.4 × 10–6 PK2 = 5.42
Fumaric acid HOOC(CH2)2COOH 9.3 × 10–4 PK1 = 3.03 3.4 × 10–5 PK2 = 4.48
Carbonic acid H2CO3 4.3 × 10–7 PK1 = 6.35 5.6 × 10–11 PK2 = 10.25
Citric acid HOOC CH2C(OH)CH2 COOH 8.7 × 10–4 PK1 = 3.06 1.8 × 10–5 PK2 = 4.74
Malic acid HOOC CH (OH) CH2 COOH 3.5 × 10–4 PK1 = 3.46 8.9 × 10–6 PK2 = 5.05
Tartaric acid (CH(OH)COOH)2 9.1 × 10–4 PK1 = 3.04 4.3 × 10–5 PK2 = 4.37
Sulphuric acid H2SO4 Strong acid K2 =1.2×10–2
PK2 =1.920
Sulphurous acid H2SO3 1.3 × 10–2 PK1 = 1.89 6.3 × 10–8 PK2 = 7.20
Phosphoric acid H3PO4 7.6 × 10–3 PK1 = 2.12 6.2 × 10–8 PK2 = 7.21 4.4 × 10–13 PK3 = 12.36
Page | 93
BASES
Bases Formula Ka pKa
Zinc hydroxide Zn(OH)2 4.4 × 10–8 PK1 = 7.35 Ammonia NH3 + H2O 1.75 × 10–5 PK1 = 4.76 Glycine H2N.CH2COOH 1.7 × 10–10 PK1 = 9.78 Methylamine (CH3)NH2 5.01 × 10–4 PK1 = 3.30 Dimethylamin (CH3)2NH 5.12 × 10–4 PK1 = 3.13 Trimethylamin (CH3)3N 8.1 × 10–5 PK1 = 4.13 Ethylamine C2H5NH2 5.6 × 10–4 PK1 = 3.25 Diethylamine (C2H5)2NH 1.26 × 10–3 PK1 = 2.90 Trielhylamine (C2H5)3N 5.65 × 10–4 PK1 = 3.20 Aniline C6H5NH2 3.83 × 10–10 PK1 = 9.39 P-Toluidine CH3-C6H4⋅NH2 1.99 × 10–9 PK1 = 8.7 Benzylamine V6H5CH2NH2 2 × 10–5 PK1 = 4.63 Diphenylamine (C6H5)2NH 6.9 × 10–14 PK1 = 13.16 Pyridine C5H5N 1.5 × 10–9 PK1 = 8.8 Piperidine C5H10NH 1.318 × 10–3 PK1 = 2.88 Urea CO(NH2)2 1.2 × 10–14 PK1 = 13.82 Thiourea C5(NH2)2 1.35 × 10–12 PK1 = 11.87 Anthranilic acid C6H4NH2COOH 1.4 × 10–12 PK1 = 11.86
Nicotine C10H14N2 7 × 10–7 PK1 = 6.49 1.4 × 10–11 PK2 = 10.86
Table No. 6
Specific Conductance of KCl in Aqueous Solutions in mhos.
Temperature oC
Æ
£Ç KCl
Æ
ÈÇ KCl
10 0.00932 0.001995 15 0.01048 0.002243 20 0.01167 0.002501 21 0.01191 0.002553 22 0.01215 0.002606 23 0.01239 0.002659 24 0.01264 0.002712 25 0.01288 0.002765 26 0.01313 0.002819 27 0.01337 0.002873 28 0.01362 0.002927 29 0.01387 0.002981 30 0.01412 0.003036 31 0.01437 0.003091 32 0.01462 0.003146 33 0.01488 0.003201 34 0.01513 0.003256 35 0.01559 0.003312
Page | 94
Table No. 7
pH of Acetic Acid Sodium Acetate Buffers
Vol. of 0.2M
Acetic Acid ml
Vol. of 0.2M
Sodium Acetate ml pH at 25
oC
9.5 0.5 3.42 9.0 1.0 3.72 8.0 2.0 4.05 7.0 3.0 4.27 6.0 4.0 4.45 5.0 5.0 4.63 4.0 6.0 4.80 3.0 7.0 4.99 2.0 8.0 5.23 1.0 9.0 5.57 0.5 9.5 5.89
Table No. 8
Acid – Base Indicators
Indicator pH
range
Color Preparation pKin
Acid Base
Cresol red (acid) 0.2-18 Red Yellow -- --
Thymol blue (acid) 1.2-2.8 Red Yellow 0.01g in 21.5ml 0.01N NaOH and dil. to 250 by water 1.7
Bromophenol blue 3.0-4.6 Yellow Blue 0.1g in 14.9ml 0.1N NaOH dil. to 250 by water 41
Methyl yellow 2.9-4.0 Red Yellow 0.1% in 90% ethanol 3.3
Methyl orange 3.1-4.4 Red Yellow 0.1% in water 3.7
Bromocresol green 3.8-5.4 Yellow Blue 0.1g in 14.3ml 0.01N NaOH dil. to 250 by water 4.7
Methyl red 4.8-6.0 Red Yellow 0.02% in 60% ethanol 5.0
Bromocresol purple 5.2-6.8 Yellow Purple 0.02% in ethanol 6.1
Bromocresol blue 6.0-7.6 Yellow Blue 0.1% in 50% ethanol 7.1
Phenol red 6.4-8.0 Yellow Red 0.1% in ethanol 7.8
Thymol blue (base) 8.0-9.6 Yellow Blue 0.1% in ethanol 8.9
Phenolphthalein 8.2-10.0 Colorless Red 1% in ethanol 9.6
Alizarin yellow R 10.2-12.0 Yellow Red 0.1% in water --
Page | 95
Table No. 9
Solubility of some Sparingly Soluble Salts in g/100 gm
Salt 20oC 25
oC 30
oC 40
oC
Barium chromate 0.0037 -- 0.0046 --
Barium sulphate 0.0024 -- 0.00285 --
Calcium sulphate -- 0.0068 -- --
Mercuric iodide -- 0.0591 -- 0.007
Mercurous bromide -- 3.9 × 10–5 -- 14.5
Mercurous chloride 0.002 -- -- 0.070
Lead chromate 7 × 10–5 -- -- --
Lead chloride 9.0 10.8 12.0 14.5
Lead iodide 0.68 0.76 0.90 1.25
Lead sulphate 0.041 0.045 0.049 0.056
Silver benzoate -- 2.61 -- --
Silver bromide 8.4 × 10–5 0.00014 -- --
Silver chloride 0.0015 0.00195 -- --
Silver chromate -- 0.029 0.036 --
Silver iodide -- -- 3 × 10–6 --
Silver thiocyanate -- 0.00017 -- --
Strontium sulphate 0.114 -- 0.114 --
Thallous chloride 3.3 3.85 4.2 5.2
Table No. 10
Viscosities of Liquids (Centi Poise)
Liquid Temp. oC Cp Liquid Temp.
oC Cp
Water 0 1.787 Chloroform 25 0.542
10 1.307 Ether 25 0.222
20 1.002 Ethylalcohol 30 1.003
25 0.8904 Glycol 20 19.9
30 0.7975 Glycerine 25 9.54
35 0.7194 Methylalcohol 25 00.547
40 0.6529 Phenol 20 12.4 Acetic acid 30 1.04 Toluene 30 0.526 Acetone 25 0.316 Benzene 30 0.564 Carbon Tetrachloride 30 0.843
Page | 96
Table No. 11
Surface Tension (Dynes/cm) of Liquids
Liquid toC S.T.
Water 25 71.97
Acetic acid 20 27.8
Acetone 20 23.7
Aniline 20 42.9
Benzene 20 28.35
Chloroform 20 27.14
Cyclohexan 20 25.5
Ethylalchol 20 22.75
Ether 20 17.01
Glycerol 20 63.4
Methylalcohol 20 22.61
Toluene 20 28.5
Page | 97
Logarithm
0 1 2 3 4 5 6 7 8 9
Mean Differences
0 1 2 3 4 5 6 7 8 9
10 0.0000 0.0043 0.0086 0.0128 0.0170 0.0212 0.0253 0.0294 0.0334 0.0374 0 4 9 13 17 22 26 30 35 39
11 0.0414 0.0453 0.0492 0.0531 0.0569 0.0607 0.0645 0.0682 0.0719 0.0755 0 4 8 12 16 20 24 28 31 35
12 0.0792 0.0828 0.0864 0.0899 0.0934 0.0969 0.1004 0.1038 0.1072 0.1106 0 4 7 11 14 18 22 25 29 32
13 0.1139 0.1173 0.1206 0.1239 0.1271 0.1303 0.1335 0.1367 0.1399 0.1430 0 3 7 10 13 17 20 23 27 30
14 0.1461 0.1492 0.1523 0.1553 0.1584 0.1614 0.1644 0.1673 0.1703 0.1732 0 3 6 9 12 15 19 22 25 28
15 0.1761 0.1790 0.1818 0.1847 0.1875 0.1903 0.1931 0.1959 0.1987 0.2014 0 3 6 9 12 14 17 20 23 26
16 0.2041 0.2068 0.2095 0.2122 0.2148 0.2175 0.2201 0.2227 0.2253 0.2279 0 3 5 8 11 14 16 19 22 24
17 0.2304 0.2330 0.2355 0.2380 0.2405 0.2430 0.2455 0.2480 0.2504 0.2529 0 3 5 8 10 13 15 18 20 23
18 0.2553 0.2577 0.2601 0.2625 0.2648 0.2672 0.2695 0.2718 0.2742 0.2765 0 2 5 7 10 12 14 17 19 22
19 0.2788 0.2810 0.2833 0.2856 0.2878 0.2900 0.2923 0.2945 0.2967 0.2989 0 2 5 7 9 11 14 16 18 21
20 0.3010 0.3032 0.3054 0.3075 0.3096 0.3118 0.3139 0.3160 0.3181 0.3201 0 2 4 7 9 11 13 15 17 19
21 0.3222 0.3243 0.3263 0.3284 0.3304 0.3324 0.3345 0.3365 0.3385 0.3404 0 2 4 6 8 10 12 14 17 19
22 0.3424 0.3444 0.3464 0.3483 0.3502 0.3522 0.3541 0.3560 0.3579 0.3598 0 2 4 6 8 10 12 14 16 18
23 0.3617 0.3636 0.3655 0.3674 0.3692 0.3711 0.3729 0.3747 0.3766 0.3784 0 2 4 6 8 9 11 13 15 17
24 0.3802 0.3820 0.3838 0.3856 0.3874 0.3892 0.3909 0.3927 0.3945 0.3962 0 2 4 5 7 9 11 13 14 16
25 0.3979 0.3997 0.4014 0.4031 0.4048 0.4065 0.4082 0.4099 0.4116 0.4133 0 2 3 5 7 9 10 12 14 16
26 0.4150 0.4166 0.4183 0.4200 0.4216 0.4232 0.4249 0.4265 0.4281 0.4298 0 2 3 5 7 8 10 12 13 15
27 0.4314 0.4330 0.4346 0.4362 0.4378 0.4393 0.4409 0.4425 0.4440 0.4456 0 2 3 5 6 8 10 11 13 14
28 0.4472 0.4487 0.4502 0.4518 0.4533 0.4548 0.4564 0.4579 0.4594 0.4609 0 2 3 5 6 8 9 11 12 14
29 0.4624 0.4639 0.4654 0.4669 0.4683 0.4698 0.4713 0.4728 0.4742 0.4757 0 1 3 4 6 7 9 10 12 13
30 0.4771 0.4786 0.4800 0.4814 0.4829 0.4843 0.4857 0.4871 0.4886 0.4900 0 1 3 4 6 7 9 10 12 13
31 0.4914 0.4928 0.4942 0.4955 0.4969 0.4983 0.4997 0.5011 0.5024 0.5038 0 1 3 4 6 7 8 10 11 13
32 0.5051 0.5065 0.5079 0.5092 0.5105 0.5119 0.5132 0.5145 0.5159 0.5172 0 1 3 4 5 7 8 9 11 12
33 0.5185 0.5198 0.5211 0.5224 0.5237 0.5250 0.5263 0.5276 0.5289 0.5302 0 1 3 4 5 7 8 9 11 12
34 0.5315 0.5328 0.5340 0.5353 0.5366 0.5378 0.5391 0.5403 0.5416 0.5428 0 1 3 4 5 6 8 9 10 11
35 0.5441 0.5453 0.5465 0.5478 0.5490 0.5502 0.5514 0.5527 0.5539 0.5551 0 1 2 4 5 6 7 9 10 11
36 0.5563 0.5575 0.5587 0.5599 0.5611 0.5623 0.5635 0.5647 0.5658 0.5670 0 1 2 4 5 6 7 8 10 11
37 0.5682 0.5694 0.5705 0.5717 0.5729 0.5740 0.5752 0.5763 0.5775 0.5786 0 1 2 4 5 6 7 8 9 11
38 0.5798 0.5809 0.5821 0.5832 0.5843 0.5855 0.5866 0.5877 0.5888 0.5899 0 1 2 3 5 6 7 8 9 10
39 0.5911 0.5922 0.5933 0.5944 0.5955 0.5966 0.5977 0.5988 0.5999 0.6010 0 1 2 3 4 6 7 8 9 10
40 0.6021 0.6031 0.6042 0.6053 0.6064 0.6075 0.6085 0.6096 0.6107 0.6117 0 1 2 3 4 5 7 8 9 10
41 0.6128 0.6138 0.6149 0.6160 0.6170 0.6180 0.6191 0.6201 0.6212 0.6222 0 1 2 3 4 5 6 7 8 10
42 0.6232 0.6243 0.6253 0.6263 0.6274 0.6284 0.6294 0.6304 0.6314 0.6325 0 1 2 3 4 5 6 7 8 9
43 0.6335 0.6345 0.6355 0.6365 0.6375 0.6385 0.6395 0.6405 0.6415 0.6425 0 1 2 3 4 5 6 7 8 9
44 0.6435 0.6444 0.6454 0.6464 0.6474 0.6484 0.6493 0.6503 0.6513 0.6522 0 1 2 3 4 5 6 7 8 9
45 0.6532 0.6542 0.6551 0.6561 0.6571 0.6580 0.6590 0.6599 0.6609 0.6618 0 1 2 3 4 5 6 7 8 9
46 0.6628 0.6637 0.6646 0.6656 0.6665 0.6675 0.6684 0.6693 0.6702 0.6712 0 1 2 3 4 5 6 7 8 8
47 0.6721 0.6730 0.6739 0.6749 0.6758 0.6767 0.6776 0.6785 0.6794 0.6803 0 1 2 3 4 5 6 6 7 8
48 0.6812 0.6821 0.6830 0.6839 0.6848 0.6857 0.6866 0.6875 0.6884 0.6893 0 1 2 3 4 5 5 6 7 8
49 0.6902 0.6911 0.6920 0.6928 0.6937 0.6946 0.6955 0.6964 0.6972 0.6981 0 1 2 3 4 4 5 6 7 8
50 0.6990 0.6998 0.7007 0.7016 0.7024 0.7033 0.7042 0.7050 0.7059 0.7067 0 1 2 3 3 4 5 6 7 8
51 0.7076 0.7084 0.7093 0.7101 0.7110 0.7118 0.7126 0.7135 0.7143 0.7152 0 1 2 3 3 4 5 6 7 8
52 0.7160 0.7168 0.7177 0.7185 0.7193 0.7202 0.7210 0.7218 0.7226 0.7235 0 1 2 3 3 4 5 6 7 8
53 0.7243 0.7251 0.7259 0.7267 0.7275 0.7284 0.7292 0.7300 0.7308 0.7316 0 1 2 2 3 4 5 6 7 7
54 0.7324 0.7332 0.7340 0.7348 0.7356 0.7364 0.7372 0.7380 0.7388 0.7396 0 1 2 2 3 4 5 6 6 7
Page | 98
55 0.7404 0.7412 0.7419 0.7427 0.7435 0.7443 0.7451 0.7459 0.7466 0.7474 0 1 2 2 3 4 5 6 6 7
56 0.7482 0.7490 0.7497 0.7505 0.7513 0.7520 0.7528 0.7536 0.7543 0.7551 0 1 2 2 3 4 5 5 6 7
57 0.7559 0.7566 0.7574 0.7582 0.7589 0.7597 0.7604 0.7612 0.7619 0.7627 0 1 2 2 3 4 5 5 6 7
58 0.7634 0.7642 0.7649 0.7657 0.7664 0.7672 0.7679 0.7686 0.7694 0.7701 0 1 1 2 3 4 4 5 6 7
59 0.7709 0.7716 0.7723 0.7731 0.7738 0.7745 0.7752 0.7760 0.7767 0.7774 0 1 1 2 3 4 4 5 6 7
60 0.7782 0.7789 0.7796 0.7803 0.7810 0.7818 0.7825 0.7832 0.7839 0.7846 0 1 1 2 3 4 4 5 6 7
61 0.7853 0.7860 0.7868 0.7875 0.7882 0.7889 0.7896 0.7903 0.7910 0.7917 0 1 1 2 3 4 4 5 6 6
62 0.7924 0.7931 0.7938 0.7945 0.7952 0.7959 0.7966 0.7973 0.7980 0.7987 0 1 1 2 3 4 4 5 6 6
63 0.7993 0.8000 0.8007 0.8014 0.8021 0.8028 0.8035 0.8041 0.8048 0.8055 0 1 1 2 3 3 4 5 6 6
64 0.8062 0.8069 0.8075 0.8082 0.8089 0.8096 0.8102 0.8109 0.8116 0.8122 0 1 1 2 3 3 4 5 5 6
65 0.8129 0.8136 0.8142 0.8149 0.8156 0.8162 0.8169 0.8176 0.8182 0.8189 0 1 1 2 3 3 4 5 5 6
66 0.8195 0.8202 0.8209 0.8215 0.8222 0.8228 0.8235 0.8241 0.8248 0.8254 0 1 1 2 3 3 4 5 5 6
67 0.8261 0.8267 0.8274 0.8280 0.8287 0.8293 0.8299 0.8306 0.8312 0.8319 0 1 1 2 3 3 4 5 5 6
68 0.8325 0.8331 0.8338 0.8344 0.8351 0.8357 0.8363 0.8370 0.8376 0.8382 0 1 1 2 3 3 4 4 5 6
69 0.8388 0.8395 0.8401 0.8407 0.8414 0.8420 0.8426 0.8432 0.8439 0.8445 0 1 1 2 3 3 4 4 5 6
70 0.8451 0.8457 0.8463 0.8470 0.8476 0.8482 0.8488 0.8494 0.8500 0.8506 0 1 1 2 2 3 4 4 5 6
71 0.8513 0.8519 0.8525 0.8531 0.8537 0.8543 0.8549 0.8555 0.8561 0.8567 0 1 1 2 2 3 4 4 5 6
72 0.8573 0.8579 0.8585 0.8591 0.8597 0.8603 0.8609 0.8615 0.8621 0.8627 0 1 1 2 2 3 4 4 5 5
73 0.8633 0.8639 0.8645 0.8651 0.8657 0.8663 0.8669 0.8675 0.8681 0.8686 0 1 1 2 2 3 4 4 5 5
74 0.8692 0.8698 0.8704 0.8710 0.8716 0.8722 0.8727 0.8733 0.8739 0.8745 0 1 1 2 2 3 4 4 5 5
75 0.8751 0.8756 0.8762 0.8768 0.8774 0.8779 0.8785 0.8791 0.8797 0.8802 0 1 1 2 2 3 3 4 5 5
76 0.8808 0.8814 0.8820 0.8825 0.8831 0.8837 0.8842 0.8848 0.8854 0.8859 0 1 1 2 2 3 3 4 5 5
77 0.8865 0.8871 0.8876 0.8882 0.8887 0.8893 0.8899 0.8904 0.8910 0.8915 0 1 1 2 2 3 3 4 5 5
78 0.8921 0.8927 0.8932 0.8938 0.8943 0.8949 0.8954 0.8960 0.8965 0.8971 0 1 1 2 2 3 3 4 4 5
79 0.8976 0.8982 0.8987 0.8993 0.8998 0.9004 0.9009 0.9015 0.9020 0.9025 0 1 1 2 2 3 3 4 4 5
80 0.9031 0.9036 0.9042 0.9047 0.9053 0.9058 0.9063 0.9069 0.9074 0.9079 0 1 1 2 2 3 3 4 4 5
81 0.9085 0.9090 0.9096 0.9101 0.9106 0.9112 0.9117 0.9122 0.9128 0.9133 0 1 1 2 2 3 3 4 4 5
82 0.9138 0.9143 0.9149 0.9154 0.9159 0.9165 0.9170 0.9175 0.9180 0.9186 0 1 1 2 2 3 3 4 4 5
83 0.9191 0.9196 0.9201 0.9206 0.9212 0.9217 0.9222 0.9227 0.9232 0.9238 0 1 1 2 2 3 3 4 4 5
84 0.9243 0.9248 0.9253 0.9258 0.9263 0.9269 0.9274 0.9279 0.9284 0.9289 0 1 1 2 2 3 3 4 4 5
85 0.9294 0.9299 0.9304 0.9309 0.9315 0.9320 0.9325 0.9330 0.9335 0.9340 0 1 1 2 2 3 3 4 4 5
86 0.9345 0.9350 0.9355 0.9360 0.9365 0.9370 0.9375 0.9380 0.9385 0.9390 0 1 1 2 2 3 3 4 4 5
87 0.9395 0.9400 0.9405 0.9410 0.9415 0.9420 0.9425 0.9430 0.9435 0.9440 0 0 1 1 2 2 3 3 4 4
88 0.9445 0.9450 0.9455 0.9460 0.9465 0.9469 0.9474 0.9479 0.9484 0.9489 0 0 1 1 2 2 3 3 4 4
89 0.9494 0.9499 0.9504 0.9509 0.9513 0.9518 0.9523 0.9528 0.9533 0.9538 0 0 1 1 2 2 3 3 4 4
90 0.9542 0.9547 0.9552 0.9557 0.9562 0.9566 0.9571 0.9576 0.9581 0.9586 0 0 1 1 2 2 3 3 4 4
91 0.9590 0.9595 0.9600 0.9605 0.9609 0.9614 0.9619 0.9624 0.9628 0.9633 0 0 1 1 2 2 3 3 4 4
92 0.9638 0.9643 0.9647 0.9652 0.9657 0.9661 0.9666 0.9671 0.9675 0.9680 0 0 1 1 2 2 3 3 4 4
93 0.9685 0.9689 0.9694 0.9699 0.9703 0.9708 0.9713 0.9717 0.9722 0.9727 0 0 1 1 2 2 3 3 4 4
94 0.9731 0.9736 0.9741 0.9745 0.9750 0.9754 0.9759 0.9763 0.9768 0.9773 0 0 1 1 2 2 3 3 4 4
95 0.9777 0.9782 0.9786 0.9791 0.9795 0.9800 0.9805 0.9809 0.9814 0.9818 0 0 1 1 2 2 3 3 4 4
96 0.9823 0.9827 0.9832 0.9836 0.9841 0.9845 0.9850 0.9854 0.9859 0.9863 0 0 1 1 2 2 3 3 4 4
97 0.9868 0.9872 0.9877 0.9881 0.9886 0.9890 0.9894 0.9899 0.9903 0.9908 0 0 1 1 2 2 3 3 4 4
98 0.9912 0.9917 0.9921 0.9926 0.9930 0.9934 0.9939 0.9943 0.9948 0.9952 0 0 1 1 2 2 3 3 4 4
99 0.9956 0.9961 0.9965 0.9969 0.9974 0.9978 0.9983 0.9987 0.9991 0.9996 0 0 1 1 2 2 3 3 4 4
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