Design of the Experiment – Revised

ENGINEERING CHEMISTRY LABORATORY DESIGN OF THE EXPERIMENTS

By DEPARTMENT OF CHEMISTRY

Maharaj Vijayaram Gajapathi Raj College of Engineering (Autonomous) (Approved by AICTE, New Delhi, and permanently affiliated to JNTUK) Chintalavalasa, Vizianagaram -535005

(A2 REGULATION)

S. No. Contents Page. No.

1 Determination of total hardness of water. 03-06

2 Determination of percentage of CaO in cement 07-16

3 Determination of acid number and saponification number of oil/ 17-25

lubricants.

4 Determination of acid strength in lead acid battery 27-32

5 Adsorption of acetic acid on activated charcoal 33-38

6 pH metric titration of strong acid using a strong base 39-52

7 Condcutometric titration of strong acid using a strong base 53-59

8 Condcutometric titration of weak acid using strong base 60-63

9 Determination of percentage of iron in cement by colorimetry 64-69

10 Potentiometric titration of Fe+2 using potassium dichromate 70-75

11 Preparation of polymer (Thiokol rubber) 76-78

12 Determination of viscosity of a polymer solution using survismeter 79-82

13 Preparation of nano materials( Zn/ Mg- ferrite) 83-87 COMPLEXOMETRY DETERMINATION OF TOTAL HARDNESS OF WATER Objective

● To know the experimentation and process involved in the determination of hardness of water

Outcomes

 By using the above procedure student can carrying out and execute, the process of determination of hardness of water experiment  Student can able to identify the type of water and it’s suitability to various applications

APPARATUS & CHEMICALS REQUIRED: Burette, pipette, volumetric flasks, funnel, conical flasks, wash bottle, with distilled water, burette stand, EDTA solution, zinc sulphate AR(solid), Eriochrome black T (EBT) indicator solution, buffer solution (NH4Cl +NH4OH) of pH 10. Theory: The total hardness of water is generally due to dissolved calcium and magnesium salts. Hardness of water is the expression for the sum of the calcium and magnesium ion concentration in a water sample. The standard way to express total hardness of water is in ppm. Ethylene diamine tetra acetic acid (EDTA) form a complexes with Ca2+ and Mg2+ ions, which are more stable at pH 10. The success of an EDTA titration depends upon the precise determination of the end point. Ordinary indicators fail hence metal ion indicators or metallochromic indicators like SBT (Solochrome Black-T), EBT (Eriochrome Black T) are used. The indicator reacts with Ca+2 and Mg+2 ions to give a wine red color. As EDTA is further added, the free Metal ions (Ca+2 and Mg+2) present in the water sample are first complexed to EDTA forming the most stable M- EDTA complex. The extra drop of EDTA added, forms a pale blue color, which indicates the end point of the titration. EBT Structure

Principle: M2+ + EBT [M-EBT] (Wine Red) M-EBT] + EDTA [M-EDTA] + EBT (Colorless)

S. No Hardness (ppm) Type of water Application 1 up to 50 Soft Industrial 2 51-100 Moderately Soft Laundry 101-150 Slightly Hard 3 151-250 Moderately Hard Agriculture 4 251-350 Hard 5 over 350 Excessively Hard Drinking *Above data is as per WHO and BIS

PROCEDURE:

PART A – Preparation of 0.01M EDTA solution: About 0.845 g of EDTA is transferred into a 250ml volumetric flask through a glass funnel. The substance is dissolved completely in a minimum amount of distilled water and the solution is made up to the mark. OBSERVATIONS:

S. No Contents in the pan Grams Milligrams Total weight(g)

1 W1

2 W2

CALCULATIONS:

W1 = Weight of weighing bottle with EDTA = g

W2 = Weight of weighing bottle after transferring the EDTA = g

Amount of the EDTA transferred = W1–W2 = g

Concentration of the EDTA = x = M

Gram molecular weight (GMW) of EDTA = 338.22 g

PART B –Determination of total hardness of water: 20.0 ml of the water sample is pipetted out into a clean conical flask to this add 3ml ammonia buffer solution and 1 or 2 drops of EBT indicator, this produces wine red color. Titrate with EDTA solution from the burette until the color changes from wine red color to pale blue.

OBSERVATIONS:

S. No Volume of water sample taken in ml Burette Reading Volume of EDTA Initial Final rundown in ml 1 Sample 1 2 Sample 2 3 Sample 3 CALCULATIONS: Volume of water sample taken = 20ml Volume of EDTA rundown =

Total hardness in CaCO3mg/lit is given by the formula: Volume of EDTA X Conc. of EDTA X100.1 X1000 Total hardness of water = ------ppm Volume of water sample titrated REPORT: The given water samples contains the following ppm (parts per million) of total hardness. 1. Sample 1 ……… ppm 2. Sample 2 ……… ppm 3. Sample 3………. ppm SIGNIFICANCE OF THE EXPERIMENT

Hardness of water has a permissible limit for use in laundering purpose and industrial applications, so this experiment assumes importance in the analysis of the water samples from time to time, so that the water has the desired quality parameters. If not, proper rectifications can be made to change the quality of water, based on the requirements. This data is relevant for knowing the quality of water for other purposes in industrial equipment such as boilers, house old equipment, heaters, water pipes etc.

Questions:

Based on the Hardness of water (ppm), would you say your water is soft, moderately hard, hard or very hard? Give a very brief explanation while addressing the following questions.

1. What is hard water? 2. Which causes hardness to water? 3. How is water classified based on the degree of hardness?

4. List the types of hardness present in water 5. State the salts responsible for temporary and permanent hardness of water 6. What is meant by softening of water? 7. What is EDTA? 8. Why is disodium salt of EDTA preferred to EDTA for estimation of hardness? 9. What is EBT ? 10. What is a buffer solution? Give an example. 11. Why is ammonium hydroxide-ammonium chloride buffer added during the determination of hardness of water? 12. Mention the disadvantages of hard water for industrial purpose. 13. List the methods of determining hardness of water. 14. What is the significance of this experiment?

DETERMINATION OF TOTAL CALCIUM OXIDE IN PORTLAND CEMENT BY RAPID EDTA METHOD

OBJECTIVES:

 To determine the amount of total calcium oxide in the given test sample of cement  To compute the percentage of total calcium oxide.  Comparison of result with the industrial standard.  Result conclusion / report writing.

OUTCOMES:

The student will be able to determine the percentage of total calcium oxide in cement sample and can compare the experimental results with industrial standards.

INTRODUCTION

Cement is a binder, a substance that sets and hardens and can bind the other materials together such as sand, bricks and other building blocks.

Cement contains oxides of Calcium, Silicon, Aluminum, Magnesium, Iron and Calcium Sulfate, Sulfur Trioxide and Alkali metals (in small amounts).

(Percentage composition may vary depending on the grade of cement)

These oxides interact with one another in the kiln at high temperature to form more complex compounds. The relative proportions of these oxide compositions are responsible for influencing the various properties of the cement.

Functions of Calcium oxide in the cement are

 It is the principal constituent of the cement.  Presence of lime in a sufficient quantity is required to form silicates and aluminates of calcium.  Excess of lime reduces the strength and makes the cement to expand and disintegrate.  Deficiency in lime causes cement to set quickly and also reduce the strength of the cement.

How do I determine the calcium oxide content in a cement sample?

Calcium oxide in the cement is determined by complexometric titration method.

What is a Complexometric titration?

It is a form of volumetric analysis, in which formation of colored complex is used to indicate the end – point of a titration. Complexometric titrations are particularly useful in determining a metal ions or mixture of metal ions in solution. (Source: Wikipedia).

These titrations are based on complexing a metal ion with a ligand (?). Ligand is an electron donating species (may be of either Organic, Inorganic and either charged or neutral).

Simple complexing agents such as ammonia are rarely used as a titrating agent because a sharp end point cannot be observed due to improper complexation. The ligands having multiple donor sites are used as complexing agents. The common ligands used in the complexometric titrations are Ethylene diamine tetraacetic acid (EDTA), 1, 2-cyclohexane tetraacetic acid (CDTA), Dimethylgyloxime, Nitrilo triacetic acid etc.

The accuracy of these titrations is high and the concentration of the metal ion can be determined even at millimole level. EDTA is one such good complexing agent.

How to determine the end – point of the reaction in a complexometric titration?

The end – point of the complexometric titration is determined by using a metal – ion indicator or complexometric indicator. Commonly used metal – ion indicators are Organic dye stuffs viz. Eriochrome Black – T, Fast Sulphon Black – F, Murexide, Calcein, Patton – Reeder’s indicator, xylenol orange etc.

How EDTA is helpful in determining the Calcium oxide content in cement?

EDTA forms a stable complex with almost all metal ions and this reaction is the basis for general analytical method for determination of these metal ions.

EDTA is ethylene diamine tetra acetic acid has four carboxylic acid groups and two amino acid groups. It is a hexadentate ligand having six donor sites. All the six donor sites will form co- ordinate covalent bonds with the metal ion and form a stable metal complex.

EDTA forms 1:1 complex with metal ion present in the test sample irrespective of the valence of the metal ion. Pure EDTA is insoluble in water and EDTA disodium salt is used instead of EDTA.

EDTA di sodium salt when in aqueous form forms a stable complex with metal ions present in the solution. The reaction is pH sensitive and has to be carried out at a pH value 10.2. Since both the reactants are colorless, a suitable metal – ion indicator is to be employed. Indicator first reacts with the metal ion and form Metal Indicator complex, which is unstable. This reaction mixture when titrated with EDTA solution, EDTA replaces the indicator and form a stable complex with the metal.

Now when the test solution is taken, the metal – ion indicator used is Patton – Reeder’s indicator. The reactions between Calcium, EDTA and Patton – Reeder’s indicator is as follows.

In the first step, Patton – Reeder’s indicator reacts with calcium ion and form an unstable wine red color complex.

In the second step, EDTA reacts with metal indicator complex, replaces the indicator with EDTA molecules.

What is the influence of pH in the above reaction?

EDTA ionizes in four stages (pK1 = 2.0, pK2 = 2.67, pK3 = 6.16 and pK4 = 10.26) and form a stable complex in alkaline solution. If pH is less than 10.26, then complete complexation of EDTA will not take place. The so formed Metal – EDTA complexes are quite stable in ammonical solutions and hence ammonical buffer solution is used to control the pH of the reaction mixture.

How to carry out the experiment?

The experiment involves a series of steps.

Step – 1: Preparation of standard ZnSO4 solution:

Around 1.45g of Zinc sulphate hepta hydrate (ZnSO4. 7H2O) is weighed accurately and transferred into a 100mL volumetric flask using a funnel. This is dissolved in a minimum amount of distilled water and finally made up to the mark.

Step – 2: Preparation of EDTA solution:

Around 4.65g of EDTA disodium salt is weighed accurately and transferred into 250mL volumetric flask using a funnel. This is dissolved in a minimum amount of distilled water and finally made up to the mark.

Step – 3: Preparation of Buffer solution:

17.42g of NH4Cl is weighed accurately and transferred into a beaker. To this 142mL of NH4OH is added, mixed properly and is finally made up to 250mL with distilled water. The pH is adjusted to 10.2 by adding either NH4Cl of NH4OH using a pH meter.

Step – 4: Preparation of 4M Sodium Hydroxide solution:

40g of NaOH is weighed and transferred into a 250mL beaker. This is dissolved in minimum distilled water and finally made up to the mark with distilled water.

Step – 5: Preparation of test Cement sample solution:

1.00g of test cement sample is weighed accurately and transferred into a 250mL beaker. This is dissolved in 20mL of concentrated hydrochloric acid and warmed until it becomes colorless. The solution obtained is filtered to remove silica and then diluted up to 250mL with distilled water.

Step – 6: Standardization of EDTA solution:

10mL of standard ZnSO4 solution is pipetted out into a 250mL conical flask. To this 10mL of distilled water is added. 5mL of pH 10.2 buffer and 2 – 3 drops of EBT indicator is added and then the reaction mixture is titrated with EDTA from the burette until the wine red colored solution changes to blue. The volume of EDTA rundown is recorded and the titration is repeated until concurrent values are obtained and the values are tabulated.

Step – 7: Determination of CaO content in Cement sample:

10mL of cement solution is pipetted out into a 250mL conical flask. 20mL of distilled water is added followed by 10mL of 4M NaOH solution to the flask. The flask is then thoroughly shaken for 2 minutes. 5mL of glycerol, 5mL of diethyl amine and a pinch of Patton - Reeder’s indicator is added. The contents are then titrated with EDTA from the burette until wine red colored solution changes to blue. The readings are then tabulated and then percentage of CaO is calculated.

COMPUTATION OF RESULTS:

1. Concentration of 0.05M ZnSO4. 7H2O Solution:

Observation table:

S. No Contents in the pan grams milligrams Total weight(g)

1 W1

2 W2

Calculations:

W1 = Weight of weighing bottle with zinc sulphate = W2 = Weight of weighing bottle after transferring the zinc sulphate = Amount of the zinc sulphate transferred = W1–W2 =

We know,

Therefore, the molarity of ZnSO4 = X M.

2. Standardization of EDTA:

Observation table:

S. No Volume of ZnSO4 Burette Reading Volume of EDTA taken in mL Initial Final rundown in mL 1 2 3

Calculations:

V1 = Volume of zinc sulphate taken = V2 = Volume of EDTA rundown = M1 = Molarity of zinc sulphate = M2 = Molarity of EDTA = n1 = No. of moles of zinc sulphate = n2 = No. of moles of EDTA =

Concentration of the supplied EDTA solution = Y M.

3. Determination of total calcium oxide:

Observation table:

S. No Volume of test Burette Reading Volume of EDTA sample taken in mL Initial Final rundown in mL 1 2 3

Calculations: V1 = Volume of test sample taken = V2 = Volume of EDTA rundown = M1 = Concentration of total CaO = M2 = Molarity of EDTA = n1 = No. of moles of Cao = n2 = No. of moles of EDTA =

Concentration of the total CaO in test sample solution = Z M.

We know,

Percentage of Total Calcium Oxide:

Discussion and significance of the experiment:

When hydrochloric acid is added to the test sample, the metal oxides are converted into their metal chlorides and they are dissolved in solution. Now, a small quantity of sodium hydroxide is added which precipitates Fe2+, Al3+ ions as their corresponding hydroxides. The Ca2+ ions left in the solution is then determined by EDTA using Patton – Reeder’s indicator.

This experiment helps in significantly determining the total Calcium oxide content in the various cement samples and also helps in grading the cement.

Chemical composition of various grades of ordinary Portland cement

S. No % of CaO Grade

1 66.67 53 2 60.41 43

3 57.5 33 4 37.5 High alumina cement

Analysis and conclusion of Results:

From the available industrial standards, the percentage of Calcium oxide present in the given sample of cement is in between 60 – 66%. The value obtained from the above computations is comparable with the industrial standards, then the cement sample is said to be of a particular grade and quality.

Mapping:

Course Objective Course Outcome Program Outcomes To verify the The student will be able to 1, 2, 4, 5, 6 fundamental concepts determine the total with experimentation. calcium oxide content. This enables him to examine the quality of the cement and can grade the cement.

References: 1. Vogel’s text book of quantitative analysis, 5Th Edition. 2. Text book of quantitative analysis, I.M. Kolthoff, 3rd Edition. 3. Standard methods of chemical analysis, I M Kolthoff and Belcher 4. ASTM – STP 985-EB.18910 – Rapid methods for chemical analysis of Hydraulic cement. 5. Engineering Chemistry, Jain and Jain, 16th edition, Dhanpat Rai Publications.

Questionnaire:

1. What is cement? Ans: Cement is a binder, a substance that sets and hardens and can bind the other materials together such as sand, bricks and other building blocks.

2. What are the constituents of cement? Ans: Cement contains oxides of Calcium, Silicon, Aluminum, Magnesium, Iron and Calcium Sulfate, Sulfur Trioxide and Alkali metals

3. What are the raw materials used in the manufacturing of cement? Ans: The raw materials used in the manufacturing of cement are Calcareous materials and argillaceous materials.

4. What is the main constituent of cement? Ans: The main constituent of the cement is calcium oxide.

5. How do you determine the total CaO content in the cement? Ans: Total CaO content is determined by complexometric titration method.

6. What is the importance of Calcium Oxide in cement? o Excess of lime reduces the strength and makes the cement to expand and disintegrate. o Deficiency in lime causes cement to set quickly and also reduce the strength of the cement.

7. Name the common ligand used in Complexometric titrations? Ans: The common ligand employed is Ethylenediamine Tetra acetic acid (EDTA).

8. Why this titration is called rapid EDTA method? Ans: This method is called rapid EDTA method because, the calcium ions are determined directly and quickly without eliminating the other metal ions.

9. What is EDTA? Give the structure of EDTA. Ans: EDTA is ethylene diamine tetra acetic acid. The structure of EDTA is

10. What is the ligancy of EDTA Ans: Ligancy of EDTA is 6. It has 6 donor sites. Four replaceable hydrogen atoms and two lone pair of electrons in two nitrogen atoms.

11. Name the indicators used in the experiment? Ans: Eriochrome Black – T (EBT) and Patton – Reeder’s indicator.

12. What is the indicator used in the standardization of EDTA? Ans: Eriochrome Black – T (EBT).

13. What is the indicator used in the determination of calcium oxide? Ans: Patton – Reeder’s indicator.

14. Why EBT is not used in the determination of Calcium Oxide? Ans: Because it forms a weak complex with calcium in the pH range of 12 – 14.

15. What is the role of glycerol? Ans: Glycerol is added to get sharp end – point.

16. What is the role of NaOH? Ans: NaOH acts as a masking agent to mask other metal ions.

17. What is the function of diethyl amine? Ans: Diethyl amine is added to maintain the pH of 12 – 14.

18. How cement solution is prepared? Ans: Cement solution is prepared by weighing accurate sample of cement, dissolved in minimum concentrated HCl, warmed and then diluted with distilled water.

19. What is the color change at the end – point of the reactions? Ans: During the standardization of EDTA – Wine red to blue During the determination of CaO – Wine red to blue.

20. What is the importance of pH in the experiment? Ans: EDTA ionizes in four stages (pK1 = 2.0, pK2 = 2.67, pK3 = 6.16 and pK4 = 10.26) and form a stable complex in alkaline solution. If pH is less than 10.26, then complete complexation of EDTA will not take place. The so formed Metal – EDTA complexes are quite stable in ammonical solutions and hence ammonical buffer solution is used to control the pH of the reaction mixture.

DETERMINATION OF ACID NUMBER AND SAPONIFICATION NUMBER OF A TEST SAMPLE OF LUBRICATING OIL

INTRODUCTION

The animal and vegetable products were used by man as lubricants for all purposes and were used in large quantities. These lubricants were rejected because they lack chemical inertness and these were substituted by petroleum products and synthetic materials. These petroleum lubricants are used because they possess the following desirable properties.

 Available in suitable viscosities  Inertness (resistance to deterioration of the lubricants)  Corrosion protection (resistance to deterioration of sliding surfaces), and  Low cost

In all working machines, the surfaces of moving or sliding or rolling parts rub against each other. Due to mutual rubbing of one part against another, a resistance is offered to their movement. This resistance is known as friction. Friction causes a lot of wear and tear of surfaces of moving parts; and a large amount of energy is dissipated in the form of heat, thereby causing a loss in efficiency of machine and also, moving parts get heated up, damaged or sometimes results in seizure (welding of the two surfaces).

Lubricants are the substances used to reduce the frictional force between moving or sliding surfaces and minimize heat generation. The process of reducing the frictional resistance between two moving or sliding surfaces, by the introduction of lubricant is known as Lubrication.

Lubrication is simply the use of a material to improve the smoothness.

Classification of Lubricants:

Lubricants are usually liquids or semi liquids, but may be solids or gases or any combination of solids, liquids and gases.

LUBRICATING OILS

ANIMAL & VEGETABLE OIL MIINERAL or PERTOELUM OIL

 These are esters of higher fatty acids like  These are obtained from distillation of stearic acid, palmitic acid, oleic acid or petroleum and length of carbon chain linolenic acid. varying between 12 to 50 carbon atoms.  They possess unsaturated compounds  They possess mostly saturated

 They possess high oiliness* compounds.

 They are costly  They possess poor oiliness

 They are available at low cost.  They undergo oxidation easily, form gummy products and get thickened when  They are available in abundance. in contact with air.  They are stable at service conditions.  They are used a blending agents with  Their oiliness is improved by the petroleum products. addition of high molecular weight compounds.

The lubricating oils used for commercial purposes are tested for the following properties and their quality is assessed by measuring them.

 Acid Number  Saponification number  Viscosity  Cloud point and pour point  Flash and fire point  Oiliness  Emulsification  Volatility  Decomposition stability  Aniline point  Precipitation number  Specific gravity  Ash content  Mechanical stability.

The acid number is used as a guide to determine the quality and age of the lubricant and the saponification number helps us to identify whether the lubricant under test contains vegetable or animal oil.

ACID NUMNER 1

Determination of acidic constituents is more common and it is referred to as “Acid Number”.

Acid Number is defined as “the number of milligrams of KOH required neutralizing the free acids in 1g of oil”.

Generally free acids are not present in the lubricants, unless refined in a faulty manner. Usually lubricating oils are blended with vegetable or animal oils. These oils when exposed to high temperatures or moisture conditions or under service conditions, easily oxidize, consequently leading to formation of oxidized products, besides gum and sludge formation.

Acidic characters are caused by numerous chemical additives such as organic and inorganic acids, esters, phenolic compounds, lactones, resins, salts of heavy metals, salts of ammonia and other weak bases, acid salts of poly basic acids and additives such as inhibitors and detergents.

The rate of change of acid number is more important. The rapid change is due to excessive degradation at high temperatures, moisture conditions or under service conditions. These oils – fats rancidify or oxidize, where the triglycerides are converted into fatty acids and alcohols causing a change (increase) in the acid number.

Lubricating oil should possess acid number less than 0.1. For example compressor and turbine oils usually run at 0.01 – 0.25 mg of KOH per gram of sample, gearbox and lube oil at 0.1 – 10 mg of KOH per gram of sample and lubricant additives at 20 – 200 mg of KOH per gram.

SIGNIFICANCE AND USE

Fresh and used petroleum products contain acidic constituents as additives or as degradation products formed during service conditions (viz. oxidation products). The relative amounts of these materials are determined by simple acid – base titration.

 Acid number is used as a guide in the quality control of the lubricating oil formulations.  This acid number is a measure of the amount of acidic substance present in the oil – under the test conditions.  It is also sometimes used as a measure of lubricant degradation in service and also used as an indicative to the age of the oil.

SAPONIFICATION NUMBER 2

It is defined as the number of milligrams of KOH required to saponify the fatty material present in one gram of oil. Saponification means to hydrolyze a fat with an alkali to form an alcohol and a salt of fatty acid.

Usually mineral oils do not saponify at all, but vegetable and animal oils do. This method helps us to ascertain whether the oil under test is animal or vegetable oil or mineral oil or a compounded oil containing mineral and vegetable oil.

SIGNIFICANCE AND USE

 Saponification value gives an estimation of non-fatty impurities present in an oil or fat i.e. the extent of adulteration.  Saponification value is used to distinguish between vegetable and animal oil, fatty and mineral oils.  Saponification value is used to identify given fatty oil, because each fatty oil has its own characteristic value.  This is also used in setting the product specifications for lubricants and additive.

DISCUSSION FOR THE DETERMINATION OF ACID NUMBER.

To determine the acid number, pre - weighed sample is taken in a conical flask and a known quantity of standard base is added. The flask along with its contents is shaken vigorously. 2 – 3 drops of phenolphthalein indicator is added and is titrated with a standard acid until the color changes from colorless to pale pink at the endpoint. Initially, the fatty acids present in the sample are neutralized by adding KOH and the excess KOH solution is back titrated with HCl. Another titration is performed in the same procedure without oil sample. The difference in the values of HCl consumed will relate to the amount of acid present in the oil sample. The chemical reaction is as follows.

DISCUSSION FOR THE DETERMINATION OF SAPONIFICATION NUMBER.

To determine the saponification number, pre – weighed oil sample is taken in a conical flask. A known quantity of standard alcoholic KOH solution is added in excess and refluxed for 30 minutes. The excess KOH solution is then back titrated with a standard HCl solution using phenolphthalein indicator until the pink colored solution changes to colorless. A blank titration is also performed in the similar procedure without oil sample. The difference in the two titrations gives the volume of KOH that is reacted.

During the reflux, a part of base is reacted with the acidic constituents that were present in lubricating oil sample and forms soap. The remaining excess base is determined by using acid base titration. The chemical reaction is as follows.

References: 1. ASTM D974-14 - Standard test methods for determination of Acid or Base Number by Color – Indicator titration. 2. ASTM D94-02 – Standard test methods for the Saponification Number of Petroleum Products. 3. Engineering Chemistry, Jain and Jain – 16th Edition - Dhanpat Rai Publications. 4. Engineering Chemistry, O . G. Palanna, 2nd edition, Mc Graw Hill Education series.

ACID NUMBER

Aim: To determine the acid number of given oil sample. Apparatus required: Conical flask, burette, water bath. Reagents required: 1) 0.1M Potassium hydroxide solution (KOH). 2) Oil sample. 3) Ethyl alcohol. 4) Phenolphthalein indicator.

Procedure: Determination of acid value A clean and dry 250mL conical flask is taken and weighed accurately. 5mL of lubricating oil is added and is again weighed. The difference in weights gives the weight of the oil sample taken. 10mL of ethyl alcohol and 2-3 drops of phenolphthalein indicator is added and titrated with 0.1M KOH until colorless solution turns to pale pink. The procedure is repeated until concurrent readings are obtained. Solution in the burette : 0.1M KOH Solution in the conical flask : Oil sample under test + 10mL of Ethyl Alcohol + 2-3 drops of Phenolphthalein Indicator used : Phenolphthalein Color change at endpoint : Colorless to pink Table:

Weight of oil sample taken Burette Readings Volume of KOH solution S. No. in g Initial Final rundown in mL

1 10.0

2 10.0

3 10.0

Calculations:

( ) Acid number of given oil sample =

SAPONIFICATION VALUE

AIM: To determine the saponification value of oil.

CHEMICALS REQUIRED: 1. 0.5M Alcoholic potassium hydroxide solution 2. 0.5M hydrochloric acid solution 3. Phenolphthalein indicator

PROCEDURE: Determination of saponification value: A clean and dry 250mL conical flask is taken and weighed accurately. 5mL of lubricating oil is added and is again weighed. The difference in weights gives the weight of the oil sample taken. 25mL of 0.5M alcoholic KOH is added and the contents in the flask are refluxed for 30 minutes. Now, to this 2 – 3 drops of phenolphthalein indicator is added and is titrated with 0.5M HCl solution until pale pink colored solution changes to colorless. The volume of HCl consumed is recorded. A blank titration is performed by taking 0.5M KOH solution in a 250 mL conical flask. 2-3 drops of phenolphthalein indicator is added and then titrated with 0.5M HCl from the burette until the pale pink colored solution changes to colorless. Solution in the burette : 0.5M HCl Solution in the conical flask : Oil sample under test + 2 + 2-3 drops of Phenolphthalein Indicator used : Phenolphthalein Color change at endpoint : Colorless to pink

CALCULATION:

Weight of the empty conical flask (W1 gm) =

Weight of the conical flask with oil (W2 gm) =

Weight of the oil taken (W1-W2 gm) = Volume of 0.5 M HCL solution used by the blank (a ml) = Volume of 0.5 M HCL solution used by the oil (b ml) =

( ) Saponification value of oil sample =

REPORT: The Saponification value of the given oil sample is ______

Mapping:

Course Objective Course Outcome Program Outcomes To verify the The student will be able to 1, 2, 4, 5, 6 fundamental concepts determine the acid number with experimentation. and saponification number. This enables him to examine the quality of lubricating oil.

Questionnaire:

1. What is lubrication? Ans: The process of reducing the frictional resistance between two moving or sliding surfaces, by the introduction of lubricant is known as Lubrication. Lubrication is simply the use of a material to improve the smoothness. 2. What is a lubricant? Ans: Lubricants are the substances used to reduce the frictional force between moving or sliding surfaces and minimize heat generation. 3. What is an oil? Ans: Oils are esters of higher fatty acids. They may be saturated or unsaturated. 4. How many types of lubricating oils are there and what are they? Ans: Lubricating oils are of two types.

a. Animal or vegetable oils. b. Petroleum or mineral oils. 5. Define acid number. Ans: Acid Number is defined as “the number of milligrams of KOH required to neutralize the free acids in 1g of oil”. 6. Why acidic impurities in oil? Ans: Oils when exposed to high temperatures or moisture conditions or under service conditions, easily oxidize, consequently leading to formation of oxidized products, besides gum and sludge formation. 7. What is the indicator used in the determination of acid number? Ans: Phenolphthalein indicator. 8. Why ethanol is used as a solvent? Ans: If your solvent is not neutral it will interfere in acidity test. The free fatty acids that are present in hydrolyzed form from triglycerides. The neutral solvents helps you find the exact acidity from free fatty acids present in a fat sample. If your solvent is not neutral it may consume more sodium hydroxide (if it acidic) and less if solution is basic, leading to wrong estimation of fee fatty acid analysis. 9. How one can analyze the quality of the oil using acid number? Ans: A lubricating oil should possess acid number less than 0.1. For example compressor and turbine oils usually run at 0.01 – 0.25 mg of KOH per gram of sample, gearbox and lube oil at 0.1 – 10 mg of KOH per gram of sample and lubricant additives at 20 – 200 mg of KOH per gram. 10. What is saponification number? Ans: It is defined as the number of milligrams of KOH required to saponify the fatty material present in one gram of oil. 11. What is saponification? Ans: Saponification is a process that involves conversion of fat or oil into soap and alcohol by the action of heat in the presence of aqueous alkali. 12. Why alcoholic KOH is used in the determination of Saponification number? Ans: KOH is used in 'anhydrous' form such as in ethanolic KOH (KOH dissolved in ethanol) when reactions are water sensitive or when performing a dehydration reaction.

13. What is the indicator used in the determination of saponification value? Ans: Phenolphthalein indicator. 14. What is the difference between the acid number and saponification number? Ans: Acid value refers to the amount of "free" fatty acids in the oil/fat. Saponification value refers to the amount of esters that can be hydrolysed and turned into soap. Both these values are given as mg KOH per g of oil or fat. 15. What is the significance of saponification number? Ans: This methods helps us to ascertain whether the oil under test is animal or vegetable oil or mineral oil or a compounded oil containing mineral and vegetable oil. Also, the magnitude of saponification number gives an idea about the average molecular weight of the fat or oil. Higher the molecular weight of the fat, the smaller is its saponification number. Saponification number also indicates the length of carbon chain of the acid present in that particular oil or fat. 16. What is purpose of ethanol during saponification process? Ans: Firstly it raises the boiling point of the mixture, which speeds up the saponification reaction. Also, it is less polar than water and helps to dissolve the nonpolar fat so that it can react with the potassium hydroxide. 17. Is saponification number same for all oils / fats? Ans: Saponification value is used to identify a given fatty oil, because each fatty oil has its own characteristic value.

DETERMINATION OF STRENGTH OF AN ACID IN Pb-ACID BATTERY

Objectives:

1. Determination of strength of an acid in lead acid battery 2. Comparison/ interpretation of results 3. Conclusions

Outcomes:

The student will be able to determine the acid strength in lead acid battery.

Theory/principal

A storage cell is that which can work both as voltaic cell as well as electrical cell. During its operation it works as a voltaic cell i.e., it supplies electrical energy due to a chemical energy and it becomes discharged. Then during its charging, it works as an electrolytic cell i.e., chemical is produced by supplying g of electrical energy. It is also called as lead storage battery.

Anode : Lead (Pb)

Cathode : Lead dioxide (PbO2)

Electrolyte : Dil. H2SO4

A simple lead acid storage cell

A lead storage battery consists of 6 cells each producing 2V output. To increase the current output of each cell, the cathode plates are joined together and the anode plates are also joined together and connected in series, we get an output of 12V as shown in figure. ….

Electrode reactions during discharge

At anode : Pb → Pb2+ + 2e-

2+ 2- Pb + SO4 → PbSO4

+ - 2+ At Cathode : PbO2 + 4H + 2e → Pb + 2H2O

2+ 2- Pb + SO4 → PbSO4 .

+ 2 Net reaction: Pb + PbO2 + 4H + 2SO4 → 2PbSO4 + 2H2O

Electrode reactions during charging

- 2- At anode : PbSO4 + 2e → Pb + SO4

- 2- + - At Cathode : PbSO4 + 2H2O → PbO2 + SO4 + 4H + 2e

+ 2- Net reaction: 2PbSO4 + 2H2O → Pb + PbO2 + 4H + 2SO4

The PbSO4 forms during discharge is a solid and sticks to the electrode s. So, it is in a position to gain or receive the electrons during electrolysis.

Such type of cell is generally used for electrical vehicles, Automobiles, railways, hospitals, power stations, in telephone exchange, UPS system etc.

An acid–base titration is a method of quantitative analysis for determining the concentration of an acid or base by exactly neutralizing it with a standard solution of base or acid having known concentration. A pH indicator is used to monitor the progress of the acid–base reaction.

When selecting an indicator for acid-base titrations, choose an indicator whose pH range falls within the pH change of the reaction. For example, in the titration of a strong acid with a strong base, the pH quickly changes from 3 to 11. Hence phenolphthalein indicator suits the best for this titration. Selection of indicator depends on the pH range. The following table illustrates different acid base indicators and their pH range.

In a lead acid battery, 20% (V/V) sulphuric acid is used as an electrolyte. Concentration of the acid has a significant effect on the voltage of the battery. The following diagram explains the concept.

From the above figure, it is noticed that, as the concentration of the acid decreases, voltage of the battery drops gradually. Hence maintaining the acid concentration is very important in lead acid batteries. Hence, the present experiment gains its importance in proper maintenance of the battery.

Design of the experiment:

Sulphuric Sodium Oxalic acid acid Hydroxide

As sulphuric acid But sodium hydroxide is a Objective is to is a strong acid, to secondary standard determine determine its solution; we have to concentration of strength we are in standardize it with an acid acid in battery a need of strong (primary standard acid) base

Conduct of experiment

Step1: preparation of oxalic acid solution Step2: standardization of sodium hydroxide solution Step3: Determination of acid strength of Pb- acid battery

Oxalic acid solution (0.025M) is prepared by dissolving 0.315g of the solid in 100 ml volumetric flask.

Sodium hydroxide solution is standardized by transferring 10ml of oxalic acid solution into a conical flask and mixed with 20 ml of distilled water and a drop of phenolphthalein indicator and titrated against sodium hydroxide solution taken in a burette. The readings are tabulated.

OBSERVATIONS:

S. No Volume of oxalic acid taken in ml Burette reading Volume of NaOH Initial Final run down in ml 1

3 CALCULATIONS:

V1 = Volume of oxalic acid taken = V2 = Volume of NaOH rundown = M1 = Molarity of oxalic acid = M2 = Molarity of NaOH = ? n1 = No. of moles of oxalic acid = 1 n2 = No. of moles of NaOH = 2

Concentration of the supplied sodium hydroxide solution = M Acid strength of lead acid battery:

The given test sample solution is made up to the mark using distilled water. 10 ml of this solution is transferred into a conical flask and mixed with 20 ml of distilled water and a drop of phenolphthalein indicator and titrated against sodium hydroxide solution taken into a burette. The readings are tabulated.

OBSERVATIONS:

S. No Volume of the acid taken in ml Burette reading Volume of NaOH Initial Final run down in ml 1

CALCULATIONS:

V1 = Volume of acid taken = V2 = Volume of NaOH rundown = M1 = Molarity of acid = M2 = Molarity of NaOH = ? n1 = No. of moles of acid = 1 n2 = No. of moles of NaOH = 2

Concentration of the supplied test sample solution = M

Conclusions:

Obtained result is compared with the true or standard value of acid concentration in batteries (4-6M). The result obtained in the present experiment is found to be in the range prescribed and hence the battery can be said as to be working efficiently.

Test your understanding:

1. Phenolphthalein is a suitable indicator for the present experiment- justify 2. What is the relation between concentration and specific gravity? 3. Can you study the effect of concentration of acid on the lead acid battery? 4. Explain the phenomenon of neutralization? 5. Write the cause of voltage drop in a battery. 6. Describe the salient features of lead acid battery.

ADSORPTION OF ACETIC ACID ON ACTIVATED CHARCOAL

Objectives

 Study the adsorption of acetic acid on activated charcoal  Study various adsorption isotherms and their patterns

Outcomes:

The student will be able to determine the extent of adsorption of acetic acid on charcoal.

Theory

Adsorption is a process that occurs when a gas or liquid solute accumulates on the surface of a solid or a liquid (adsorbent), forming a molecular or atomic film (adsorbate). It is different from absorption, in which a substance diffuses into a liquid or solid to form a solution. The term sorption encompasses both processes, while desorption is the reverse process. Adsorption is operative in most natural physical, biological, and chemical systems, and is widely used in industrial applications such as activated charcoal, synthetic resins and water purification. Similar to surface tension, adsorption is a consequence of surface energy. In a bulk material, all the bonding requirements (be they ionic, covalent or metallic) of the constituent atoms of the material are filled. But atoms on the (clean) surface experience a bond deficiency, because they are not wholly surrounded by other atoms. Thus it is energetically favorable for them to bond with whatever happens to be available. The exact nature of the bonding depends on the details of the species involved, but the adsorbed material is generally classified as exhibiting physisorption or chemisorption.

Physisorption or physical adsorption is a type of adsorption in which the adsorbate adheres to the surface only through Van der Waals (weak intermolecular) interactions, which are also responsible for the non-ideal behavior of real gases. Chemisorption is a type of adsorption whereby a molecule adheres to a surface through the formation of a chemical bond, as opposed to the Van der Waals forces which cause physisorption. Adsorption is usually described through adsorption isotherms, that is, functions which connect the amount of adsorbate on the adsorbent, with its pressure (if gas) or concentration (if liquid). One can find in literature several models describing process of adsorption, namely Freundlich isotherm, Langmuir isotherm, BET isotherm, etc. We will deal with Langmuir isotherm in more details: Langmuir isotherm in 1916, Irving Langmuir published an isotherm for gases adsorbed on solids, which retained his name. It is an empirical isotherm derived from a proposed kinetic mechanism. It is based on four hypotheses:

1 The surface of the adsorbent is uniform, that is, all the adsorption sites are equal. 2 Adsorbed molecules do not interact. 3 All adsorption occurs through the same mechanism.

4 At the maximum adsorption, only a monolayer is formed: molecules of adsorbate do not deposit on other, already adsorbed, molecules of adsorbate, only on the free surface of the adsorbent.

For liquids (adsorbate) adsorbed on solids (adsorbent), the Langmuir isotherm (Figure 1) can be expressed by:

kc kcA m = 1 max [mol g-1, resp. mol.kg-1] (1) where: m – is the substance amount of adsorbate adsorbed per gram (or kg) of the adsorbent, the unit is mol g-1, resp. mol.kg-1

Amax – is the maximal substance amount of adsorbate per gram (or kg) of the desorbent, the unit is mol g-1, resp. mol.kg-1 k – is the adsorption constant (mol-1 dm3)

c – is the concentration of adsorbate in liquid (mol dm-3)

In practice, activated carbon is used as an adsorbent for the adsorption of mainly organic compounds along with some larger molecular weight inorganic compounds such as iodine and mercury.

Activated carbon, also called activated charcoal or activated coal, is a general term that includes carbon material mostly derived from charcoal. For all three variations of the name, “activated" is sometimes substituted by "active." By any name, it is a material with an exceptionally high surface area. Just one gram of activated carbon has a surface area of approximately 500 m² (for comparison, a tennis court is about 260 m²). The three main physical carbon types are granular, powder and extruded (pellet). All three types of activated carbon can have properties tailored to the application. Activated carbon is frequently used in everyday life, in industry, food production, medicine, pharmacy, military, etc. In pharmacy, activated charcoal is considered to

34 be the most effective single agent available as an emergency decontaminant in the gastrointestinal tract. It is used after a person swallows or absorbs almost any toxic drug or chemical.

Task: Determination of adsorption isotherm of acetic acid on activated charcoal.

Determine the adsorption constant (k) and the maximal amount of acetic acid per gram of charcoal (Amax)

Equipment and chemicals:

6 boiling flasks (250 ml), 6 Erlenmayer’s flasks (250 ml), 6 funnels, 3 burettes (50 ml), 10 titrimetric flasks, 3 pipettes, holders for funnel, holders for burettes, filtering paper, vessels for weighing coal, spoon, rubber stoppers, solution of acetic acid (c = 1 mol dm-3), solution of NaOH (c = 0.2 mol dm-3), activated charcoal, phenolphthalein

Method 1 Prepare aqueous solutions of acetic acid into numbered flasks following the scheme given in the table Table 1. The total volume of each solution is 60 ml. Use flasks fitted with stoppers.

Flask No. V (acetic acid) in ml V Distilled water in ml

Table.1 scheme of acetic acid dilution

2 Transfer 10 ml of the solution from each flask into numbered titrimetric flask. Final volume of acetic acid solution is VA = 50 ml per flask.

3 Determine the actual concentration of acetic acid in flasks by titration in this way: For titration, modify the volume in each titrimetric flask. Take away defined volume of the solution, to obtain in each flask the volume as given in the Table 2.

Titrimetric flask No. volume of solution V in ml Table.2 Volume of acetic acid solutions (V) used for the titration before and after adsorption

4 Add 1-2 drops of phenolphthalein and titrate by NaOH.

5 Record the value of base consumed for each titration in the Table 3.

6 Calculate the actual concentration of acetic acid before adsorption ( ) in the flask No.1-6, respectively according Equation 2:

7. Using practical balance and vessels for weighing coal, weigh 6 portions of activated charcoal, each portion 5 g. The accuracy of weighing must be with accuracy 0.01 g.

8 Put activated charcoal into numbered flasks with stoppers (1 portion per flask).

9 Plug up the flasks, and shake them. Wait for 20 minutes, the process of adsorption is in progress. Mix the mixtures for several times by flasks shaking within this period. Remark: The process of adsorption is a function of time too. It is important to put charcoal into flasks at the same time, to provide adsorption for the same period in each flask.

10 Filter the mixtures into clean and dry flasks. To avoid disturbing effect of adsorption of acetic acid into filtering paper, remove away the first portion of filtration, app. 5 ml.

11 Determine the final concentration of acetic acid ( ) in each of the flasks after adsorption: From each solution, transfer the asked volume into clean and dry titrimetric flask, again following Table 2.

12 Repeat points 4-6, and from the consumed base ( ) from Table 3 determine the concentration of acetic acid ( ) after adsorption according Equation 3:

Finishing experiment, wash carefully used flasks, pipettes, etc.

0 0 Flask No Xi (ml) ci (mol/ l) Xi ci mi 1/ci 1/mi (moles/l) (mmol/g) (l/mol) (g/mmol)

Table.3 Experimental data

References:

KOPECKÝ, F., at al.: Laboratory manual for physical chemistry, Pharmaceutical faculty, Comenius University in Bratislava (1996). - http://en.wikipedia.org/wiki/Adsorptionhttp://www.chemvironcarbon.com - OREMUSOVÁ, J., GREKSÁKOVÁ, O.: Manual for laboratory practice in physical chemistry for students of pharmacy.

Test your understanding

1. Illustrate the application of this experiment to real life examples 2. How do you classify various types of adsorption 3. Collect data on various adsorbents and their use in water treatment 4. Differentiate between Langmuir and Freundlich isotherms 5. What significant information you deduce from isotherms 6. Narrate the effect of temperature on adsorption process 7. Discuss the important applications of activated charcoal 8. Can you prepare charcoal from biological materials? Can they be used for adsorption processes? 9. Name adsorption isotherms other than Langmuir and Freundlich. 10. What happens if the adsorbent is varied?

pH METRIC TITRATIONS

Objective of the experiment:

a. Determination of concentration of a strong acid using a strong base. b. Comparison of the result obtained with true value.

Outcome of the Experiment:

The student will be able to determine the end point of the acid – base reactions and pH of a test sample by using the instrumental method of analysis.

Introduction

What is pH?

The concept of pH was given by Sorensen in 1909. It is a scale to measure the acidity or alkalinity of an aqueous solution. It is defined as the negative logarithm (to base 10) of activity + of hydronium ion (H3O ) concentration of a solution.

i.e., pH = - log a [Hydronium Ion]

At low concentrations, the activity of the hydronium ions is approximately equal to the numerical value of their molarity. Hence the determination of pH is an indication of the hydronium ion concentrations. pH may also be defined as the negative logarithm (to base 10) of the hydrogen ion concentration at low concentrations.

i.e., pH = -log [H+]

According to Arrhenius theory, H+ causes acidic nature and OH- causes alkalinity. Pure water in solution state has a tendency to dissociate into H+ and OH- ions, equal in their number, thus making water neutral. The ionic product of water at 22oC has been found to be 10-14,

[H+][OH-] = 10-14

Since, each water molecule dissociated to give hydrogen and hydroxyl ions, each of their concentration in pure neutral water shall be 10-7.

[H+] = [OH-] = 10-7 Thus for all neutral solutions, their hydrogen (or hydroxyl) ion concentration shall be 10-7 and the corresponding pH value calculated is 7. For acid solutions, the concentration of hydrogen ions shall be greater than 10-7 and hence their corresponding pH value shall be less than 7. Similarly for alkaline solutions, the hydrogen concentration shall be less than 10-7 and hence their corresponding pH value shall be greater than 7 up to 14.

Thus in pH scale from 0 to 14, 0 to 7 indicates acidity, 7 indicates neutrality and 7 to 14 indicates alkalinity of an aqueous solution. The pH values of strong acids lie in 0 to 1 region while that of strong bases lie in 13 to 14 regions.

How to measure pH?

There are several methods with indicator solutions and indicator papers for the instant determination of pH. Even though quick, they are approximate. The best and widely used accurate method for the determination of pH is the measurement of EMF produced by H+ ions around an electrode in the instrument pH meter, it is a suitably modified potentiometer calibrated in terms of pH.

Here the basic principle is that when two electrodes are kept in two solutions having different hydrogen ion concentration and the electrodes are connected to a potentiometer and the solutions via salt bridge, a small potential difference (EMF) can be observed. If the standard EMF produced by hydrogen ion species of a known concentration is known, the EMF can be measured by comparison and calibration. The pH and EMF are related by the equation,

pH = EMF/ 0.00019837 T Where T is the absolute temperature.

How to measure the EMF produced by Hydrogen ion species?

The measurements of pH are carried out with a glass electrode. A glass electrode consists of a thin glass membrane that is responsive to changes in H+ activity. The glass used for construction of the glass electrode is a special glass of approximate composition of SiO2 (72%), Na2O (22%) and CaO (6%). These satisfactorily work over the pH range 0-10. If Na2O is replaced by Li2O, glass electrode can measure the pH values from 0 to 14. If the glass electrode is coupled with another reference electrode such as calomel or silver/sliver chloride electrode as a combined electrode, the hydrogen ion concentration and the pH of an unknown solution can be directly measured as EMF of the cell constituted by the two electrodes.

Contact between the test solution and reference electrode is usually achieved by means of a liquid junction, which forms part of the reference electrode. The pH meter is a high impedance electrometer, calibrated in terms of pH. The glass electrode as a half-cell contains a very thin glass membrane with constant hydrogen ion activity on one side (0.1M HCl) and the variable hydrogen ion activity on the other side (of the unknown solution). The glass electrode is applicable for use in all solutions including those containing oxidizing and reducing agents. The solutions are not contaminated and the sensitivity is up to 0.01%. However, since glass membrane is attacked by strong alkalis like NaOH or KOH, it may not function well beyond pH 10 and loses its sensitivity in such a case.

Glass electrode pH Combination electrode

How does pH meter look like? The pH meter is basically a high impedance meter amplifier that accurately measures the minute electrode voltages and displays the resistants’ directly in pH units. The block diagram the pH meter is as follows.

Calibration of the instrument (why the instrument has to be calibrated?*)

 Changing Characteristics – Over time, aging and coating of pH electrodes can cause changes in their characteristics, and even the most stable electrodes cannot be produced with exactly the same characteristics. Calibration helps to match the current characteristics of the pH meter with the pH sensor in use, compensating for any difference between a pH electrode’s behavior in theory and reality.  Higher Accuracy – A minimum of three standards are required for a calibration curve, and a pH meter cannot be calibrated without a standardized buffer. If pH meter calibration has not been performed properly, the resulting measurements are likely to be inaccurate.  Reduced Drift – Measurement drift is a common issue with pH meters, as with any other instruments using electrodes. While drift from calibrated settings can’t be eliminated or prevented, regular equipment calibration helps you maintain accuracy in measurement results.  Sample Differences – Multiple samples of the same substance can have different characteristics, and calibrating pH meters against standardized buffers helps to prevent membrane-related issues such as differences in ionic strength.

*- https://www.rscal.com/what-is-a-ph-meter-and-why-is-ph-meter-calibration-important/

Instrument calibration for pH meters is generally conducted in two ways:

1. Two-Point Calibration – In this method, a microprocessor-based pH meter calculates the real slope and offset error for the pH electrode. Based on this information, the meter’s mV/pH- equation is then adjusted to match the characteristics of the pH electrode in use. The two calibration points bracket the range of values to be measured, so this method is also known as bracketing calibration. Readings that go past the calibrated range may be displayed with slight deviations from the true value, since they are extrapolated by the pH meter assuming linearity.

2. Multi-Point Calibration – With some pH meters, calibration can be performed for more than two pH values on both sides of the zero point, which in this case is pH 7.00. Calibration at three or more pH values increases the measurement range of the device without recalibration being required.

This is generally done by measuring different buffer solutions with standardized, well- defined values, and then adjusting the pH meter based on any deviations from the buffer’s known pH value.

Calibration of the instrument:

The instrument is calibrated by using two point calibration method. Two buffers of pH = 4.01 and 9.14 (at 30°C) are used for calibrating the instrument.

Methodology:

Model Graphs:

Normal Graph Derivative graph

pH METRIC TITRATION OF STRONG ACID (HCL) AND STRONG BASE (NAOH)

Aim: Determination of concentration of a strong acid solution, hydrochloric acid with a strong base, sodium hydroxide solution, by a pH metric titration method.

APPARATUS & CHEMICALS REQUIRED: pH meter with electrode, burette (50ml), volumetric flask (100 ml), conical flask (100 ml), pipette (10 ml/25 ml), beakers (100 ml), stirrer / glass rod, hydrochloric acid solution, sodium hydroxide solution, oxalic acid AR solid.

PRINCIPLE: The neutralization reaction between electrolytic solutions of HCl and NaOH is + - + - + - H + Cl + Na +OH Na + Cl + H2O

A measured volume of HCl solution is taken in a 100 ml beaker in which the combined electrode is immersed without touching the bottom and NaOH solution is added from the burette, when the above reaction occurs, forming feebly ionized water molecules by the combination of H+ and OH- ions. With the progress of the titration, as the NaOH solution is gradually added, highly mobile H+ ions disappear and are replaced by less mobile Na+ ions causing a decrease in the potential of the solution. This trend of decrease continues till the end point is reached, when all the H+ ions present in solution due to HCl are neutralized by combination with OH- ions and disappear, due to addition of NaOH. Beyond end point, further addition of NaOH solution results in gradual change in potential due to unreacted OH- ions and Na+ ions. At the point of neutralization, the pH value of the solution is constant. A plot of pH on Y-axis measured at each addition versus volume of NaOH solution added, on X-axis gives a smooth curve and the sudden inflexion (jump in pH) indicates the end point. The corresponding value of NaOH added with respect to the inflexion gives the titer value.

How to carry out the experiment?

The following series of steps are involved in the experiment.

Step – 1: Preparation of Standard Oxalic acid (0.1 M) solution:

Around 3.15g of oxalic acid (H2C2O4. 2H2O) is weighed accurately and transferred into a 100mL volumetric flask using a funnel. This is dissolved in a minimum amount of distilled water and finally made up to the mark.

Step – 2: Standardization of Sodium Hydroxide solution supplied:

The stoichiometric equation of the neutralization reaction between sodium hydroxide and oxalic acid is H2C2O4 + 2 NaOH Na2C2O4 + 2H2O

According to the above equation, two moles of sodium hydroxide react with one mole of oxalic acid.

10mL of standard oxalic acid solution is pipetted out and transferred carefully into a 250mL conical flask. To this 10mL of distilled water and 2 drops of phenolphthalein indicator is added. The contents of the conical flask are then titrated with sodium hydroxide solution from the burette until the color changes from colorless to pink. The volume of NaOH rundown is recorded and the procedure is repeated until we get the concurrent readings.

Step – 3: pH metric titration of hydrochloric acid solution with standardized sodium hydroxide solution.

pH metric titrations are usually carried out by taking a test sample solution in a beaker, in which the glass electrode is immersed without touching the bottom. The titrant is added from the

44 burette, with continuous stirring and the stable value of pH is measured after each addition of sodium hydroxide. The pH meter should be switched on 15 minutes before the start of the experiment for a warm up. The pH meter is calibrated by immersing the electrode in a buffer solution of known pH value (usually 4.0/9.0) or it can be calculated by using proper electronic probe of a definite known pH value. 10mL of HCl test sample is pipetted out into a 100mL beaker. Now, to this 40mL of distilled water is added. The contents are then titrated with NaOH by measuring the changes in pH using a pH meter. Two titrations are performed namely, pilot titration and accurate titration. In pilot titration, the NaOH is added 1mL at a time by simultaneous measurement of pH at each time of addition of NaOH. The sudden change in pH from acidic range to basic range is noted and the corresponding volumes are recorded. In the accurate titration, the titrimetric procedure is repeated as above, 0.1mL of NaOH is added from the volumes where a change in pH value is noted. At the equivalence point, the rate of change of pH is high and next 0.1mL of NaOH added will give the end – point. The average value of the end point is calculated by plotting a graph between the volume of NaOH taken on X – axis and the corresponding pH values on Y – axis. For more accurate results, a first derivative graph is plotted between change in volume of NaOH on X – axis and corresponding change in pH values on Y – axis.

Observations and Calculations:

Step – 1: Preparation of Standard Oxalic acid solution:

S. Contents in the pan grams milligrams Total weight(g) No

1 W1

2 W2

CALCULATIONS:

W1 = Weight of weighing bottle with oxalic acid = g W2 = Weight of weighing bottle after transferring the oxalic acid = g Amount of the oxalic acid transferred = W1–W2 = g

W1 - W2 1000 Concentration of the oxalic acid = x = = M GMW of Oxalic acid 100

Gram molecular weight (GMW) of oxalic acid = 126g

Step – 2: Standardization of sodium hydroxide solution:

S. No Volume of oxalic acid taken in ml Burette reading Volume of NaOH Initial Final rundown in ml 1 2 3

CALCULATIONS: V1 = Volume of oxalic acid taken = V2 = Volume of NaOH rundown = M1 = Molarity of oxalic acid = M2 = Molarity of NaOH = ? n1 = No. of moles of oxalic acid = 1 n2 = No. of moles of NaOH = 2

Concentration of the supplied sodium hydroxide solution = M

Step – 3: pH metric titration of hydrochloric acid solution with standardized sodium hydroxide solution.

OBSERVATIONS:

Volume of HCl solution taken = 10 ml Temperature of the solution = oC

Pilot Table:

S.No Volume of NaOH added,ml pH measured 1 2 3 4 5 6 7 8

Accurate titration

S.No Volume of NaOH pH value Change in Change in ∆pH/∆v added in mL volume (∆V) pH value (∆pH) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39

CALCULATIONS:

V1 = Volume of HCl taken = 10 ml V2 = Volume of NaOH required = M1 = Molarity of HCl solution = ? M2 = Molarity of NaOH = n1 = No of moles of HCl = 1 n2 = No. of moles of NaOH = 1

Concentration of the supplied HCl solution = M

The amount of hydrochloric acid present in supplied solution in the 100 ml volumetric flask = Molarity of HCl x 36.5/10 = g

Discussion and significance of the experiment:

pH metric titration is very valuable at low titrant concentrations of the range of 0.0001 M, at which the normal titrimetric methods fail. This method can be used with much diluted solutions or with colored or turbid solutions, in which end point cannot be seen with naked eye. This method can be used in reactions where there is no suitable indicator and has many applications, i.e. it can be used for acid- base titrations. Further, choice of an indicator for titrations weak acids, weak bases and sometimes both, is difficult to be carried out and pH metric titrations are accurate alternative for such experiments. Since the end point is located graphically, any errors creeping in the operation shall stand minimized or averaged out. In acid-base visual titrimetry, location of end point accurately is very important and any error requires repetition of the entire process. For obtaining concurrent values, the experiment may have to be repeated several times. However, in pH metry, no observation of end point is necessary and one single experiment is sufficient as there is no visual location of the end point, as it is located graphically by extrapolation.

APPLICTIONS OF pH METRY:

1. Agriculture 2. Brewing 3. Corrosion prevention 4. Dyeing 5. Jam and Jelly manufacturing 6. Printing 7. Pharmaceutical industry. 8. Dissociation constants of various indicators can be calculated.

Mapping: Course Objective Course Outcome Program Outcomes To verify the The student will be able 1, 2, 4, 5, 6 fundamental to determine the end concepts with point of the acid – base experimentation. reactions and pH of a test sample by using the instrumental method of analysis.

References:

1. Vogel’s text book of quantitative analysis, 5Th Edition, Pearson Education, Singapore. 2. Text book of quantitative analysis, I.M. Kolthoff, 3rd Edition. 3. Standard methods of chemical analysis, I M Kolthoff and Belcher, Van Nostrand Company, USA 4. Text book of Quantitative chemical analysis, D C Harris, W. H. Freeman and Company, New York, USA. 5. Engineering Chemistry, Jain and Jain, 16th edition, Dhanpat Rai Publications

Questionnaire:

1. What is pH? Ans: pH is the negative logarithm of the hydrogen ion concentration.

2. What is the effect of temperature on pH? Ans: The pH value of the solution increases with the increase in temperature.

3. What is the pH of pure water at 25°C? Ans: 7

4. What is the pH of the acidic solution? Ans: Less than 7

5. What does pH of a solution signify? Ans: It signifies the hydronium ion concentration in moles per liter.

6. What is ionic product of water? + - Ans: Kw = [H3O ] [OH ]

7. Does the addition of the salt having a common ion onto weak acid change the pH of the solution? Ans: Yes, the pH of the solution increases.

8. What is the pH scale range? Ans: The pH scale given by Sorensen, ranges from 1 to 14, 1 to 7 is acidic, 7 to 14 is basic and 7 being neutral.

9. How the end point of the titration is obtained in pH metry? Ans: The end point of the reaction is obtained from graph.

10. How do you measure the pH of a solution? Ans: The pH of a given test solution is measured by using a glass electrode.

11. What is the composition of the glass membrane in glass electrode? Ans: The composition of glass electrode is SiO2 (72%), Na2O (22%) and CaO (6%). These satisfactorily work over the pH range 0-10. If Na2O is replaced by Li2O, glass electrode can measure the pH values from 0 to 14.

12. What is the primary standard used in this experiment? Ans: Oxalic acid. 13. Calculate the pH of 10-8 M hydrochloric acid solution 14. Calculate the pH of 0.01 M NaOH solution

CONDUCTOMETRIC TITRATION

CONDUCTOMETRIC TITRATION OF STRONG ACID USING STRONG BASE

Objectives:

a. Determination of concentration of a strong acid using a strong base b. Comparison of the result obtained using true value.

Outcome:

The student will be able to explain the use of instrumental methods in chemical analysis and can determine the concentration of acid using an instrumental method.

Theory/ Principle:

What is conductance? How do you categorize conductance? What are its units?

Conductance (L) is the reciprocal of the electric resistance R: (1) L=1/ R

Conductance unit is 1 −Ω (also called Siemens, S). Conductance of aqueous electrolyte solutions is, as rule in the range of mS or µS.

In order to do conductance measurements, two metal electrodes (Pt) should be immersed in the solution and connected to an alternating current (AC) voltage source (Fig. 1).

Why AC in conductometer?

Describe the advantage of using Pt electrode in conductivity cell.

Under the effect of the potential difference between electrodes, each ion in solution moves to the electrode of opposite charge. According to Ohm’s law, the electric current in this circuit

(including the solution) is: (2)

L depends on both solution properties (i.e. conductivity, k) and geometrical parameter of the cell (i.e. electrode area, A, and distance between electrodes, l, Fig.1) according to the following equation:

L= k (A/l)

What is an electrolyte? How many types of electrolytic conductance are there? What are their units? Which type of electrolytic conductance we measure in these titrations? Narrate the various factors that influence K? The ratio (A/l) (in cm) is termed as the cell constant and can be determined by measuring the conductance (L) of a solution with a known conductivity. Conductivity depends on solution properties, such as ion concentrations, ion charges and ion mobilities1. It is worth noting that H+ and OH- have a much higher mobility compared to other ions. Conductivity is widely used for estimating the overall ion content in various sample of practical interest (Table 2). But conductivity values cannot indicate the concentration of a specific ion in the sample. Ion concentration can be determined by means of condcutometric titration. The reaction between the ion of interest and the added reagent (i.e. neutralization, precipitation, or formation of complex compound) brings about a strong modification of solution conductivity. The reagent should be added in the form of a standard solution and the conductance (or conductivity) is measured as a function of the added volume. This procedure is a condcutometric titration. With a properly selected reaction, the end point of the titration is indicated by a particular point on the titration curve (i.e. conductance vs. added reagent volume).

Sample type Conductivity range 25o C High pressure boilers <0.1 µS/cm to 0.2 µS/cm Demineralized water 1 µS/cm 80 µS/cm Drinking water 100 µS/cm to 1 mS/cm Waste water 85 µS/cm to 9 mS/cm Surface water 100 µS/cm to 10 mS/cm Industrial process water 8 mS/cm to 130 mS/cm Concentrated acids and dyes 85 mS/cm to >1000 mS/cm Table.2 Conductivity measurements of various samples

Describe the significance of Table.2 in your own words.

Cell calibration

250 ml standard 0.1000 M KCl solution should be prepared and used to adjust the measuring instrument according to the KCl solution conductivity at the working temperature (TABLE 3).

HCl titration against NaOH

HCl is a strong acid which dissociates completely in solution:

HCl → H+ + Cl-

Neutralization occurs when NaOH is added:

(H+;Cl- ) + (Na+;OH-)  H2O + Cl- + Na+

Titration of HCl by NAOH leads therefore to a gradual replacement of the fast H+ ions by slower Na+ ions (see Table 1). Accordingly, the conductivity decreases until the end point is reached (Fig. 1). Further, the excess of NaOH brings about a gradual increase of conductivity. The equivalence volume lies at the intersection of the two linear branches of the titration curve. Points in the region of the end point slightly deviates from the linear trend and should be disregarded. Equivalence volume determined according to Fig. 2 will be employed for calculating the amount of HCl in the delivered sample.

Explain the significance of the plot? What additional information do you obtain from the plot?

Analyze the pattern of the plot in your words.

Design of the experiment:

Sodium Strong acid Oxalic acid Hydroxid

As the test sample As sodium hydroxide is a Objective is to is a strong acid, to secondary standard determine determine its solution; we have to concentration of strength we have standardize it with an acid acid in test chosen sodium (primary standard acid) sample hydroxide

Conduct of experiment

 Step-1: Prepare standard oxalic acid solution  Step-2: Standardize sodium hydroxide solution  Step-3: Determination of concentration of strong acid

OBSERVATIONS:

S.No Volume of oxalic acid taken in ml Burette reading Volume of NaOH Initial Final rundown in ml 1

3 CALCULATIONS:

V1 = Volume of oxalic acid taken = V2 = Volume of NaOH rundown = M1 = Molarity of oxalic acid = M2 = Molarity of NaOH = ? n1 = No. of moles of oxalic acid = 1 n2 = No. of moles of NaOH = 2

Concentration of the supplied sodium hydroxide solution = M

Determination of concentration of strong acid by condcutometric titration

The given test Made up to the

sample solution mark using distilled water

25 ml of test sample (HCl) solution+ 25 Titration of strong acid with strong base ml of distilled water using conductometer

The conductance measurements were tabulated. A plot of volume of NaOH versus corrected conductance is plotted. The intersection point of the two straight lines gives the end point for the titration.

Why the volumes of acid and water are equal? What happens if changed?

Analysis of data

S. No Volume of Conductance measured, mho NaOH added, Conductance(C) Corrected Conductance (C1) Why do we use volume correction? ml 1 C (V + 50) C = 50 Explain corrected 1 conductance. Why 2 we need to calculate 3 C’? 4 What happens if C’ is 5 not computed? 6 7 8 9 10 11 12 13 14 15 16 17 18 19

CALCULATIONS:

V1 = Volume of HA taken = 10 ml V2 = Volume of NaOH required =

M1 = Molarity of HA solution = ? M2 = Molarity of NaOH = n1 = No of moles of HA =1 n2 = No. of moles of NaOH = 1

Concentration of the supplied strong acid solution = M Interpretation of results:

The conductance values first decreased and then increased as hydrogen ions of the acid get neutralized readily by the strong base used. When volume of base is plotted against C’, two straight lines with intersection is the result. The extrapolation of the intersection point gives the volume of base consumed for the neutralization of the acid taken.

Conclusions:

The result obtained is compared with the standard/ true value. And deviations from the true value are recorded.

Test your understanding

1. Illustrate the following plot in your own words

2. Try to design an experiment for the above plot.

CONDCUTOMETRIC TITRATION OF WEAK ACID WITH A STRONG BASE

Design of the experiment:

Sodium Weak acid Oxalic acid Hydroxide

As the test sample But sodium hydroxide is a Objective is to is a acid, to secondary standard determine determine its solution; we have to concentration of strength we are in standardize it with an acid acid in test sample a need of strong (primary standard acid) base.

Conduct of experiment

 Step-1: Prepare standard oxalic acid solution  Step-2: Standardize sodium hydroxide solution  Step-3: Determination of concentration of weak acid

OBSERVATIONS:

S. No Volume of oxalic acid taken in ml Burette reading Volume of NaOH Initial Final run down in ml 1

3 CALCULATIONS:

V1 = Volume of oxalic acid taken = V2 = Volume of NaOH rundown = M1 = Molarity of oxalic acid = M2 = Molarity of NaOH = ?

59 n1 = No. of moles of oxalic acid = 1 n2 = No. of moles of NaOH = 2

Concentration of the supplied sodium hydroxide solution = M

Determination of concentration of strong acid by condcutometric titration

The given test Made up to the

sample solution mark using distilled water

25 ml of test sample CH3COOH) Titration of weak acid with strong base solution+ 25 ml of distilled water using conductometer

The conductance measurements were tabulated. A plot of volume of NaOH versus corrected conductance is plotted. The intersection point of the two straight lines gives the end point for the titration. Why the volumes of acid and water are equal? What happens if changed?

Analysis of data

S. No Volume of Conductivity measured, mho NaOH added, Conductance Corrected Conductance (C1) Explain corrected conductance. Why ml 1 C (V + 50) (C) C = we need to calculate 50 C’? 1 2 What happens if C’ is not computed? 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 CALCULATIONS:

V1 = Volume of CH3COOH taken = 10 ml V2 = Volume of NaOH required =

M1 = Molarity of CH3COOH solution = ? M2 = Molarity of NaOH = n1 = No of moles of CH3COOH =1 n2 = No. of moles of NaOH = 1

Concentration of the supplied strong acid solution = M Interpretation of results:

The conductance values gradually increases and then increased as weak acid possess low dissociation in turn low values of C. When volume of base is plotted against C’, two straight lines with intersection is the result. The extrapolation of the intersection point gives the volume of base consumed for the neutralization of the acid taken.

Conclusions:

The result obtained is compared with the standard/ true value. And deviations from the true value are recorded.

Test your understanding

1. Differentiate between classical method of titration and instrumental method. 2. Narrate the applications of this experiment. 3. What happens if DC source is used for the instrument? 4. Discuss the principle behind working of conductometer. 5. Explain the effect of temperature on electrolytic conductance. 6. Give the advantage of condcutometric titration. 7. Study the effect of dilution on conductance measurement 8. Write various applications of conductometry. 9. As an engineer why this experiment is significant. 10. Can you calculate conductance of acid at a given volume say 1.26 ml? 11. Illustrate the following plot in your own words

DETERMINATION OF PERCENTAGE OF IRON IN CEMENT SAMPLE BY COLORIMETRY

Experimental Design Process

Define a problem

Identification of objectives

Theory/ principal

Experimental set up

Conduct of experiment and collection of data

Analyze data

Interpret results

Verification of results with standards

Objective (s):

1. Determination of amount of iron in the given test sample 2. Computation of percentage of iron 3. Comparison of the result with true value 4. Report writing/ conclusions

Outcome:

After conducting the experiment, the student will be able to determine the percent of iron cement and categorize different graded cement based on standard grades of cement

Theory / principle:

Colorimetric method is an instrumental method by which concentration of any test solution (colored solutions) [what if not colored?] can be determined by measuring its absorbance at a specific wavelength [why specific wavelength?]. [How concentration and absorbance are related to each other?]. For illustration, a series of solutions of

varying concentration of KMnO4 may be prepared and the relation between concentration and intensity of color can be established (Fig.1).

Fig.1 Relation between concentration and color intensity

This relation has been explained by Beer- Lamberts law. The law states the concentration of a chemical is directly proportional to the absorbance of a solution. The relation may be used to determine the concentration of a chemical species in a solution using a colorimeter.

Equation for Beer's Law

Beer's Law may be written simply as:

A = εbc

Where A is absorbance (no units) ‘ε’ is the molar absorptivity with units of L mol-1 cm-1 (formerly called the extinction coefficient) ‘b’ is the path length of the sample, usually expressed in cm ‘c’ is the concentration of the compound in solution, expressed in mol/ L

(Principles of Instrumental Analysis, Skoog, West and Holler)

[Do Beer’s law valid for all conditions? What are its limitations?]

Test your understanding:

A sample is known to have a maximum absorbance value at 490 nm. Its molar absorptivity is 8400 M-1cm-1. The width of the cuvette is 1 cm. A spectrophotometer finds A = 0.70. What is the concentration of the sample?

[Try to study the effect of path length (b) on the absorbance measurements]

Fig.2 Schematic of digital colorimeter

In the present experiment, concentration of iron in a given test sample of cement is to be determined using colorimeter. Beer’s law is valid only for very dilute solutions. When a highly concentrated solution is used for the study, depending on the nature of the test sample, either association or dissociation of the chemical takes places results in the deviations in absorbance measurements. Hence always very dilute solutions are preferred (ppm range) for colorimetric determinations. As we have to determine the concentration of iron and as it is a colorless solution, in carrying out colorimetric experiment we are in need of a chromogen (color generating reagent). It was found in literature (Text Book of quantitative analysis, Vogel 5th Edition, pp 690) potassium thiocyanate {KSCN} forms

an instantaneous, stable, clear red colored compound. Strong acids {HCl or HNO3 of concentration 0.05-0.5M} should be present to suppress the hydrolysis of iron. [Why not sulphuric acid?]

+3 - 3-n Fe + n SCN [Fe(SCN)n]

Design of the experiment:

 Preparation of a stock solution of iron, potassium thiocyanate solution(20%) and a series of varying concentrated solutions of iron, by transferring 2.5ml, 5ml, 7.5ml, 10ml of the stock solution into 25 ml volumetric flasks and each of the flask mixed

with 3ml of HNO3 and 4 ml 20% of KSCN. The resulting solution is made up to the mark.  Preparing a calibration curve  Determination of iron in test sample

Conduct of experiment. Absorbance measurements of each of the red color compound developed, as mentioned above, were recorded and tabulated. Collect your test sample (cement sample solution),

mix with 3ml of HNO3 and 4 ml 20% of KSCN. The resulting solution is made up to the mark using distilled water. Record the absorbance for the test sample also and tabulate the reading.

S.No Concentration Absorbance 1 c1 0.1 0.40

0.35 2 c2 0.2 0.30 3 c3 0.3 0.25 4 c4 0.4

absorbance 0.20 5 test sample 0.25 0.15

0.10

1.0 1.5 2.0 2.5 3.0 3.5 4.0 concentration in mol/L

From the plot concentration of iron in the test sample (cement) can be determined.

Amount of iron in the sample is concentration X gmw X 25/1000

% of iron in the test sample is calculated using the formula

(Amount of iron/ weight of cement sample) X100

Interpretation of results

The following Table gives the details of chemical composition of cement

Ingredient Percentage in cement

Lime 60-65

Silica 17-25

Alumina 3-8

Magnesia 1-3

Iron oxide 0.5-6

Calcium Sulfate 0.1-0.5

Sulfur Trioxide 1-3

Alkaline 0-1 Source: Indian standards ( IS 2296), Engineering Chemistry, Jain and Jain, 16th Edition

From the standard table, iron (iron oxide) in cement sample must be in the range 0.5-6.

Iron in cement sample provides

 Iron oxide imparts color to cement.  It acts as a flux.  At a very high temperature, it imparts into the chemical reaction with calcium and aluminum to form tricalcium alumino-ferrite.  Tricalcium alumino-ferrite imparts hardness and strength to cement.

In the light of the above facts, the present experiment gains importance.

S. No % of Iron in Cement Grade 1 4.94 53 2 3.56 43 3 3.4 33 4. 13.1 High alumina cement Chemical composition of different grades of cement

The role of iron oxides (FeO, Fe20 3) is discussed extensively in the special literature. It is decisive in the development of the melt phase of the clinker and in reducing the

temperature of clinker-forming reactions. The CaO-Al2O3- Fe2O3 system includes many compounds affecting composition and development of Portland clinker. From the viewpoint of clinker chemistry only the crystal compound series likely to develop

between the incongruous dicalcium ferrite (C2F) and different calcium aluminates,

primarily C2Al, is of interest. Among these crystal compounds the brown millerite (C

4AF) holds a notable place. In the development of the cement quality, ferric compounds are only of significance by facilitating calcination and reducing the burning temperature. This feature of ferric compounds gets an increasing actuality by improving the energy balance of clinker combustion.

Conclusions by student:

From the standards it is noticed that, the presence of 0.5-6% of iron oxide in cement is significant for hardness and its strength. In the present experiment, the results showed that the test sample has 4% (for example) of iron oxide in it. As this value is in the range given by standards, it can be concluded that the test sample is of good quality.

Program specific applications:

S. No Name of the Program Application 1 CIVIL, CHEM and MECH This method (colorimetry) can be applicable for the determination of iron in various ores, metals and alloys (steel)

Mapping of the experiment outcome with PO:

The student will be able to apply knowledge of science in conducting and designing experiments; analyze and interpret data; solve engineering problems and communicate effectively.

References:

1. ESTIMATION OF FERRIC ION IN THE GIVEN SOLUTION.mp3 2. Vogel’s text book of quantitative analysis, 5Th Edition. 3. Text book of quantitative analysis, I.M. Kolthaff, 3rd Edition. 4. IS Standards 2269, BIS, India. 5. Engineering Chemistry, Jain& Jain, 16th Edition

POTENTIOMETRIC TITRATION OF Fe2+ WITH POTASSIUM DICHROMATE

Objectives

After performing this experiment you will be able to: • Discuss how the potential changes with relative concentration of oxidised/reduced from, • perform a redox titration of ammonium iron (II) sulphate using potassium dichromate as oxidizing agent, • determine the equivalence point of the redox titration by plotting titration curve using potential change values and amount of oxidizing agent added during titration, • estimate the strength of iron (II) ions in the given solution, • state the advantages of potentiometry in redox titrations, and • practice the precautions while performing a potentometric titration experiment.

Theory/ Principle:

This is an example of redox titration and is based on the oxidation-reduction reaction between the titrand and the tirant. Here the end pint is detected using a potentiometer. From the Nernst equation, you know that the potential of a given reaction will depend on the relative concentration of oxidised/reduced from. During their titration, the solution potential changes due to the change in the concentration of oxidised/reduced form. At one stage, where either of the form is absent i.e. at the end point, there is a sharp change in potential. Potassium dichromate is an oxidising agent and in acid medium; it follows the half reaction to give Cr (III) as the reduction product.

Typical circuit diagram for potentiometer

Two electrodes are used in this experiment namely calomel electrode ( reference electrode) and platinum electrode (Indicating electrode)

Calomel electrode is a type of half-cell in which the

electrode is mercury coated with calomel (Hg2Cl2) and the electrolyte is a solution of potassium chloride and saturated calomel.

Typical calomel and platinum electrodes

Potentiometric titration of Mohr’s salt solution using potassium dichromate and its first derivative plot

Procedure 1. Using 100 ml volumetric flasks prepare of 0.02 M potassium dichromate solution and 0.10 M mmonium iron (II) sulphate solution. You may have to add sufficient amount of dilute acid to prepare ammonium iron (II) sulphate solution. 2. Take 25 ml of given Fe2+ solution and add 25 cm3 dilute sulphuric acid and 50 ml of distilled water in a 250 ml beaker. 3. SCE is used as the Reference electrode. Platinum metal foil, dipped in Fe2+ solution is used as the indicator electrode. 4. Carry out necessary connections as shown in Fig. 3.1. 5. Standardize the potentiometer using a standard cell before replacing it with the working cell. 6. Add 2 cm3 of 0.02 M potassium dichromate solution from burette, operate the magnetic stirrer for 2 minutes, stop it, wait it for 1 minute and measure the E.M.F. 7. Repeat the above step, each time adding two more cm3 of potassium dichromate at a time and go on noting the E.M.F. after each addition. 8. When the volume reached near about 1 cm3 of the expected equivalence point (approximate), add the solution from burette in 0.5 cm3 installments and note the potential each time. 9. Continue adding these installments even after the equivalence point (This can be easily observed from the change in measured potential). The change becomes very small. Continue for another 5-6 additions. Note the potential readings. 10. Record the observations in the Observation Table 1.Follow the same procedure for plotting the graphs and locate the equivalence point as given in Experiment No. 3.

Volume of the sample of ferrous ammonium sulphate taken = 10ml Temperature = oC

Pilot table:

S. No Volume of Potassium dichromate in mL EMF (V)

Accurate Table:

S.No Volume of potassium dichromate in ml EMF ∆E ∆V (V) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

17. 18. 19. 20. 21. 22. 23. 24. 25.

A graph is plotted to the volume of potassium dichromate added on X-axis and corresponding potential in mV measured on Y-axis to obtain a smooth curve through the obtained points. The steep upward jump in potential in the curve causing a change of 200 – 300mV in the EMF is the point of neutralization and the corresponding volume of potassium dichromate gives the volume required to oxidize the 10ml of ferrous ammonium sulphate taken. The concentration of ferrous ammonium sulphate is calculated as below.

CALCULATIONS:

M1= Concentration of ferrous iron solution = ?

V1 = Volume of ferrous iron solution pipetted out = 10.0 ml

n1 =No. of moles of ferrous Iron = 6

M2 =Concentration of potassium dichromate solution =

V2 = Volume of dichromate solution rundown =

n2 =No. of moles of dichromate = 1

Concentration of ferrous iron present in given 100 ml of solution (M1) = Amount of ferrous iron [Iron (II)] present in given 100ml solution =

Atomic weight of iron = 55.86

RESULT: The amount of ferrous iron [Fe (II)] present in a given sample 100 ml = g

Test your understanding:

a. What happens if the titrant and titrate are reversed? b. Explain the plot in your own words. c. Plot the second derivative plot of the same titration. d. Describe the significance of the experiment in engineering. e. How do you calibrate a potentiometer? f. Why we add acid in the titration mixture? g. Can you do the same titration using potassium permanganate? Predict the chemical reaction and explain. h. Is it possible to do the same titration by using indicator? If so name the indicator and color change at end point. i. You are given with a sample of iron ore. Can you determine the amount of iron in the sample using this method?

PREPARATION OF THIOKOL RUBBER

Objectives: By doing this lab the student will able to

 Demonstrate the synthesis of a polysulfide rubber  Calculate the weight %-yield  Determine the density of product,  Determine the melting point of product  Determine the solubility of product in various solvents,  Write the repeat unit of the product,  Describe the end product uses of this polymer

Theory: Thiokol rubber (poly sulphide rubber) is an elastomer which is obtained by copolymerization of 1,2-dichloroethane and sodium tetra sulphide. This rubber cannot undergo vulcanization and it cannot form a hard rubber. It possesses extremely good resistance to mineral oils, fuels, oxygen, solvents, ozone, and sunlight and possesses low abrasion resistance

Properties: The properties of the polymers depend on the length of the aliphatic groups and the number of sulphur atoms present in it. The polymer behaves like elastomer when four sulphur atoms, or greater, are present per monomer; lower amounts of sulfur give a hard, brittle resin. Some of the important properties are:

(i) Thiokol is resistant to the action of oxygen and ozone. (ii) It is also resistant to the action of petrol lubricants and organic solvents. (iii) Thiokols outstanding resistance to swelling by organic solvents but benzene and its derivatives cause some swelling. (iv) Thiokol films are impermeable to gases to a large extent. (V) But Thiokol has poor heat resistance and low tensile strength. The odour of thiokol is very bad. It tends to lose its shape under continuous pressure.

Uses (i) Thiokol mixed with oxidizing agents is used as a solid fuel in rocket engine. (ii) It is used to engine gaskets and other such products that come into contact with oil. (iii) Thiokols are used for hoses and tank lining for the handling and storage of oils and solvents.

Aim: To prepare the sample of thiokol rubber by co-polymerization. Apparatus required: 100 ml Breaker, 100 ml Measuring cylinder, glass rod. Reagents required: 1,2-Dichloro ethane (ethylene dichloride), Sodium hydroxide, Powdered sulfur.

Procedure: Dissolve 2g of NaOH in 50-60 ml of warm water taken in a 100 ml beaker. Boil the solution and add 4g of sulfur powder in lots with constant stirring till the yellow solution turns deep red color. Cool it to 60-70oC and add 10 ml of 1,2-dichloroethane with stirring. Stir the solution for another 20 min till the rubber polymer gets separated out as a lump. Eliminate the liquid from the beaker to obtain thiokol rubber. Wash it water under the tap and dry the sample.

Record the physical properties of the product

Find the dry mass of the product and determine the %-yield

Determine the density of the polymer

(optional) Measure the melting point of the product.

Labeled 5 test tubes: acetone, conc. nitric acid, conc. sulfuric acid, 6 M sodium hydroxide,and methyl alcohol. Place 1 mL of the appropriate solvent into each test tube. Add a smallamount of your polymer product. Shake gently to mix. Record your observations

Result: The yield of thiokol rubber obtained is ……………………gm.

Turn in the remainder of the sample to your instructor along with your results and the answers to the questions.

STUDENT REPORT:

1. Will this material bounce? Do its elastic properties resemble that of a rubber band? 2. Does the polymer have an objectionable feature? 3. There are two types of Thiokol produced. Identify and describe these two types Which type did you produced? 4. What is the repeat unit of your polymer product? 5. What can be said about the solubility of this polymer? 6. Under what conditions the product can be used as a coating or sealant for tanks and pipelines? Explain your reasoning 7. Based on your observations (and research), what end product applications would this material fit? 8. What is copolymerization? 9. What is vulcanization? 10. Why thiokol rubber does not undergo vulcanization? 11. Give the properties of thiokol rubber? 12. What is Thiokol rubber also called as? 13. Give the reaction for synthesis of thiokol rubber? 14. Industrial and domestic applications of Thiokol rubber

References:

1. Budavari, Susan, Ed. The Merck Index, Twelfth Edition; Merck Research Laboratories: Whitehouse Station, 1996; p. 1593.

2. Shakhashiri, B.Z. Chemical Demonstrations, Volume 1; The University of Wisconsin Press: Madison, 1983; pp. 245-246

DETERMINATION OF VISCOSITY OF POLYMER SOLUTION USING SURVISMETER OBJECTIVES

 To measure viscosity of a polymer solution using Survismeter

AIM: Determination of viscosity of a polymer solution using Survismeter

APPARATUS AND CHEMICALS REQUIRED: Man Singh Survismeter, polymer solution, stop watch

INTRODUCTION

Currently a survismeter was fabricated based on 2-in-1 technique for surface tension and viscosity data measurement. As the innovations keep going on for betterment in experimental techniques for reducing the consumptions of the time, efforts and materials used in. Survismeter resolve hurdles in obtaining such data for biological solutions (very expensive and get spoiled on exposure to atmosphere ) for the study of the atherosclerosis with minimum operational steps.

MATERIAL AND METHODS

The triple distilled water was used to determine the survismeter constant at desired temperatures. The methanol, ethanol, glycerol, ethyl acetate and n-hexane each of analytical reagent (AR, Merck, India) grade were distilled and used for the measurements. The tetrahydrofuran (THF), dimethylformamide (DMF), dimethylsulfoxide (DMSO) and acetronitrile (AR, Merck, India) are distilled in the lab; for purity, their solution were prepared w/w, in ultra-pure water. The deionized water was triply distilled in KMnO4 and KOH (AR, Merck, India) for removing dissolved CO2 and boiled off for further removal of the dissolved gases, its conductivity was maintained to be 1 ×10— 6 V— 1 cm— 1. Handling and operation

The survismeters were washed very cautiously with freshly prepared chromic acid followed with ordinary and distilled water, the final was made with aqueous acetone. It was dried in oven for 24 h at 120 oC. The reference liquid and solutions were separately filled in the bulb number 8 of both the survismeters with utmost care; sucking of the liquids from bulb number 8 to the bulbs 7, 6 and 5 for flow times and bulbs number 7 and 9 for drop numbers were made very cautiously. The attention was paid to avoid the formation of the air bubble in the liquid at any step during sucking operation as it disrupts the smooth flow. The sucking was made with 100 × 10-3 dm3 capacity syringe connecting the needle of the syringe and the respective upper end of the survismeter limb.

Fig. 1. The schematic diagram of survismeter type I ,

Handling

Usually, survismeters are calibrated with any standard solution, however ultra-pure water is preferred, and their calibration constants are known as survismeter constants and are found in the range of 0.1 m2 s— 2. Unit numbers 4 and 2 are calibrated separately for viscosity and surface tension measurements, respectively.

Measurement of Viscosity

About 15 × 10—3 dm3 of solution of known viscosity value is taken in the 8th bulb through 3rd limb keeping the other ends of the limbs open for pressure control. The survismeter was mounted on the stainless steel stand and kept in thermostat at desired a temperature with F0.018C control, read with Beckman thermometer. The solution was sucked upward from bulb number 8 through syringe. With needle syringe, one end of Teflon/PVC tube is attached and another end is attached to a movable stopper with hole. The stopper is fixed in the socket of the 4th limb and, pulling out the plunger, sucks up the solution to bulb number 5; once filled up, the stopper is removed. During sucking of the solution, the socket of limb 3 is kept open, while that of the 2nd and 1st are blocked with stoppers; for down flow of solution, the syringe and stopper of limb number 1 are taken

79 back. The flow time between upper and lower marks of bulb number 6 is counted. The flow times were recorded for four to five times to ensure the reproducibility in values; similarly, the time data are collected for solutions.

After taking the sample in the reservoir bulb, close the pressure limb and the surface tension limb with the operational stoppers. Place the glass adaptor of the syringe pump over the viscosity limb and close it air tightly. Now pull the plunger of the syringe pump slowly and you will see the sample getting into the control pressure unit. Continue pulling the plunger and you will observe that the sample starts to fill into the viscosity unit without going to the surface tension limb. Pulling the plunger slowly helps in maintaining homogeneous pressure inside the Survismeter unit and maintains uniform pressure distribution among the constituents of the liquid sample. The viscosity limb contains two bulbs: the lower bulb is the functional bulb and the upper one is the buffer bulb. Fill the viscosity limb until the buffer bulb is half filled. The role of the buffer bulb is to give the user an extra time to start the analysis. This also gives the constituents of the sample to achieve proper orientation, since extra pressure used to lift the sample in the viscosity limb increased the overall kinetic energy of the system. This matters a lot when the sample is protein or surfactant solution and this buffer zone helps the constituents to attain the same energy state with ordered orientation as they were, before the start of the analysis. After the attainment of equilibrium, open the glass adaptor of the syringe from the viscosity limb. There will be no formation of viscous flow seen in the control pressure unit till now. To start the viscous flow formation, open the operational stopper from the pressure limb. The viscous flow formation is continuous because the flow of the sample is not stopped in the midway and the sample solution continuously flows due to gravity. When the sample solution cross the upper mark located between the functional bulb and the buffer bulb of the viscosity limb, start the digital timer and stop the timer once the sample crosses the lower mark located below the functional bulb and record the viscous flow time taken between the two marks. Repeat the experiment again without taking out the sample from the reservoir bulb to find out the repeatability associated with the experiment. For viscosity calculation, in addition to the viscous flow time recorded for the sample solution, the viscous flow time record between the two marks of the viscosity limb for a reference liquid is also required at the experimental temperature. The solvent in which the sample solution is prepared is generally selected as the reference liquid. For the viscosity limb, at a given temperature, the viscous flow time for a particular reference liquid will always be constant and hence with every sample there is no need to find the viscous flow time for the reference liquid, as it can the taken from previous experiment at that temperature. For example, suppose the experimental temperature is 25 degree Celsius and the sample is 0.1 % weight by weight aqueous solution of any water soluble protein. The reference liquid will be water. The viscous flow time for water can be taken from any previous experiments where the viscous flow time was recorded for water at 25 degree Celsius. Now, to calculate the viscosity of the sample, the viscous flow time of the sample and the reference liquid are now used in the formula as:

Test your understanding

1. Define viscosity. Give its units. 2. Describe the importance of viscosity. 3. Explain the effect of temperature on viscosity 4. What is the importance of a survismeter? 5. Can you compare your result with standard values? Prepare a chart of standard viscosity values of different polymer solutions.

Nanomaterials

Definition: The nanochemistry is explained as the study of the synthesis of materials and properties of materials in the range of nanoscale.

Therefore, 1nm = 10-7 cm OR 1nm = 10-9 m Note: Surface of a material increases with decrease in the particle size Classification of nanomaterials: The types of nanomaterials are explained in the below figure:

1. One dimensional nanomaterial: This type of material has one dimension arrangement of atoms in the nanoscale range. The examples for one dimension nanomaterial are surface coatings and thin films.

2. Two-dimensional nanomaterial: This type of material has two dimension arrangements of atoms in the nanoscale range. The examples for two dimension nanomaterial are biopolymers, nanotubes, and nanowires.

3. Three-dimensional nanomaterial: This type of material has three dimension arrangements of atoms in the nanoscale range. The examples for three dimension nanomaterial are fullerenes.

The significance of nanoscale:

It is observed that the properties are distinct on the nanoscale from those at the larger scale. The properties like quantum, mechanical and thermodynamical become vital at the nano level and these are not seen at the macroscopic level. The nanoscience is based on the fact that along with the physical dimensions of the materials the properties of the materials also change. The transformation in the properties of the materials in these confined spaces is because of the

82 changes in the electronic structure of the materials. The properties of the materials are distinct at the nano level due to the following two reasons:

. Increased surface area . Quantum confinement effect

Properties of nanomaterials: The properties of the nanomaterials depend on the crystal structure and size and few properties of the nanomaterials are explained below:

. Physical properties . Mechanical properties . Magnetic properties . Optical properties . Thermal properties . Surface area . Catalytic activity . Electrical properties . Semiconductors . Superconductors 1. Physical properties: In the nanoscale range, surface area to volume enhances. This enhancement transforms the surface pressure and results in a change in the interatomic spacing. This feature is explained in the below figure: 2. Mechanical properties: The nanomaterials have the application on both the low temperatures and high temperatures. So, the mechanical properties of the nanomaterials are studied at low temperatures and also at the high temperatures. 3. Magnetic properties: The magnetic behavior of the nanomaterials is size dependent. In the tiny ferromagnetic particles, the magnetic properties are distinct from that of the very large material. In the range of nanoscale, magnetic material has a single magnetic domain. As an outcome, the remanent magnetization and coercive field exhibit a strong dependence on the size of a particle. 4. Optical properties: If the particles of a semiconductor are made tiny then the effects of quantum come into play which limits the energies at which the electrons and holes can exist in the particles. 5. Thermal properties: The nanocrystalline materials are expected to have a low conductivity of thermal when compared with the conventional materials. 6. Surface area: Nanomaterials have large surface area and the nanocrystal’s surface area is inversely proportional to the crystal’s size. 7. Catalytic activity: Nanomaterials can be availed as effective catalysts in many reactions.

8. Electrical properties: Nanomaterials exhibit electrical properties in between semiconductors and metals depending on the chirality of the molecules and diameter of the molecules. 9. Semiconductors: Most of the nanomaterials play the role of a semiconductor and these nanowires which have the feature of being a semiconductor are used in making distinct components. 10. Superconductors: Few nanomaterials behave as the superconductors. Applications of nanomaterials: The nanomaterials have its applications in distinct fields and few of them are as follows:

. Material technology . Information technology . Biomedicals . Energy storage . The carbon nanotubes and nanowires are utilized in making many components Synthesis of nanomaterials: The processes used for the synthesis of nanomaterials are explained in the below figure:

Few methods of a top-down process are explained below:

. Ball milling . Plasma arching . Laser sputtering . Vapour destination

Few methods of a bottom-up process are explained below:

. Sol-gel . Colloidal . Electrodeposition

. Solution phase reductions Preparation of zinc ferrite/magnesium ferrite nano particles

Aim: to synthesize zinc/magnesium ferrite nano particles by using sol-gel method

Materials: Magnesium nitrate [Mg (NO3)2.6H2O], Zn(NO3)2.6H2O Ferric nitrate [Fe(NO3)3.9H2O], PVA(Polyvinyl alcohol)PEG(poly ethylene glycol) and Citric acid [C6H8O7]

Principle: Zn(NO3)26H2O + 2 Fe(NO3)2 9 H2O → ZnFeO4 + Volatiles(H2O + CO2 + NO2)

Procedure:

Step-1: Calculation of the molecular weight of the required product

Ex: ZnFe2O4(Zinc ferrite)

Zn → 65.380 x 1 ( No. of atoms in the formula) = 65.38 Fe → 55.840 x 2 = 111.68 O → 15.999 x 4 = 63.996 241.056

Step-2: Calculation of the amount of required metal nitrate of the respective metal ferrite for the desired quantity of metal ferrite

Where 10 is the desired quantity of metal ferrite

Example: Amount of zinc nitrate required for the preparation 10 g of ZnFe2O4 (Zinc ferrite)

[Zinc nitrate hexahydrate (Zn(NO3)2.6H2O)= 297.4815]

= 8.1046 grams

Step-3: Combustion

Dissolve metal nitrates individually in slightly excess volume of solvent than their respective amounts and stir these solutions for 30 minutes for the complete dissolution then add PVA (poly vinyl alcohol) equal amount to metal nitrate, and heat the mixture to 50 oC until it dissolves then mix the solutions and continue the heating at 65oC for 15-30 minutes and 135 oC for further 45 minutes.

Result: The amount of ferrite nano particles obtained:

Test your understanding

1. What is nano material. 2. Explain the classification of nano materials? 3. Why nano materials gain prominence in these days? 4. Explain the applications of nano materials and nano ferrite materials. 5. Mention any two properties of nano materials.