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Engineering Chemistry

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Molecule

The universe is made up of matter. The matter is anything that occupies space and has mass or weight. The matter is made up of fundamental particles called elements. To further understand chemistry the basic concepts are essential. An Atom is the smallest particle of an element which can take part in chemical reactions. It may or may not be capable of independent existence.

A Molecule is defined as the smallest particle of matter (element or compound) which can exist independently. It may be made up of two or more atoms of the same or different elements. For example Molecules of Chlorine, Oxygen, Nitrogen contain only two atoms.

Molecules of sulphur dioxide, carbon dioxide etc., are built up by more than two atoms and involve the combinations of atoms of different elements. Thus in the molecule of carbon di oxide, one atom of carbon and two atoms of oxygen have united.

H O H O H2 O2

1.1.2 Molecular Formula

Example:

Significance of molecular formula

1.1.3 Molecular Mass

Calculation of Molecular Mass

Molecular formula is the short form of symbolic representation of one molecule of an element or a compound.

Molecular formula of Oxygen is O (element). Molecular formula of Water is H O (compound)

1. Molecular formula represents one molecule of an element or a compound.

2. It shows the elements present in one molecule.

3. It gives the number of atoms of each element present in one molecule.

4. It helps to calculate the molecular mass.

5. It gives the ratio between the masses of the elements to form the substances.

Molecular mass of an element or a compound is the ratio between the mass of one molecule of the element or the compound and the mass of 1/12 part of a carbon Atom. Molecular mass can be calculated as the sum of total atomic mass of each element present in one molecule of an element or a compound. 1/12 part bymass of a carbon atom Mass of one molecule of an element or a compound a compound of an element or Molecular Mass

Example

1.1.4 Mole

Molecular mass of O = Atomic mass x No. of atoms = 16x2=32

Molecular mass ofNH = (14 x 1) + (1 x 3) = 17

If the molecular mass is expressed in grams, then it is called gram molecular mass or mole.

Example

Problem: 1

Molecular mass of O = 32 Gram molecular mass of O (or)

1 mole of O = 32 g

Problem: 2

How many moles are present in 8.5g of Ammonia?

How many moles are represented by 4.4 g. of CO ? 

Atomic mass of Carbon 12

Atomic mass of Oxygen 16

Molecular mass of CO (12 x 1) (16 x 2 )

12 32 44 Mass

No of moles Molecular mass 4.4

1.1.5 Avogadro's Hypothesis

Applications of Avogadro's Hypothesis

How many grams of SO are present in 0.4 moles of SO?

What is the mass of 3 gram atoms of Bromine?

1 gram atom of Bromine = 80g

3 gram atom of Bromine = 80 x 3 = 240g.

How many moles of Carbon atoms are present in three moles of C H ?

1 mole of C H has 3 moles of carbon 3 moles of C H will have 9 moles of carbon atoms.

Avogadro's Hypothesis states that “Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules”

1. It helps in deduction of the relationship between vapour density and molecular mass of gas.

2. Avogadro's Hypothesis helps in determining the relationship between weight and volume of gases.

3. It enables us to calculate the density of a gas.

4. Avogadro's Hypothesis has established the truth about Dalton's atomic theory by making a clear distinction between atoms and molecules.

5. Avogadro's Hypothesis helps in determining the atomicity of gases.

6. It helps in the calculation of molecular mass of hydrocarbons.

1.1.6 Relationship between molecular mass and vapour density

Vapour density Vapour density of a gas is the ratio between the mass of certain volume of the gas and the mass of the same volume of hydrogen under the same conditions of temperature and pressure. Let there be 'n' molecules in certain volume of a gas Let n = 1 Mass of 1 molecule of hydrogen = 2 2 x vapour density = Mass of one molecule of a gas Therefore, Molecular mass = 2 xVapour Density Molar volume is the volume occupied by one gram molecular mass or one mole of a gas at S.T.P. Molar volume of the gas at S.T.P is 22.4 litres, i.e. the molar volume is the same for all gases at S.T.P and it is equal to 22.4 litres. According to Avogadro's hypothesis,

i) one molecular mass (i.e. one mole) of every gas occupies 22.4 litres at S.T.P (or N.T.P)

ii) equal volumes of all gases at S.T.P contain equal number of molecules.

Hence it follows that 1 mole of every gas contains the same number of molecules. This number is called the Avogadro's number or Avogadro's Constant. It is denoted by. It has been found to be equal to 6.023 x 10. It can be defined as “the no of atoms or molecules present in one mole of an element or a compound respectively”

1.1.7 Molar volume

1.1.8 Avogadro's number

N 23 Mass of certain volume of a gas at S.T.P Vapour Density

Mass of same volume of hydrogen at S.T.P _ Mass of nmolecules of hydrogen

Mass of n molecules of a gas Vapour Density _ Mass of 1 molecule of a gas Vapour Density

Mass of 1 molecule of hydrogen

Mass of 1molecule of a gas Vapour Density

TEST YOUR UNDERSTANDING

1.2. EQUIVALENT MASS

1.2.1 Introduction:-

1.2.2 Definition

Elements combine among themselves in a definite ratio by mass to form compounds. The term equivalent mass expresses the combining capacity of an element or a compound in terms of mass with reference to some standard.

Consider the formation of hydrogen chloride.

H + Cl HCl Atomic mass 1.008 35.45.

Here the mass of chlorine combining with 1.008 g of hydrogen is 35.45g. Consider the formation of water

2H + O H O. 2 x 1.008 16

Here the mass of oxygen combining with 1.008g of hydrogen is 8g. In the above two cases 35.45g of chlorine and 8g of oxygen are equivalent as they combine with the same mass of H i.e., 1.008g. Expressed without units, the numbers 35.45 and 8 are the equivalent masses of chlorine & oxygen respectively. So, 1.008 parts by mass of hydrogen is taken as a standard. Certain metals which do not combine with hydrogen displace it from dilute acids. Then the equivalent mass is the number of parts by mass of the metal which displaces 1.008 parts by mass of hydrogen. Many metals do not either combine with hydrogen or displace hydrogen from acids. But almost all elements combine with oxygen and chlorine. Hence 8 parts by mass of oxygen and 35.45 parts by mass of chlorine are also chosen as standards.

Equivalent Mass of an element is the number of parts by mass of the element which combines with or displaces 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.45 parts by mass of chlorine or one equivalent mass of any other element.

When the equivalent mass of sodium is 23, it means that 23 parts by mass of sodium combines with or displaces 1.008 parts by mass of H or 8 parts by mass ofO or 35.45 parts by mass of Cl Relationship betweenAtomic Mass and Equivalent Mass Equivalent mass is mere a number. When equivalent mass is expressed in grams, it is called gram equivalent mass or gram equivalent. Equivalent mass of oxygen = 8

Gram Equivalent mass of oxygen = 8g.

In other words, 1 gram equivalent of oxygen= 8g

Therefore 3 grams equivalents of oxygen = 8 x 3 = 24g.

How many gram equivalent mass are present in 36g of Magnesium?

Equivalent mass ofMg = 12.

Gram equivalent mass ofMg = 12g.

(i.e) 12g ofMg = 1 gram equivalent mass

36 g ofMg = 36/12 = 3 gram equivalents mass.

The equivalent mass of an element could be determined by displacement methods and combination methods. The following are the methods to determine the equivalent mass.

1. Hydrogen displacement method.

2. Oxide method.

3. Chloride method.

4. Metal displacement method.

5. Electrolysis method.

Oxide Method

Direct Oxide Method

Principle:

Procedure:

Calculation

There are two methods to determine the equivalent mass of substances by oxide method. They are direct oxidation and indirect oxidation methods. In these methods, a metal is converted into its oxide either directly or indirectly. The process of conversion of metal to its oxide is called “oxidation”. A known mass of the metal is directly converted to its oxide by heating in air. From the mass of oxide formed, the mass of metal which combines with 8 parts by mass of oxygen is calculated and it gives the equivalent mass of the metal. This method is useful to determine the equivalent mass of elements like Magnesium and Calcium.

i) Aclean dry crucible with lid is weighed (W g).

ii) Apiece of the metal is placed in the crucible and weighed again (W g).

iii) The lid is kept slightly open and the crucible is heated.

iv) The metal burns to form the oxide.

v) When the metal is completely burnt, the crucible is allowed to cool to the room temperature.

vi) A drop of water is added and once again heated strongly.

vii) The crucible is allowed to cool in a dessicator and weighed. The heating and cooling are repeated until a constant mass (W g) is obtained.

Mass of the empty crucible = W g.

Mass of the crucible + metal = W g.

Mass of the crucible + metal oxide = W g .

Mass of the metal = (W W ) g.

Mass of the oxygen combined = (W –W ) g

(W –W ) g of oxygen combines with = (W W ) g of metal.

Thus, equivalent mass of the metal , = Mass of oxygen = 48g

Equivalent mass of oxygen = 8

No. of gram equivalents of oxygen =

No of gram equivalents of chlorine = 3

Equivalent mass of chlorine = 35.45

Mass of 3 gram equivalents of chlorine = No of g equivalents x Equivalent mass = 3 x 35.45 = 106.35.

Weight of the metal oxide = 0.80g

Weight of metal = 0.48g

Weight of oxygen = (0.80 - 0.48) =0.32g 

1.3.4 Theories Of Acids and Bases

(1) Arrhenious Theory (1827)

Depending upon the ions produced in the solution, the substances are classified into acids and bases. The acidic and basic nature of matter is more essential to study the chemical reactions. An acid is defined as a substance that gives hydrogen ions (H ) in aqueous solution. A base is defined as a substance that gives hydroxyl ions (OH ) in aqueous solution. The acidic and basic nature of substances can be easily explained by the following theories

1. Arrhenius theory

2. Lowry-Bronsted theory

3. Lewis theory.

According to this theory, an acid is a substance that gives hydrogen ions in aqueous solution. A Base is a substance that gives hydroxyl ions in aqueous solution. Strength of an acid (or) base depends upon the extent to give H (or) OH during ionization.

1. It is valid only in aqueous solution.

2. It is difficult to explain the basic properties of non-hydroxyl compounds like NH and Na CO An Acid is a substance which donates proton (H ) to any other substance. Hence, it is a proton donor.

A Base is substance which accepts proton (H ) from any other substance. Hence, it is a proton acceptor.

1. It explains the acidic and basic characters of a substance only if a solvent like water is present.

2. It does not explain the neutralization of acidic oxides (CO ,SO ) by basic oxides (CaO, BaO) because there are no H ions.

Lewis concept is known as electronic concept because it involves electron pair transfer during base formation.

According to this theory, an acid is s substance that accepts a pair of electrons.

Abase is a substance that donates a pair of electrons. So, base is an electron.

Limitations

(2) Lowry-Bronsted Theory (1923) (Protonic concept) Drawbacks

(3) Lewis Theory (G.N.LEWIS – 1923) (Electronic concept ) electronpair acceptor pair donor

ACID H B

Proton donor Proton acceptor

15 + -2+ -2

NaOH Na + OH

Ca (OH) Ca + 2OH

Here, OH and NH donate pair of electrons. So, they are bases.

H and BF accepts a pair of electrons. So, they are acids.

1. It explains the acid base reactions by way of electron transfer.

2. It gives an idea about the acidic character of substances that do not contain hydrogen. The acidic nature of the following substances like BF FeCl AlCl can be explained by this concept.

3. The neutralization of acidic oxides and basic oxides can also be explained by this concept. CaO +SO CaSO Acid Base

4. The concept can be easily understood. pH is a convenient way of expressing acidity or the concentration of H in solution. It helps to avoid the use of large of negative power of10. Higher the concentration ofH , lower will be the value of pH.

Advantages:

1.3.5 Concept of pH and pOH (S.P.L Sorenson)  

Definition

pH Scale.

Ionic product of water.

Points to be remembered

pH is defined as the negative logarithm to the base 10 of the hydrogen ion concentration [H ] in a solution. pH = -log [H ].

Similarly, pOH is defined as the negative logarithm to the base 10 of the hydroxyl ion concentration [OH ] in a solution. pOH = -log [OH ]

In a neutral aqueous solution [H ] [OH ] = 10 g ions/litre

Then, log [H ] + log [OH ] = log 10 = -14.

-log [H ] - log [OH ] = 14. So, pH + pOH=14

The product of [H ] and [OH ] is known as ionic product of water. The value is 1 x 10.

K = [H ] [OH ] = 1x10 x1x10 = 1 x 10 g ions/litre

1. For pure water, pHandpOH-value is 7

2. The sum of pH andpOHis equal to 14 pH + pOH = 14.

1. Calculate the pH of a solution whose hydrogen ion concentration is 4.6 x 10 g- ions/litre.

[H ] = 4.6 x10 g ions/litre

pH = -log [H ] = -log [4.6 x 10 ]

pH = 3.3373

2. Calculate the pH of 0.003M HCl solution. The hydrogen ion concentration is 0.003MpH = - log [H ] = -log [1 x 10 ]pH = 3.

3. Calculate the hydrogen ion concentration of a solution whose pH is 4.45. pH = 4.45 pH = - log [H ] 4.45 = - log [H ] log [H ] = - 4.45 [H ] = Antilog of -4.45

[H ] = 3.548 x 10 gram ions/litre 4. If the pH of a solution is 5.25, calculate the hydrogen ion concentration of the solution. pH = 5.25 pH = - log [H ] 5.25 = - log [H ] log [H ] = - 5.25 [H ) = antilog of -5.25 = 5.623 x 10 g ions/litre.

5. Calculate the pH of 0.1MNaOH solution.

0.1MNaOHmeans = 0.1g ions/litre.

pOH = - log [0H ] = - log [0.1] = - log [1 x 10 ] = - [- 1] = 1.

pH +pOH = 14

pH + 1 = 14

pH = 14 – 1 = 13

An indicator is a substance which has one colour in acid solution and a different colour in alkaline solution. Example: Phenolphthalein Methyl Orange Methyl Red Indicators are used in titrations to determine the end point. During the acid-base titration, if certain pH is reached, the indicator changes its colour. Phenolphthalein Colourless to Pink 8 - 9.5 (Acid) (Base) Litmus Red to Blue 5.5 – 7.5 (Acid) (Base) Methyl Orange Red to Yellow 4.5 – 6.5 (Acid) (Base) Methyl Red Pink to Yellow 3.5 – 4.5 (Acid) (Base) Phenol Red Yellow to Red 6.8 – 8.4 (Acid) (Base) pOHofNaOH

1.3.5 Indicators

Indicator Colour Change pH Range

It is very clear from the study of the above table that phenolphthalein becomes colourless when pH is 8. This indicates that the solutions is acidic. Methyl orange shows a yellow alkali colour when pH is 4.5. This indicates that the solution is acidic. Therefore, selection of indicators is more important for acid-base titrations. It depends upon the nature of acid and the base involved in the titrations. Maintaining of pH is more important in many industries. For that purpose ,buffer solution is needed. Buffer solution is one which maintains a constant pH even when small amount of acid or alkali are added to the solution. Buffer solution is classified into two types.

1. Acidic buffer

2. Basic buffer is obtained by mixing a weak acid with a salt of the same weak acid .It maintains the pH between 0 and 7 Example : Mixture of acetic acid and sodium acetate

 

CH COOH + CH COONa Acetic acid sodium acetate is obtained by mixing a weak base with a salt of the same weak base. It maintains the pH between 7and 14.

Example: Mixture of ammonium hydroxide and ammonium chloride

NH OH + NH Cl

Ammonium Ammonium Hydroxide chloride.

Maintenance of pH is more important in many industries to get good yield and the quality. pHplays a vital role in industries.

1. Textile Industry: In textile industry, the pH of dye has to be maintained otherwise dyeing will not be uniform and permanent.

1.3.7 Buffer solution

Acidic buffer 

Basic buffer

1.3.8 Industrial Applications of pH

2. Sugar Industry: The pH of the sugarcane juice should be maintained at 7. Otherwise crystallization of sugar will not be better .The quality and yield will be poor.

3. Leather Industry: The pH of solution for Tanning purpose should be maintained between 2.5 to 3.5. If not, the hides will putrify.

4. The pH of liquid chlorine is more effective in the pH range of 5 to 6.5 in water treatment.

5. The pH of the soil should be maintained for plants to grow and to get better yield.

6. The pH of human blood is 7.2. If not, it causes coagulation of blood which leads to death.

7. The pH of a gastric juice is 1.4 to 2. Otherwise it may cause vomiting and stomach disorder.

8. Here are the industries where pH plays a vital role: Paper industry, Alcohol industry, Medicine and Chemical industry, Food production industry etc. By electronic concept, the oxidation and reduction can be explained as below. Oxidation is a process that involves removal of electrons. Here Na, KandHare oxidised into Na ,K andH respectively. Reduction is a process that involves addition of electrons. Here Cl, F and Br are reduced into Cl , F and Br respectively.

1.3.9 Electronic concept of Oxidation and Reduction Oxidation

Part-A

Part-B

PROBLEMS

In this chapter, Students have understood the acidic and basic nature of substances based on their electronic configuration which is a fundamental property. Students have also learnt the importance of pH in various fields.

1. What is Arrhenius theory of acids and bases?

2. What is Lowry- Bronsted theory of acids and bases?

3. What is Lewis theory of acids and bases?

4. Define pH?

5. Define pOH?

6. What is ionic product of water?

7. What is an Indicator?

8. What is a Buffer solution?

9. Give any two examples for indicators?

1. Explain the Lowry- Bronsted theory of acids and bases with a suitable example.

2. Explain the Lewis concept of acids and bases with examples. Mention its advantages also.

3. Define pH. Calculate the pH of a solution whose Hydrogen ion concentration is 1x10 gramions / litre.

4. Write a note on indicators.

5. Write a note on applications of pH in industries.

6. Explain oxidation and reduction by electronic concept.

1. The hydrogen ion concentration of a solution is 2x10 gramions/litre Calculate the pH of the solution.

2. The hydroxyl ion concentration of a solution is 1x10 gram ion/litre. Calculate the pH of the solution.

3. The pH of a solution is 4.28.Calculate the hydrogen ion concentration of the solution.

4. The pH of a solution is 11.5. Calculate the hydrogen ion concentration of the solution. Try to apply the Lewis theory and explain the following compounds as acid or base. Al Cl ZnCl

TEST YOUR UNDERSTANDING

(1) Ionic bond

Example : Formation of Sodium Chloride

Explanation:

When two atoms in a molecule strongly tend to remain together, they are said to be in chemical bonding with each other. In other words, it is said that a chemical bond has been established between the two atoms. “A chemical bond may be defined as an attractive force which holds together the constituent atoms in a molecule” According to Kossel and G.N.Lewis (1916) who put forward the octet theory of valency, assumed that all atoms have a tendency to acquire a stable grouping of 2 or 8 valence electrons as the elements in the zero group (Noble gases). Thus it may be concluded that it is the tendency of the atoms to acquire a stable configuration or to complete their outermost orbit which is the cause of the chemical combination between them.

(I) Ionic bond (or) Electrovalent bond or Polar bond

(ii) Covalent bond or Non-Polar bond

(iii) Co-ordinate covalent bond or Dative bond

(iv) Metallic bond.

This type of bond is formed as a result of the complete transfer of one or more electrons from one atom to other. This bond is generally present in inorganic compounds The atomic number of sodium is 11.

The electronic configuration is 1s , 2s , 2p , 3s (2,8,1). The electron dot formula is Na Sodium has only one electron in its outermost orbital. 2 2 6 1

The atomic number of chlorine is 17.

The electronic configuration is 1s , 2s , 2p , 3s , 3p (2,8,7)

The electron dot formula is Sodium has one electron in excess of the stable Neon configuration (2,8). While chlorine is one electron short of the stable Argon configuration (2,8,8). By transferring one electron to chlorine, sodium acquires a unit positive charge. The chlorine atom after gaining one electron acquires a unit negative charge. These charged ions are held together by electrostatic force of attraction and form a neutral molecule of sodium chloride. Compounds formed in this way are called electrovalent or ionic compounds and the bond is called Ionic bond or electrovalent bond. This type of bond is formed by the of a pair of electrons between two atoms. The shared electrons are contributed by the atoms. The covalent bond is indicated by a line called ‘single bond.( - ).’ 2 2 6 2 5

 

To get stable electronic configuration, Nitrogen shares its three electrons with electrons of 3 Hydrogen atoms. Thus Ammonia Molecule is formed by 3 covalent bonds. It is yet another type of linkage by virtue of which atoms acquire a stable configuration. Both transfer as well as sharing of electrons is involved in this mode of bond formation. The “shared pair” of electrons is supplied by one atom only and the other atom simply takes part in sharing. Thus “co-ordinate linkage is one in which the electron pair is contributed by one atom only and the sharing is done by both combining atoms”. The atom which provides the shared pair of electrons (called lone pair) is termed the and the atom which accepts this pair for the purpose of forming the molecule is called the atom. The coordinate linkage is shown by an arrow mark ( ). The direction of the arrow points to the acceptor atom.

Example : Formation of AmmoniumIon [NH ]

Explanation:

(4) Metallic Bonding: (Electron Sea Model)

Explanation

When Ammonia (NH ) reacts with Hydrogen ion (H ), Nitrogen of Ammonia donates its unshared lone pair of electrons to Hydrogen ion which needs two electrons to get stable configuration. [Thus the pair of electrons is shared between nitrogen atom & hydrogen ion]. The donor atom which contributes its lone pair becomes positive and the acceptor atom becomes negative. The co-ordinate linkage is also called semipolar linkage. The Ammonium ion is formed as follows. Metallic bond may be defined as the force which binds metal kernels to a number of electrons within its sphere of reference. (The atom without valence electrons is called kernal). In every metal atom there can be a complete free movement of valence electrons in the vacant valence orbits around the nucleus. In other words the valence electrons can move from one kernel to another in the metallic crystal. As a result there may be number of positively charged

Structure of Metallic crystal

Kernels in the metallic crystal. Therefore a metal crystal consists of an assembling of positively charged kernels immersed in a sea of mobile valance electrons. The positively charged kernels are stationary. There is considerable electrostatic force of attraction between positively charged kernels and the negatively charged mobile valence electrons. This force of attraction between the metal atom is called metallic bond. In this chapter, students have studied the different types of bonding including the nature of bonding in metals.

1. What are the types of bonding?

2. Define ionic bond.

3. Define covalent bond.

4. Define co-ordinate covalent bond.

5. What is metallic bond?

1. Explain electrovalent bond with an example.

2. Explain covalent bond with an example.

3. Explain co-ordinate covalent bond with an example.

4. Explain metallic bond by electron sea model theory.

5. Predict the nature of the bond in the following and give explanation.

i) NaCl II)NH III)NH Try to find the type of bonds inCO andSO

UNIT II SOLUTIONS, TECHNOLOGY OF WATER AND SOLID STATE

2.1 SOLUTIONS

2.1.1Introduction

Definition Solute Solvent Units for measuring concentration of solutions

(1) Percentage by mass

Eg:

In this chapter we are going to study about true solutions and the different modes of expressing the concentrations of chemical solutions. Study of different units used to measure the concentration of solutions is essential, since in various estimations of chemicals in different fields these concentration units are employed. Solution is a homogeneous mixture of solute and solvent whose proportion varies within certain limits. Solute is a substance which is present in small quantity in a solution. Solvent is a substance which is present in large quantity in a solution. Action It is defined as the number of grams of solute present in 100grams of solution. A 20% solution of NaOH by weight contains 20 parts by mass of NaOH dissolved in 80 parts by mass of water. Generally this unit is used to prepare solutions of approximate concentration.

1. Percentage by mass

2. Normality

3. Molarity

4. Molality

5. Mole-fr

(2) Normality (N)

Normal solution (Normality – 1NDecinormal Solution (Normality = 0.1N) Problem Mole

(3) Molarity (M)

Normality is the number of gram equivalents of solute present in1000 ml of solution. Normality is represented by the symbol 'N') Anormal solution contains one-gram equivalent of solute present in one litre of solution. A decinormal solution contains one tenth of a gram equivalent of solute in one litre of solution. Calculate the normality of an oxalic acid solution containing 3.2g of oxalic acid in 2 litres of solution. Mass of oxalic acid = 3.2g Volume of oxalic acid = 2000ml ( 2litres ) Equivalent mass of oxalic acid= 63= 0.025N Mole or gram mole is the molecular mass of a substance expressed in grams. Molarity is the number of moles of solute present in 1000 ml or one litre of solution. Molarity is represented by the symbol 'M'. Molar solution (Molarity =1M)

Mass of Solute 1000 Normality N Equivalent mass of solute Volume of solution Molecular mass of solute Mass of Solute in grams Number of moles _ Mass of Solute 1000Normality N

Equivalent mass of solute Volume of solution A Molar solution contains one mole of solute in one litre of solution.We can also have another formula which relates Molarity and Normality Calculate the molarity of a solution containing 40g of sugar (C H O) in 200 ml of solution. Molecular mass of sugar = 12x12+22x1+16x11=342

Mass of sugar = 40g

Volume of solution = 200ml

Find the normality of a solution of sulphuric acid whose molarity is equal to 4M. Molecular mass of sulphuric acid = 98

Equivalent mass of sulphuric acid = 49

Molarity of the solution = 4M

4 98 = Normality 49

Molality is the number of moles of solute present in 1000g. (or) 1Kg of the solvent. Molarity is represented by the symbol 'm'. In this lesson students have learnt the meaning of percentage by mass , Normality , Morality , Molality and Mole-fraction. Problems were worked out to explain the concept.

1. Calculate the Molarity of a solution containing 5.2 g. of calcium bromide (CaBr ) in 200ml.

2. Calculate the Molarity of a solution containing 6.516 g. of sodium chromate (Na CrO ) in 100ml of solution.

3. Calculate the weight of potassium hydroxide (KOH) required to prepare 400ml of 0.082Msolution.

4. Find the mass of urea (Molecular mass = 60) required to prepare a decimolar solution of it, in 250 ml of solution.

5. Calculate the Molality of a solution containing 5.6 g of potassium hydroxide in 250 g of water.

6. Find the mass of sugar required to prepare m/10 solution of it in 300ml of water.

7. Find the Molality of a solution of 2 g of sodium chloride in 450 g of water.

8. 64g of methyl alcohol (CH OH) is dissolved in 144g of water. Calculate the mole-fraction of methanol and water.

9. An aqueous solution contains 10% glucose by weight. Find the molefraction of both solute and solvent.

10. Find the mole-fraction of both solute and solvent in a solution containing 1.12g of potassium hydroxide in 179.64 g of water.

11. A solution of sodium carbonate contains 1.06 g of Na CO dissolved in 100 ml of water. Calculate its concentration in normality. 2 2 4 3 2 3

12. An aqueous solution of potassium permanganate contains 0.316 g of KMnO in 200 ml of solution. Calculate its concentration in normality.

13. Find the mass of oxalic acid present in 500 ml of a decinormal solution of oxalic acid.

14. Find the normality of 0.5M sulphuric acid solution.

15. Find the molarity of a 2Nsolution of sulphuric acid solution.

1. A20% solution of sulphuric acid (by weight) has a density of 1.14g/ml. Calculate molality and molarity of the solution.

2. Find the mole fractions of both solutes and solvent in a solution containing 5%sodium chloride and5%sodium hypochlorite.

 

2.2.3 Rain water Harvesting

Water is the most essential compound for all living matter on the earth. It plays an important role in human living and industries. The two important sources of water are (1) surface water and (2) underground water. The water available on the earth's surface is called as surface water. Surface water includes rainwater, river water, lake water and seawater. Underground water includes water present between the rocks in the earth crust, spring water, well water etc. The decrease in the quantum of underground water is depletion of water. Depletion of water is mainly caused by,

1. Modernization, industrialization and population growth

2. Global warming causing excess evaporation of surface water

3. Deforestation

4. Decrease in rainfall caused by seasonal changes and

5. Effluents from the industries spoiling the ground water source.

To meet out this depletion of ground water sources, it is essential to find alternate plans using water management techniques to recharge the ground water sources. One of the techniques adopted is rainwater harvesting. Rainwater harvesting (RWH) is collection of rainwater for useful purposes . The methods employed are

1. Roof top harvesting

2. Open space harvesting

2.2.5 Hardness of water

Rainwater is directly used for recharging open wells and bore wells by directing them into it. It can also be stored in sumps or over head tanks and used directly. Open spaces around the buildings are used for rainwater harvesting as follows

1. With percolation/recharge pits

2. Recharge trenches

3. Recharge wells

The recharge method used depends on the soil condition. There are three types of impurities present in water. They are

(i) Suspended and colloidal impurities.

(ii) Dissolved salts.

(iii) Micro - organisms.

There are two types of water. They are (i) soft water and (ii) hard water.

(i) Soft water readily gives lather with soap.

(ii) Hard water does not give lather with soap.

There are two types of hardness in water. They are:

(i) Temporary Hardness: (Carbonate hardness) It is due to the presence of calcium bicarbonate [Ca(HCO ) ] and

magnesium bicarbonate [Mg (HCO ) ]. Temporary Hardness can be removed by boiling.

(ii) Permanent Hardness: (Non-Carbonate hardness) It is due to the presence of chloride and sulphate salts of calcium and magnesium. (CaCl , CaSO , MgCl MgSO ).It cannot be removed by boiling. Hence it is known as permanent hardness. 3 2 3 2 2 4 2, 4

Disadvantages of a hard water sample

2.2.6 Degree of Hardness

Units for measuring hardness

Note

Hard water cannot be used for drinking purpose. It cannot be used for cooking purposes. It cannot be used for bathing and washing purposes as it does not give lather with soap. Hard water cannot be used in laboratories as it gives unwanted chemical reactions. Hard water cannot be used in boilers in steam raising. It cannot be used in sugar and paper industries. Hard water cannot be used in textile and leather industries.

 

1.mg/litre of CaCO 

2.parts per million of CaCO Usually ,the hardness of water is expressed in terms of calcium carbonate equivalents The formula used to convert the mass of hardness producing salt to mass ofCaCO equivalents is given below Molecular masses of hardness producing salts are given below.

CaSO 136

MgSO 120

CaCl 111

MgCl 95

Ca(HCO ) 162

Mg(HCO ) 146

CaCO 100

Hardness producing salt Molecular Mass

3 Mass of salt Molecular mass of CaCO Calcium carbonate equivalents Molecular mass of salt

2.2.7 Estimation Of Hardness OfWater – EDTA Method

PRINCIPLE

 

PROCEDURE

 

A water sample contains 48 mg of MgSO per 200ml of water. Calculate the hardness in terms ofCaCO equivalent in mg/litre ofCaCO Mass ofMgSO = 48mg Molecular mass ofMgSO = 120 Mass ofCaCO present in200 ml of water = 40mg Therefore, mass ofCaCO present in 1000ml of water = 200mg Hardness of water = 400mg/litre ofCaCO

EDTA method is used to determine the hardness of a sample of water.

EDTA refers to Ethylene diamine tetra acetic acid. This method is also called as Modern method.

This is a volumetric method based on the principle of formation of complexes. Ethylene diamine tetraacetic acid (E.D.T.A.) forms colourless complexes with Ca and Mg ions present in water. Similarly Eriochrome Black-T, another dye, also forms wine red coloured complexes with Ca and Mg ions. Pure Eriochrome Black-T is blue in colour. At the pH range of 9 to 10, the Eriochrome complexes are less stable when compared to E.D.T.A. complexes. Thus when E.D.T.A. solution is added to Eriochrome-Ca or Mg complexes it displaces pure Eriochrome to form E.D.T.A-Ca or Mg complexes. Thus at the end point E.D.T.A. frees the total Eriochrome Black-T to change the colour of the solution from wine red to steel blue.

Eriochrome-Ca +E.D.T.A.---------> .D.T.A-Ca + Eriochrome Black-T The burette is filled with the standard E.D.T.A. solution. A 50-ml pipette is washed with distilled water and rinsed with the sample of hard-

Calciumcarbonate equivalents water. Exactly 50 ml of hard-water is pipetted out into a conical flask and 5ml of NH Cl - NH OH buffer solution is added. A pinch of Eriochrome black-T indicator is added. The colour of the conical flask solution changes into wine red. The water sample is titrated against the E.D.T.A. solution taken in the burette. The colour changes from wine red to steel blue. This is the end point of the titration. The burette reading is noted. Titrations are repeated until two consecutive values agree. From the volume of E.D.T.A. the hardness of the sample of water is calculated. In the estimation of hardness of water, a standard data relating the mass of CaCO and volume of 0.01M EDTAsolution .is given below Let,the Volume of water taken =50ml

A sample of 100 ml of water consumed 12.5 ml of 0.01 M EDTA solution. In another titration 100 ml of the same sample, after boiling for half an hour consumed 8.2 ml of the same EDTA solution. Calculate the carbonate and non-carbonate hardness of the sample of water. 12.5ml of 0.01M EDT Asolution 12.5mgofCaCO 12.5ml of 0.01M EDTA solution 100 ml of hard water Mass of CaCO present in 100 ml of hard water = 12.5mg Mass ofCaCO present in 1000ml of hard water = 125mg Hence, Total hardness of water = 125 mg/litre of CaCO 8.2ml of 0.01M EDTA solution 8.2mgofCaCO 8.2ml of 0.01M EDTA solution 100 ml of hard water Mass ofCaCO present in 100 ml of hard water = 8.2mg Mass of CaCO present in 1000ml of hard water = 82mg Hence, Non-carbonate hardness of water = 82 mg/litre of CaCO Therefore, Carbonate Hardness =Total hardness – Non-carbonate Hardness = (125 – 82) = 43 mg/litre of CaCO

Problem- 2

Total hardness

Non-carbonate Hardness

2.2.8 Methods of softening hard water

(1) Ion exchange method

Softening Process

In this method the hard water is first passed through an acidic resin (RH ) to remove the captions [Ca , Mg ] and then it is passed through a basic resin [R'(OH) ] to remove the anions. Thus both types of ions are totally removed. Acidic resin is represented by RH . Basic resin is represented by R'(OH) When the hard water sample is passed through the I-Cylinder (acidic resin) calcium and magnesium ions are replaced by hydrogen ions of the acidic resin.

RH + Ca RCa + 2H

RH + Mg ----------> RMg + 2H.When this water is passed through the II-Cylinder (basic resin) chloride, bicarbonate and sulphate ions are replaced by the hydroxide

ions of the basic resins. R'(OH) + 2Cl ---------->R'Cl + 2OHR'(OH) +2HCO ---------> R'(HCO ) + 2OHR'(OH) +SO ----------> R'SO + 2OHThus all the ions responsible for hardness are removed from water. TheH andOH ions combine together to form water.

H +OH ----------> H O The quality of water obtained by this method is equivalent to distilled water.

Reverse Osmosis

Method

When a hydrostatic pressure greater than the osmotic pressure is applied on the concentrated side, solvent molecules move from concentrated side to the dilute side across the membrane. This is called reverse osmosis. This principle is used in Reverse Osmosis plants to soften hard water. In this method hard water and soft water are taken in two different chambers separated by a semi permeable membrane. When a hydrostatic pressure greater than the osmotic pressure is applied on the hard waterside, the water molecules move from hard waterside to soft waterside leaving the impurities on the membrane due to reverse osmosis. Thus hard water is converted to soft water by Super filtration or hyper filtration. The semi permeable membrane is made of polysulphone or cellulose acetate or polyamide.

Water for Drinking purpose ( Potable water )

1) In this method ionic, non-ionic, colloidal, and organic particles are removed from water.

2) The semi permeable membrane can be replaced and reused.

3) There is no wastage of water. Municipal water is mainly used for drinking purposes and for cleaning, washing and other domestic purposes. The water that is fit for drinking purposes is called potable water

(1)Characteristics of Potable water

1.It should be colourless, odourless and tasteless.

2.It should be free from turbidity and other suspended Impurities.

3.It should be free from germs and bacteria.

4.It should not contain toxic dissolved impurities.

5. It should be moderately soft.

6. It should not be corrosive to the pipe lines.

7. It should not stain clothes.

 (3)Water quality standards in India The three stages involved in purifying a water sample for drinking purpose are 

1. Sedimentation

2. Filtration

3. Sterilisation

Water from river or lake is taken in the big tank called sedimentation tank. Here the insoluble matter settles down at the bottom of the tank as sediments. In this tank the colloidal impurities are converted into precipitate by adding Alum. The clear water from the top layer is sent to the next tank, called Filtration tank. In filtration tank, the suspended impurities and the microorganisms are removed. In all types of filtration, the filter bed used is constructed as follows.

Sedimentation

The filter bed consists of a layer of fine sand, followed by a layer of coarse sand, which is then followed, by a layer of gravel. There is a drain at the bottom to remove the filtered water. The layer of fine sand acts as the filtering unit and the other two beds support the fine sand layer. Generally filtration is done due to the gravitational force. The filtered water is then taken to the sterilization tank. In industrial areas where large amount of drinking water is required in short period, Pressure filters are used in which water is sent through filter beds using external pressure.) Sterilization is destroying of bacteria. It is done by Chlorination.

Chlorination is addition of chlorine. Chlorine is added to water in the pH range of 6.5 to 7. When chlorine is added to water, it forms HCl and HOCl. The hypochlorous acid enters into the living cells of bacteria and destroy them.

H O+Cl ---------- >HCl + HOCl

Hypochlorous acid Other sterilizing agents used are chloramines, bleaching powder etc. The advantage of using chloramines is that it does not evaporate out easily and can be carried over to a longer distance along with the water. Ultra-violet rays can also be used for sterilizing purpose.

Water is used in boilers, steam engines etc., to raise steam. When a sample of hard water is used in boiler to prepare steam, the following problems will occur.

1. Scale formation

2. Corrosion of boiler metal

3. Caustic Embrittlement and

4. Priming and foaming.

When hard water is used in boilers to get steam, the impurities that are present in the hard water will settle down on the sides of the boiler. This residue in due course will adhere to the boiler vessel surface in the form of a sludge or scale. This is called as boiler scale. The following calcium salts are responsible for the formation of boiler scale.

CaSO ,CaCO CaSiO ,Ca (OH) Mg(OH) ,etc

1. The salt deposit formed is a poor conductor of heat. Therefore, fuel is wasted in raising the temperature of the boiler.

2. Due to the increase in the temperature, the plates may melt. This may lead to explosion of boiler.

3. At higher temperature, more oxygen may be absorbed by the boiler metal, which causes corrosion of boiler metal.

4. The sudden spalling of the boiler scale exposes the hot metal suddenly to super-heated steam, which causes corrosion of boiler. The two types of methods employed to prevent scale formation are,

1. Internal conditioning method

2. External conditioning methods.

1. Internal conditioning methods involve addition of complexing agents like Calgon to boiler feed water. Another method of internal conditioning method is Phosphate conditioning. In this method sodium phosphate is added to boiler feed water which forms non-sticky Calcium

2.2.10 Boiler feed water

(1) Boiler scale formation Disadvantages of Boiler scale

4 3, 3 2, 2 and Magnesium Phosphate which can be removed by blow down operation.

2. In external conditioning methods water is purified either by Zeolite process or by ion-exchange method before being fed into boilers. The impurities such as dissolved oxygen, dissolved Carbon dioxide, mineral acids, dissolved salts of calcium and magnesium, organic matter etc. are responsible for the corrosion of boilers.

The dissolved matter undergoes hydrolysis and forms acids. The acid slowly attacks the inner part of the boiler. The dissolved oxygen attacks iron at high temperature. TheCO and H Oform carbonic acid (H CO ), which slowly attacks the metal.

1. By using proper water treatment procedures.

2. By degasification to remove the dissolved gases like oxygen, CO , etc.,

3. The dissolved CO can be removed by the addition of limewater.

4. Adding calculated amount of base could neutralize the mineral acids. Sometimes cracks appear inside the boiler parts, particularly at the places, which are under stress. Metal becomes brittle at these places. It is due to the high concentration of caustic soda (NaOH) and a little amount of silica in water. This is called as caustic embrittlement. Caustic soda is formed by the hydrolysis of Na CO .

Na CO +H O----------> 2NaOH +CO

Removal of Na CO present in water can prevent caustic embrittlement.

This can be done by the following methods.

1. By adding sulphuric acid.

2. By addingCaSO and CaCl to boiler water

3. By adding Na SO

4. By adding trisodium phosphate.

(2) Corrosion of Boiler metal Prevention of Boiler Corrosion

Foaming is nothing but the formation of foam. Bubbles of water will enter the surface of water inside the boilers and results in the formation of foam. Foam comes out of the boiler along with the steam. Hence the steam becomes wet and the heat content of the steam is reduced considerably. This type of wet steam spoils the machine parts where it is used. The main cause for foaming is the presence of dissolved salts in water. Hence soft water should be used in boilers to avoid foaming. Priming is violent and rapid boiling of water inside the boiler. Due to priming the water particles mix up with the steam when it comes out of the boiler. Like foaming, priming also reduces the heat content of the steam and reduces the efficiency of the steam. Main reasons for Priming

a) Defective design of the boiler.

b) Presence of large quantities of dissolved salts, oily matter, alkaline and suspended matter.

1. Priming can be controlled by proper design of the boiler

2. By uniformly heating the water in the boiler.

3. By using a better sample of water.

Students have learnt about rain water harvesting, estimation of hardness, methods of softening and bad effects of hard water in boilers.

1. Define hard and soft water.

2. List the salts that cause Carbonate hardness in a water sample.

3. List the salts that cause Non-carbonate hardness in water.

4. What is rain water harvesting?

5. Mention any two disadvantages of hard water.

6 List any two methods of softening of hard water.

2.3 SOLID STATE

All solids are classified into two types, based on the arrangement of particles present in them. They are

(1) Crystalline solids & (2) Amorphous solids.

In crystalline solids the particles present are arranged in a regular three-dimensional way. Acrystalline solid is made up of number of layers in which the particles are arranged regularly in two dimension. These layers are called 'plane surfaces'. In a big crystal a representative unit is again and again repeated. This representative unit is called as 'unit cell'. Therefore a unit cell is the smallest arrangement of a model of a crystal.

When a big crystal is cut with a sharp edged tool, it breaks into two smaller crystals of the same shape. All crystalline solids have sharp melting points. Generally, when the molten substance is cooled slowly, we get crystalline solids. When the molten substance is cooled slowly, the particles present get sufficient time to arrange themselves in the crystal lattice.

(For eg.) When molten sulphur is cooled slowly, we get needle sulphur and rhombic sulphur.

In amorphous solids the particles are arranged in an irregular manner. Amorphous solids do not have sharp melting point. When the molten substance is cooled suddenly, we get amorphous solids. When the molten substance is cooled suddenly, the particles present in them do not find sufficient time to arrange themselves properly.

(E.g.) When molten sulphur is cooled suddenly by pouring it in ice cold water, we get clay of sulphur.

(1) CRYSTALLINE SOLIDS

(2) AMORPHOUSSOLIDS

TYPES OF CRYSTALLINE SOLIDS

3.MOLECULAR SOLIDS

Crystalline solids are classified into four types based on the type of particles present in them and the force of attraction operating between these particles. They are

1. Ionic solids

2. Covalent solids

3. Molecular solids and

4. Metallic solids.

In ionic solids the particles present are ions and the force of attraction operating between these particles is 'positive-negative attraction' or 'Electrostatic force of attraction.'

NaCl, KCl, CsCl etc. For example in NaCl, the particles present are Na and Cl ions and the force of attraction operating between these ions is 'Electrostatic force of attraction'. In NaCl each sodium ion is surrounded by six chloride ions and each chloride ion is surrounded by six sodium ions. In covalent solids all particles are connected to each other by covalent bonds throughout the crystal. Diamond,graphite etc. In diamond each carbon atom is connected to four other carbon atoms by covalent bonds throughout the crystal. Therefore diamond isthe hardest substance in the world. In graphite carbon atoms form layers of hexagonal rings (honey comb structure) where each carbon atom is surrounded by three other carbon atoms satisfying three valences of carbon and the fourth valence is satisfied by the loosely bound ' 'electrons between the layers. In molecular solids the particles present are molecules and the force attraction operating between these particles is 'Vander wall's force of attraction' or 'dipole-dipole attraction'.

Example: - Dry ice and Ice

4.METALLIC SOLIDS  

Dry ice is solid carbon dioxide. In dry ice the particles present are CO molecules and the force of attraction operating between these particles is 'Vander wall's force of attraction'. In ice the particles present are H O molecules and the force of attraction operating between H Omolecules is dipole-dipole attraction. All metal crystals are metallic solids. Metals show the following properties.

a. They have high melting point and density.

b. They conduct electricity.

c. On heating they emit electrons. This property is called 'Thermionic emission'. A new type of bonding called 'metallic bonding' explains all these properties. The three type of packing generally present in metal crystals are

(I) Body centered cube (BCC)

(ii) Face centered cube (FCC)

(iii) Hexagonal Close Packing (HCP)

This type of packing is present in metals like sodium, potassium etc.

The diagram of the unit cell is as follows.

 (I) Body Centered Cube (BCC Diagram

In the unit cell 8 atoms are placed at 8 corners of a regular cube and one atom is placed at the centre. Therefore the total number of atoms present in a unit cell inBCCpacking is calculated as follows. Number of atoms present in a unit cell = 1 + 8 X / = 2 atoms. The packing density of the bcc crystal is found to be 68%. In this type of packing the atoms are packed in the least efficient way. Therefore BCC metals are very soft in nature. They can be cut even with a knife. These metals have low melting point and density. This type of packing is present in metals like copper, silver and gold. The model diagram of the unit cell is as follows.

In the unit cell 8 atoms are placed at 8 corners of a regular cube and six atoms are placed at the centre of the six faces. The total number of atoms present in a unit cell of aFCCcrystal is calculated as follows.

Number atoms present in a unit cell = 8 X / + 6 X = 1+3=4.

The packing density of the FCC crystal is calculated as 74%.

Therefore FCC metals are harder than BCC metals. They have high melting point and density. These metals are malleable and ductile. This type of packing is present in metals like Titanium (Ti), 1 1 8 8

(ii) Face centre cube (FCC)

(iii)Hexagonal Close Packing (HCP)  

1. Explain the four types of crystalline solids.

2. Describe BCC packing in metal crystal with a neat diagram. How this type of packing explains the properties ofBCCmetals.

3. Explain FCC packing of metal atoms with a neat diagram. How it reflects the properties of FCC metals.

4. What is HCP packing of metals? Explain with a neat diagram with examples.

5. What are ionic and molecular solids? Distinguish between them.

1. Both ice and dry ice are molecular solids. But ice has got comparatively high melting point. Why?

2. Calculate the packing density of BCC and FCC crystals.

Unit III

COLLOIDS, NANO PARTICLES AND

PHOTO CHEMISTRY

3.1 COLLOIDS

3.1.1 Introduction

 

Definition

 

An aqueous solution of salt or sugar is homogeneous and it contains the solute particles as single molecules or ions. This is called a true solution. The diameter of the dispersed particles ranges from 1A to 10 [1A = 10 cm]; whereas in a suspension of sand stirred in water, the diameter of the dispersed particles will be more than 2000A . The particles which are larger than a molecule and smaller than a suspended particle are said to be colloids and such solutions are called colloidal solution or sol. Moleculer size < colloids < suspension

 

(1A - 10A ) (10 - 2000 ) (More than 2000A ) A colloidal system is made up of two phases. The substance distributed as colloidal particles is called the dispersed phase (analogous to solute) and the phase where the colloidal particles are dispersed is called the dispersion medium (analogous to solvent). A colloidal solution can form eight different types based upon the physical state (solid, liquid, gas) of dispersed phase / dispersion medium. The common example of colloids are milk, curd, cheese, clouds, paint etc. The properties of these

 

colloidal solution are in many ways different from that of true solution.

 

 

 

Table 3.1. Differences between true solution and colloidal solutions Colloidal solutions in which the dispersed phase has very little affinity for the dispersion medium are termed as lyophobic colloids [Lyosolvent; phobic-hate]. eg. colloidal solutions of metals and sulphur in

 

water.

 

Table 3.2 Distinction between lyophilic and lyophobic colloids

 

3.1.2 Types of colloids

 

Lyophobic colloids

 

 

 

Lyophilic colloids

 

3.1.3 Properties of colloids

 

(1) Brownian movement (Mechanical / kinetic property).

 

Colloidal solutions in which the dispersed phase has strong affinity for the dispersion medium are called lyophilic colloids e.g. solution, gum, protein etc..The lyophilic and lyophobic colloids have different characteristics, which is given inTable 3.2.

 

Colloids exhibit certain exclusive properties. They are:

 

(i) Brownian movement (Mechanical / kinetic property)

 

(ii) Optical property

 

(iii) Electrical property

 

When the colloidal particles are seen through an ultramicroscope, it is found that the colloidal particles are found to be in constant zig-zag, chaotic motion. This was first observed by Brown and so this random movement of colloidal particles is called Brownian movement. This movement is due to the collision of colloidal particles with the molecules of the dispersion medium. The motion becomes more rapid when the temperature of the dispersion medium is high and less viscous. When a beam of light is passed through a true solution, and observed at right angles to the direction of the beam, the path of the light is not clear. At the same time, if the beam of light is passed through a colloidal solution, the path of the light is quite distinct due to scattering of light by the colloidal particles.

 

If an electric potential is applied across two platinum electrodes immersed in a colloidal solution, the colloidal particles move in a particular direction, depending upon the charge of the particles.

 

Fig.3.1. Brownian movement

 

(2) Tyndall effect (Optical property)

 

(3) ( )

 

The phenomenon of scattering of light by the colloidal particles is known as “Tyndall effect”.

 

Thus the migration of colloidal particles under the influence of electric field is called

 

electrophoresis.

 

Fig.3.2. Tyndall effect

 

Electrophoresis Electrical property

 

Brownian movement Colloidal Particle Temperature is high Particles of dispersion medium

 

Powerful beam of light True Solution Path is invisible Mircroscope Colloidal solution Path is Visible. This phenomenon can be demonstrated by placing a layer of arsenic sulphide solution under two limbs of a U-tube. When current is passed through the limbs, it can be observed that the level of the colloidal solution decreases at one end of the limb and rises on the other end. The entire colloidal particles are electrically charged; all are positively charged or negatively charged. Therefore every colloidal particle repel each other and remains stable. In order to coagulate a colloid, these charges have to be nullified. This can be done in three

 

ways:

 

(I) By adding a double salt (electrolyte)

 

(ii) By introducing an electrode of opposite charge

 

(iii) By introducing another colloid of opposite charge

 

After nuetralising the charges, the colloidal particles are brought together and they are large enough to settle down. Smoke is a colloidal suspension of carbon particles in air. The smoke is first introduced into a chamber and subjected to a very high voltage. The particles get deposited in one of the electrodes and the hot air alone is let out through the chimney.

 

The suspended impurities of the water cannot be filtered. So it is better to coagulate them. This is done by adding potash alum.

 

Fig.3.3. Electrophoresis

 

(4) Coagulation of colloid

 

3.1.4Industrial applications of colloids

 

(1)Smoke precipitation (cottrell's method)

 

(2)Purification of drinking water

 

Thus the process of precipitating a colloidal solution is called coagulation.

 

 

 

Part-B

 

The dirt particles stick to the cloth or body by the greasy oily substance. It forms an emulsion with soap. The dirt particles get detached from the cloth / body and washed away along with soap with excess of water.

 

Animal hides are colloidal in nature. When a hide, positively charged particles, soaked in tannin, a negatively charged particle, mutual coagulation takes place. This results in hardening of leather. The process is called tanning. Chromium salts are used as tannin.

 

Sewage dirt particles are electrically charged. So the sewage is allowed to pass through disposal tanks. It is then subjected to high potential. The sewage particles lose the charges and coagulated. Clean water is recycled or used for gardening. Sludge is used as manure.

 

1. What is a colloid?

 

2. Give any two examples for colloid.

 

3. What are the two types of colloids?

 

4. Give any two example for lyophilic colloid.

 

5. Define Tyndall effect.

 

6. What is called Brownian movement?

 

7. What is meant by electrophoresis?

 

8. Define coagulation of colloid.

 

1. Distinguish between true solution and colloidal solution.

 

2. What are the differences between lyophilic and lyophobic colloids?

 

3. Write notes on (i) Brownian movement (ii)Tyndall effect.

 

4. Write notes on (i) Electrophoresis (ii) Coagulation of colloids.

 

5. Write down any five applications of colloids.

 

 

 

TEST YOUR UNDERSTANDING

 

1. Why is silver iodide powder is sprinkled over clouds for artificial rain?

 

2. What is the difference between soap and detergent?

 

 

 

3.2. NANO PARTICLES

 

3.2.1 Introduction

 

3.2.2 Characterization

 

3.2.3 Application of Nano particle technology in medicine

 

Diagnostics:

 

Drug Delivery:

 

3.2.4 Tissue Engineering

 

3.2.5 Application of nanotechnology in electronics

 

Nano technology is the study of matter on an atomic and molecular scale. One nanometer (nm) is one billionth or 10 m. The carbon-carbon bond length is in the range of 0.12-0.15 nm and the DNAdouble helix has a diameter of 2nm and the bacteria will be around 200nm. So particles of nanometer size are called Nano particles Materials reduced to nanometer scale show unique characteristics. For instance, opaque substance become transparent (copper); stable materials turn combustible (aluminium); insoluble materials become soluble (gold). Therefore materials on nanoscale find wide applications in the field of medicine, electronics and in all fields of engineering. The biological and medical research communities have utilized the properties of nano particles for various applications: Integration of nano materials with biology led to the development of diagnostic devices and drug delivery vehicles.

 

Gold nano particles tagged with DNA can be used for the detection of genetic sequence.

 

Drug can be delivered for specific cell using nano particles. This may replace todays conventional treatment like organ transplants/artificial inplants. Advanced forms in tissue engineering may lead to life extension. It can repair damaged tissue.

 

(I) Today’s solar cells utilize only 40% of solar energy. Nano technology could help to increase the efficiency of light conversion using nanostructures.

 

 

 

.

 

 (ii) The efficiency of internal combustion engine is about 30-40%. Nanotechnology could improve combustion by designing catalysts with maximised surface area.

 

(iii) Nano porous filters may reduce pollutants.

 

(iv) The use of batteries with higher energy content is possible with nano materials.

 

(v) Nano technology has already introduced integrated circuits in nanoscale (50nm) in CPU's and DRAM devices.

 

(vi) Carbon nano-tubes based cross bar memory called Nano-Ram has been developed. Food and bio processing industry for manufacturing high quality of safe food can be solved using nano technology. Bacteria identification and food quality monitoring using bio-sensors are examples of application of nano technology. Anano composite coating act as anti microbial agents. Natural bone surface is 100nm across; if the artificial bone implant is smooth, the body rejects it; so nano sized finishing of hip and knee would help the body to accept the implant.

 

1. What are nano particles?

 

2. Mention few unique characteristics of nano particles.

 

1. Mention few applications of nano technology in engineering.

 

2. Explain the applications of biomaterials.

 

3. How come the nano technology becomes useful in the field of medicine?

 

1. Imagine few innovative applications of nano technology.

 

2. What could be the ill effects of nano technology?

 

 

 

3.3 PHOTO CHEMISTRY

 

3.3.1 Introduction

 

Absorption and emission of photon by sodium atom

 

3.3.2 Important terms used in Photo chemistry

 

(1) Charge transfer / transition:

 

(2) Electronic energy migration

 

Ground state

 

Excited State:

 

Emission:

 

Frequency:

 

(3) Fluorescence

 

Photochemistry is the interaction of light and matter. When a sodium atom absorb a photon it goes to an excited state. After a short while, the excited sodium atom emits a photon of 589 nm light and falls back to its ground state. The atom can be excited by a flame, called flame test. An electronic transition in which a large fraction of electronic charge is transferred from one region to the other region. The movement of electronic excitation energy from one molecular entity to another. : The lowest energy state of a chemical entity (atom) Ahigher energy state of a chemical entity after absorbing energy.

 

Deactivation of excited state; transfer of energy from molecular entity to electromagnetic field. ( or ) The number of wave periods per unit time. When a beam of light is incident on a substance, it emits visible light back; but as and when the incident light is cut off, they stop emitting. This phenomenon is called fluorescence and those substance which emit such light are called fluorescent substances. The other name for fluorescence is “cold light”. This happens due to the absorption of energy by the electron. They move from ground state to higher energy level (excited state).

 

Photon 589nm Photon 589nm

 

Na ground state_ Na ground state Na _excited state

 

This is produced in three ways. When current flows through a medium the moving electrons collide with molecules in the medium. The energy make them excited to emit light e.g. Neon lights, fluorescent light, incandescent bulb. Electricity discharges through a gas (mercury vapour) and causing emission of u.v. light. Burning copper, excite electrons to give green light. Fluorine, naphthalene, bibhenyl etc.When light is incident on certain substances, they emit light radiations continuously even after the incident light is cut off. This is called phosphorescence. Those substances are called phosphorescent substances. A fluorescent paint glows under u.v.lamp, but stops glowing as the lamp is turned off. A phosphorescent paint keeps glowing for a while. Phosphorescent substance have the ability to store up light and release it gradually. Ground state molecules absorb photons and go to excited

 

i. Using electricity:

 

ii. Using ultraviolet light:

 

iii.Use of heat:

 

Somefluorescent substances

 

 

 

Most of them come back to the ground state causing fluorescence. Few of them come to an intermediate state, called a metastable state by non-radiative process from the metastable state. It comes down to the ground state and slowly give out light. Some of the phosphorescent substances are zinc sulphide, calcium sulphide, etc. A chemical reaction leading to the expelling light is called chemiluminescence.

 

x + y ––––––> [x+y] ––––––> products + light A common example of chemiluminescence is the reaction between a fuel and oxidant producing excited products that emit light. Nitrogen monoxide reacts with ozone to produce nitrogen dioxide in an excited state. [NO ] returns to lower energy state and emits light. E.g. Sea divers use chemiluminescence for want of light under water. A photoelectric cell is a device that is activated by electromagnetic energy in the form of light waves. Light is a form of energy. When light strikes certain chemical substances such as selenium or silicon, its energy causes a push on the electrons. Based on the nature of photo electric effect there are three types of cells. (i) photo conductive cell (ii) photo emissive cell (iii) photo voltaic cell. These cells are very popular for triggering the automatic opening of doors. This type of cell contains a wire, as anode and a semi cylindrical cathode with an emitting surface. These are seated in an evacuated or gas filled bulb. The cathode surface is coated with cesium, potassium or rubidium.As light incident upon cathode, the surface emits electrons. The emitted electrons are attracted to the positive anode as photo current. This is properly designed in a electronic circuit.

 

Photo synthesis is a process by which green plants absorb light energy from sunlight, it then prepares glucose and carbohydrates by the combination of carbondioxide and water. The byproduct of photo

 

(5) Chemiluminescence

 

(6) Photo Electric Cell

 

(7) Photo Emissive cell

 

(8) Photo synthesis

 

**2 synthesis is oxygen which is essential for all living organisms. All atmospheric oxygen has originated only from photo synthesis. 6H O+6CO –––>C H O +6O

 

Green plants use the energy absorbed in the form of light to make organic compounds from inorganic compounds. The simple form of photosynthesis can be written as

 

Light CO +H O–––––––> [CH O] +O Carbohydrate

 

It was proved that oxygen come only by the photolysis of water molecule. Therefore two molecules of water must be involved to release one oxygen molecule. Therefore,

 

Light CO + 2H O–––––––> [CH O] +O +H O

 

A pigment may be defined as a substance that absorbs light. The colour which is not absorbed appears as the colour of the pigment. Chlorophyll absorbs all the wavelength except green.

 

Photosynthesis is a two stage process. The first stage called light dependent process is called light reactions. It is followed by light independent process called dark reactions. In the light reactions, light strikes chlorophyll-a and excite electrons to a higher energy. In a series of reaction the energy is converted to ATP (Adenosine triphospate) and NADPH (Nicotinamide-Adenine Dinucleotide Hydrogen Phosphate). Water is split in the process releasing oxygen as byproduct of the reaction. During the day time plants make their own food in the form of glucose. It is then stored in the form of starch.

 

2 2 6 12 6 2

 

2 2 2 n 2

 

2 2 2 n 2 2

 

General chemical reactions in photo synthesis

 

Chlorophyll andAccessory pigments

 

Mechanism of photosynthesis

 

(9) Light reactions:

 

 

 

(10) Dark reactions

 

3.3.3 PHOTOSYNTHESISANDACID RAIN

 

QUESTIONS

 

Part-A

 

Carbon fixing reactions are known as dark reactions. The control feature of dark reactions in photo synthesis is the carbon reduction cycle called Calvin Benson cycle. The ATP and NADPH are used to make C-C bonds in the absence of light to form glucose from stored starch. Plants help to remove carbon-di-oxide from the atmosphere and oceans. Plants also produce carbon dioxide during respiration but that is again used for photosynthesis. Plants also convert light energy into chemical energy to form C-C bonds. Animals are carbon-di-oxide producers and obtain their energy from carbohydrates and other products from plant kingdom. The balance between the carbon dioxide removal by the plants and the carbon dioxide produced by animals is equalised by the formation of carbonates in the oceans.

 

CO (from animal kingdom) +CO (Plant kingdom) = CO (taken for photo synthesis) +CO (Carbonates in ocean) The CO from animal kingdom includes CO generated by fossil

 

fuels (coal, petrol, diesel etc.) and industrial outlet. The equation is not balanced due to

 

(i) Deforestation [Remove lessCO ]

 

(ii) Industrialisation [Add moreCO ]

 

This leads to global warming. The increase in CO and other pollutants like NO and SO leads to acid rain.

 

1. What is called ground state of an atom?

 

2. Define excited state of a chemical entity.

 

3. Give any two example of fluorescent light.

 

4. Name any two fluorescent substances.

 

5. What are the three types of photo electric cell?

 

6. What is coated on the cathodic area of a photo emissive cell?

 

7. Define photosynthesis.

 

8. What is the role of chlorophyll- in a green leaf?

 

9. Mention the disadvantage of deforestation.

 

10. Define photo chemistry.

 

1. Write notes on (i) flurescence (ii) Phosphorescence.

 

2. Explain chemi luminescence.

 

3. Explain the mechanism of photo synthesis.

 

4. Explain the importance of photosynthesis in preventing acid rain.

 

1. What makes the sky appear blue?

 

2. Think about a town/city-plan that is free from deforestation.

 

Part-B

 

TEST YOUR UNDERSTANDING

 

 

 

UNIT IV

 

ELECTROCHEMISTRY, CELL AND

 

BATTERIES

 

4.1 ELECTROCHEMISTRY

 

4.1.1 Introduction

 

Chemical effects producing electricity

 

4.1.2 Electrolyte

 

Non-electrolytes

 

Electrochemistry is a branch of chemistry which deals with the relationship between electrical energy and chemical energy. Electrochemical reactions find applications in many industries. Electrochemistry broadly discusses about electrical effects on passing electricity through a solution. Electrolysis comes under this category and finds applications in

 

I) Metallurgy

 

ii) Electroplating

 

iii) Chemical manufacturing processes including medicines. Electrochemical cells including Dry cells, Daniel cells, Laclanche cells, rechargeable batteries are used in day to day life in torchlight, transistors, wall clocks, automobiles and cell phones. The basics of electrolysis and its applications in electroplating are discussed in the following sections.

 

An electrolyte is a substance, which conducts electricity both in solution and in fused state.

 

Example: Sodium chloride, hydrochloric acid, copper sulphate solution, etc.

 

A non-electrolyte is a substance which does not conduct electricity either in solution or in fused state. Example: Sugar, urea, alcohol, etc.

 

 

 

Strong electrolytes

 

Weak electrolytes

 

4.1.3 Electrolysis

 

4.1.4 Mechanism of electrolysis

 

The electrolytes, which ionize completely in solution, are called strong electrolytes.

 

Example: Sodium hydroxide, potassium chloride, sodium chloride, etc.

 

Electrolytes which do not ionize completely in solution are called weak electrolytes.

 

Example: Acetic acid, oxalic acid, etc.

 

Decomposition of a substance by passing electric current is called electrolysis. During electrolysis, electrical energy is converted into chemical energy.

 

Example: Electrolysis of hydrochloric acid. Hydrochloric acid contains H ions and Cl ions. During electrolysis, H ions move towards the cathode (-ve electrode). So, H ions are called cations. Similarly Cl ions move towards the anode (+ve electrode). So, Clions are called anions.

 

Anodic reaction: At the anode, Cl ions get oxidized to chlorine atoms by the loss of

 

electrons.

 

Cl Cl + e (oxidation)

 

2Cl Cl (gas)

 

Chlorine gas is liberated at the anode.

 

Anode Cathode

 

+ -

 

HCI

 

Solution

 

Cathodic reaction: At the cathode, H ions get reduced to hydrogen atoms by gain of

 

electrons. H + e H (reduction) 2H H (gas) Hydrogen gas is liberated at the cathode. Thus, hydrochloric acid decomposes into hydrogen and chlorine. Electrolysis depends on the following factors: (i) Nature of electrodes used and (ii) Physical nature of electrolytes

 

used.

 

Electrolysis is applied in

 

(i) Electroplating

 

(ii) Anodization ofAluminium

 

(iii) Electrolytic refining of metals.

 

Electroplating is coating of a more noble metal over a less noble metal by electrolysis. Electroplating is done for the following purpose.

 

(a) To make the surface corrosion resistant.

 

(b) To improve the surface appearance.

 

In electroplating, The metal which is to be electroplated (base metal) is taken as cathode; the metal to be coated on (coat metal) is taken as anode. A salt solution of coat metal is taken as electrolyte. Example: Chrome plating, silver plating, copper plating, gold plating etc. It is essential to clean the article thoroughly before applying a coating. The cleaning of the article is called as 'preparation of surface'. First, a surface is buffed with emery sheet to get a polished (cleaned) surface.

 

4.1.5 Industrial Applications of Electrolysis

 

4.1.6 Electroplating

 

4.1.7 Preparation of surface

 

The surface is then washed with solvents like acetone to remove oil

 

and grease. It is then washed with tri-sodium phosphate (TSP) to remove any oil

 

and dirt. It is finally dipped in 3N hydrochloric acid for few minutes to remove any

 

oxide impurities. In between the above operations, the article is washed with water.

 

Finally it is washed thoroughly with demineralised water. The nature, stability and thickness of the coating depends on the following factors:

 

1. Nature of the electrolyte.

 

2. Nature of the electrode.

 

3. Solubility of the electrolyte.

 

4. Concentration of electrolyte solution.

 

5. Temperature.

 

6. Voltage applied (low).

 

7. Current density (high).

 

8. Time for which the current is passed.

 

9. pH of electrolyte solution.

 

Coating of chromium over nickel or copper (coated mild steel) is called chrome plating.

 

Process:

 

1) The nickel or copper coated iron article (base metal) is placed at the cathode.

 

2) Alead-antimony rod is used as the anode.

 

3) A solution of chromic acid and sulphuric acid (100:1) is used as the electrolyte.

 

4) Temperature of the electrolyte solution is maintained at 40 Cto 50 C.

 

5) Acurrent density of 100 – 200 mA/cm is used.

 

6) Sulphate ions act as catalyst for coating.

 

7) When electric current is passed, electrolysis takes place and chromium is deposited over the base metal.

 

4.1.8 Factors affecting the stability of the coating

 

4.1.9 Chrome plating

 

Aschematic diagram of coating of chromium is given below.

 

Electroless plating is a technique of depositing a noble metal (from its salt solution) on a catalytically active surface of a less noble metal by employing a suitable reducing agent without using electrical energy. The reducing agent causes reduction of metallic ions to metal which gets plated over the catalytically activated surfaces giving a uniform thin coating.

 

Metal ions + Reducing agent Metal + Oxidized products (deposited)

 

Example: Electroless nickel plating.

 

Procedure: The pretreated object (example: Stainless steel) is immersed in the plating bath containing NiCl and a reducing agent, sodium hypophosphite for the required time. During the process, Ni gets coated over the object.

 

Anodic reaction:

 

H PO +H O H PO +2H +2e Cathodic reaction:

 

Ni + 2e Ni

 

 

 

4.1.11 Advantages of Electroless plating over electroplating

 

4.1.12 Applications of Electroless plating

 

SUMMARY

 

QUESTION

 

PART-A(1 Mark)

 

·         No electricity is required for electroless plating.

 

·         Electroless plating on insulators (like plastics, glass) and semiconductors can be easily carried out.

 

·         Complicated parts can also be plated uniformly in this method.

 

·         Electroless coatings possess good mechanical, chemical and magnetic properties.

 

·         Electroless nickel plating is extensively used in electronic appliances.

 

·         Electroless nickel plating is used in domestic and in automotive fields.

 

·         Electroless nickel coated polymers are used in decorative and functional applications.

 

·         Electroless copper and nickel coated plastic cabinets are used in digital as well as electronic instruments.

 

·         Electroless copper plating is used in manufacture of double sided and multilayered printed circuits boards (PCB).

 

In this lesson, types of electrolytes, mechanism of electrolysis, industrial applications of electrolysis, preparation of surface, factors affecting coating, electroplating, electro less plating, its advantages and applications are discussed.

 

1. What is an electrolyte?

 

2. Give two examples for strong electrolytes.

 

3. Give two examples for weak electrolytes.

 

4. Define strong electrolyte.

 

5. Define weak electrolyte.

 

6. Define electrolysis.

 

7. Give any two industrial applications of electrolysis.

 

8. What is electroplating?

 

9. Mention any two factors affecting the stability of coating.

 

10. What is chrome plating?

 

11. What is the anode and electrolyte used in chrome plating?

 

12. What is electroless plating?

 

13. Give any two advantages of electroless plating over electroplating.

 

14. Give any two applications of electroless plating.

 

1. Explain electrolysis with a suitable example.

 

2. What are the steps involved in preparation of surface?

 

3. What are the factors affecting the stability of coating?

 

4. Explain electroplating with an example.

 

5. Describe chrome plating with a neat diagram.

 

6. Explain electroless plating with an example.

 

7. I) What are the advantages of electroless plating over electroplating?

 

ii) Give the applications of electroless plating.

 

 

 

4.2 CELL

 

4.2.1 Introduction

 

4.2.2 Electrochemical Cell

 

4.2.3 Single electrode potential

 

4.2.4 Galvanic Cell

 

Asystem in which two electrodes are in contact with an electrolyte is called as cell. There are two types of cells,

 

I) Electrolytic Cell

 

ii) Electrochemical cell.

 

Electrolytic cell is a device which produces chemical change by supplying electric current from outside source. Here, electrical energy is converted into chemical energy. Electrochemical cell is a device in which chemical energy from a redox reaction is utilized to get electrical energy. Here, chemical energy is converted into electrical energy.

 

Example: Daniel cell.

 

The measure of tendency of a metallic electrode to lose or gain electrons when in contact with a solution of its own salt in a half cell of an electrochemical cell is called as single electrode potential. The tendency of an electrode to lose electrons is called oxidation potential while the tendency of an electrode to gain electrons is called reduction potential. Galvanic cells are electrochemical cells in which the electrons transferred due to redox reaction, are converted into electrical energy. A galvanic cell consists of two half-cells with each half-cell contains an electrode. The electrode at which oxidation takes place is called anode and the electrode at which reduction occurs is called cathode. The electrons liberated to the electrolyte from the metal leaves the metal ions at anode. The electrons from the solution are accepted by the cathode metal ion to become metal. Galvanic cell is generally represented as

 

follows.

 

M /M ||M /M orM / (Salt ofM ) ||M / (Salt ofM ) 1 1 2 2 1 1 2 2 + +

 

Where, M & M are Anode and Cathode respectively and M & M are the metal ions in respective electrolyte. The symbol || denotes salt bridge. The above representation of galvanic cell is known as galvanic cell diagram.

 

Example: The typical example for galvanic cell is Daniel cell.

 

This cell consists of a zinc rod as anode dipped in zinc sulphate solution (electrolyte) in a glass tank and copper rod as cathode dipped in copper sulphate (electrolyte) in another glass tank. Each electrode is known as half cell. The two half cells are inter-connected by a salt bridge and zinc and copper electrodes are connected by a wire through voltmeter. The salt bridge contains saturated solution of KCl in agar-agar gel. The cell diagram of Daniel cell is

 

Zn / Zn ||Cu / Cu or Zn / ZnSO || /CuSO Cu Redox reaction occurs at Daniel cell:

 

At anode Zn Zn + 2e (Oxidation) At cathode Cu + 2e Cu (Reduction) Overall Cell reaction

 

Zn+Cu Cu+Zn 1 2 1 2 4 4

 

+ +

 

2+

 

2+ -

 

2+ -

 

2+ 2+

 

 

 

4.2.6 Electrochemical series

 

Electrochemical series

 

4.2.7 Significance and applications of electrochemical series

 

When various metals as electrodes are arranged in the order of their increasing values of standard reduction potential on the hydrogen scale, then the arrangement is called electrochemical series.

 

1. Calculation of standard EMFof a cell Standard electrode potential of any cell can be calculated using this series.

 

2. Relative ease of oxidation and reduction Higher the value of standard reduction potential (+ve value) greater is the tendency to get reduced. Thus, metals on the top having more negative (–ve) values are more easily ionized (oxidized).

 

3. Displacement of one element by another

 

Metals which lie higher in the series can displace those elements which lie below them in the series.

 

4. Determination of equilibrium constant for the reaction The equilibrium constant for the cell can be calculated from the standard electrode potential.

 

5. Hydrogen displacement behaviour Metals having more –ve potential in the series will displace hydrogen from acid solutions.

 

6. Predicting spontaneity of redox reactions Spontaneity of redox reaction can be predicted from the standard electrode potential values of the complete cell reaction.

 

The cell which produces electrical energy by transfer of a substance from the solution of higher concentration to the solution of lower concentration is called concentration cell.

 

This is also an electrochemical cell. The difference in concentration may be brought about by the difference in concentration of the electrodes or electrolytes.

 

The concentration cells are classified into two types.

 

I) Electrode concentration cell

 

ii) Electrolyte concentration cell.

 

Two identical electrodes of different concentrations are dipped in the same electrolytic solution of the electrode metal in a cell is called electrode concentration cell.

 

Example: Amalgam concentration cells. Amalgam electrodes are produced by mixing various proportions of lead and mercury. It is represented as,

 

Hg – Pb(C ) /PbSO || Hg-Pb(C ) Where,C &C are concentrations of electrolytes Two identical electrodes of same concentrations are dipped in the electrolytic solutions of different concentration in a cell is called electrolyte concentration cell.

 

 

 

4.2.8 Concentration cell

 

Electrode concentration cell:

 

Electrolyte concentration cell:

 

Example: Silver ion concentration cell

 

The diagram of an electrolytic concentration cell is

 

Ag / Ag (C ) || Ag (C ) / Ag (C >C )

 

Diluted Concentrated In this lesson, electrochemical cells, single electrode potential, galvanic cell, construction and working of Daniel cell, significance and applications of electrochemical series and two types of concentration cells are discussed.

 

1. What is an electrochemical cell?

 

2. Give two examples for electrochemical cell.

 

3. Define single electrode potential.

 

4. What is galvanic cell?

 

5. Write an example for a galvanic cell.

 

6. What is Daniel cell?

 

7. How will you write a short representation of a Daniel cell?

 

8. Define electrochemical series.

 

9. Write any two applications of electrochemical series.

 

10. Define concentration cell.

 

11. What are the types of concentration cells? Give examples.

 

12. Give an example for electrode concentration cell.

 

13. Give an example for electrolyte concentration cell.

 

 

 

4.3 STORAGE BATTERIES

 

4.3.1 Introduction

 

Primary battery

 

Secondary battery

 

Fuel battery or Flow battery

 

4.3.2 Dry Cell

 

A device that stores chemical energy and releases it as electrical energy is called as battery or storage battery. A battery is an electrochemical cell which is often connected in series in electrical devices as a source of direct electric current at a constant voltage.

 

Batteries are classified as follows,

 

I) Primary battery

 

ii) Secondary battery

 

iii) Fuel battery or Flow battery

 

Primary battery is a cell in which the cell reaction is not reversible. Thus, once the chemical reaction takes place to release the electrical energy, the cell gets exhausted. They are use and throw type. Example: Dry cell, Laclanche cell etc. Secondary battery is a cell in which the cell reaction is reversible. They are rechargeable cells. Once the battery gets exhausted, it can be recharged. Example: Nickel-Cadmium cell, Lead-acid cell (storage cell), etc. Flow battery is an electrochemical cell that converts the chemical reaction into electrical energy. When the reactants are exhausted, new chemicals replace them. Example: Hydrogen-oxygen cell,Aluminium-air cell, etc. In Aluminium-air cell, when the cell is exhausted, a new aluminium rod is used and the solution is diluted with more water as the electrochemical reaction involves aluminium and water. Acell without fluid component is called as dry cell.

 

Example: Daniel cell, alkaline battery.

 

Construction and working: The anode of the cell is zinc container containing an electrolyte consisting of NH Cl, ZnCl and MnO to which starch is added to make it thick paste-like so that is less likely to leak. A graphite rod serves as the cathode, which is immersed in the electrolyte in the centre of the cell. The electrode reactions are given below.

 

The dry cell is a primary battery, since no reaction is reversible by supplying electricity. Dry cell is very cheap to make. It gives voltage of about 1.5V.

 

Matal cap (positive)

 

Insulating washer

 

Collar to keep rod in

 

Zinc cup (negative)

 

Carbon rod

 

Metal cover (negative)

 

Mixture of manganese (iv)

 

oxide. graphite, ammonium chloride and zinc chloride

 

 

 

But, it has few demerits: i) When current is drawn rapidly, drop in voltage occurs. ii) Since the electrolyte is acidic, Zn dissolves slowly even if it is not in use.

 

 

 

Uses

 

Dry cells are used in flash-lights, transistor radios, calculators, etc. The typical example for storage cell is Lead-acid storage cell. It is a secondary battery which can operate as a voltaic cell and as an electrolytic cell. When it acts as a voltaic cell, it supplies electrical energy and becomes run down. When it is recharged, the cell operates as an electrochemical cell.

 

Construction and Working: Alead – acid storage cell consists of a number of voltaic cells (3 to 6) connected in series to get 6 to 12 V battery. In each cell, a number of Pb plates used as anodes are connected in parallel and a number of PbO plates used as cathodes are connected in parallel. The plates are separated by insulators like rubber or glass fibre. The entire combinations are immersed in dil.H SO .

 

The cell is represented as Pb |PbSO ||H SO |PbO |Pb When the lead-acid storage battery operates, the following cell reactions occur.

 

Anodic reaction:

 

Lead is oxidized to Pb ions, which further combines with SO

 

forms insoluble PbSO .

 

Pb (s) +SO PbSO (s) + 2e

 

Cathodic reaction:

 

PbO is reduced to Pb ions, which further combines with SO

 

forms insoluble PbSO .

 

PbO (s) + 4H +SO + 2e PbSO (s) + 2H O

 

Overall cell reaction during discharging:

 

Pb (s) +PbO (s) + 2H SO (aq) PbSO (s) + 2H O+ Energy

 

From the above cell reactions, it is clear that PbSO is precipitated at both the electrodes and the concentration of H SO decreases. So, the battery needs recharging. Overall cell reaction during recharging: The cell can be charged by passing electric current in the opposite direction. The electrode reaction gets reversed. As a result, Pb is deposited on anode and PbO on the cathode. The concentration of H SO also increases.

 

2PbSO (s) + 2H O+ Energy— Pb (s) +PbO (s) + 2H SO (aq)

 

Advantages of Lead – acid batteries:

 

1. It is made easily.

 

2. It produces very high current.

 

3. The self discharging rate is low.

 

4. It works effectively even at low temperatures.

 

Uses:

 

1. Lead – acid batteries are used in cars, buses and trucks etc.

 

2. It is used in gas engine ignition, telephone exchanges, power stations

 

etc.

 

Anickel – cadmium storage cell consists of Cadmium as anode and

 

NiO paste as cathode andKOHas the electrolyte.

 

 

 

Cd | Cd(OH) ||KOH(aq) | NiO |Ni

 

Construction andWorking:

 

When the nickel battery operates, Cd is oxidized to Cd ions at

 

anode and the insoluble Cd(OH) is formed. NiO is reduced to Ni ions

 

which further combines with OH ions to form Ni(OH) . It produces about

 

1.4 V. The following cell reactions occur.

 

Anodic reaction:

 

Cd (s) +2OH Cd(OH) (s) + 2e

 

Cathodic reaction:

 

NiO (s) + 2H O+ 2e Ni(OH) (s) +2OH

 

Overall cell reaction during discharging:

 

Cd (s) + NiO (s) + 2H O Cd(OH) (s) + Ni(OH) (s) + Energy From the above cell reactions, it is clear that Cd(OH) and Ni(OH) are deposited at both the anodes and cathodes respectively. So, this can be reversed by recharging the cell.

 

Overall cell reaction during recharging: The cell can be charged by passing electric current in the opposite direction. The electrode reactions get reversed. As a result, Cd is deposited on the anode and NiO on the cathode. Cd(OH) (s) + Ni(OH) (s) + Energy Cd (s) + NiO (s) + 2H O Advantages of Ni-Cd battery:

 

1. It is portable and rechargeable cell.

 

2. It has longer life than lead – acid battery.

 

3. It can be easily packed like dry cell since it is smaller and lighter.

 

Uses:

 

1. It is used in calculators.

 

2. It is used in gas electronics flash units.

 

3. It is used in transistors, cordless electronic appliances, etc.

 

A typical example of pollution free cell is H -O fuel cell in which the

 

fuel is hydrogen and the oxidizer is oxygen.

 

A full cell converts the chemical energy of the fuels directly to

 

electricity. The essential process in a fuel cell is

 

Fuel + Oxygen Oxidation products + Electricity

 

Construction and working:

 

Hydrogen – oxygen fuel cell consists of two porous electrodes made

 

up of compressed carbon coated with small amount of catalysts (Pt, Pd,

 

Ag) andKOHorNaOHsolution as the electrolyte.

 

During working, Hydrogen (the fuel) is bubbled though the anode

 

compartment, where it is oxidized. The oxygen (oxidizer) is bubbled

 

though the cathode compartment, where it is reduced. The following cell

 

reactions occur.

 

Anodic reaction:

 

2H +4OH 4H O+ 4e

 

Cathodic reaction:

 

O + 4H O+ 4e 4OH

 

Overall cell reaction:

 

2H (g) +O (g) 4H O(l)

 

Electrolyte

 

Porous carbon

 

electrodes

 

Anode Cathode

 

From the above cell reactions, hydrogen molecules are oxidized to water. When a large number of fuel cells are connected in series, it is called fuel battery.

 

Advantages of fuel cells:

 

1. Fuel cells are efficient and take less time for operation.

 

2. No harmful chemicals are produced in fuel cells.

 

Uses

 

1. It is used as auxiliary energy source in space vehicles, submarines etc.

 

2. It is used in producing drinking water for astronauts in the space.

 

Adevice which converts the solar energy (energy obtained from the sun) directly into electrical energy is called 'Solar cell'. This is also called as 'Photovoltaic cell'.

 

Principle:

 

The basic principle involved in the solar cells is based on the photovoltaic (PV) effect. When sun rays fall on the two layers of semiconductor devices, potential difference between the two layers is produced. This potential difference causes flow of electrons and thus produces electricity. Example: Silicon solar cell Construction: Solar cell consists of a p-type (such as Si doped with boron) and a ntype (such as Si doped with phosphorous). They are in close contact with each other.

 

4.3.6 Solar cell

 

e- e-

 

n-type semiconductor

 

p-type semiconductor

 

93

 

Working:

 

When the solar rays fall on the top layer of p-type semiconductor, the electrons from the valence band get promoted to the conduction band and cross the p-n junction into n-type semiconductor. Thereby potential difference between two layers is created, which causes flow of electrons (i.e. electric current). The potential difference and hence current increases as more solar rays falls on the surface of the top layer. Thus, when this p- and n- layers are connected to an external circuit, electrons flow from n-layer to p-layer and hence current is generated. Applications of solar cells:

 

1. Solar cells are used in street lights.

 

2. Water pumps are operated by using solar batteries.

 

3. They are used in calculators, watches, radios and TVs.

 

4. They are used for eco-friendly driving vehicles.

 

5. Silicon Solar cells are used as power source in space crafts and satellites.

 

6. Solar cells can even be used in remote places and in forests to get electrical energy without affecting the atmosphere. In this lesson, various types of batteries, construction, working with cell reactions of storage batteries like, dry cell, lead - acid cell, Ni - Cd cell,

 

H -O fuel cell, solar cell and their uses are discussed.

 

1. Define a storage battery.

 

2. What is a primary battery? Give example.

 

3. What is a secondary battery? Give example.

 

4. Define a fuel cell.

 

5. What is dry cell? Give an example.

 

6. Write short representation of lead - acid storage cell.

 

7. Give any two fuel batteries.

 

8. What is a green fuel cell? Why is it called so?

 

9. Give any two use of lead – acid battery.

 

10. What is a solar cell?

 

11. Give any two applications of solar cells.

 

 

 

Part -B

 

1. Explain construction and working of dry cell with example.

 

2. Explain the construction and working of lead – acid battery.

 

3. Describe a nickel – cadmium battery with cell reactions.

 

4. What is green fuel cell? Explain its working.

 

5. Explain the construction and working of flow battery with example.

 

6. Write a note on solar cell.

 

7. Explain the uses of solar cells.

 

 

 

 

 

UNIT V

 

CORROSION ENGINEERING

 

5.1 CORROSION

 

5.1.1 Introduction

 

5.1.2 Types of Corrosion:

 

Corrosion is a 'billion dollar thief'. Even though it is a natural phenomenon in which the gases present in the atmosphere react chemically with metals to convert them into their salts, it results in loss of material and money. Metals have a strong crystalline structure and when they are converted into their salts they lose the metallic strength resulting in the damage to machineries in which they are used. Thus corrosion causes damage to metals and there by to the society. The estimate of loss due to corrosion is approximately 2.5 billion dollars / annum all over the world. Hence it is necessary to understand the mechanism of corrosion. In this lesson we are going to study about the causes and the mechanism of corrosion so that we can find ways to prevent this social enemy. Corrosion is defined as the deterioration of a metal by chemical or electro chemical reactions with its environment. Due to corrosion the useful properties of a metal like malleability, ductility, electrical conductivity and also the surface appearance are lost. The most familiar example of corrosion is rusting of iron when exposed to atmospheric conditions. Another example is the formation of green film or basic Carbonate

 

[CuCO +Cu(OH) ] on the surface of copper when exposed to moist air containing CO .O

 

1. Chemical Corrosion or Dry Corrosion

 

2. Electro chemical Corrosion orWet Corrosion

 

3 2

 

2 2

 

 

 

(1) Chemical Corrosion or Dry Corrosion

 

(2) Electro Chemical Corrosion or wet corrosion

 

5.1.3Galvanic Cell Formation Theory

 

Anodic reaction anode.

 

The direct chemical action of atmospheric gases like oxygen, halogen, H S etc in a dry environment on metals, a solid filim of the Corrosion produced is formed on the surface of the metal. This is known as chemical Corrosion. A solid film of the corrosion product is formed on the surface of the metal which protects the metal from further corrosion. If a soluble or volatile corrosion product is formed, then the metal is exposed to further attack. 'For example, chlorine attack silver generating a protective film of silver chloride on the surface. 2 Ag+Cl 2AgCl. Wet corrosion occurs due to the electrochemical action of moisture and oxygen on metals. Corrosion of very important metal, iron takes place due to electrochemical attack.

 

There are two theories proposed to explain rusting of iron.

 

1) Galvanic cell formation theory

 

2) Differential aeration theory (concentration cell theory)

 

When iron piece with impurity is exposed to atmosphere, a mini galvanic cell is formed.

 

Iron with atmosphere forms one electrode. Impurity (copper, tin, dirt, etc) with atmosphere forms another electrode.

 

Iron which is more electropositive acts as Therefore iron is oxidized to ferrous ions (Fe ions) by the removal of electrons. Ferrousions combine with hydroxide ions to form ferrous hydroxide by atmosphere. Finally ferric hydroxide decomposes to form ferric oxide,

 

which is nothing, but rust.

 

The electrons released at the anode are absorbed at the cathode to form either hydrogen or water or hydroxide ion depending on the nature of the atmosphere.

 

This completes the formation of the cell, which favours rusting of iron.

 

Depending on the nature of the atmosphere the following reactions take place at the cathode.

 

When the atmosphere is acidic and contains no oxygen, hydrogen will be given out.

 

2H +2e H When the atmosphere is acidic and contains more oxygen water will be given out.

 

2H + [O]+2e H O When the atmosphere is basic or neutral and contains no oxygen, OH andH will be given out. 2H O+2e H 2OH When the atmosphere is basic or neutral and contains more oxygen OH will be given out. 2H O+O +4e 4OH

 

According to this theory, corrosion occurs due to the development of concentration cell formed by varying concentration of oxygen or any electrolyte on the surface of the metal.

 

Thus,

 

Cathodic reaction

 

(2) Differential aeration theory or concentration cell formation

 

theory

 

The less oxygenated area acts asAnode (gets corroded) The more oxygenated area acts as the Cathode ( Protected from Corrosion)

 

Reaction At anode (less oxygenated area)

 

Fe Fe + 2e (oxidation or corrosion)

 

At the cathode (more oxygenated area)

 

2H o +O +4e 4OH (Reduction)

 

Fe +2OH Fe(OH)

 

Fe(OH) is further oxidized to Fe(OH) . Since the anodic area is small and the cathodic area is large, corrosion is more concentrated at the anode. Thus, a small hole is formed on the surface of the metal.

 

1. Corrosion noted on the barbed wire fencing, In a wire fence, the areas where the wires cross are less accessible to air than the rest of the fence and hence corrosion takes place at the wire crossings which are anodic.

 

2. Corrosion noted in the iron water tanks near the water level water line corrosion.

 

In iron water tanks , iron portion inside the water is less expose to the oxygen when compared to other portions. Thus a concentration cell is formed and iron rusting takes place at the water level. (the place where the anode and cathode meet). The rust spreads when the water evaporates. This type of corrosion is called water line corrosion.

 

3. When a drop of water or salt solution is placed over an iron piece corrosion occur at the ridge of the water drop Areas covered by droplets, having less access of oxygen become anodic with respect to the other areas which are freely exposed to air. The factors that affect he rate of corrosion are

 

1) Factors connected with the metal and its surface

 

2) Factors connected with the atmosphere

 

3) Factors connected with the corrosion product

 

The type of impurity present in it and its electropositive nature decides the corrosion of a metal. For example when iron has impurities like copper, tin, etc. iron corrodes since iron is more electropositive than metals like copper and tin. On the other hand when iron is coupled with zinc, zinc corrodes since zinc is more electropositive than iron.

 

Generally pure metal does not corrode, as there is no cathode spot available to induce corrosion. A rough surface corrodes readily as it collects more dirt and provides more cathode spot for corrosion. A polished surface does not corrode easily.

 

 

 

4) Stress corrosion

 

5) Anode to cathode area ratio.

 

6) Physical state of a metal.

 

Factors connected with the atmosphere

 

Examples

 

Stress in a metal surface is produced by mechanical workings such as quenching, pressing, bending, and riveting, improper heat treatment etc. The portion subjected to more stress acts as anode and other portion act as cathode. This leads to the formation of stress corrosion. Stress corrosion is noted in fabricated articles of certain alloys like high zincbrasses and nickel brasses. Caustic embrittlement noted in boilers is a typical example for stress corrosion, which is due the attack of alkali present in water on stressed boiler metal. When a bigger cathode area covers a smaller anode area, severe corrosion is noted in the anode spot. This is called erosion. It is frequently encountered in piping agitators, condenser tubes etc. where turbulent flow of gases and vapors remove the coated surfaces resulting in differential cells. Removal of surface coatings can also be caused by rubbing or striking activities of solids on the coated surfaces. The rate of corrosion is influenced by grain size, Orientation of crystals, stress etc.The smaller the grains size of the metal greater the rate of corrosion.

 

1) Nature of the atmosphere

 

2) Temperature of the atmosphere

 

3) pH of the atmosphere

 

4) Amount of moisture in the atmosphere

 

5) Amount of oxygen in the atmosphere

 

6) Amount of chemical fumes in the atmosphere etc.

 

1. Buried pipelines and cables passing from one type of soil to another suffer corrosion due to differential aeration.

 

2. Lead pipe lines passing through clay and then through sand.

 

3. Lead pipe line passing through clay get corroded because it is less aerated than sand.

 

 

 

TESTYOURUNDERSTANDING

 

a. In some causes the corroded product sticks to the surface and absorbs more moisture. This induces further corrosion. E. g Rusting of iron,As rust formed over iron absorbs more moisture, rate of rusting of iron increases.

 

b. In some cases the corroded product acts as the protective coating which prevents further corrosion.

 

E g Aluminium oxide formed over the surface of aluminium prevents further corrosion and act as a protective coating. This is the basic principle of anodization.

 

c. In some other cases the corroded product falls out of position exposing the fresh metal surface for further corrosion.

 

E g Magnesium Oxide formed over the surface of Magnesium falls out of position exposing a fresh surface for further corrosion. In this lesson various types of corrosion, theories explaining corrosion and factors influencing corrosion are explained.

 

·         What is corrosion?

 

·         What is dry corrosion?

 

·         What is wet Corrosion?

 

·         What type of corrosion takes place in a metal when anode is small and cathode is large. Why ?

 

·         Explain the Galvanic cell formation theory.

 

·         Explain the differential aeration theory with suitable examples.

 

·         What are the factors affecting the rate of corrosion.

 

·         Why corrosion often takes place under metal washers.

 

·         Welded joints are better than riveted joints. Why?

 

 

 

5.2 METHODS OF PREVENTION OF CORROSION

 

5.2.1 Modifying the Environmental Conditions.

 

5.2.2 Alloying:

 

5.2.3 Surface Coating:

 

Objectives of Coating Surfaces

 

The corrosion rate can be reduced by modifying the environment.

 

The environment can be modified by the following:

 

(a) Deaeration: The presence of increased amounts of oxygen is harmful since it increases the corrosion rate. Deaeration aims at the removal of dissolved oxygen. Disolved oxygen can be removed by deaeration or by adding some chemical substance like Na CO .

 

(b) Dehumidification: In this method, moisture from air is removed by lowering the relative humidity of surrounding air. This can be achieved by adding silica gel which can absorb moisture preferentially on its surface.

 

(c) Inhibitors: In this method, some chemical substance known as inhibitors are added to the corrosive environment in small quantities. These inhibitors substantially reduce the rate of corrosion. Both corrosion resistance and strength of many metals can be improved by alloying, e-g. Stainless steels containing chromium produce a coherent oxide film which protects the steel from further attack. The other non-corrosive alloys are German silver, Aluminium bronze, Nickel bronze, Duralumin etc

 

Corrosion of metal surfaces is a common phenomenon. To protect a metal surface from corrosion, the contact between the metal and the corrosive environment is to be cut off. This is done by coating the surface of the metal with a continues, non-porous material, insert to the corrosive atmosphere. Such a coating is referred to as surface coating or protective coating. In addition to protective action, such coatings also give a decorative effect and reduce wear and tear.

 

1. To prevent corrosion.

 

2. To enhance wear and scratch resistance.

 

3. To increase hardness

 

4. To insulate electrically

 

5. To insulate thermally

 

6. To impart decorative colour.

 

Surfacing coatings made up of metals are known as metallic coatings. These coatings separate the base metal from the corrosive environment and also function as an effective barrier for the protection of base metals.

 

The metal which is coated upon is known as the base metal.

 

The metal applied as coating is referred to as coat metal.

 

The different methods used for metal coating are.

 

1. Hot dipping

 

(a) Galvanization

 

(b) Tinning

 

2. Metal spraying.

 

3. Cladding.

 

4. Cementation

 

(a) Sherardizing – Cementation with Zinc powder is called Sherardizing.

 

(b) Chromizing - Cementation with 55% Chromium powder & 45% Alumina is called chromizing

 

(c) Calorizing – Cementaion withAluminium andAlumina powder is called Calorizing

 

5. Electroplating or electrodeposition. In the process of hot dipping, the metal to be coated is dipped in the molten bath of the coating metal. Such hot dip coatings are generally nonuniform. The common examples of hot dip coatings are galvanizing and tinning. The process of coating a layer of zinc on iron is called galvanizing. The iron article is first pickled with dilute sulphuric acid to remove traces of rust, dust, etc. at 60-90'c for about 15 -20 minutes. Then this metal is dipped in a molten zinc bath maintained at 430'c.

 

 

 

5.2.4 Metallic Coating:

 

1.Hot dipping

 

(a) Galvanizing:

 

104

 

The surface of the bath is covered with ammonium chloride flux to prevent oxide formation on the surface of molten zinc. The coated base metal is then passed through rollers to correct the thickness of

 

the film. It is used to protect roofing sheets, wires, pipes, tanks, nails, screws, etc. The coating of tin on iron is called tin plating or tinning. In tinning, the base metal is first pickled with dilute sulphuric acid to remove surface impurities. Then it is passed through molten tin covered with zinc chloride flux. The tin coated article is passed through a series of rollers immersed in a palm oil bath to remove the excess tin. Tincoated utensils are used for storing foodstuffs, pickles, oils, etc. Galvanizing is preferred to tinning because tin is cathodic to iron, whereas zinc is anodic to iron. So, if the protective layer of the tin coating has any cracks, iron will corrode. If the protective layer of the zinc coating has any cracks, iron being cathodic does not get corroded. The corrosion products fill up the cracks, thus preventing corrosion.

 

(b) Tinning:

 

5.2.5Differences between Galvanizing and Tinning.

 

Galvanising Tinning

 

1.A process of covering iron with a thin coat of 'Zinc' to prevent it from rusting.

 

2.Zi n c p r o t e c t s t h e i r o n sacrificially.(Zinc undergo corrosion)

 

3.Zinc continuously protects the base metal even if broken at

 

some places.

 

4.Galvanized containers cannot be used for strong acidic food stuffs as Zinc becomes toxic in acidic medium. Tin is non-toxic in nature of any

 

medium. A process of covering iron with a thin coat of 'tin' to prevent it from corrosion. Tin protects the base metal with out undergo any corrosion (non sacrificially) A break in coating causes rapid corrosion of base metal.

 

105

 

5.2.6Electroplating:

 

Objectives of Electroplating:

 

Process

 

Electroplating is process in which the coat metal is deposited on the base metal by passing a direct current through an electrolytic solution.

 

a. To increase corrosion resistance.

 

b. To get better appearance.

 

c. To get increased hardness.

 

d. To change the surface properties of metals and non - metals.

 

In electroplating, the cleaned base metal is made as the cathode and the coat metal is taken as the anode.Asolution of the coat metal salt is taken as the electrolyte. The electrodes are connected to a battery and DC current passed. Now electrolysis takes place and the coat metal deposited over the base metal. The nature of coating depends on 1) the current density 2) time 3) temperature and 4) the concentration of the electrolyte. For example, to coat silver on copper material, the copper material is taken as the cathode. A silver plate is taken as the anode. Silver thiocyanate solution is the electrolyte. When the electrodes are connected to aDCsource of electricity, silver is deposited over the copper

 

material. The following electrolytes are used for coating other metals.

 

Copper sulphate – Copper

 

Nickel sulphate – Nickel

 

Chromic acid – Chromium

 

 

 

Factors affecting electroplating

 

5.2.7 Anodizing:

 

The following are the factors affecting electroplating:

 

1. Cleaning of the article is essential for a strong adherent electroplating.

 

2. Concentration of the electrolyte is a major factor in electroplating.

 

3. Low concentration of metal ions will produce uniform coherent metal deposition.

 

4. Thickness of the deposit should be minimized in order to get a strong adherent coating.

 

5. Additives such as glue and boric acid should be added to the electrolytic bath to get a strong adherent and smooth coating.

 

6. The electrolyte selected should be highly soluble and should not undergo any chemical reaction.

 

7. The pH of the electrolytic bath must be properly maintained to get the deposition effectively. Anodizing is the process of coating the base metal with an oxide layer of the base metal.

 

This type of coating is produced on metals like Al, Zn, Mg and their alloys by anodic oxidation process, by passing direct electric current though a bath in which the metal is suspended from anode. Here the base metal behaves as an anode. The electrolytes are sulphonic, chromic, phosphonic, oxalic or boric acid. Anodised coating have more corrosion resistance due to thicker coating.

 

'Aluminium oxide coatings” are formed by the oxidation taking place on the aluminium surface at moderate temperatures (35 to 40 c) and moderate current densities. The formed oxide film is initially thin and gain thickness by the continous oxidation of aluminium anode. The surface of oxide film contains pores, which may cause corrosion. The pores can be sealed by exposing to boiling water, when the oxide is converted into monohydrate (Al O .H O). This process is called sealing process.

 

 

 

5.2.8 Phosphating:

 

5.2.9 Cathodic Protection:

 

(a) SacrificialAnodic Method

 

Phosphate coatings are produced by the reaction between base metal and aqueous solution of phosphoric acid with accelerators (copper salt).Accelerators are used to enhance the rate of the reaction. Such coatings are applied on iron, steel, zinc, aluminium, cadmium and tin. After the reaction the surface film consists of crystalline zinc iron or manganese iron phosphates. Phosphate coatings do not prevent corrosion completely, they are principally used as an adherent base primer-coat for paint, lacquers, oils etc.

 

The principle involved in cathodic protection is to force the metal behave like a cathode. Since there will not be any anodic area on the metal, corrosion does not occur. There are two types of cathodic protection.

 

(a) Sacrificial anodic method.

 

(b) Impressed voltage method.:

 

In this technique, a more active metal is connected to the metal structure to be protected so that all the corrosion is concentrated at the more active metal and thus saving the metal structure from corrosion. This method is used for the protection of sea going vessels such as ships and boats. Sheets of zinc or magnesium are hung around the hull of the ship. Zinc and magnesium being anodic to iron get corroded. Since they are sacrificed in the process of saving iron (anode), they are called

 

Chromic acid Anode Lead cathode sacrificial anodes. The corroded sacrificial anode is replaced by a fresh one, when consumed completely. Important applications of sacrificial anodic protection are as follows:

 

(I) Protection from soil corrosion of underground cables and pipelines.

 

(ii) Magnesium sheets are inserted into domestic water boilers to prevent the formation of rust.

 

In this method, an impressed current is applied in an opposite direction to nullify the corrosion current and converting the corroding metal from anode to cathode. This can be accomplished by applying sufficient amount of direct current from a battery to an anode buried in the soil and connected to the corroding metal structure which is to be protected. The anode is in a backfill (composed of gypsum) so as to increase the electrical contact with the soil. Since in this method current from an external source in impressed on the system, this is called impressed current method. Buried pipe made cathode

 

(protected) This type of protection is given in the case of buried structures such as tanks and pipelines.

 

 

 

5.3 ORGANIC COATINGS

 

5.3.1 Introduction

 

5.3.2 Paint

 

5.3.3Components of paints and their functions

 

1. Pigment:

 

Functions:

 

Organic coatings include paints and varnishes. In the this lesson we are going to study about paint and its components A little introductions to special paints used also discussed. further we are going to study about varnish, its types and their preparation .A preliminary idea is also given

 

about dyes. Paint is a dispersion of a pigment in medium oil (vehicle). When paint is applied on a surface, the medium oil saves the surface from corrosion. The pigment saves the medium oil from the ultra violet light given by the sun.

 

The important constituents of paint are as follows.

 

1. Pigment

 

2. Drying oil or medium oil or vehicle

 

3. Thinner

 

4. Drier

 

5. Filler or extender

 

6. Plasticizer

 

7. Antiskinning agent

 

Apigment is a solid and colour-producing substance in the paint.

 

The following are the functions of the pigment:

 

(a) A pigment gives opacity and colour to the film.

 

(b) A pigment gives strength to the film.

 

(c) It protects the film by reflecting the destructive ultraviolet rays.

 

(d) It covers the manufacturing defects

 

 

 

2. Drying oil or medium oil or vehicle.

 

Functions:

 

4. Thinner:

 

Functions:

 

4. Drier:

 

Functions:

 

5. Filler or extender:

 

The liquid portion of the which the pigment is dispersed is called a medium or vehicle. E.g. linsed oil, dehydrated castor oil, soyabean oil and fish oil.

 

(a) Vehicles hold the pigment particles together on the surface.

 

(b) They form the protective film by evaporation or by oxidation and polymerization of the unsaturated constituents of the oil.

 

(c) Vehicles give better adhesion to the surface.

 

(d) They impart water repellency, durability and toughness to the film.

 

Thinners are added to paints to reduce the viscosity of the paints so that they can be easily applied to the surface. E.g. turpentine and petroleum sprit.

 

(a) Thinners reduce the viscosity of the paint to render it easy to handle and apply to the surface.

 

(b) They dissolve the oil, pigments etc. and produce a homogeneous mixture.

 

(c) Thinners evaporate rapidly and help the drying of the film.

 

(d) They increase the elasticity of the film.

 

(e) Thinners increase the penetrating power of the vehicle.

 

Driers are used to accelerate the drying of the oil film.

 

Eg Naphthenates of lead and cobalt, Resonates of lead and cobalt.

 

Driers act as oxygen carrier catalysts which help the absorption of oxygen and catalyze the drying of the oil film by oxidation, polymerization and condensation.

 

Fillers are added to reduce the cost and increase the durability of the paint, E.g. talc, Gypsum, mica, asbestos etc.

 

Functions:

 

6. Plasticizers:

 

7. Antiskinning agent:

 

5.3.4Varnishes

 

There are two types of varnishes

 

Preparation of oilVarnish

 

(a) Fillers serve to fill the voids in the film.

 

(b) They reduce the cracking of the paints.

 

(c) Fillers increase the durability of the paints.

 

(d) They reduce the cost of the paint.

 

Plasticizers are chemicals added to paints to give elasticity to the film and to prevent cracking of the film, e.g. triphenyl phosphate, tertiary amyl alchol,Tributyl Phthalate. They are chemicals added to the paint to prevent skinning of the paint, E.g. polyhydroxy phenols.

 

Varnish is a homogenous colloidal dispersion of natural or synthetic resin in oil or thinner or both. It is used as a protective and decorative coating to surfaces. It provides a hard, transparent, glossy, lustrous and durable film to the coated surface.

 

1. Oil Varnish

 

2. Spirit Varnish

 

The resin is taken in an aluminium vessel, known as a kettle, and heated The resin melts and the temperature is slowly increased to about 300 . This process, is known as gum running, Some cracking or depolymerization of the resin takes place and about 25 per cent of the resin is lost in the form of pungent fumes. The required quantity (about 25 per cent of the weight of the resin) of boiled oil or linsed oil along with driers is separately heated to 200 to 220 and is slowly added to the heated resin with constant stirring until thorough combination has taken place. This operation is known as cooking. The kettle is removed from the furnace and allowed to cool, white spirit, which is a thinner is added to varnish when the temperature of the varnish is below the flash point of the thinner with constant stirring during the addition. The varnish is stored in tanks for some days for maturing. Filtered or packed for marketing.

 

Preparation of sprit varnish

 

Difference between paint and varnish

 

5.3.5 Special paints

 

1. Luminescent paints

 

2. Heat Resistance paints

 

Sprit varnish is obtained by dissolving a resin in a sprit. E.g. shellac resin dissolved in methylated sprit, a solution of shellac in alcohol, etc. Resins of trees like Manila, Damar, etc are also used for making varnish. In addition to the normal ingredients some special chemicals are incorporated to paints for some specific purposes. They are commonly known as special paints.

 

Luminescent paints contains luminophor pigments are used for visibility in the dark. They find application in inks, advertising signboards, road marks, number plates of vehicles, watch dials, etc. The active components in luminous paint are specially prepared phosphorescent materials like CaS, ZnS, etc. They absorb light radiations and emit them in the dark. For colour effect in luminous paints, certain chemicals like copper salts (green), silver salts (blue), cerium and uranium salts

 

(yellow), etc. are used. When the surfaces are exposed to high temperatures such as in chimneys, exhaust pipes, furnaces, oil stills, etc. Oil paints tend to decompose or get charred, they being organic in nature. Then the surfaces become liable for corrosion. To overcome this problem, a suspension of graphite or lamp black in small amounts of drying oils and more thinners can be used. But more recently, silicone paints are used for heat resistance.

 

 

 

3. Fire-retardant paints

 

4. Antifouling paints

 

5. Cement Paint

 

6.Aluminium Paint

 

These are paints containing chemicals which are fire-resistant in nature. In other words, they produce gases like CO NH , HCl,HBr on heating which are themselves non-combustible and do not support

 

combustion, there by minimizing the rate of burning or extinguishing the fire.

 

Oil paints are liable for attack by living organisms because of the organic content in them. So, in places where living organisms are handled or are present, such paints cannot be used. For use in breweries and biochemical laboratories, the paint is mixed with compounds having fungicidal properties. The active ingredients employed are HgO, Cu O, Hg Cl ,DDT, pentachlorophenol, etc.

 

Such paints are called Antifouling paints.

 

Cement paint is the coating, which is applied on plastered brickwork, concrete work, etc. The ingredients are

 

1. White cement (about 70%)

 

2. Hydrated lime [Ca(OH) ]

 

3. Pigment ( a colouring agent)

 

4. Very fine sand (an inert filler) and

 

5. Water- repellent compound

 

Such paints of different colors are marked in powder form (eg Snowcem, Smocem). The powder is mixed with a suitable quantity of water to get a thin slurry, and applied on surfaces. For good results, a 1.5% to 2% aqueous solution of sodium silicate and Zinc sulphate is applied as primer coat.

 

The base material in aluminium paint is a fine powder of aluminium. The finely powder of aluminium is suspended in either spirit varnish or in an oil-varnish depending on the requirement. When paints is

 

applied, the thinner evaporates and oil, if any, undergoes oxidation and polymerization. A bright adhering film of aluminium is obtained on the painted surface.

 

Advantages of aluminium paint:

 

7. Distempers

 

Advantages

 

5.3.6DYES

 

1. It possesses a good covering power.

 

2. It imparts very attractive appearance to the surface.

 

3. It has fairly good heat-resistance.

 

4. It has very good electrical resistance

 

5. The painted film is waterproof.

 

6. The electrical surface is visible even darkness.

 

7. Corrosion protection for iron and steel surface is better than all other paints.

 

Distempers are water paints. The ingredients of distemper are

 

·         Whitening agent or chalk powder (the base)

 

·         Glue or casein (the binder)

 

·         Colouring pigment and

 

·         Water (the solvent or thinner).

 

·         Distempers are cheaper than paints and varnishes

 

·         They can be applied easily on plasters and wall surfaces in the interiorof the buildings.

 

·         They are durable.

 

·         They give smooth and pleasing finish to walls.

 

·         They have good covering power.

 

Dyes are coloured organic compounds capable of colouring various

 

materials and articles.

 

According to commercial classification of dyes which is based on the

 

application of the dyes are as follows.

 

1. Direct dyes

 

2. Basic dyes

 

3. Mordant dyes etc.

 

 

 

Acid Dyes:

 

Basic Dyes.

 

Mordant or Adjective dyes:

 

Questions

 

Part -A

 

Part -B

 

Salts of organic acids are called acid dyes. It donate colour cellulose fibres. They can readily dye only animal fibres. Eg: Methyl red, Methyl Organe etc. They are salts of colour bases with hydrochloric acid or Zinc chloride. Basic dyes can dye animal fibres directly and vegetable fibres after the fibres are mordanted with tannin. Basic dyes are mostly used for dyeing silk and cotton. Eg: Magenta, Para-rosaniline dye,AnilineYellow. Amordant is any substance, which is fixed to the fibre before dyeing.

 

Commonly used mordants are hydroxides or basic salts of chromium, aluminium or iron.Tannic acid is also used as mordant for basic dyes. Generally the fabric is dipped in a mordant and then in a solution of dye. The dye coat thus formed is insoluble and does not fade on washing.

 

E.g. Alizarin andAnthraquinone dyes.

 

1. Define paint.

 

2. Give examples for heat resistant paint.

 

3. What is Varnish?

 

4. What is the function of the drier in paint?

 

5. What are Fire-retardant paints?

 

6. Define Dyes.

 

 

 

 

 

1. What are the components present in the paint. Explain their functions.

 

2. How is oil varnish prepared ?

 

3. Write a short note on special paints.

 

4. write short notes on Dyes.

 

 

 

TESTYOURUNDERSTANDING

 

1. What are toners?

 

2. What is an enamel?

 

 

 

SEMESTER- I

 

PRACTICAL- I

 

VOLUMETRIC ANALYSIS

 

The method to determine the exact amount of the substance in a given sample is termed as quantitative analysis volumetric analysis is a branch of quantitative analysis involving accurate measurement of volumes of reacting solutions. The volumetric analysis is very much in use due to simplicity rapidity accuracy and wide applicability. The reacting substances are taken in the form of solutions and made to react. The concentration of one solution is determined using another suitable solution whose concentration is accurately known. A known volume of one solution is measured with a pipette and taken in a conical flask. The other solution is taken in a burette and run into the first solution till the chemical reaction is just complete. The volume of the second solution is read from the burette and the two volumes are compared. The process of adding one solution from the burette to another in the conical flask in order to complete the chemical reaction is termed titration. It is the exact stage at which chemical reaction involved in the titration is just complete

 

It is a substance which will show the end point of the reaction by change of colour. For example phenolphthalein and methyl orange are indicators used in acid alkali titrations. Potassium permanganate itself acts as an indicator in potassium permanganate titrations.

 

Acidimentry refers to the titration of alkali with a standard acid and alkalimetry refers to the titration of an acid with a standard alkali.

 

Various terms used in volumetric analysis are given below:

 

Titration

 

Endpoint

 

Indicator

 

Acidimetry andAlkalimetry Titration:

 

Permanganimetry Titration:

 

Normality:

 

Standard solution:

 

Decinormal Solution:

 

Law of volumetric analysis:

 

The titration involving KMnO is called permanganimetry titration. In presence of dilute H SO KMnO oxidizes ferrous sulphate to ferric sulphate and oxidizes oxalic acid toCO andH O.

 

The strength of a solution is expressed in terms of normality.

 

Normality is the number gram equivalent mass of solute dissolved in one litre of solution.

 

A solution of known strength (Normality) is called a standard solution.

 

A solution having the strength of 0.1 N is called decinormal solution. Whenever two Substances react together, they react in the ratio of their equivalent mass. One litre of a normal solution of a substance will react exactly with same volume of a normal solution of another substance. In other words equal volumes of equal normal solutions will exactly react with each other. This result is stated in the form law of volumetric analysis If V ml of a solution of strengthN is required or complete reaction by

 

V ml of the second solution of strengthN then V N =V N

 

If any three factors (V V & N ) are known, the fourth factor N can be calculated. The following are the important formula used in all volumetric estimations

 

Mass of solute per litre of the solution = Equivalent mass x Normality

 

Equivalent mass of some important compounds

 

Name of the compound

 

Equivalent

 

1 ESTIMATION OF SULPHURIC ACID

 

EX.NO

 

 

 

Procedure

 

Titration I:

 

Standardisation of Sodium hydroxide

 

Titration II:

 

Standardisation of Sulphuric acid

 

 

 

To estimate the amount of Sulphuric acid present in 400 ml of the given solution. You are provided with a standard solution of oxalic acid of normality ..............N and an approximately decinormal solution of Sodium hydroxide. (Test solution should be made up to 100 ml) The titration is based on the neutralisation reaction between oxalic acid and Sodium hydroxide in titration I and Sulphuric acid and Sodium hydroxide in titration II. The burette is washed with water, rinsed with distilled water and then with the given oxalic acid.It is filled with same acid up to zero mark. The initial reading of the burette is noted. A 20ml pipette is washed with water,rinsed with distilled water and then with the given Sodium hydroxide solution.20 ml of Sodium hydroxide is pipetted out in to a clean conical flask.Two drops of phenolphthalein indicator is added into the flask.The solution becomes pink in colour. The solution is titrated against oxalic acid taken in the burette.The end point of the titration is the disappearance of pink colour to give colourless solution.The titration is repeated to get the concordant value. From the titre value,the normality of Sodium hydroxide is calculated.

 

The given Sulphuric acid solution is made upto 100 ml in a 100 ml standard flask.The solution is thoroughly shaken to get a uniformly concentrated solution.The burette is washed with water,rinsed with distilled water and then with the given Sulphuric acid from the standard flask. It is filled with same acid upto zero mark. The initial reading of the burette is noted.Exactly 20 ml of the sodium hydroxide is pipetted out in to a clean conical flask.To this solution two drops of phenolphthalein indicator is added. The solution becomes pink in colour.The solution is titrated against Sulphuric acid taken in the burette.The end point of the titration is the disappearance of pink colour to give colourless solution.The titration is repeated to get the concordant value.From the titre value,the normality of Sulphuric acid and the amount of sulphuric acid present in 400 ml of the given solution is calculated.

 

Normality of Sodium hydroxide =...........................N

 

Normality of Sulphuric acid =...........................N

 

Amount of Sulphuric acid present in

 

400ml of the given solution =...........................g

 

Volume of oxalic acid (V ) =

 

Normality of oxalic acid (N ) =

 

Volume of sodium hydroxide (V ) = 20 ml

 

Normality of sodium hydroxide (N ) = ?

 

By the principle of volumetric analysis, V N = V N

 

Normality of sodium hydroxide (N ) = ___________ N

 

Result

 

Titration-I : Sodium hydroxide Vs Oxalic acid

 

Concordant value= Calculation:

 

 

 

SHORT PROCEDURE

 

Volume of sulphuric acid (V ) =

 

Normality of sulphuric acid (N ) = ?

 

Volume of sodium hydroxide (V ) = 20ml

 

Normality of sodium hydroxide (N ) =

 

By the principle of volumetric analysis, V N = V N

 

Normality of sulphuric acid (N )= ____________N

 

Amount of sulphuric acid present

 

400 ml of the given solution =Equivalent mass x Normality

 

of sulphuric acid x 400 / 1000

 

 

 

2. ESTIMATION OF SODIUM HYDROXIDE

 

 

 

Titration I:

 

Standardisation of Sulphuric acid

 

Titration II:

 

Standardisation of Sodium hydroxide

 

To estimate the amount of Sodium hydroxide present in one litre of the given solution.You are provided with a standard solution of Sodium carbonate of normality ..............N and an approximately decinormal solution of Sulphuric acid.(Test solution should be made upto 100 ml )

 

Principle: The titrat ion is based on the neutralisation reaction between

 

Sulphuric acid and Sodium carbonate in titration I and Sulphuric acid and

 

Sodium hydroxide in titration II.

 

The burette is washed with water, rinsed with distilled water and then with the given Sulphuric acid. It is filled with same acid upto zero mark. The initial reading of the burette is noted. A 20ml pipette is washed with water, rinsed with distilled water and then with the given Sodium carbonate solution.20 ml of Sodium carbonate is pipetted out in to a clean conical flask. Two drops of methyl orange indicator is added into the flask.The solution becomes pale yellow in colour. The solution is titrated against Sulphuric acid taken in the burette. The end point of the titration is the change in colour from yellow to permanent pale pink. The titration is repeated to get the concordant value. From the titre value, the normality of Sulphuric acid is calculated.

 

The given Sodium hydroxide solution is made upto 100 ml in a 100 ml standard flask. The solution is thoroughly shaken to get a uniformly concentrated solution. A 20 ml pipette is washed with water, rinsed with distilled water and then with the given Sodium hydroxide. Using the rinsed pipette, exactly 20 ml of the made-up solution is transferred into a clean conical flask. To this solution two drops of methyl orange indicator is added. The solution becomes pale yellow in colour. The solution is titrated against Sulphuric acid taken in the burette. The end point of the titration is the change in colour from yellow to permanent pale pink. The titration is repeated to get the concordant value. From the titre value, the normality of Sodium hydroxide is calculated.

 

(i) Normality of Sulphuric acid =......................N

 

(ii) Normality of Sodium hydroxide =......................N

 

(iii) Amount of Sodium hydroxide present

 

in one litre of given solution =......................g

 

Volume of sulphuric acid (V ) =

 

Normality of sulphuric acid (N ) = ?

 

Volume of sodium carbonate (V ) = 20ml

 

Normality of sodium carbonate (N ) =

 

By the principle of volumetric analysis,V N =V N

 

Normality of sulphuric acid (N )= ___________N

 

 

 

3. COMPARISON OF STRENGTHS OF TWO ACIDS

 

 

 

Titration I:

 

Standardisation of hydrochloric acidA

 

Titration II:

 

Standardisation of hydrochloric acidB

 

To compare the strengths of two hydrochloric acids solutions in bottles A and B and estimate the amount of hydrochloric acid present in 250 ml of the weaker solution. You are provided with a standard solution of sodium hydroxide of normality ..............N.

 

The experiment is based on the neutralisation reaction between hydrochloric acid A and Sodium hydroxide in titration I and hydrochloric acid Band Sodium hydroxide in titration II.

 

The burette is washed with water, rinsed with distilled water and then with the given hydrochloric acid in bottle A. It is filled with same acid upto zero mark. The initial reading of the burette is noted. A 20ml pipette is washed with water, rinsed with distilled water and then with the given

 

Sodium hydroxide solution.20 ml of Sodium hydroxide is pipetted out in to a clean conical flask. Two drops of phenolphthalein indicator is added into the flask. The solution becomes pink in colour. The solution is titrated against hydrochloric acid A taken in the burette. The end point of the titration is the disappearance of pink colour to give colourless solution.

 

The titration is repeated to get the concordant value. From the titre value, the normality of hydrochloric acidAis calculated. The burette is washed with water, rinsed with distilled water and then with the given hydrochloric acid in bottle B. It is filled with same acid upto zero mark. The initial reading of the burette is noted.20 ml of standardised Sodium hydroxide is pipetted out in to a clean conical flask. Two drops of phenolphthalein indicator is added into the flask. The solution becomes pink in colour. The solution is titrated against hydrochloric acid B taken in the burette. The end point of the titration is the disappearance of pink colour to give colourless solution. The titration is repeated to get the concordant value. From the titre value, the normality of hydrochloric acid Bis calculated

 

1. Normality of hydrochloric acid A =......................N

 

2. Normality of hydrochloric acid B =......................N

 

3. Hydro in bottle--------

 

is weaker then

 

4. Amount of hydrochloric acid present in

 

250 ml of the weaker solution =......................g

 

Volume of hydrochloric acidA (V ) =

 

Normality of hydrochloric acidA (N ) = ?

 

Volume of sodium hydroxide (V ) = 20 ml

 

Normality of sodium hydroxide (N ) =

 

By the principle of volumetric analysis, V N = V N

 

Normality of Hydrochloric acidA=........N

 

 

 

4. COMPARISON OF STRENGTHS OF TWO BASES

 

Titration I:

 

Standardisation of Sodium hydroxide (A)

 

Titration II:

 

Standardisation of Sodium hydroxide (B)

 

To compare the normalities of Sodium hydroxide solutions supplied in bottles A and B and to estimate the amount of Sodium hydroxide present in 500 ml of the stronger solution. You are provided with a standard solution of Oxalic acid of normality ..............N. The titration is based on the neutralisation reaction between Oxalic acid and Sodium hydroxide. The burette is washed with water, rinsed with distilled water and then with the given oxalic acid. It is filled with same acid upto zero mark. The initial reading of the burette is noted.A20ml pipette is washed with water, ,rinsed with distilled water and then with the given Sodium hydroxide solution in bottle A.20 ml of Sodium hydroxide A is pipetted out in to a clean conical flask. Two drops of phenolphthalein indicator is added into the flask. The solution becomes pink in colour. The solution is titrated against Oxalic acid taken in the burette. The end point of the titration is the disappearance of pink colour to give colourless solution. The titration is repeated to get the concordant value. From the titre value and normality of Oxalic acid. The normality of Sodium hydroxide in bottleAis calculated. 20 ml of Sodium hydroxide from bottle B is pipette out into a clean conical flask using clean rinsed pipette. To this solution two drops of phenolphthalein indicator is added. The solution becomes pink in colour.

 

The solution is titrated against Oxalic acid taken in the burette. The end point of the titration is the disappearance of pink colour to give colourless solution. The titration is repeated to get the concordant value. From the titre value, the normality of Sodium hydroxide(B) is calculated.

 

 

 

5. ESTIMATION OF MOHR'S SALT

 

 

 

Titration I:

 

Standardisation of Potassium permanganate

 

Titration II:

 

Standardisation of Mohr's salt (ferrous ammonium sulphate)

 

To estimate the amount of crystalline ferrous ammonium sulphate present in 100 ml of the given solution. You are provided with a standard solution of crystalline ferrous sulphate of normality ..............N and an approximately decinormal solution of potassium permanganate. (Test solution should be made upto 100 ml ) The titration is based on the oxidation and the reduction reaction.

 

The oxidising agent i.e Potassium permanganate oxidises the reducing agent ferrous sulphate and ferrous ammonium sulphate in acidic medium to ferric sulphate.

 

The burette is washed with water, rinsed with distilled water and then with the given Potassium permanganate solution. It is filled with same solution upto zero mark. The initial reading of the burette is noted.A20ml pipette is washed with water, rinsed with distilled water and then with the given ferrous sulphate solution.20 ml of ferrous sulphate solution is pipetted out in to a clean conical flask. One test tube full of dilute sulphuric acid (20 ml) is added to it. It is titrated against Potassium permanganate taken in the burette. Potassium permanganate acts as the self indicator. The end point of the titration is the appearance of permanent pale pink colour. The titration is repeated to get the concordant value. From the titre value, the normality of Potassium permanganate solution is calculated. The given Mohr's salt solution is made upto 100 ml in a 100 ml standard flask. The solution is thoroughly shaken to get a uniformly concentrated solution. A 20 ml pipette is washed with water, rinsed with distilled water and then with the given Mohr's salt solution .Using the rinsed pipette,exactly 20 ml of the made-up solution is transferred into a clean conical flask. To this solution one test tube full of dilute sulphuric acid (20 ml) is added. The solution is titrated against standardised potassium permanganate taken in the burette. The end point of the titration is the appearance of permanent pale pink colour. The titration is repeated to get the concordant value. From the titre value, the normality of given Mohr's salt solution and the amount of Mohr's salt in 100 ml of the given solution is calculated.

 

 

 

6. ESTIMATION OF FERROUS SULPHATE

 

 

 

Titration I:

 

Standardisation of Potassium permanganate

 

Titration II:

 

Standardisation of ferrous sulphate

 

To estimate the amount of crystalline ferrous sulphate present in 500 ml of the given solution. You are provided with a standard solution of crystalline ferrous ammonium sulphate of normality ..............N and an approximately decinormal solution of potassium permanganate. The titration is based on the oxidation and the reduction reaction. The oxidising agent i.e Potassium permanganate oxidises both ferrous sulphate and ferrous ammonium sulphate in acidic medium to ferric sulphate. The burette is washed with water, rinsed with distilled water and then with the given Potassium permanganate solution. It is filled with same solution upto zero mark. The initial reading of the burette is noted.A20ml pipette is washed with water, rinsed with distilled water and then with the given ferrous ammonium sulphate solution.20 ml of ferrous ammonium sulphate solution is pipetted out in to a clean conical flask.One test tube full of dilute sulphuric acid is added to it. It is titrated against Potassium permanganate taken in the burette. Potassium permanganate acts as the self indicator. The end point of the titration is the appearance of permanent pale pink colour. The titration is repeated to get the concordant value. From the titre value, the normality of Potassium permanganate solution is calculated. The given ferrous sulphate solution is made upto 100 ml in a 100 ml standard flask. The solution is thoroughly shaken to get a uniformly concentrated solution. A 20 ml pipette is washed with water, rinsed with distilled water and then with the given ferrous sulphate solution .Using the rinsed pipette, exactly 20 ml of the made-up solution is transferred into a clean conical flask. To this solution one test tube full of dilute sulphuric acid (20 ml) is added. The solution is titrated against standardised potassium permanganate taken in the burette. The end point of the titration is the appearance of permanent pale pink colour. The titration is repeated to get the concordant value. From the titre value,the normality of given ferrous sulphate solution and the amount of ferrous sulphate in 500 ml of the given solution is calculated.

 

1.Normality of Potassium permanganate =......................N

 

2.Normality of ferrous sulphate =......................N

 

3.Amount of ferrous sulphate present in 500 ml of given solution=...g

 

Volume of FAS (V ) =20 ml

 

Normality of FAS (N ) =

 

Volume of Potassium permanganate (V ) =

 

Normality of Potassium permanganate (N ) = ?>

 

By the principle of volumetric analysis, V N =V N

 

Normality of Potassium permanganate (N ) =

 

Result

 

Titration-I : Potassium permanganate Vs Mohr's salt (FAS)

 

Concordant value= Calculation:

 

 

 

7. ESTIMATION OF TOTAL HARDNESSS OF WATER

 

Procedure:

 

Titration- I

 

Standarisation ofEDTAsolution

 

Tiration- II

 

Standarisation of hard water

 

Result:

 

To estimate the total hardness of the given sample of water by EDTA titration.

 

The total harness of water can be determined by titrating a known volume of hard water against EDTA solution using Erriochrome Black – T indicator. The estimation is based on the complexometric titration. 50 ml of the given calcium chloride solution is pipetted into a clean conical flask. Half a test tube of ammonia buffer solution is added into the conical flask. A pinch of Erriochrome Black - T indicator is added into the conical flask. The solution turns wine red in colour. It is tirated against EDTA solution taken in a clean, rinsed burette. The end point is the change in colour from wine red to steel blue. The tiration is repeated to get concordant values. From the titre values and the molarity of calcium chloride solution, the molarity of EDTAis calculated. 50 ml of the given sample of hard water is pipetted into a clean conical flask. Half a test tube of ammonia buffer solution is added into the conical flask. A pinch of Erriochrome Black - T indicator is added into the

 

conical flask. The solution turns wine red in colour. It is tirated against EDTA solution taken in a clean, rinsed burette. The end point is the change in colour from wine red to steel blue. The tiration is repeated to get concordant values. From the titre values, the hardness of the given sample of water is calculated in ppm. The total harness of the given sample of water = …………….....Ppm

 

 

 

8. DETERMINATION OF pH AND CALCULATION OF

 

HYDROGEN ION CONCENTRATION

 

Procedure:

 

Result:

 

[H ]CONCENTRATION

 

1. The pH of given solutions in bottlesA,B,C ,D and E

 

2. To calculate hydrogen ion concentration of the solutions.

 

The pH of the solution can be directly measured using a pH meter.

 

Acids give hydrogen ions in solution. The acidic nature of the solution depends on the hydrogen ion concentration which is expressed as gram ions per litre. The pH of the solution varies with concentration of hydrogen ions.

 

PH= - log [H ]

 

Exactly 50 ml of the given five sample solutions are taken in five 150 ml beakers and labeled as A,B,C,D and E. The pH meter is standardized using a known buffer solution. The electrodes are then washed with distilled water and then immersed in the solution taken in the beaker.

 

The pH readings are noted. The pH of all the other solutions are to be determined similarly. The electrodes are washed well with distilled water before the electrodes are immersed in next solution. The amount of hydrogen ions present in the solutions are then calculated from the pH. The amount of pH and hydrogen ion concentration of the given five sample solutions are

 

pH and

 

(1) SampleA……………..g ions / lit

 

(2) Sample B……………..g ions / lit

 

(3) SampleC……………..g ions / lit

 

(4) SampleD……………..g ions / lit

 

1.1.3 Air Pollution

 

Harmful Effects of Air Pollutants

 

1.1.4 Acid Rain

 

“Excess discharge of unwanted harmful substances into the atmospheric air” is known as Air Pollution. Gaseous pollutants like hydrogen sulphide, sulphur dioxide, carbon monoxide, and Nitrogen dioxide etc., are known as the most important primary air pollutants.

 

It means that rain water contains more acids. It is due to the dissolved oxides of sulphur and nitrogen. The gases like SO and NO from industries dissolves in water and form respective acids.

 

Causes respiratory disease, eye irritation and throat troubles, damage to agriculture.

 

Cement industry, mines, glass industry, ceramic industry

 

Combustion of fuels, explosive industry, acid manufacture.

 

Phosphates, fertilizer industry, aluminium industry, brick pottery From industries like petroleum, paper and leather.

 

From petroleum industry, thermal power station, sulphuric acid manufacturing plants.

 

Causes eye irritation severe throat pain, headache, corrosion of metals.. Causes irritation, bone, tooth disorders respiratory diseases.

 

Causes respiratory illness, eye irritation, affects plant growth, smog formation

 

Causes respiratory diseases, affects lungs, affects

 

agriculture, accelerates corrosion.

 

4 Automobile industry, oil refineries, cigarette smoke, etc.

 

Carbon monoxide

 

(CO) Causes Headache, visual difficulty, paralysis.

 

 

 

Harmful effects ofAcid Rain

 

1.1.5 Green House Effect

 

1.1.6 GlobalWarming

 

1. Acid rain makes the soil more acidic thereby reduces the fertility of the soil.

 

2. It affects the growth of crops, plants etc.

 

3. It affects the survival of fishes and reduces the population of aquatic species.

 

4. It badly makes damage to buildings, vehicles, structural materials etc.,

 

5. It affects human being's life system and organs like skin, lungs and hair.

 

6. It damages the memorable monuments, buildings etc., The famous

 

'Taj Mahal' is being affected severely.

 

The earth surface gets warmed due to the blanketing effect of

 

Pollutants like CO2 present in the atmosphere. It is known as Green house effect.

 

Several radiations like UV, visible and infra-red from the sun come in contact with the ozone layer. UV radiation is screened by the ozone layer. The visible and infra-red radiation reaches the earth surface and produce heat energy. The pollutants likeCO and other gases which form a blanket around the earth prevent the heat energy to escape from the surface of the earth. Hence the earth gets warmed. It is similar to green glass houses where heat radiation cannot escape from that

 

Gases that cause Green house effect are mainly CO , Methane, Water Vapour, and Chloro Fluoro Carbon (CFC). They are called Green

 

house gases. The increase in the temperature of the earth surface due to the effect

 

of greenhouse gases is termed as Global Warming.

 

 

 

Effects of Global warming

 

1.1.7 Ozone Layer

 

Importance of ozone layer

 

Causes for depletion of Ozone Layer

 

1. Evaporation process of surface water is enhanced very much due to increase in temperature of earth surface which leads to drastic seasonal change.

 

2. Sea level is increased due to melting of glaciers. Hence low lying land areas will be submerged under sea water.

 

3. Food production is mainly affected and it leads to draught.

 

4. It spreads some diseases like malarial fever, Cholera etc.,

 

5. It causes drastic change in seasons. Hence human beings and animals are mostly affected by climatic change.

 

6. Natural calamities like Cyclones, Hurricane, Typoons and Tsunami may occur frequently and strongly. One of the best gifts given by the nature is ozone layer. It is present about 20 km above the earth's surface. Oxygen is converted into ozone by photo chemical change as follows.

 

The region in whichO density is high is called ozone layer.

 

Ozone layer covers the earth surface and prevents the entry of harmful UV radiation. It saves the lives of human beings and animals. If not so, no life is found on the earth.

 

For depletion of ozone layer chlorine plays a vital role. Chlorine converts the ozone molecules into oxygen in the presence of UV radiation as follows.

 

It is noteworthy that one molecule of chlorine may convert very huge number of molecules of ozone into oxygen. The main source for chlorine is CFC which is released by Aircrafts, Jet planes, Refrigerators, Air- conditioners etc., Other gases which cause ozone layer depletion areNO

 

And NO .

 

1. They affect human beings and cause skin disease and eye defects.

 

2. Due to ozone layer depletion, the harmfulUVradiation may enter freely into the earth's surface and affect the lives in earth.

 

3. They reduce the population of aquatic species.

 

4. They affect growth of plants and vegetables.

 

5. They affect the Eco-system very badly.

 

“Prevention is better than cure” Similarly it is better to control the air pollutants at its source itself. The following are the steps to be taken to control the air pollution.

 

1. The exhaust gases from automobiles and vehicles should be minimized by the use of catalyst.

 

2. Chimneys may be used to reduce the concentration of pollutants at the ground level.

 

3. Smoke may be removed by cottrell's electrostatic precipitator.

 

4. Dust Particles can be removed by the use of bag filters and dust separators.

 

5. Plants take carbon dioxide during photosynthesis and release oxygen to environment. Hence more trees should be planted.

 

6. Non –polluting fuels can be used for energy production. In this lesson air pollution, acid rain, green house effect, global warming, ozone layer depletion, their causes and harmful effects and methods to control air pollution are discussed.

 

 

 

TESTYOURUNDERSTANDING

 

·         What are air pollutants?

 

·         Define air pollution.

 

·         What is acid resin?

 

·         What is green house effect?

 

·         What is global warming?

 

·         What is the reason for ozone layer depletion?

 

·         Mention the sources of air pollutants and their causes.

 

·         Give the harmful effects of acid rain.

 

·         Explain green house effect and give the causes.

 

·         Give the harmful effects of global warming.

 

·         Give the importance of ozone layer.

 

 What are the causes for depletion of ozone layer?

·         What are the effects of ozone layer depletion?

·         Write a note on control of air pollution.

 

 

 

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