X CBSE Science Tutorials

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Table of Contents – Chemistry

CHAPTER 01: CHEMICAL REACTIONS AND EQUATIONS ... 1

CHAPTER MAP: ... 1

INTRODUCTION: ... 1

Chemical Equations: ... 1

What is a word-equation? ... 2

Writing a Chemical Equation: ... 2

Types of Chemical Reactions:... 4

Combination Reaction: ... 4

Decomposition Reaction: ... 4

Displacement Reaction: ... 5

Double Displacement Reaction: ... 5

Oxidation and Reduction Reactions: ... 6

APPENDIX:NCERTACTIVITIES ... 7

REVIEW QUESTIONS: ... 10

NCERTEXERCISE: ... 11

MULTIPLE CHOICE QUESTIONS:PRACTICAL (THEORY) ... 13

ANSWERS: ... 14

CHAPTER 02: ACIDS, BASES AND SALTS... 16

Properties of Acids and Bases: ... 16

Comparative properties of all Acids and Bases: ... 17

Bleaching Powder: ... 21

Baking Soda (Sodium Bicarbonate): ... 21

Plaster of Paris: ... 23

REVIEW QUESTIONS: ... 29

MULTIPLE CHOICE QUESTIONS:PRACTICAL THEORY ... 34

ANSWERS: ... 41

CHAPTER 03: METALS AND NON–METALS ... 43

METALS: ... 44

NON–METALS: ... 44

Metalloids ... 44

Comparative Properties – Metals and Non Metals:... 44

Reactions of Metals: ... 45

Activity Series of Metals: ... 47

METALLURGY: ... 50

Conversion to Metal Oxide: ... 52

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Universal Tutorials – X CBSE (2012–13) – Chemistry

Refining of Metals (Purification of Metals): ... 55

Corrosion: ... 55 Alloys: ... 56 Steel: ... 57 Alloying of Gold: ... 57 NON METALS: ... 58 Importance:... 58

Chemical Properties of Non Metals: ... 58

REVIEW QUESTIONS:CLASS WORK ... 61

REVIEW QUESTIONS:HOME WORK ... 64

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Table of Contents – Biology

CHAPTER 06: LIFE PROCESSES ... 77

CHAPTER MAP: ... 77 LIFE PROCESSES: ... 78 NUTRITION: ... 79 Modes of Nutrition: ... 79 Nutrition in Animals: ... 82 Nutrition in Amoeba: ... 82

Human digestive system: ... 82

APPENDIX: ... 85 RESPIRATION: ... 87 Types of Respiration: ... 87 Respiration in Plants: ... 88 Respiration in Animals: ... 88 Aquatic animals: ... 88 Terrestrial animals: ... 88

Human Respiratory System: ... 89

Mechanism of Breathing: ... 90

Exchange of Gases in Tissues: ... 90

APPENDIX: ... 91

TRANSPORTATION: ... 92

Transportation in Humans: ... 92

Structure and Function of Heart and Blood Vessels: ... 93

Blood Vessels: ... 94

Lymphatic System: ... 95

Transportation in Plants: ... 96

Raw Materials and Source: ... 96

Transportation of Food and other Substances: ... 97

APPENDIX: ... 98

EXCRETION: ... 99

Excretion in Humans: ... 99

Excretion in Plants: ... 101

PREVIOUS BOARD QUESTIONS: ... 102

MISCELLANEOUS EXERCISES:... 105

CHAPTER 07: CONTROL AND CO-ORDINATION ... 119

CO–ORDINATION IN PLANTS AND ANIMALS: ... 119

Coordination in Animals: ... 119

Human Nervous System: ... 120

Function of Human Nervous System: ... 120

Neuron - Functional Unit of the Nervous System: ... 120

Components of Nervous System: ... 122

Central Nervous System (CNS): ... 122

Reflex Action: ... 124

Hormones: ... 124

Hormonal control by feedback Mechanism: ... 125

Coordination in Plants: ... 125

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Universal Tutorials – X CBSE (2012–13) – Biology Volume NCERTEXERCISE: ... 127 PREVIOUS BOARD QUESTIONS: ... 128 MISCELLANEOUS EXERCISE: ... 129

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Table of Contents – Physics

CHAPTER 12: ELECTRICITY ... 131

CHAPTER MAP: ... 131 INTRODUCTION: ... 131 CURRENT ELECTRICITY: ... 132 Classification of Substances: ... 132

Electric Potential and Potential Difference: ... 132

ELECTRIC CIRCUITS: ... 133

OHM’S LAW: ... 134

Verification of Ohm’s Law: ... 134

RESISTANCE: ... 134

Concept of Resistance: ... 135

Methods of joining Resistors: ... 136

Resistors in Series: ... 136

Resistors in Parallel: ... 136

HEATING EFFECT OF ELECTRIC CURRENT: ... 137

Joule’s law for heating effect of electric current: ... 137

Applications of heating effect of current: ... 138

Electric bulb: ... 138 Electric Fuse: ... 138 Electric Power: ... 138 Useful Conversions: ... 139 APPENDIX: ... 139 SOLVED EXAMPLES:... 142 REVIEW QUESTIONS: ... 146 NCERTQUESTIONS: ... 151

ANSWERS TO UNSOLVED EXERCISES: ... 156

CHAPTER 13: MAGNETIC EFFECTS OF CURRENT ... 158

MAGNETIC EFFECTS OF ELECTRIC CURRENT: ... 159

Oersted’s Experiment: ... 159

Magnetic field due to a straight wire carrying electric current: ... 159

Factors affecting the magnetic field produced due to current: ... 160

Magnetic field due to a Solenoid: ... 160

EFFECTS OF EXTERNAL FIELD ON A WIRE CARRYING CURRENT: ... 161

Fleming’s Left hand Rule:... 161

Factors to increase the force: ... 162

ELECTRIC MOTOR: ... 162

ELECTROMAGNETIC INDUCTION: ... 163

Fleming’s Right Hand Rule:... 163

Factors to increase the strength of induced current:... 163

Domestic Electric Circuits:... 165

APPENDIX: ... 165

SOLVED EXAMPLES:... 168

REVIEW QUESTION: ... 169

NCERTQUSESTIONS: ... 173

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CHAPTER 14: SOURCES OF ENERGY ... 177

A good source of energy: ... 178

Conventional Sources of Energy:... 178

Fossil Fuels: ... 178

Disadvantages: ... 178

Pollution Control: ... 179

Traditional sources of energy: ... 179

Thermal Power Plant: ... 179

Hydro Power Plants: ... 179

Improvements in Technology for using Conventional Sources of Energy: ... 180

Bio-Mass: ... 180

Wind Energy: ... 181

Non–conventional Sources: ... 181

Solar Energy: ... 182

Energy from Sea: ... 184

Forms of energy from Sea: ... 184

Geothermal Energy: ... 185

Nuclear Energy: ... 185

Environmental Consequences: ... 185

How long will an energy source last us? ... 186

REVIEW QUESTION: ... 186

NCERTQUESTIONS: ... 188

PREVIOUS YEARS’BOARD QUESTIONS ... 189

APPENDIX: ... 190

ANSWERS TO THE UNSOLVED EXERCISE: ... 192

PRACTICAL SKILLS IN SCIENCE AND TECHNOLOGY ... 193

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Chapter 01: Chemical Reactions and Equations

Chapter Map:

Introduction:

 Chemical reactions are taking place very frequently in our day to day life. E.g. mangoes changing their colour from green to yellow (broadly ripening of all fruit), milk left at room temperature during summer (it curdles) an iron nail or tawa/ pan exposed to humid air (rusts, oxidizes) grapes fermenting, food being cooked and digested, respiration of living organisms etc.

 In all the above examples the characteristics of all the substances taking part in the reaction (physical and chemical) have changed along with their identity. When their physical form changes we say that a physical reaction has taken place. However when their chemical nature/ properties changes and a new substance/s are formed we say that a chemical reaction has taken place.  When a magnesium ribbon is burnt in oxygen, it gets converted to magnesium oxide. This is a

word statement. However, it can be written as a chemical equation as 2Mg + O2 → 2MgO.

Chemical Equations:

 Observations to determine whether a chemical reaction has taken place. When a chemical reaction takes place we observe the following:

 Change in state of reactants  Change in colour of reactants  Evolution of a gas as a product

 Change in temperature of the entire reaction.

 The word-equation for the reaction when Magnesium reacts with oxygen to give us Magnesium oxide would be

Magnesium + Oxygen → Magnesium oxide

(Reactants) (Product)

→ Chemical Equations

→ Writing a Chemical Equation → Balancing Chemical Equations → Types of Chemical Reactions

→ Combination Reactions → Decomposition Reactions → Displacement Reactions

→ Double Displacement Reactions → Oxidation and Reduction Reactions

→ The effects of Oxidation Reactions in Everyday Life → Corrosion

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 The substances that undergo chemical change in the above reaction i.e. magnesium and oxygen, are the reactants. The new substance, magnesium oxide, formed during the reaction is the product.

What is a word-equation?

 A chemical reaction written in words showing change of reactants to products by an arrow placed between them is called a word equation. The reactants are written on the left-hand side (LHS) with a plus sign (+) between them if they are more than one. Similarly, products are written on the right-hand side (RHS) with a plus sign (+) between them if they are more than one. The arrowhead points towards the products, and shows the direction of the reaction.

Writing a Chemical Equation:

 A chemical equation represents a chemical reaction and is written using symbols for element and formulae for compounds. The reaction of magnesium with oxygen to form magnesium oxide can be written as

2Mg + O2 → 2MgO

 The number of atoms of each element on the LHS and RHS of the arrow must be the same on both the sides. If not, then the equation is unbalanced as the mass of the elements is not the same on both sides of the equation.

Balancing Chemical Equations:

 The law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction. That is, the total mass of the elements present in the products of a chemical reaction has to be equal to the total mass of the elements present in the reactants. The number of atoms of each element remains the same, before and after a chemical reaction. Hence, we need to balance a chemical equation. The word–equation for the reaction of zinc with sulphuric acid is

Zinc + Sulphuric acid → Zinc sulphate + Hydrogen

 The above word-equation may be represented by the following chemical equation, Zn + H2SO4 → ZnSO4 + H2

 Let us count the number of atoms of different elements on both sides of the arrow.

Element Number of atoms in

reactants (LHS) Number of atoms in products (RHS) Zn 1 1 H 2 2 S 1 1 O 4 4

 As the number of atoms of each element is the same on both sides of the arrow. The above equation is a balanced chemical equation. Let us try to balance the following chemical equation,

Fe + H2O → Fe3O4 + H2

 Step I: List the number of atoms of different elements present in the unbalanced equation

Element Number of atoms in

reactants (LHS) Number of atoms in products (RHS) Fe 1 3 H 2 2 O 1 4

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 Step II: Start balancing with the compound that contains the maximum number of atoms. It may be a reactant or a product. We select Fe3O4 and the element oxygen in it. There are four oxygen atoms on the RHS and only one on the LHS.

 To balance the oxygen atoms:

Atoms of oxygen In reactants In products Initial 1(in H2O) 4 (in Fe3O4)

To balance 1 × 4 4

 To equalise the number of atoms, we cannot alter the formulae of the compounds or elements involved in the reactions. For example, to balance oxygen atoms we can put coefficient ‘4’ as 4H2O and not H2O4 or (H2O)4. Now the partly balanced equation becomes: Fe + 4H2O → Fe3O4 + H2 [partly balanced equation]

 Step III: Fe and H atoms are still not balanced. Let us balance hydrogen atoms in the partly balanced equation. To equalise the number of H atoms, make the number of molecules of hydrogen as four on the RHS.

Atoms of hydrogen In reactants In products Initial 8(in 4 H2O) 2(in H2)

To balance 8 2 × 4

 The equation would be,

Fe + 4 H2O → Fe3O4 + 4 H2 [partly balanced equation]

 Step IV: Now balance Fe in the equation.

Atoms of iron In reactants In products

Initial 1 (in Fe) 3(in Fe3O4)

To balance 1 × 3 3

 To equalise Fe, we take three atoms of Fe on the LHS. 3Fe + 4H2O → Fe3O4 + 4 H2 [balanced equation]

 The number of atoms of elements on both sides of the equation are equal. This equation is now balanced. This method of balancing chemical equations is called hit-and-trial method as we make trials to balance the equation by using the smallest whole number coefficient.

 Step V: Finally, to check the correctness of the balanced equation, we count atoms of each element on both sides of the equation.

3Fe + 4H2O → Fe3O4 + 4H2.

 Step VI: Writing Symbols of Physical States: To make a chemical equation more informative, the physical states of the reactants and products are mentioned along with their chemical formulae. The gaseous, liquid, aqueous and solid states of reactants and products are represented by the notations (g), (l), (aq) and (s) respectively. The word aqueous (aq) is written if the reactant or product is present as a solution in water. The balanced equation above becomes

3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)

 Sometimes the reaction conditions, such as temperature, pressure, catalyst etc., for the reaction are indicated above and/or below the arrow in the equation. For example:

CO(g) + 2H2(g) 340atm→ CH3OH(l) 6CO2(aq) + 6H2O(l) l Chlorophyl Sunlight__  →  C6H12O6(aq) + 6O2(aq) (Glucose)

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Types of Chemical Reactions:

 We have learnt in Class IX that during chemical reactions atoms of one element do not change into those of another element. Nor do atoms disappear from the mixture or appear from elsewhere, chemical reactions involve the breaking and making of bonds between atoms to produce new substances.

 Chemical change or chemical reaction is always associated with change in energy.

 In a reaction where a large amount of heat is evolved the reaction mixture is warm. Reactions in which heat is released along with the formation of products are called exothermic chemical reactions.

 Examples of exothermic reactions are:

 Burning of natural gas, CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + heat  Respiration is an exothermic process.

 We all know that we need energy to stay alive and work. We get this energy from the food we eat. During digestion, food is broken down into simpler substances.

 For example, rice, potatoes and bread contain carbohydrates. Carbohydrates are broken down to form glucose. This glucose combines with oxygen in the cells of our body and provides energy. The special name of this reaction is respiration.

C6H12O6(aq) + 6O2(aq) → 6CO2(aq) + 6H2O(l) + energy

(Glucose)

 The decomposition of vegetable matter into compost is also an example of an exothermic reaction.

 Reactions in which energy is absorbed are known as endothermic reactions.  Examples of endothermic reactions are:

 Reaction of barium hydroxide with ammonium chloride to give barium chloride and ammonium hydroxide.

 Calcium carbonate on heating breaks up into calcium oxide and carbon dioxid

Combination Reaction:

 Calcium oxide reacts vigorously with water to produce slaked lime (calcium hydroxide) releasing a large amount of heat.

CaO(s) + H2O(l) → Ca(OH)2(aq)

(Quick lime) (Slaked lime)

 In this reaction, calcium oxide and water combine to form a single product, calcium hydroxide. Such a reaction in which a single product is formed from two or more reactants is known as a combination reaction. Some more examples of combination reactions are:

 Burning of coal, C(s) + O2(g) → CO2(g)

 Formation of water from H2(g) and O2(g); 2H2(g) + O2(g) → 2H2O(l)

Decomposition Reaction:

 The decomposition reactions require energy either in the form of heat, light or electricity for breaking down the reactants.

 Splitting of the compounds by heat energy is called thermal decomposition.

 2FeSO4(s)  →Heat Fe2O3(s) + SO2(g) + SO3(g)

(Ferrous sulphate) (Ferric oxide)

 In this reaction you can observe that a single reactant breaks down to give simpler products. This is a decomposition reaction. Ferrous sulphate crystals (FeSO4⋅7H2O) lose

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water when heated and the colour of the crystals changes. It then decomposes to ferric oxide (Fe2O3), sulphur dioxide (SO2) and sulphur trioxide (SO3). Ferric oxide is a solid, while SO2 and SO3 are gases.

 Decomposition of calcium carbonate to calcium oxide and carbon dioxide on heating is an important decomposition reaction

CaCO3(s)  →Heat CaO(s) + CO2(g)

(Limestone) (Quick lime)

 Another example of a thermal decomposition reaction is 2Pb(NO3)2(s)  →Heat 2PbO(s) + 4NO2(g) + O2(g)

(Lead nitrate) (Lead oxide) (Nitrogen oxide) (Oxygen)

 Splitting of a compound by light energy is called photochemical reaction.

 White silver chloride turns grey in sunlight. This is due to the decomposition of silver chloride into silver and chlorine by light.

2AgCl(s) Sunlight→ 2Ag(s) + Cl2(g)

 Silver bromide also behaves in the same way. 2AgBr Sunlight→ 2Ag(s) + Br2(g)

 The above reactions are used in black and white photography.

 Decomposition brought about by electrical energy is known as electrolytic decomposition or electrolysis.

 2H2O(l) electriccurrent→ 2H2↑ + O2↑

(acidified water)

Displacement Reaction:

 Consider the reaction, Fe + CuSO4 → FeSO4 + Cu

 The iron nail become brownish in colour and the blue colour of copper sulphate solution fades  In this reaction, iron has displaced or removed another element copper, from copper sulphate

solution. This reaction is known as a displacement reaction.  Other examples of displacement reactions are

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

(Copper sulphate) (Zinc sulphate)

Pb(s) + CuCl2(aq) → PbCl2(aq) + Cu(s)

(Copper chloride) (Lead chloride)

 As zinc and lead are more reactive elements than copper. They displace copper from its compounds.

 A chemical reaction in which less reactive atom or group of atoms is displaced by another more reactive atom or group of atoms to form a new substance is called a displacement reaction.

Double Displacement Reaction:

 In the reaction below

Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)

(Sodium sulphate) (Barium chloride) (Barium sulphate) (insoluble) (Sodium chloride) (soluble)

 A white substance, which is insoluble in water, is formed. This insoluble substance formed is known as a precipitate. Any reaction that produces a precipitate can be called a precipitation reaction. A double displacement reaction always produces a precipitate.

 The white precipitate of BaSO4 is formed by the reaction of SO

− 2

4 and Ba

2+

. The other product formed is sodium chloride which remains in the solution (soluble)

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 A chemical reaction in which the cations of two compounds mutually exchange places to form two new compounds one of which is a precipitate is called a double displacement reaction.

Oxidation and Reduction Reactions:

 In this reaction, 2Cu + O2  →

Heat

2CuO

 The surface of copper powder becomes coated with black copper (II) oxide. This is because oxygen is added to copper and copper oxide is formed. If hydrogen gas is passed over this heated material (CuO), the black coating on the surface turns brown as the reverse reaction takes place and copper is obtained.

CuO + H2  →

Heat

Cu + H2O

 If a substance gains oxygen during a reaction, it is said to be oxidised. If a substance loses oxygen during a reaction, it is said to be reduced. During the above reaction the copper(II) oxide is losing oxygen and is being reduced. The hydrogen is gaining oxygen and is being oxidised. In other words, one reactant gets oxidised while the other gets reduced during a reaction. Such reactions are called oxidation-reduction reactions or redox reactions.

 Some other examples of redox reactions are:

i) ii)

 In the first reaction carbon is oxidised to CO and ZnO is reduced to Zn. In the second reaction HCl is oxidised to Cl2 whereas MnO2 is reduced to MnCl2. From the above examples we can say that if a substance gains oxygen or loses hydrogen during a reaction, it is oxidised. If a substance loses oxygen or gains hydrogen during a reaction, it is reduced.

 Oxidation reaction in everyday life:

 Iron articles are shiny when new, but get coated with a reddish brown powder when left for some time. This process is commonly known as rusting of iron. Some other metals also get tarnished in this manner.

 When a metal is attacked by substances around it such as moisture, acids, etc., it is said to corrode and this process is called corrosion.

 The black coating on silver and the green coating on copper are other examples of corrosion.

 Corrosion causes damage to car bodies, bridges, iron railings, ships and to all objects made of metals, especially those of iron. Corrosion of iron is a serious problem. Every year an enormous amount of money is spent to replace damaged iron.

 When fats and oils are oxidised, they become rancid and their smell and taste changes. Usually substances which prevent oxidation (antioxidants) are added to foods containing fats and oils. Keeping food in air tight containers helps to slow down oxidation.

Oxidation CuO + H2  →Heat Cu + H2O Reduction Oxidation ZnO + C  →Heat Zn + CO Reduction Oxidation MnO2 + 4HCl → MnCl2 + 2H2O + Cl2 Reduction

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APPENDIX: NCERT ACTIVITIES

Activity 1.1 (Experiment):

Aim: To show the reaction between magnesium and oxygen present in the air.

Apparatus: Mg ribbon, watch – glass, pair of tongs, burner, sand paper

Procedure: Clean a magnesium ribbon about 2 cm long by rubbing it with sandpaper. Hold it with a pair of tongs Burn it using a burner and collect the ashes so formed in a watch – glass as shown in the fig.

while burning the magnesium ribbon keep it as far as possible from your eyes.

Observation: magnesium ribbon burns with a dazzling white flame and changes into a white powder. This powder is magnesium oxide

Conclusion: MgO is formed due to the reaction between magnesium and oxygen present in the air. Equation, 2Mg + O2 → 2MgO

Aim: To show change in state and colour when a chemical reaction takes place.

Apparatus: lead nitrate solution, potassium iodide solution, test tube, and a conical flask. Procedure: Take lead nitrate solution in a test tube Add potassium iodide solution to this. Observation: lead nitrate reacts with potassium iodide to give lead iodide and potassium nitrate.

The colour of the products is different from that of the reactants.

Conclusion: Pb(NO3)2 + 2KI → PbI2 + 2KNO3. The change in colour is because lead iodide and potassium nitrate have been formed after the reaction. PbI2 is a yellow ppt. KNO3 is a colourless solution.

Activity 1:3 (Experiment):

Aim: To show the interaction between zinc granules and hydrochloric acid or sulphuric acid (formation of H gas)

Apparatus: Zinc granules, conical flask, test tube, hydrochloric acid / sulphuric acid

Procedure: Take a few zinc granules in a conical flask or a test tube. Add dilute hydrochloric acid or sulphuric acid to this. Touch the conical flask or test tube.

Observation: A gas is evolved and the conical flask is warm

Conclusion: From this we conclude that when a chemical reaction takes place there is a change in the state, colour, and temperature.

Zn + 2HCl → ZnCl2 + H2↑ + heat Zn + H2SO4 → ZnSO4 + H2↑ + heat

Aim: Formation of slaked lime by the reaction of calcium oxide with water Apparatus: Calcium oxide, beaker, and water

Procedure: Take a small amount of calcium oxide or quick lime in a beaker. Slowly add water to this. Touch the beaker

Observation: The beaker becomes hot (exothermic reaction). It reacts vigorously with water.

Conclusion: Slaked lime is formed by the reaction of the combination of calcium oxide with water

CaO(s) + H2O(l) → Ca(OH)2(aq) + heat

(quick lime) (Slaked lime) (exothermic reaction)

This is a combination reaction, and also an exothermic reaction.

Burner Tongs Magnesium ribbon Watch glass Magnesium oxide

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Aim: To show a decomposition reaction

Apparatus: ferrous sulphate crystals, dry boiling tube and burner

Procedure: Take about 2g ferrous sulphate crystals in a dry boiling tube. Note the colour of the ferrous sulphate crystals. Heat the boiling tube over the flame of a burner or spirit lamp observe the colour of the crystals after heating

Observation: The green colour of the ferrous sulphate crystals has changed there is also the characteristic odour of burning sulphur.

Conclusion: This is a Decomposition reaction, shown by the reaction 2FeSO4(s)  →Heat Fe2O3 + SO2(g) + SO3(g)

Aim: Heating of lead nitrate and emission of nitrogen dioxide

Apparatus: lead nitrate powder, boiling tube, pair of tongs and burner Procedure: Take about 2 g lead nitrate powder in a boiling tube. Hold the boiling tube with a pair of tongs and heat it over a flame. Observe the change if any.

Observation: We observe the emission of brown fumes.

Conclusion: These fumes are of nitrogen dioxide (NO2). The reaction that takes place is

2Pb(NO3)2(s)  →

Heat

2PbO(s) + 4NO2(g) + O2(g)

(Lead nitrate) (Lead oxide) (Nitrogen dioxide) (Oxygen)

Aim: To show that water is a compound containing two atoms of hydrogen and 1 atom of oxygen

Apparatus: Plastic mug, rubber stoppers, carbon electrodes, 6 volt battery, water, dilute sulphuric acid, burning candle

Procedure: Take a plastic mug. Drill two holes at its base and fit rubber stoppers in these holes. Insert carbon electrodes in these rubber stoppers as shown in fig then connect these electrodes to a 6 volt battery. Fill the mug with water such that the electrodes are immersed.

Add a few drops of dilute sulphuric acid to the water. Take two test tubes filled with water and invert them over the two carbon electrodes. Switch on the current and leave the apparatus undisturbed for some time. Observe the formation of bubbles at both the electrodes. These bubbles displace water in the test tubes. The volume of the gas collected is not the same in both the test tubes. Once the test tubes are filled with the respective gases, remove them carefully. Test these gases one by one by bringing a burning candle close to the mouth of the test tubes. Observation: We observe the formation of bubbles at both the electrodes. These bubbles

displace water in the test tubes. The volume in one test tube is twice that in the other test tube. The test tube containing hydrogen gas has double the volume of the test tube containing oxygen gas. Hydrogen gas burns with a light blue flame with a pop sound.

Conclusion: Water is a compound containing two atoms of hydrogen and 1 atom of oxygen.

Wafting gas gently towards nose Do not point the mouth of boiling tube at your neighbours or yourself Boiling tube Ferrous sulphate crystals Burner Burner Boiling tube Lead nitrate Test tube holder

6V Battery Plastic mug Hydrogen Water Water Rubber stopper Cathode Switch Graphite rod Test tube Oxygen Anode

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Aim: To show how silver chloride is affected by sunlight Apparatus: silver chloride and a china dish.

Procedure: Place this china dish containing silver chloride in sunlight for some time. Observe the colour of the silver chloride after some time.

Observation: White silver chloride turns grey in sunlight.

Conclusion: This is due to the decomposition of silver chloride into silver and chlorine by light 2AgCl(s) Sunlight→ 2Ag(s) + Cl2(g)

Aim: To show displacement reaction

Apparatus: two iron nails, sand paper, copper sulphate solution, thread and two test tubes. Procedure: Take two iron nails and clean them by rubbing them with sand paper. Take two test

tubes marked as (A) and (B). In each test tube, take about 10 ml copper sulphate solution. Tie one iron nail with a thread and immerse it carefully in the copper sulphate solution in test tube A for about 20 minutes. Keep one iron nail aside for comparison. After 20 minutes, take out the iron nail from the copper sulphate solution. Compare the intensity of the blue colour of copper sulphate solutions in the test tubes (A) and (B). Also, compare the colour of the iron nail dipped in the copper sulphate solution with the one kept aside.

Observation: the iron nail dipped in the copper sulphate solution in test tube A becomes brownish in colour and the blue colour of copper sulphate solution fades in test tube A. While the blue colour of copper sulphate solution in test tube B remains the same.

Conclusion: iron has displaced or removed another element, copper, from copper sulphate solution in test tube A. This reaction is known as displacement reaction.

Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)

[copper sulphate] [Iron sulphate]

Aim: To show double displacement reaction

Apparatus: 3 ml of sodium sulphate, 2 test tubes, 3 ml of barium chloride solution.

Procedure: Take about 3 ml of sodium sulphate solution in a test tube. In another test tube, take about 3 ml of barium chloride solution, Mix the two solutions.

Observation: A white substance, which is insoluble in water, is formed. This insoluble substance formed is known as a precipitate. Any reaction that produces a precipitate can be called a precipitation reaction.

Conclusion: This is a double displacement reaction shown by

the equation below where Na+ being more reactive than Ba+2 displaces Ba+2 from its compound BaCl2 and takes its place to form NaCl.

Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)

(sodium sulphate) (Barium chloride) (Barium sulphate) (Sodium chloride)

Test tube containing solution of sodium sulphate Test tube containing solution of barium chloride China dish Silver chloride Sunlight

Iron nail Iron nail taken out from copper Sulphate solution Copper Sulphate Solution (test tube A) Test tube Test tube stand Reaction Mixture (test tube B) Test tube Thread Copper sulphate solution Iron nail Stand

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Aim: To show oxidation and reduction reaction Apparatus: China dish 1g copper powder, burner,

wire gauze, tripod stand and Hydrogen gas. Procedure:

 Heat a china dish containing about 1 g copper powder.

 Now pass hydrogen gas over this heated material.

Observation:

 The surface of copper powder becomes coated with black copper oxide.

 Hydrogen gas is passed over this heated material [CuO] the black coating on the surface turns brown as the reverse reaction takes place and copper is obtained.

Conclusion: In the first case copper, was oxidized to copper oxide by the following reaction 2Cu + O2  →Heat 2CuO this is an oxidation reaction

(black coating)

In the second case when hydrogen gas was passed over heated [CuO] copper metal was obtained because copper oxide got reduced to copper metal by the following reaction

CuO + H2  →Heat Cu + H2O

REVIEW QUESTIONS:

CW Exercise:

1) Balance the following reactions wherever required.

a) SO2 + H2O → H2SO3 b) SO3 + H2O → H2SO4 c) Na2O + H2O → NaOH d) K + H2O → KOH + H2 e) NaOH + HCl [dil.] → NaCl + H2O f) Zn + HCl [dil.] → ZnCI2 + H2 g) NaCl + H2SO4 (conc.) <200°C→ NaHSO4 + HCl

h) NaNO3 + H2SO4 (conc.) >200°C→ Na2SO4 + HNO3

2) Translate the reactions into word equations after balancing them a) NaCl + H2SO4 (conc.) <200°C→ NaHSO4 + HCl(g)

b) NaCl + H2SO4 (conc.) >200°C→ Na2SO4 + HCl(g)

c) HCl(g) H2 + Cl2 d) NH3 + HCl(g) → NH4Cl e) Zn + HCl(g) → ZnCl2 + H2 f) Fe + HCl(g) → FeCl2 + H2

g) Mg + HCl [dil.] → MgCl2 + H2 h) CaO + HCl [dil.] → CaCl2 + H2O

i) NH4OH + HCl [dil.] → NH4Cl + H2O j) Na2CO3 + HCl [dil.] → NaCl + H2O + CO2

3) Write balanced equations for the following reactions and identify the types of reactions. a) CuO + H2SO4 [dil.] → ______ + H2O b) Fe + H2SO4 [dil.] → _____ + H2

c) Fe + Cl2 → _____ d) Pb(NO3)2 + NaCl → _____ + _____ e) CaCl2 + Na2CO3 → _____ + _____ f) ZnCO3 <400°C→ ZnO + CO2 g) Al2O3 . 2H2O + Na2CO3 →∆ NaAlO2 + H2O + CO2 h) Al(OH)3 1100°C→ Al2O3 + H2O i) C + O2 → CO2 + ∆ j) CO2 + C → CO – ∆ k) CaCO3 → CaO + CO2 – ∆ Wire gauz China dish containing130 copper powder Tripod stand Burner >500° C

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HW Exercise:

1) Balance the following reactions wherever required. a) NaHCO3 + H2SO4 [dil.] → Na2SO4 + H2O + CO2 b) Na2CO3 + H2SO4 [dil.] → Na2SO4 + H2O + CO2

c) Cu(OH)2 + Na2SO4 [dil.] → CuSO4 + NaOH d) Al(s) + H2O → Al2O3(s) + H2(g) e) Fe + H2SO4 (dil.) → FeSO4 + H2 f) Fe + Cl2 → FeCl3

g) Zn + S → ZnS h) Pb(NO3)2 + NaCl → NaNO3 + PbCl2 i) CaCl2 + Na2CO3 → NaCl + CaCO3

2) Translate the reactions into word equations after balancing them

a) NaHCO3 + HCl [dil.] → NaCl + H2O + CO2 b) NaHCO3  →Heat Na2CO3 + H2O + CO2 c) NaHSO3 + HCl [dil.] → NaCl + H2O + SO2 d) FeS + HCl [dil.] → FeCl2 + H2S

e) AgNO3 + HCl [dil.] → AgCl↓ + HNO3 f) Pb(NO3)2 + HCl [dil.] → PbCl2↓ + HNO3 g) MnO2 + HCl [conc.] → MnCl2 + H2O + Cl2 h) PbO2 + HCl [conc.] → PbCl2 + H2O + Cl2 i) Pb3O4 + HCl [conc.] → PbCl2+ H2O + Cl2

3) Write balanced equations for the following reactions and identify the types of reactions. a) CaO + SiO2 → CaSiO3 b) Fe2O3 + CO → Fe + CO2

c) Zn + O2 → ZnO d) Zn + S → ZnS

e) Zn + NaOH → Na2ZnO2 + H2 f) Zn + H2SO4 [dil.] → ZnSO4 + H2 g) Fe + O2 → Fe3O4 h) Fe + H2O  Fe3O4 + H2

i) Fe + Cl2 → FeCl3 j) Fe + S → FeS

NCERT EXERCISE:

1) Why does the colour of copper sulphate solution change when an iron nail is dipped in it? 2) Give an example of a double displacement reaction other than the one given in the notes.

3) Identify the substances that are oxidised and the substances that are reduced in the following reactions.

i) 4Na(s) + O2(g) → 2Na2O(s) ii) CuO(s) + H2(g) → Cu(s) + H2O(l)

4) Why should a magnesium ribbon be cleaned before burning in air? 5) Write a balanced equation for the following chemical reactions.

i) Hydrogen + Chlorine → Hydrogen chloride

ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride iii) Sodium + Water → Sodium hydroxide + Hydrogen

6) Write a balanced chemical equation with state symbols for the following reactions.

i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.

ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.

7) A solution of a substance ‘X’ is used for white washing. i) Name the substance ‘X’ and write its formula.

ii) Write the reaction of the substance ‘X’ named in (i) above with water.

8) Why is the amount of gas collected in one of the test tubes in the decomposition of water double the amount collected in the other? Name this gas.

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Which of the statements about the reaction below are incorrect? 9) 2PbO(s) + C(s) → 2Pb(s) + CO2(g)

a) Lead is getting reduced b) Carbon dioxide is getting oxidised c) Carbon is getting oxidized d) Lead oxide is getting reduced i) (a) and (b) ii) (a) and (c) iii) (a), (b) and (c) iv) all 10) Fe2O3 + 2Al → Al2O3 + 2Fe. The above reaction is an example of a

a) combination reaction b) double displacement reaction c) decomposition reaction d) displacement reaction

11) What happens when dilute hydrochloric acid is added to iron fillings? Tick the correct answer. a) Hydrogen gas and iron chloride are produced

b) Chlorine gas and iron hydroxide are produced c) No reaction takes place

d) Iron salt and water are produced

12) What is a balanced chemical equation? Why should chemical equations be balanced? 13) Translate the following statements into chemical equations and then balance them.

a) Hydrogen gas combines with nitrogen to form ammonia

b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide

c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate

d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas 14) Balance the following chemical equations, if necessary.

a) HNO3 + Ca(OH)2 → Ca(NO3)2 + H2O b) NaOH + H2SO4 → Na2SO4 + H2O c) NaCl + AgNO3 → AgCl + NaNO3 d) BaCl2 + H2SO4 → BaSO4 + HCl 15) Write chemical equations for the following reactions. Balance them if required.

a) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water b) Zinc + Silver nitrate → Zinc nitrate + Silver

c) Aluminium + Copper chloride → Aluminium chloride + Copper

d) Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride

16) Write the balanced chemical equation for the following and identify the type of reaction in each case.

a) Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s) b) Zinc carbonate(s) → Zinc oxide(s) + Carbon dioxide(g)

c) Hydrogen(g) + Chlorine(g) → Hydrogen chloride(g)

d) Magnesium(s) + Hydrochloric acid(aq) → Magnesium chloride(aq) + Hydrogen(g) 17) What does one mean by exothermic and endothermic reactions? Give examples. 18) Why is respiration considered an exothermic reaction? Explain.

19) Why are decomposition reactions called the opposite of combination reactions?

20) Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.

21) What is the difference between displacement and double displacement reactions? Write equations for these reactions.

22) In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved.

23) What do you mean by a precipitation reaction? Explain by giving examples. 24) Explain the following in terms of gain or loss of oxygen with two examples each.

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25) A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.

26) Why do we apply paint on iron articles?

27) Oil and fat containing food items are flushed with nitrogen. Why? 28) Explain the following terms with one example each:

a) Corrosion b) Rancidity

MULTIPLE CHOICE QUESTIONS: PRACTICAL (THEORY) 1) When CuSO4 reacts with Fe the product formed will be:

a) FeSO4 b) Fe2(SO4)3 c) Fe d) FeS

2) The substance which loses electrons is called as:

a) Oxidising agent b) Reducing agent c) Catalyst d) None of these 3) Which of the following reactions is a double displacement reaction?

a) Zn + CuSO4 → ZnSO4 + Cu b) NaOH + HCl → NaCl + H2O c) CaCO3 → CaO + CO2 d) CaO + H2O → Ca(OH)2

4) An arrow pointing downwards in an equation indicates:

a) Evolution of gas b) Formation of precipitate c) An aqueous solution d) Both (a) and (b)

5) Name one compound of hydrogen which is very essential for existence of life on the earth

a) H2O b) H2O2 c) CH4 d) H3O

+

6) Find X in the reaction, Zn + H2SO4→ X + H2

a) ZnS b) ZnSO4 c) ZnO d) ZnSO3

7) Which one of the following is a decomposition reaction?

a) Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) b) BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq) c) CaCO3(s) → CaO(s) + CO2(g) d) C(s) + O2(g) → CO2(g)

8) CaCO3  →

heat

CaO + CO2. This reaction is an example of:

a) Electrolysis b) Thermal decomposition

c) Displacement d) Combination

9) Which of the following is not a chemical reaction?

a) Souring of milk b) Rusting of iron

c) Dissolution of sugar in water d) Digestion of food in our body

10) a Mg3N2 + b H2O → c Mg(OH)2 + d NH3. When the equations is balanced, the coefficients a,

b, c, d respectively are

a) 1, 3, 3, 2 b) 1, 6, 3, 2 c) 1, 2, 3, 2 d) 2, 3, 6, 2 11) Which of the following reaction has not been correctly represented?

a) BaCI2(aq) + H2S04(aq) → BaSO4↓ + 2 HCl b) Zn(s) + H2SO4(aq) → ZnSO4↓ + H2(g) c) C(s) + O2(g) → CO2(g) + Heat d) 2KCIO3(s)  →heat 2 KCI(s) + 3O2↑

12) Which of the following is not a combination reaction?

a) Fe + S → FeS b) CaO + CO2 → CaCO3

c) NH3 + HCI → NH4Cl d) AgNO3 + NaCI → AgCI + NaNO3

13) Which of the following is not a thermal decomposition reaction?

a) 2H2O → 2 H2 + O2 b) 2 FeSO4 → Fe2O3 + SO2 + SO3 c) ZnCO3 → ZnO + CO2 d) 2 KClO3 → 2 KCI + 3 O2

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14) Which of the following reactions will not occur?

a) Mg + H2SO4 → MgSO4 + H2 b) Cu + H2SO4 → CuSO4 + H2 c) 2 Al + 6 HCl → 2 AICl3 + 3 H2 d) Fe + 2 HCl → FeCI2 + H2

15) Which of the following reactions will occur?

a) 2 Ag + Cu(NO3)2 → 2 AgNO3 + Cu b) Cu + ZnSO4 → CuSO4 + Zn c) 2 Ag + H2SO4 → Ag2SO4 + H2 d) 2 Al + 3 FeSO4 → Al2(SO4)3 + 3 Fe

16) In the reaction, 2H2S + SO2→ 3S + 2H2O

a) H2S has been oxidized b) SO2 has been oxidized c) H2S is the oxidizing agent d) SO2 is the reducing agent

17) Which of the following reactions is a redox reaction as well as a displacement reaction? a) 2 HgCI2 + SnCl2 → Hg2CI2 + SnCI4 b) ZnO + C → Zn + CO

c) 2 Al + 6 HCI → 2 AlCl3 + 3 H2 d) H2S + Cl2 → 2 HCI + S

18) The correct formula of rust is

a) Fe2O3 b) Fe3O4 c) Fe2O3 . x H2O d) Fe3O4 . x H2O

19) Galvanisation of iron means coating iron with

a) Chromium b) Nickel c) Zinc d) Tin

20) Which of the following metals is protected by a layer of its oxide?

a) Copper b) Silver c) Iron d) Aluminium

21) The term ‘rancidity’ represents

a) Acid rain b) Oxidation of fatty food

c) Rotting of fruit d) Fading of coloured clothes in the sun ANSWERS: CW Exercise: 1) c) Na2O + H2O → 2NaOH d) 2K + 2H2O → 2KOH + H2 f) Zn + 2HCl(dil) → ZnCl2 + H2 2) b) 2NaCl + H2SO4 (conc.) → ° >200C Na2SO4 + 2HCl(g) e) Zn + 2HCl(g) → ZnCl2 + H2 f) Fe + 2HCl(g) → FeCl2 + H2

g) Mg + 2HCl [dil.] → MgCl2 + H2 h) CaO + 2HCl [dil.] → CaCl2 + H2O j) Na2CO3 + 2HCl [dil.] → 2NaCl +H2O +CO2

3) a) CuO + H2SO4 [dil.] → CuSO4+ H2O [Neutralization]; double displacement. b) Fe + H2SO4 [dil.] → FeSO4+ H2 [simple displacement]

c) Fe + Cl2 → FeCl2 [Direct combination] d) Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3 [Double displacement] e) CaCl2 + Na2CO3 → CaCO3 + 2NaCl [Double displacement] f) ZnCO3 → ° <400C ZnO + CO2 [Decomposition] g) Al2O3 . 2H2O + Na2CO3 → ∆

2NaAlO2 + 2H2O + CO2 [displacement and combination] h) 2Al(OH)3 →

°C 1100

Al2O3 + 3H2O [Decomposition]

i) C + O2 → CO2 + ∆ [Combination and oxidation] j) CO2 + C → 2CO – ∆ [Combination and reduction] k) CaCO3 →∆ CaO + CO2 – ∆ [Decomposition and endothermic]

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HW Exercise:

1) a) 2NaHCO3 + H2SO4 [dil.] → Na2SO4 + 2H2O + 2CO2 c) Cu(OH)2 + Na2SO4 [dil.] → CuSO4 + 2NaOH

d) 2Al(s) + 3H2O → Al2O3(s) + 3H2(s) f) 2Fe + 3Cl2 → 2FeCl3

h) Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 i) CaCl2 + Na2CO3 → 2NaCl + CaCO3 2) b) 2NaHCO3  →Heat Na2CO3 + H2O + CO2 d) FeS + 2HCl [dil.] → FeCl2 + H2S

f) Pb(NO3)2 + 2HCl [dil.] → PbCl2↓ + 2HNO3 g) MnO2 + 4HCl [conc.] → MnCl2 +2H2O +Cl2 h) PbO2 + 4HCl [conc.] → PbCl2 + 2H2O + Cl2 i) Pb3O4 + 8HCl [conc.] → 3PbCl2+ 4H2O + Cl2 3) a) CaO + SiO2 → CaSiO3 [Combination]

b) Fe2O3 + 3CO → 2Fe + 3CO2 [Reduction – oxidation] c) 2Zn + O2 → 2ZnO [Combination and oxidation]

d) Zn + S → ZnS [Combination]

e) Zn + NaOH → Na2ZnO2 + H2 [Combination and displacement] f) Zn + H2SO4 [dil.] → ZnSO4 + H2 [Displacement]

g) 3Fe + 2O2 → Fe3O4 [Combination] h) 3Fe + 4H2O  Fe3O4 + 4H2 [Redox] i) 2Fe + 3Cl2 → 2FeCl3 [Combination]

j) Fe + S → FeS [Combination]

NCERT Exercise:

1) Displacement reaction 2) CaCl2 + MgSO4 → CaSO4 + MgCl2 3) (i) Na oxidized, (ii) CuO is reduced, H2 is oxidized to water.

4) Mg gets oxidized to MgO when exposed to air at ordinary temperature. Hence it has to be cleaned / brushed to get Mg.

5) (i) H2 + Cl2 = 2HCl (ii) 3BaCl2 + Al2(SO4)3 → 3BaSO4 + 2AlCl3 (iii) 2Na + 2H2O → 2NaOH + H2 6) (i) BaCl2(aq) + Na2SO4(aq) → BaSO4(s)↓ + 2NaCl(aq) (ii) NaOH(aq) + HCl(aq) → NaCl(l) + H2O(l) 7) i) Calcium oxide, CaO ii) CaO + H2O → Ca(OH)2

8) Water consists of two parts of H2 and one part of O2. Hence the above statement the gas whose quantity is double is hydrogen.

9) 1 10) d 11) a

13) a) 3H2 +N2 → 2NH3 b) 2H2S + 3O2 → 2H2O + 2SO2 c) 3BaCl2 + Al2(SO4)3 → 3BaSO4↓ + 2AlCl3 d) 2K + 2H2O → 2KOH + H2↑

14) a) 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O b) 2NaOH + H2SO4 → Na2SO4 + 2H2O c) NaCl + AgNO3 → AgCl + NaNO3 d) BaCl2 + H2SO4 → BaSO4 + 2HCl 15) a) Ca(OH)2 + CO2 → CaCO3 + H2O b) Zn + 2AgNO3 → Zn(NO3)2 + 2Ag c) 2Al + 3CuCl2 → 2AlCl3 + 3Cu d) BaCl2 + K2SO4 → BaSO4 + 2KCl 16) a) 2KBr + BaI2 → 2KI + BaBr2 [double displacement]

b) ZnCO3 → ZnO + CO2 [decomposition]

c) H2 + Cl2 → 2HCl [Combination]

d) Mg(s) + 2HCl (aq) → MgCl2(aq) + H2(g)↑ [displacement] 22) 2AgNO3 + Cu → Cu(NO3)2 + 2Ag↓

24) a) gaining of O2 or loss of H2 b) loss of O2 or gaining of H2

25) X = Cu; black color CuO 26) To prevent rusting/ oxidation of Fe

Note: Not all questions have been solved from the exercises behind the chapter. Students are

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Chapter 02: Acids, Bases and Salts

Chapter Map:

What is an acid?

 An acid is a compound which on dissolving in water gives hydronium ions as the only positively charged ions.

What is a base?

 Any substance which gives hydroxyl ions on dissolving in water or reacts with acids to form salt and water only or neutralizes an acid is called a base.

Properties of Acids and Bases:

Substance Acids Bases

Methyl orange Reddish orange Yellow

Red litmus Red Blue

Blue litmus Red Blue

Phenolphthalein Colourless Pink

Distinguish between acids and Bases:

Acids Bases

Sour to taste Bitter to taste

Corrosive to touch Soapy to touch

Turn blue litmus red Turn red litmus blue

Produce H+ ions when mixed with water Produce (OH)– ions when mixed with water → Chemical properties of Acids and Bases

→ Acids and Bases in the Laboratory

→ How do Acids and Bases react with Metals?

→ Reaction of Acids with Metal Carbonates and Metal Hydrogencarbonates → Reaction of Acids and Bases with each other

→ Reaction of Metallic oxides with Acids → Reaction of Non–metallic Oxides with Bases → Common properties of Acids and Bases

→ Behaviour of Acids and Bases in water → Strength of Acidic and Basic solutions → pH of solutions

→ Importance of pH in Everyday Life → Salts

→ Family of salts

→ Chemical compounds from Common salt

(NaCl, NaOH, CaOCl2, NaHCO3, Na2CO3.10H2O) → Crystals of salts are not dry → Plaster of Paris

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What is litmus solution?

 This solution is a natural indicator. It is neither basic nor acidic its colour is purple. However, this colour changes to red in an acidic medium and to blue in a basic medium.

 There are other indicators as well eg Turmeric, methyl orange, methyl red, phenolphthalein.

 Some naturally occurring materials like red cabbage leaves, coloured petals of some flowers eg. Hydrangea, Petunia, Geranium also indicate the presence of an acid or a base in a solution.

What is a Salt?

 A salt is an ionic compound containing a positive ion other than hydrogen ion and a negative ion other than a hydroxyl ion.

 A salt is a substance having different tastes. It has a mild irritating sensation when touched.

Reaction of Acids with metal carbonates & Metal Hydrogencarbonates:

 Acids react with metal carbonates to give salt, water and CO2. Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

 Acids react with metal hydrogen carbonates to give salt, water and CO2. NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)

Reaction of metallic oxides with Acids:

 When acids react with metallic oxides a salt and water are formed. This reaction is similar to the reaction when acids and bases react with each other. This proves that metallic oxides are basic in nature.

Na2O + 2HCl → 2NaCl + H2O

Reaction of non–metallic oxides with bases:

 When bases react with non–metallic oxides a salt and water are formed. This reaction is similar to the reaction when acids and bases react with each other. This proves that non– metallic oxides are acidic in nature.

Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l)

(Lime water) (White precipitate)

Comparative properties of all Acids and Bases:

 All Acids generate hydrogen gas on reacting with metals, so hydrogen seems to be common to all acids. Hence, when we pass electricity through an acidic solution the electric current is carried through the solution by ions.

 Since the cation present in acids is H+

, this suggests that acids produce hydrogen ions, H+(aq), in solution, which are responsible for their acidic properties.

 When we dissolve a base like sodium hydroxide in water; (OH–

) ions (anions) are generated in water.

 When HCl is added to water the following reaction takes place HCl + H2O → H3O

+ + Cl–  The separation of H+

ions from HCl molecules cannot occur in the absence of water.

 Hydrogen ions cannot exist alone, but they exist after combining with water molecules. Thus hydrogen ions must always be shown as H+(aq) or as an hydronium ion (H3O

+ ).  Acids give H3O

+

or H+(aq) ions in water. Now, when a base is dissolved in water, OH –

ions are generated in water. Bases which are soluble in water are called alkalis. Eg.

NaOH(s)  → O H2 _Na+ (aq) + OH – (aq) KOH(s)  → O H2 _K+ (aq) + OH – (aq) Mg(OH)2(s)  →H2O Mg 2+ (aq) + 2O H – (aq)

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 All bases do not dissolve in water.

 Now as we have identified that all acids generate H+

(aq) and all bases generate OH –

(aq), we can view the neutralisation reaction as follows,

Acid + Base → Salt + Water H X + M OH → MX + HOH H+(aq) + OH

–

(aq) → H2O(l)

 The process of dissolving an acid or a base in water is a highly exothermic one. Care must be taken while mixing concentrated nitric acid or sulphuric acid with water. The acid must always be added slowly to water with constant stirring.

 If water is added to a concentrated acid, the heat generated may cause the mixture to splash out and cause burns. The glass container may also break due to excessive local heating.

What is dilution of an acid or a base?

 Mixing an acid or a base with water results in decrease in the concentration of ions (H3O

+

/OH–) per unit volume. Such a process is called dilution and the acid or the base is said to be diluted.

Strength of acidic or basic solutions:

 Acid-base indicators are used to distinguish between an acid and a base.

 On dilution or decreasing the concentration of H+ or OH– ions in solutions, per unit volume we can quantitatively find the amount of these ions present in a solution.

 Thus, we judge how strong a given acid or a base is. We can do this by making use of an universal indicator, which is a mixture of several indicators. The universal indicator shows different colours at different concentrations of hydrogen ions in a solution.

 A scale for measuring hydrogen ion concentration in a solution, called pH scale has been developed. The ‘p’ in pH stands for ‘potenz’ in German, meaning power. This scale was introduced by Sorensen.

Definition:

 pH of a solution is defined as the negative logarithm of hydrogen ion concentration or pH = – log [H+]

 If hydrogen ion conc. is expressed as 10 raised to its negative power numerical value then the numerical value written over 10 is called pH of the solution.

i.e. if [H+] = 10–X then pH = X.

 Every aqueous solution whether neutral, acidic or basic contains both H+

and OH– ions.  The product of [H+

] [OH–] for any solution is always 1 × 10–14 at 25°C.  If H+

concentration is more, the solution is acidic and if OH– concentration is more, the solution is basic.

 Thus a convenient scale for pH was devised; termed as pH scale.

In pure water [H+] = [OH–] = 10–7mol / litre

 On the pH scale we can measure pH from 0 (very acidic) to 14 (very alkaline). pH should be thought simply as a number which indicates the acidic or basic nature of a solution. Higher the hydronium ion concentration, lower is the pH value.

[H+] = 10–7

pH = – (–7) log 10 (pH = –log [H+]) pH = 7(neutral) [H+] 100 10–1 10–2 10–3 10–4 10–5 10–6 10–7 10–8 10–9 10–10 10–11 10–12 10–13 10–14

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 Solutions with pH from 0 to 6.9 are acids while solutions with pH from 7.1 to 14 are bases.  Solutions with pH 0 to 3 are strong acids while those with pH 4 to 6 are weak acids.  Solution with pH 7.1 to 10 are weak bases

 Solutions with pH 11 to 14 are strong bases.

Variation of pH with the change in concentration of H+(aq) and OH–(aq) ions

S. No. Solution Colour of Universal paper Approx. pH value Nature of substance

1) Saliva (before meals) Green 7 Neutral

2) Saliva (after meals) Red < 7 Acidic

3) Lemon juice Red < 7 Acidic

4) Colourless aerated drink Red < 7 Acidic

5) Carrot juice Red < 7 Acidic

6) Coffee Red < 7 Acidic

7) Tomato juice Red < 7 Acidic

8) Distilled water Green 7 Neutral

9) 1 M NaOH Blue 14 Basic

10) 1 M HCl Red < 7 Acidic

Strong Acids / Weak Acids:

 Strong acids are those acids, which undergo complete dissociation on dissolving in water and hence release large number of hydrogen ions in aqueous solutions. Eg. HCl, H2SO4

e.g. HCl(aq) → H +

(aq) + Cl –

(aq)

 Weak acids do not undergo complete dissociation and hence release few hydrogen ions. e.g. Acetic acid, Oxalic acid.

e.g. CH3COOH(aq) → CH3COO –

(aq) + H +

(aq)

Strong Bases / Weak Bases:

 Strong bases are those bases, which release large number of hydroxyl ions by undergoing complete dissociation in aqueous solutions.

E.g. Sodium hydroxide, potassium hydroxide

 Weak bases release few hydroxyl ions, as they do not dissociate completely. E.g. Ammonium Hydroxide.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 Gastric juice (about 1.2) Pure water, blood (7.4) Lemon juice (about 2.2) Milk of magnesia (10) Sodium hydroxide solution (about 14)

Acidic nature increasing Basic nature increasing Neutral

7 ₁₄

OH−

Decrease in H+ ion concentration Increase in H+_{ion concentration}

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Importance of pH in Everyday Life:

 Our body works within the pH range of 7.0 to 7.8. Living organisms can survive only in a narrow range of pH change.

 When rain water dissolves SO2 and NO2 gases present in the atmosphere its pH is less than 5.6. This is called acid rain. When this acid rain flows into the rivers and seas it lowers the pH of the water there, thus making aquatic life difficult.

pH of the soil in your backyard:

 Plants require a specific pH range for their healthy growth i.e. between 6.0 – 7.5.

pH in our digestive system:

 Our stomach produces hydrochloric acid. It helps in the digestion and preservation of the food stored in the stomach during digestion of food without harming the stomach. During indigestion the stomach produces too much acid and this causes pain and irritation.

 To get rid of this pain, people use bases called antacids. These antacids neutralise the excess acid. Magnesium hydroxide (Milk of magnesia), a mild base, is often used for this purpose.

pH change is the cause of tooth decay:

 Tooth decay starts when the pH of the mouth is lower than 5.5. Tooth enamel, made up of calcium phosphate is the hardest substance in the body.

 It does not dissolve in water, but is corroded when the pH in the mouth is below 5.5. Bacteria present in the mouth produce acids by the degradation of sugar and food particles that remain in the mouth after eating.

 The best way to prevent this is to clean the mouth after eating food. Using toothpastes, which are generally basic, for cleaning the teeth can neutralise the excess acid and prevent tooth decay.

Self defence by animals and plants through chemical warfare:

 Bee-sting leaves an acid; formic acid which causes pain and irritation. Use of a mild base like baking soda on the stung area gives relief. Stinging hair of nettle leaves inject methanoic acid causing burning pain.

More about salts:

 Salts of a strong acid and a strong base are neutral with pH value of 7. On the other hand, salts of a strong acid and a weak base are acidic with pH value less than 7 and those of a strong base and a weak acid are basic in nature, with pH value more than 7.

Chemicals from Common Salt:

 Common salt is formed by the combination of hydrochloric acid and sodium hydroxide solution and is called sodium chloride. This is the salt that is used in food, it is a neutral salt.

 Seawater contains many salts dissolved in it. Sodium chloride is separated from these salts. Deposits of solid salt are also found in several parts of the world.

 These large crystals are often brown due to impurities. This is called rock salt. Beds of rock salt were formed when seas of bygone ages dried up. Rock salt is mined like coal.

 Common salt – A raw material for chemicals.

 The common salt thus obtained is an important raw material for various materials of daily use, such as sodium hydroxide, baking soda, washing soda, bleaching powder and many more.

Sodium hydroxide:

 When electricity is passed through an aqueous solution of sodium chloride (called brine), it decomposes to form sodium hydroxide.

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 The process is called the chlor-alkali process because of the products formed – chlor for chlorine and alkali for sodium hydroxide.

2NaCl(aq) + 2H2O(l) electricity→ 2NaOH(aq) + Cl2(g) + H2(g)

 Chlorine gas is given off at the anode, and hydrogen gas at the cathode. Sodium hydroxide solution is formed near the cathode.

 The three products produced in this process are all useful. Figure below shows the different uses of these products.

Bleaching Powder:

 Formula: CaOCl2

 Industrial method of preparation is the action of chlorine on dry slaked lime.

Ca(OH)2 + Cl2 → CaOCl2 + H2O  It is also called as chloride of lime

Uses:

 Used for bleaching cotton, linen in textiles industry, wood pulp in paper factories

 As an oxidizing agent. For disinfecting drinking water.

Baking Soda (Sodium Bicarbonate):

 Common Name: Baking Soda Formula: NaHCO3

 It is produced using sodium chloride as one of the raw materials. NaCl + H2O + CO2 + NH3 → NH4Cl + NaHCO3

(Ammonium chloride) (Sodium hydrogencarbonate) at Cathode

at Cathode at Anode

Fuels, margarine ammonia for fertilisers

Water treatment, swimming pools,

PVC. disinfectants, CFCs. pesticides detergents, paper making, artificial fibres De-greasing metals, soaps and HYDROCHLORIC ACID

For: cleaning steel, ammonium chloride, medicines, cosmetics

BLEACH For: household bleaches,

Bleaching fabric

Hasenclever Plant

Waste gases Slaked lime _Hopper

Chlorine

Bleaching powder

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Properties:

 Crystals are white in color, sparingly soluble in water, this solution is alkaline

 If solution of sodium hydrogencarbonate is boiled or heated, CO2 is given off. So it is used as baking powder to aerate the dough.

2NaHCO3 Heat→ Na2CO3 + H2O + CO2

 It is a mild non-corrosive base. The following reaction takes place when it is heated during cooking:

2NaHCO3  →heat Na2CO3 + H2O + CO2

(Sodium hydrogencarbonate) (Sodium carbonate)

 Sodium hydrogencarbonate has got various uses in the household. Uses:

 The soda commonly used in the kitchen for making tasty crispy pakoras is baking soda. Sometimes it is added for faster cooking.

 As an antacid.

 As an additive in food and drinks.

 In fire extinguishers.

Note: Baking powder contains tartaric acid, which neutralises sodium carbonate and thus prevents

the cake from tasting bitter

Uses of sodium hydrogencarbonate (NaHCO

3

):

 For making baking powder, which is a mixture of baking soda (sodium hydrogencarbonate) and a mild edible acid such as tartaric acid. When baking powder is heated or mixed in water, the following reaction takes place

NaHCO3 + H +_{→ CO}

2 + H2O + Sodium salt of acid

(From any acid)

 Carbon dioxide produced during the reaction causes the dough for the bread or cakes to rise making them soft and spongy.

Washing soda

 Another chemical that can be obtained from sodium chloride is Na2CO3.10H2O (washing soda).

 Common name: Washing soda

 Formula: Na2CO3 . 10 H2O

 A Molecule of sodium carbonate contains 10 molecules of water of crystallization. Anhydrous sodium carbonate is commonly known as soda ash.

 Sodium carbonate can be obtained by heating baking soda. On recrystallisation of sodium carbonate we get washing soda. It is also a basic salt.

Na2CO3 + 10H2O → Na2CO3 . 10H2O

(Sodium carbonate)

 When Na2CO3 10H2O is strongly heated then the following reaction takes place.

 Na2CO3 10H2O strong.heating.→ Na2CO3 + 10H2O

(soda ash)

Uses:

 In the manufacture of glass, soap, paper and other sodium compounds such as borax, caustic soda etc.

 for removing permanent hardness of water

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 As a cleansing agent for domestic purposes

 As a laboratory reagent.

Na2CO3 . 10H2O is an efflorescent salt i.e. at room temperature it gives out 9 molecules of its water of crystallization and dissolves in this water. This phenomenon is called efflorescence Na2CO3 . 10H2O room.temp.→ Na2CO3 . H2O + 9H2O

That is why washing soda is sold as soda ash in grocer shops.

Crystals of Salts are not dry:

 Blue copper sulphate crystals which seem to be dry contain water of crystallisation. When we heat the crystals, this water is removed and the salt turns white.

 If you moisten the crystals again with water, you will find that the blue colour of the crystals reappears.

 Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt. Five water molecules are present in one formula unit of copper sulphate. Chemical formula for hydrated copper sulphate is CuSO4 . 5H2O.

 One other salt, which possesses water of crystallization is gypsum. It has two water molecules as water of crystallization. It has the formula CaSO4.2H2O.

Plaster of Paris:

 Formula: CaSO4 ½H2O hemihydrated salt of calcium sulphate.

Preparation:

 It is prepared by controlled heating of gypsum at 373 K in a kiln. CaSO4 • 2H2O  →373K (CaSO4) • ½H2O + 1½H2O

Note: If heating is not controlled anhydrous calcium sulphate is formed. This is called burnt gypsum.

Properties:

 Plaster of Paris is a white powder. When it is mixed with water, crystals of gypsum are produced again, but this time they are set to form a hard solid mass.

CaSO4 ½H2O + 1½ H2O → CaSO4 2H2O

Plaster of Paris Gypsum

 Half a water molecule is shown to be attached as water of crystallisation.

 It is written in this form because two formula units of CaSO4 share one molecule of water.

 About 5000 years ago, Egyptians obtained a powder by heating gypsum (calcium sulphate) in open air fires.

 This powder was used for cementing blocks of their monuments. The powder is called plaster of Paris because it is made by using gypsum, which was mainly found in Montmartre in Paris.

APPENDIX: NCERT ACTIVITIES

Activity 2.1:

Aim: To show how the following solutions behave in the presence of different indicators

Apparatus: HCl, H2SO4, HNO3, CH3COOH, NaOH, Ca(OH)2, KOH, Mg(OH)2, NH4OH, watch – glass, phenolphthalein, Methyl orange, red litmus, blue litmus.