chemistry notes(ruther,heisen).pdf

EIJE _ Class Note _ Applied Chemistry_103(N)_AR Class 1 Topic: Rutherford model of atom, Bohr's theory and hydrogen spectrum explanation based on Bohr's model of atom, Heisenberg uncertainty principle Scientist Dulton...

EIJE _ Class Note _ Applied Chemistry_103(N)_AR Class 1 Topic: Rutherford model of atom, Bohr's theory and hydrogen spectrum explanation based on Bohr's model of atom, Heisenberg uncertainty principle Scientist Dulton first give the idea of atom in 1808 in his atomic theory. According to this theory, atom is the smallest indivisible particle of the matter that takes part in the chemical reaction. Later different discoveries reveal that an atom is made up of several smaller particles (~35 sub-atomic particles) of which electron, proton and neutron are the fundamental ones. In 1911. Scientist Rutherford first give the scientific idea about the structure of atom depending upon his famous α scattering experiment. Rutherford’s α particle scattering experiment Experiment A narrow beam of α (42He2+) particles are allowed to bombard with high speed on a very thin gold foil of thickness 0.0004 mm which is kept kept in a vacuum tube. Scattering of α particles are detected by the scintillations, α particles made on a semi circular ZnS coated screen. Observations i) Most of the α particles pass through the foil without any deflection from their own paths. ii) A few α particles get deflected from their paths making small angles. As the atomic weight of the target element increases, the angle of deflection also increases. iii) A very few α particles (1 in 20000) rebound tracing its incidental path. Inferences i) Most of the space inside an atom is empty. ii) As the mass of electron is negligible compared to the mass of an α particle, it is impossible for an electron to deflect an α particle from its path. Hence it can be concluded from above observations that the total mass and +ve charge of an atom are concentrated in a very small space compared to the size of atom. iii) The negatively charged electrons remain in the extra-nuclear part. Rutherford Atomic Model Based on the above experiment Rutherford gave the following model: i) Total positive charge and total mass of an atom are concentrated in the centre of the atom. It is called nucleus. Its diameter (10-13cm) is 1/10000th of diameter of the atom (10-8 cm). Positively charged protons and neutral neutrons remain in this small area. ii) In the extra-nuclear part of the atom, negatively charged electron revolves around the nucleus in a circular path or orbit. Now as the atom is neutral, the number of the protons in the nucleus is same as the number of electrons. iii) The coulombic force (electrostatic force) of attraction acting between the oppositely charged nucleus and electron is equal and opposite to the outward centrifugal force originating due to the rotation of electron around the nucleus. As a result the electrons do not fall on the nucleus. Thus Rutherford explainedthe stability of the atom. Drawbacks of Rutherford ‘s Model i) According to Maxwell’s electromagnetic theory an accelerating charged particle emits energy continuously. Hence negatively charged electron while revolving around the nucleus following a circular path emits energy continuously in the form of electromagnetic wave. The electron losing its energy approaches the nucleus in a spiral path of decreasing radius and finally falls on the nucleus. Hence the atom becomes unstable. So, Rutherford’s model cannot explain the stability of the atomic system. 1  EIJE _ Class Note _ Applied Chemistry_103(N)_AR ii) Rutherford’s model predicts the continuum spectrum of atom. But in actual case, a discontinuous line spectrum is produced. Bohr’s Atomic Model Niels Bohr in 1913 modified the drawbacks of Rutherford’ model with the help of Planck’s Quantum Theory and explained the stability and line spectrum of the atom. Bohr considered the simplest atom hydrogen (1 electron and 1 proton) as his model and made following postulates which are known as Bohr’s Postulates. Bohr’s Postulates i) Electrons can not rotate around the nucleus in any circular orbit. Rather the electron rotates around the nucleus only in some selected circular orbits of fixed radii such that the angular momentum (mvr) of electron in this circular orbit is an integral multiple of h/2π, where h is Planck’s constant. nh mvr= —— [Here, m= mass of electron, v= velocity of electron, r = radius of the orbit] 2π Here, n = 1,2,3,4…. n is called principal quantum number. It indicates the orbits in which the electron rotates around the nucleus. ii) While rotating around the nucleus in such selected orbits, the electron does not emit or absorb any energy. That means the energy of electron in these selected orbits remain fixed. Hence these orbits are called stationary orbits. iii) When an electron is transferred from one stationary orbit to another, the electron emits or absorbs energy in the form of electromagnetic wave. But in any condition, the electron can never exist in between the two successive stationary states. The energy absorbed or emitted during the transfer of electron from one stationary orbit (Ei) to another stationary orbit (Ef) is ΔE = Ef ~ Ei = hν, here ν is the frequency of the radiation. If Ef > Ei i.e. if electron is transferred from lower energy orbit to higher energy orbit*, then energy is absorbed. If Ei > Ef i.e. if electron is transferred from higher energy orbit to lower energy orbit, then energy is emitted. Success of Bohr Model i) The electrons can revolve around the nucleus only along certain selected orbit. An electron does neither absorb nor emit any radiation as long as it moves in such a selected orbit. As the energy of an electron in such orbits remain unaltered, such orbits known as ‘stationary orbit’. Thus an electron can move only in those orbits for which its angular momentum is integral multiple of h/2 p.That means angular momentum is quantized. Radiation is emitted or absorbed only when transition of electron takes place from one quantized value of angular momentum to another. Therefore, Maxwell’s electromagnetic theory does not apply here that is why only certain fixed orbits are allowed. Thus the stability of atom can be explained. ii) Using Bohr’s model, the discrete line spectrum of the one electron system like H, He+,Li2+,Be3+ can be explained. The origin of discontinuous line spectrum can be When an electron jumps from one such selected orbit to another, the difference 2  EIJE _ Class Note _ Applied Chemistry_103(N)_AR of energy between the two states is emitted or absorbed in accordance with the quantum theory of radiation. The frequency ν of this emitted radiation will be given by the relation: Ef − Ei = hν. Thus the appearance of discrete lines in the spectrum of atomic hydrogen can be explained in the terms of jumps of electrons between selected orbits, radiations of various definite frequencies are emitted thereby. (Picture from internet) iii) Using Bohr’s model, the radius and the energy of a stationary orbit can be calculated. iv) Bohr first give the idea of Quantum number. Drawbacks of Bohr Model i) Bohr’s model can not be applied for two or multielectron systems. ii) Fine and hyperfine splitting of line spectrum of one electron system in the presence the magnetic and electric field can not be explained by Bohr’s model. iii) According to Bohr’s theory, atom is two-dimensional. But in actual case atom is three dimensional. iv) Bohr’s theory directly opposes Heisenberg’s uncertainty principal. v) Bohr’s theory fails to explain the wave-particle dual nature of electron Heisenberg’s Uncertainty Principle: Werner Heisenberg a German physicist in 1927, stated uncertainty principle which is the consequence of dual behaviour of matter and radiation. It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Mathematically, it can be given as in equation. Δx. Δp ≥ h/2p or, Δx. Δ(mv) ≥ h/2p or, Δx. Δv ≥ h/2pm where Δx is the uncertainty in position and Δp (or Δv) is the uncertainty in momentum (or velocity) of the particle. If the position of the electron is known with 100% accuracy (Δx = 0 or very small), then the velocity of the electron will be uncertain [Δv is large]. On the other hand, if the velocity of the electron is known precisely, then the position of the electron will be uncertain ( Δx will be large). The effect of Heisenberg Uncertainty Principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects. Additional Reading: Planck’s Quantum theory:The energy is absorbed or emitted in the form of discrete packets of energy known as quanta (plural, quantum in singular). Each quantum is associated with definite amount of small energy which is related to the frequency of absorbed or emitted radiation i.e. energy E = hν. Here h is Planck’s constant. Its value is 6.626 ×10-34 J s. The absorsorption or emission of 3  EIJE _ Class Note _ Applied Chemistry_103(N)_AR energy is done in the integral multiple of hν. i.e 1hν, 2hν, 3hν …. This is known as Quantisation of energy. In case of light, the quantum is called photon. Orbits do not have their own energies. Rather the energy of the rotating electron in a particular orbit is the energy of that orbit. Dual Nature of Electron — de Broglie Hypothesis In Rutherford’s and Bohr’s theory, electron has been considered as negatively charged particle. But with the particle nature only, all the properties of an electron can not be explained properly. In 1924, de Broglie proposed the wave-particle dual nature for subatomic particles like electron. According to his hypothesis, a microscopic particle in motion like a moving electron exhibits both its particle and wave nature. If the wavelength of the wave associated with moving electron with momentum p is λ, then h h λ = — = —— [m= mass of electron, v = velocity of electron and h = Planck’s constant] p mv Although de Broglie hypothesis is applicable for all particles be it small or large, it is only significant for microscopic particles like subatomic and atomic particles. For macroscopic particles, the momentum of the particle is very large. So the de Broglie wavelength is insignificant for them and macroscopic particles can not show wave nature. ILO: 1. State Rutherford’s and Bohr’s atomic model and its success and failures. 2. Discuss Heisenberg’s uncertainty principle Questions: 1. Discuss Rutherford’s model and discuss its success and failure 2. State Bohr’s Postulates and its success and fail 3. State and explain Heisenberg Uncertainty Principle. Class 2 Topic: Quantum numbers Quantum Numbers The numbers which are required to identify an electron and to define its energy in an atom precisely are called quantum numbers. Four (4) quantum numbers are required to identify an electron in an atom. These quantum numbers are (i) Principal Quantum Number (n) (ii) Azimuthal Quantum Number (l) (iii) Magnetic Quantum number (m) (iv) Spin Quantum Number (s) Principal Quantum Number: Its concept comes from Bohr’s theory. According to Bohr’s model, electrons rotate around the nucleus in some definite orbits of fixed energy, known as energy shell or main energy levels of an atom. These energy levels are denoted by principal quantum number n. The energy levels are identified by K, L, M, N … etc. alphabets for the values of n =1, 2, 3, 4… As the value of principal quantum number n increases, the distance of the energy level from the nucleus increases and the size and energy associated with the energy level also increase. The order of the energy levels in increasing energy is K(n=1) < L (n =2) (lone-pair – bond-pair) of electrons > (bond-pair – bond-pair) of electrons. (Bond pair : two electrons participating in bond formation, lone pair: electron pair that doesn’t participate in bond formation.) During molecule formation, these hybrid orbitals of one atom participate in bonding with the atomic orbitals or hybrid orbitals of other atom to produce sigma-bond. sp3 Hybridisation In this hybridization 1 s orbital and 3 p orbitals of an atom participate and 4 identical hybrid orbitals are formed. Each sp3 hybrid orbital has 25% s character and 75% p character. The sp3 hybrid orbitals are oriented towards the for vertices of a regular tetrahedron.The bond angle is 109 o28'. Examples of sp3 hybridised molecule is CH4, H2O, NH3 etc. CH4 NH3 In NH3, Three sp3 hybrid orbitals of N with one electron each, overlap with the 1s orbitals of three H-atoms individually and forms 3 sigma-bonds. The 4th hybrid orbital contains a lone pair of electrons. As the repulsion between (lone-pair – bond-pair ) of electrons is greater than the (bond-pair –bond-pair) of electrons, the bond angle H-N-H deviates from the normal angle of sp3 hybridisation 109 o 28' to 106 o45' and the shape of the molecule is pyramidal. H2O In H2O molecule, 2 sp3 hybrid orbitals of O-atom with one electron each, overlap with the 1s orbitals of two H-atoms, individually and forms two sigma bonds. Rest two sp3 orbitals coantain two lone pairs of electrons. As the order of repulsion between different 14  EIJE _ Class Note _ Applied Chemistry_103(N)_AR types electron pair is (lone-pair – lone-pair) of electrons > (lone-pair – bond-pair) of electrons > (bond-pair – bond-pair) of electrons. Due to lone-pair-lone pair repulsions, the bond angle H-O-H is 104.5o and H2O molecule is V-shaped. sp2 Hybridisation In this hybridization, one s orbital and 2 p orbitals of an atom participate to produce 3 identical sp2 hybrid orbitals. Each sp2 hybrid orbital has 33.33% (1/3) s character and 66.66% p (2/3) character. These three sp 2 hybrid orbitals remain in the same plane making 120o bond angle with each other. BCl3 In BCl3, the central atom B is sp2 hybridised. 3 sp2 hybrid orbitals with B is hybridized with 3p orbital of 3 Cls, individually. So BCl3 molecule is planar trigonal (triangular shape with B atom at centre of triangle) and the bond angle Cl-B-Cl is 120o. sp Hybridisation In sp hybridization, one s orbital and one p orbital of same atom participate to produce 2 identical sp hybrid orbitals. Each sp hybrid orbital 50% s character and 50% p character. The bond angle between two sp-hybrid orbitalsis 180o (bonds are in a straight line). This type of hybridisation is known as diagonal hybridisation. BeCl2 In BeCl2, the central atom Be is sp-hybridised. Two sp hybrid orbital overlap with 3p orbitals of 2 Cl-atoms, indivudually. The molecule is linear. The bond angle Cl-Be-Cl is 180o. Sigma Bond: Sigma bond is formed due to the headon overlap between two atomic orbitals. It can be formed between the head on overlap of two s orbitals, one s and one p orbital or between two p orbitals. The order of strength is as follows (s-s) σ-bond < (s-p) σ-bond < (p-p) σ- 15  EIJE _ Class Note _ Applied Chemistry_103(N)_AR bond. If there is a single bond between two atoms, then the bond must be σ-bond. σ-bond even can be formed by the head on overlap of one pure atomic orbital with one hybrid orbital or overlap of two hybrid orbitals. Pi Bond Pi Bond is formed by the sideways overlap of two parallel p orbitals. The strength of π-bond is much less than the strength of σ- bond, as σ-bond is formed by the headon overlap of two atomic orbitals and π-bond is formed by the sideways overlap of atomic orbitals. Structure of Diamod Every C-atom in diamond is sp3 hybridised. Each carbon atom is attached to another 4 carbon atoms through the covalent bonds tetrahedrally and thus a three dimensional macromolecular structure of diamond is formed. In diamond, C–C–C bond angle is 109o28' and C–C bond strength is 1.54 Å. Properties of Diamond from its Structure (i) All 4 valance electrons of each C–atom of diamond participate in the formation of covalent bonds and there is no free electron to conduct electricity. Hence diamond is a bad conductor of electricity. (ii) Since the carbon-carbon bonds of diamond are very strong covalent bonds and the diamond possess a macromolecular structure, hence large amount of energy would be required to break those bonds. Hence diamond is very hard and can be used as abrasive. (iii) The refractive index of diamond is very high. As a result light gets internally reflected several times. Hence diamond is very bright and used as precious stone. Structure of Graphite (i) Each C atom in graphite is sp2 hybridised. Each C atom is bonded to 3 other C atoms forming a hexagonal planar structure which finally results in two dimensional sheet or layer. (ii) These layers remain parallel to each other and are held together by weak van der Waals attractive force. (iii) Each carbon–carbon bond length in the planar hexagonal ring within a layer is 1.42Å and the distance between two successive layers is 3.35Å. (iv) As the each C atom in graphite is sp2 hybridised, the 4th electron on C atom is free and mobile. Properties from the Structure of Graphite (i) As the each C atom in graphite is sp2 hybridised, the 4th electron on C atom is free and mobile. As a result graphite is a good conductor of electricity. (ii) As the two consecutive parallel layers are held by weak van der Waals force, hence one layer can slide over the another layer As a result graphite is soft and can act as solid lubricant. (iii) Carbon – carbon bonds within a layer are strong covalent bonds and hence graphite has high melting point. ILO: Discuss about ionic, covalent and coordinate bonding and hybridisation with examples Questions: 1. “Diamond is one of hardest substance where as graphite is soft”--- Explain. 2. ‘Graphite is used to make electrodes whereas diamond is bad conductor of electricity’---- Explain 3. Though CH4, NH3, H2O are all sp3 hybridized but their shapes are different. --- Explain 4. What kind of hybridization are present in BeCl2, BCl3, H2O, NH3, CH4. Discuss about their shape. 5. Discuss about which chemical bond is present in Al2O3, MgCl2, O3, H2SO4, O2, NaCl 6. Compare the properties of ionic and covalent compounds. Class 7 Topic: Coordinate bond, Hydrogen bond, Metallic bond Coordinate Bond 16  EIJE _ Class Note _ Applied Chemistry_103(N)_AR If the electron pair, that is shared mutually by both atoms to fulfill the octate, comes from one atom only, then the bond formed between two atoms are called coordinate bond. The atom which donates the electron pair is called donor atom and the atom which accepts the electron pair is called acceptor atom. Hence donor atom must have atleast one lone pair of electron for donation and the acceptor must have atleast one vacant orbital to accept elevtron pair.The coordinate bond formed between two atoms is expressed by using → symbol. Formation of H3O+ ion: Addition compound between NH3 and BF3 Formation of NH4+ ion Ozone Name a compound which has electrovalent, covalent and coordinate bonds present in it. NH4Cl, NH4NO3, [Cu(NH3)2]SO4 etc. [In NH4Cl NH4+ ion and Cl- ion are paired by covalent bond. NH4+ ion is formed by the coordinate bonding between NH3 and H+ where NH3 donates its lone pair of electron to H+. N combines with H via covalent bond to form NH3. Hence all three bonds are present in NH4Cl.] Hydrogen Bond (H-bond) When hydrogen atom forms a covalent molecule with strong electronegative atoms like N, O or F, the electronegative atom pulls the boned pair of electrons towards it and becomes partially negatively charged while the H atom attached to it becomes partially positively charged. Hence the H atom of one molecule is bonded to the electronegative atom (N, O or F atom) of the other molecule via weak electrostatic force of attraction. This weak electrostatic force is called H-bond. It is not like normal chemical bond and much weaker than the ionic and covalent bonds. Hence it is shown by dotted line. Types of Hydrogen Bond Intermolecular H-Bond This kind of H-bond is formed between same or different types of molecules. 1. H-bond in Water Oxygen atom in H2O molecule is strongly electronegative, as a result O atom gains partial negative charge where as H atom gains a partial positive charge. As a result, a large number of H 2O molecules get associated through the intermolecular H bonding. Large amount of energy is required to break these H-bonds. Hence boiling point of water increases and water is liquid at normal temperature. 17  EIJE _ Class Note _ Applied Chemistry_103(N)_AR S atom being larger in size and much less electronegative than oxygen, no intermolecular H-bond takes place in H2S. Only a weak van der Waals force of attraction acts among the H 2S molecules. Hence H2O is liquid but H2S is gaseous at normal temperature. 2. H-bond in HF Due to the presence of strongly negative F atom, HF bond becomes polar, where H atom becomes partially positively charged and F atom becomes partially negatively charged. As a result HF molecules are associated with intermolecular H- bonding. For this reason HF is less acidic than HCl. Due to the the intermolecular H-bonding, HF molecules undergoes weak ionization to produce smaller number H+ ions. On the other hand, Cl being larger in size and much less electronegative than F atom, HCl does not participate in intermolecular H-bonding and HCl in aqueous solution undergoes complete ionization. That is why HF is weaker acid than HCl. 3. HF forms basalt though it is a monobasic acid. F being strongly electronegative, HF bond becomes polar, where H atom becomes partially positively charged and F atom becomes partially negatively charged. Hence, F– ion forms strong H-bond with HF molecule [F– ---Hδ+––Fδ-] and forms HF2– ion (bifluoride ion). That’s why HF forms bisalt like KHF2 though it is a monobasic acid. Cl being larger in size and much less electronegative than F atom, H bonding is not possible between Cl – ion and HCl to form HCl2– ion. That’s why, bisalts are possible for HF, not for HCl. 4. Ice floats on Water. In the crystal of ice, each H2O molecule is tetrahedrally surrounded by other 4 H2O molecules through H-bonds to form open cage structure which results in intermolecular spaces in the ice crystal. When ice melts, water molecules are associated via H-bonds to produce chain like structure. Hence the water molecules approach each other more closely and there is no such intermolecular space in water molecules. So, the density of ice is much less than that of the water. This is the reason solid ice floats on liquid water. At 4oC temperature, water molecules approach each other as close as possible. Hence the density of water is maximum at 4oC. 5. Molecular weight of both ethanol and formic acid are same but the boiling point of formic acid is more than that of ethanol. This is because of the following facts: Formic acid and ethanol both undergoes H-bonding, but the H-bonding in formic acid is much stronger than that of ethanol. Along with this O atom of C=O group also participates in H bonding. Hence the H-bond in HCOOH is much stronger than that of C2H5OH. 18  EIJE _ Class Note _ Applied Chemistry_103(N)_AR In vapour phase formic acid forms stable dimer through the H bonding. As a result, the effective molecular weight of formic acid increases, as a result boiling point increases. Because of these two reasons, the boiling point of HCOOH is higher than the ethanol, though they have same molecular weight. 6. Dimethyl ether is more volatile than ethanol though they have same molecular weight. This is because of the fact that molecules of ethanol (C2H5OH) remains associated through the H-bonds. But dimethyl ether does not participate in H- bonding. Hence the boiling point of C2H5OH is more than that of dimethyl ether. So, dimethyl ether is more volatile than ethanol. Due to the intermolecular H-bonding, the molecules of a compound remain associated. Hence the boiling point and the melting point of the compound increase and the compound becomes less voltatile. Diethyl ether is less soluble in water than the ethanol. Due to the greater extent of H-bonding between ethanol and water, the solubility of ethanol is more in water. But the extent of H-bonding of diethyl ether with water is much less. Hence diethyl ether is less soluble. Intramolecular H-bond In this case H-bond is formed in the molecule itself. The boiling point of p-nitrophenol is higher than that of o-nitrophenol. O-nitrophenol undergoes intramolecular H bonding and exist as single molecule. On the other hand, p-nitrophenol molecules are associated via intermolecular H bonding. That is why the boiling point of p-nitrophenol is higher than that o-nitrophenol. Metallic Bonds In metallic solids the metal atoms are held together by metallic bonds. Example: Any metals like Au, Ag, Cu etc. 19  EIJE _ Class Note _ Applied Chemistry_103(N)_AR Metallic bond is not a chemical bond. Metal atoms form metal ions after losing one or more valance electrons. These free and mobile electrons do not belong to any particular ion, rather they belong to the metallic crystal as a whole. Metallic ions may be considered to be in a sea of free electrons. Now these mobile electrons can migrate from one metal atom to another freely. As a result a strong bond is formed among the metal atoms. This bond is called metallic bond. Properties: A metallic solid is a good conductor of electricity and heat as it contains free and mobile electrons. Metallic solids have very high melting and boiling points. As the number of valance electrons of the metals increases, the metallic bonds become stronger. Hence their melting point also increases. Metallic solids are ductile, malleable and have metallic lusture. ILO: Demonstrate coordinate bonding, hydrogen bonding and metallic bond with examples. Questions 1. Which type of chemical bonds are present NH4Cl, H3O+, NH4+? 2. Why does solid ice float on liquid water? 3. Why dimethyl ether is more volatile than ethanol? 4. Write short note on metallic bonding. Class 8 & 9 Topic: Solution A Solution is a homogenous mixture of two or more components. Solvent: The component that is present in largest quantity is called solvent. It determines the physical state of solution. Solute: One or more components present in solution other than solvent is called solute. Binary solutions: Solution consisting of two components only. ILO: solve the simple numerical problems related to solution Questions: numerical problems as discussed in class. 20  EIJE _ Class Note _ Applied Chemistry_103(N)_AR Class 10 & 11 Topic: Hardness of water, Numerical problems related to hardness GRAPHICAL PRESENTATION OF WATER DISTRIBUTION ON EARTH Bar Diagram Pie diagram Soft Water and Hard Water Based on action of soap with it, water is classified in two categories: Soft Water: Water which easily forms lather with soap is called soft water. E.g. distilled water, rain water etc. Hard Water: Water which does not form lather easily with soap is called Hard water. On reaction with the soap, hard water first produces white scum. Hard water produces lather only after the consumption of considerable amount of soap. E.g. sea water, spring water, tube well water etc. Causes of Hardness: Hardness of water is due to the presence of dissolved bicarbonate, sulphate and chloride salts of Ca, Mg and Fe like Mg(HCO3)2, Ca(HCO3)2, Fe(HCO3)2, MgCl2, CaCl2, MgSO4 etc. Hardness of water is normally caused by the presence of dissolved salts of any metal except Na and K. Anions do not play any role to cause hardness. 21  EIJE _ Class Note _ Applied Chemistry_103(N)_AR Soaps Soaps are Na or K salts of higher fatty acids like stearic acid, palmitic acid or oleic acid. E.g. Sodium Stearate (C17H35COONa), Sodium Palmitate (C15H31COONa) etc. These salts are soluble in water and produces lather after dissolving in water. Additional Reading Detergents which are used for the household purposes are not the salts of fatty acids like ordinary soap. They are derivatives of linear alkyl benzene sulphonic acid or aryl alkyl sulphonic acid. Hence these detergents produce lather even ith hard water. So these detergents are called soap less soap. Action of Soaps on Hard Water Soaps are Na or K salts of higher fatty acids like stearic acid, palmitic acid or oleic acid. Ca, Mg and Fe salts, present in hard water, undergo exchange reaction with the soaps and form the Ca, Mg and Fe salts of the fatty acids which are insoluble in water and separate as white precipitate. C17H35COONa + Ca-salt = C17H35COOCa ↓+ Na-salt Insoluble Scum When all the Ca, Mg and Fe salts are removed from water as white scum, the soap produces lather. In this way, after the consumption of large amount of soap finally the lather is produced. Thus hard water is not suitable for washing purpose. Types of Hardness in Water Depending upon the nature of dissolved salts in water, the hardness is divided into following two classes: Temporary Hardness: This type of hardness can be removed easily by simple processes like boiling etc. That’s why it is called temporary hardness. This type of hardness is caused by the presence of dissolved bicarbonate salts of Ca, Mg and Fe. It is also called carbonate hardness. Permanent hardness: Permanent hardness is caused by the presence of dissolved salts of sulphate and chloride of Ca, Mg and Fe. This type of hardness can not be removed easily by simple processes like boiling etc., but special chemical treatments are to needed to remove hardness from water. It is also known as non-carbonate hardness. Permanent hardness can not be removed by boiling. More over due to boiling amount of water decreases and the concentration of the salts in water increases. As a result, degree of hardness increases. Disadvantages of Using Hardness of Water (i) Hard water is not suitable for the use of household purposes. It causes wastage of soap in washing. Use of hard water causes deposition of an insoluble and thermally insulating layer of metal salt on the inner surface of cooking utensil. As a result, wastage of fuel increases and the lifetime of the utensil decreases due to overheating. Hard water is not also suitable for drinking, cooking and bathing purposes. (ii) Hard water is not suitable to use in boiler. In boiler, the water is boiled constantly to form water vapour. Boiling of hard water in the boiler causes the formation of sludge and scale. A layer of salts, mainly CaCO3 and CaSO4, are deposited on the inner wall of the boiler. This layer is called scale. The scale may cause following problems -- (a) This layer acts as thermal insulator and greater amount of fuel is required for heating. (b) The lifetime of the boiler decreases due to overheating. (c) Due to the unequal expansion of the deposited scale and the metallic body of the boiler, the boiler may burst causing an accident. (d) Large amount of dissolved matters in water on boiling may cause foaming. (e) Sometimes the hydrolysis of dissolved salts in water produces HCl which corrodes the all of the boiler. MgCl2 + H2O = Mg(OH)2 + HCl So, the water used in boiler must be soft and free from dissolved or suspended matters to avoid scale formation, corrosion, foaming and accidental bursting. In boiler, the water is boiled constantly to form water vapour. When hard water is used in boiler, the concentrations of dissolved salts within in it gradually increase due to the evaporation and when the concentration crosses the saturation limit of the salts, the salts precipitate out from the solution. Because of this phenomenon, sludge and 22  EIJE _ Class Note _ Applied Chemistry_103(N)_AR scale are formed with in boiler. Sludge is loose, non-adherent precipitates with in boiler where as scale is a hard, insoluble, sticky, adherent and hard to remove deposit on the inner wall of the boiler. Scale acts as thermal insulator. As a result, wastage of fuel increases and the lifetime of the boiler decreases due to overheating. (iii) Hard water can not be used in paper industry, textile industry, tanneries, laundries etc. Units of Hardness The temporary and the permanent hardness are normally present simultaneously in a sample of natural water. The total hardness of water = (the temporary hardness + the permanent hardness) present in water. Hardness of water is measured in terms of the amount of the CaCO 3. If other salts are present in water, then the hardness is expressed as the equivalent amount of CaCO3 of these salts. Degree of hardness of a sample of hard water is defined as the parts of CaCO 3 or its equivalent of other salts present million (106) parts of water of that sample of water. The unit of the hardness of water is parts per million or ppm. To express the weight of a salt, that causes hardness in water, as the equivalent of CaCO 3, we shall use the concept of chemical equivalence. For example let us consider w gm of MgSO4 is present in a sample of water Molecular weight of CaCO3 = 100, so the equivalent weight = 100/2 = 50 Molecular weight of MgSO4 = 120, so the equivalent weight = 120/2 = 60 Now, 1 equivalance of MgSO4 ≡ 1 equivalance of CaCO3 or, gm-equivalent weight of MgSO4 ≡ gm-equivalent weight of CaCO3 or, 60 mg of MgSO4 ≡ 50 mg of CaCO3 50 × w or, w mg of MgSO4 ≡ ———— mg of CaCO3 60 CaCO3 equivalent can be generalized as Weight of hardness producing salt × Chemical equivalent of CaCO 3 = ————————————————————————————— Chemical equivalent of hardness producing salt The unit of CaCO3 equivalent is same as the unit of the weight of the hardness producing salt. Hardness of water is also expressed in milligram per litre (mg/lt) unit. 1 mg/lt of water = 1 mg CaCO3 in 1 lt of water = 1 mg CaCO3 in 1000 ml of water = 1 mg CaCO3 in 1000 gm of water (as density of water = 1mg/ml) = 1 mg CaCO3 in 1000 × 1000 mg of water = 1 ppm 1 mg/lt = 1 ppm ILO: Solve the simple numerals related to hardness of water. Questions 1. What is the reason of hardness of water? 2. What is soap? What is the action of soap on hard water? 3. Compare between soft water and hard water. 4. Why hardwater can not be used in boiler? 5. What are the units of hardness of water. 6. Numerals related to hardness of water as discussed in class. Class 12 & 13 Topic: Water Softening Techniques Processes to Remove Permanent and Temporary hardness: Soda-Lime Process: 23  EIJE _ Class Note _ Applied Chemistry_103(N)_AR In this process, calculated amount of Ca(OH)2 and Na2CO3 are added to remove the hardness of water. Both permanent and temporary hardness of water can be removed in this process. CaCl2 + Na2CO3 = CaCO3 ↓+ 2NaCl CaSO4 + Na2CO3 = CaCO3 ↓+ 2Na2SO4 MgCl2 + Na2CO3 = MgCO3 ↓+ 2NaCl MgCO3 + Ca(OH)2 = Mg(OH)2↓ + CaCO3 ↓ Mg(HCO3)2 + 2Ca(OH)2 = 2CaCO3 ↓+ Mg(OH)2 ↓+ 2H2O We know that MgCO3 is partially soluble in water. In this process usage of calculated amount of Ca(OH)2 and Na2CO3 both helps to remove the hardness completely. Zeolite Process/Permutit Process Permutit is the trade name of the artificially prepared compound sodium aluminium silicate which is analogous to natural mineral zeolite (NaAlSiO4.3H2O). In Zeolite process, Na+ ions present in the permutit are exchanged with the Ca 2+, Mg2+ etc. ions in water. That’s why this process is also known as base exchange process. In this process both temporary and permanent hardness are removed. When the hard water is passed through the permutit bed, Na+ ions in Zeolite are replaced with the equivalent amount of dissolved Ca2+, Mg2+ etc. ions in water and insoluble Ca/Mg-Zeolite is formed. So, in water only Na+ cations are present and the water becomes soft. Ca/Mg—salt + Na—Zeolite = Ca/Mg— Zeolite ↓+ Na—salt The water obtained from Zeolite process is soft and degree of hardness is zero. After prolonged use, Zeolite loses its activity. Then the exhausted Zeolite is regenerated by passing 10% NaCl solution through it. Ca/Mg— Zeolite + 2NaCl = Na— Zeolite + CaCl2/ MgCl2 Ion-Exchange Process Ion exchange process, ion-exchange resins are used to soften the hard water. Except the distillation process, it is possible to obtain deionised or demineralised water only by the ion-exchange process. The water obtained is free from all electrolytes and both temporary and permanent hardness can be removed by this process. For deionization, the hard water is first passed through the cation exchange resins and then through the anion exchange resins. Resins are synthetic polymeric organic molecules. The cation exchange resin contains –SO3H, –COOH etc. as active groups. They contain exchangeable H+ ions which are exchanged with all dissolved the cations which may or may not be responsible for hardness of water (e.g. Ca2+, Mg2+, K+, Na+ etc.). Cation exchange resins are represented by RH (R = alkyl group). Anion exchange resin is represented by R4N+OH- (R = alkyl group or H). This OH- ions are exchanged with all anions present in water. The hard water is passed through the cation exchange resin. As a result all dissolved cations in water such as Ca 2+, Mg2+, K+, Na+ etc. are exchanged with equivalent amount of H+ ions of resins. The effluent water contains only H+ cation and the water becomes acidic. The anions present in water remains unaffected by cation exchange resins. 2RH +CaCl2 = R2Ca + 2HCl 2RH + MgSO4 = R2Mg + H2SO4 2RH + Ca(HCO3)2 = R2Ca + 2H2CO3 RH + NaCl = RNa + HCl Then the water is passed through the anion exchange resins and the anions present in water such as Cl-, SO42- etc. are exchanged with equivalent amount of OH- ions of the resins. R4N+OH- + HCl = R4N+Cl- + H2O R4N+OH- + H2SO4 = (R4N+)2 SO42- + H2O In this way, the obtained water contains no dissolved cations except H+ ion and no dissolved anions except OH- ion. H+ and OH- form undissociated H2O. Thus the water obtained in this way is free from all cations and anions and is called deionised or demineralised water. After prolonged use, the ion-exchange resins lose their activity. Then the exhausted cation exchange resin is regenerated by passing dil. H2SO4 through the resin. R2Mg + H2SO4 = RH + MgSO4 The anion exchange resin is regenerated by passing dil. NaOH through it. R4N+Cl- + NaOH = R4N+OH- + NaCl 24  EIJE _ Class Note _ Applied Chemistry_103(N)_AR Deionised or demineralised water is a sample of soft water which contains no dissolved cations except H+ ion and no dissolved anions except OH- ion. H+ and OH- form undissociated H2O. Its degree of hardness is zero. It is obtained by using cation and anion exchange resins. Deionised Water and Distilled Water Deionized water or demineralised water is obtained by removing all the cations except H+ ion and all the anions except OH- ion by using cation and anion exchange resins. On the other hand, distilled water is obtained in distillation process in which the water vapour, obtained by boiling, is condensed back to water. There is as such no difference between deionised water and distilled water. In both cases degree of hardness is zero. But distilled water is purer than the deionised water. In deionised water, non-electrolyte materials such as organic impurities may be present in dissolved condition. Both deionised and distilled water contain little amount of CO 2 and Silica. Estimation of Total Hardness by E.D.T.A. Method Hardness of water is estimated using the standard solution of EDTA by complexometric titration. EDTA (Fig a) is Ethylene Di-amine Tetra-acetic Acid. Di-sodium salt of EDTA (Fig b) is actually used for the estimation of hardness. Among the four replaceable Hs of EDTA, two are replaced by the Na. EDTA forms complex with Ca 2+, Mg2+ ,Fe2+ ions present in water (Fig c). Ca2+ + EDTA —————> Ca2+—EDTA complex Mg2+ + EDTA —————> Mg2+—EDTA complex (Fig of EDTA (Fig a), its di-sodium salt (Fig b) and its calcium salt (Fig c) in Bengali version: Please insert) In this titration Eriochrome Black T (EBT) is used as indicator. EBT indicator forms wine-red coloured complex with Ca2+, Mg2+, Fe2+ ions present in water. Ca2+ + EBT —————> Ca2+—EBT complex Mg2+ + EBT —————> Mg2+—EBT complex M2+—EDTA complex is more stable than M2+—EBT complex (M=Ca, Mg, Fe etc.). During titration at first EBT indicator is added to the water sample where EBT indicator forms a wine-red complex with M2+ ions (M=Ca, Mg, Fe etc.), Now EDTA is added drop wise. EDTA first forms complexes with the free metal ions present in water. On gradual addition of EDTA all free metal ions form complex with it. Then on further addition of EDTA, M 2+—EBT complex being less stable than M2+—EDTA complex , M2+—EBT complex is broken to form M2+—EDTA complex and EBT becomes free. As a result at end point the colour solution changes from wine red to steel blue (colour of free indicator). M2+—EDTA complex is stable in the pH range 8-10. to maintain this pH, NH4Cl-NH4OH buffer solution is used. Procedure of Titration: A sample of 100 ml hard water is taken in a 500 ml conical flask. 5 ml NH 4Cl-NH4OH buffer solution is added to it. Followed by 3-4 drops of EBT solution (indicator). The colour of the solution is turned to wine-red. The standard EDTA solution of strength 0.01 (M) (taken in burette) is then added dropwise to the conical flask with constant shaking until the colour of the solution is changed to steel- blue colour. Let us consider that the mean volume of EDTA required = V ml 1000 ml 1(M) EDTA ≡ 1000 ml 1(M) CaCO3 ≡ 100 gm CaCO3 (1 molecule of EDTA foms complex with 1 Ca2+ ion ) So, 1000 ml 1(M) EDTA ≡ 100 gm CaCO3 or, 1000 ml 0.01 (M) EDTA ≡100 × 0.01 gm CaCO3 100 × 0.01 × V or, V ml 0.01(M) EDTA ≡ ———————— gm CaCO3 1000 V = ——— gm CaCO3 1000 100 ml water = 100 gm water (density of water is 1 gm/ml ) V 100 gm water contains = ——— gm CaCO3 1000 V × 106 6 10 gm water contains = ————— gm CaCO3 1000 × 100 = 10V gm CaCO3 The hardness of water sample is 10V ppm. 25  EIJE _ Class Note _ Applied Chemistry_103(N)_AR Total Solids in water: They occur in water either in solution (Total dissolved solid: TDS) or in suspension (Total suspended solid: TSS). These 2 types of solids are identified by using a glass fiber filter. Suspended solids are retained on the top of the filter and dissolved solids pass through the filter with the water. If the filtered portion of the water sample is placed in the small dish and then evaporated, the residue solid is called total dissolved solids. Total Solid (TS) = TSS + TDS Freshwater: 5000 mg/L TDS Municipal Water Supply 1. Settling of water in a large tank to remove suspended impurities. 2. Coagulation of colloidal particles by adding suitable coagulants like alum [K 2SO4, Al2(SO4)3.24H2O], green vitriol [FeSO4.7H2O] etc. Positive charge from coagulants neutralizes the negative charge of dirt and other dissolved particles and the neutralized particles bind with each other to form slightly large particles which undergoes sedimentation. 3. Flocculation: gentle mixing of the water to form larger, heavier particles called flocs. Often, water treatment plants will add additional chemicals during this step to help the flocs form. 4. Sedimentation: Coagulated or sedimented particles settled at the bottom of water as they are heavier than water. 5. Filtration: Once the heavier particles settled down at the bottom, the clear water from the top is filtered to separate additional solids from water. The filter can be of different pore sizes and are made of different materials like sand, gravel, charcoal etc. These filters remove dissolved particles and germs, such as dust, chemicals, parasites, bacteria, and viruses. Activated carbon filters also remove any bad odours. Water treatment plants can use a process called ultrafiltration in addition to or instead of traditional filtration. During ultrafiltration, the water goes through a filter membrane with very small pores. This filter only lets through water and other small molecules. Reverse osmosis is another filtration method that removes additional particles from water. Water treatment plants often use reverse osmosis when treating reused water or salt water for drinking. 6. Disinfection/ Sterilization: i) By using bleaching powder or by direct chlorination. HOCl produced kills the pathogenic bacteria present in the water. Ca(OCl)Cl + H2O ---→ Cl2+Ca(OH)2 Cl2+H2O ---→ HOCl +HCl ii) Sterilization by chloramine iii) Sterilization by ozonization iv) Boiling, exposure to sunlight, using UV light. ILO: Discuss different water softening methods Discuss Municipal water treatment process. Questions: 1. How the water is softened using Zeolite process? (Chemical reactions only) 2. How the water is softened using ion-exchange resin process? (Chemical reactions only) 3. Discuss municipal water treatment process. 4. Discuss how the hardness of water is estimated using EDTA? Project: 1. Bring water sample from your home (tap water/ RO water/ normal filtered water/pond/river water etc.) and calculate hardness in lab class, prepare a report on class basis. 2. Prepare a chart based on water quality standard of India. Class 14 &15 Topic: Metallurgy 26  EIJE _ Class Note _ Applied Chemistry_103(N)_AR Minerals and Ore Minerals:The Minerals are the naturally occurring inorganic substances in which the metals are found in either free state or combined form of different metallic compounds along with impurities. The chemical composition of a mineral is always fixed. Ores: Ores are those minerals from which the metals can be easily and economically extracted. Example: Haematite (Fe2O3) and Iron Pyrites (FeS2) both are the minerals of Fe. But Haematite (Fe2O3) is the ore of Fe. Iron Pyrites has high S % and sulpher can not be removed completely from the extracted iron. Presence of sulpher makes the iron brittle. Hence Iron Pyrites is not regarded as an ore of Fe. ‘All ores are minerals but all minerals are not ore.’-Explain Minerals are the naturally occurring compounds of the metals. But from all minerals metals can not be extracted efficiently and easily. Ores are those minerals from which the metals can be extracted easily and economically profitably. Hence all ores are minerals but all minerals are not ore. For example, Haematite (Fe2O3) and Iron Pyrites (FeS2) both are the minerals of Fe. But haematite (Fe2O3) is the ore of Fe. Metallurgy The processes including easy and efficient extraction of metal, purification of metal and analysis of metallic properties and preparation of alloys are called metallurgy. In ores, metals are present in cationic forms, During metal extraction, metallic ions accepting electrons form metals. Thus the metal extraction process is a reduction process. The oxidation number of free metal is 0. Gangue: Unwanted substances like clay,silica, silicates etc. are present as impurities in the ore along with the crucial metallic compounds. These impurities are called gangue or matrix. Methods Used in Extraction of Metals i) Crushing and Grinding of Ore: The metal ores are obtained in the forms of rocks. First these ores are crushed in pieces and ground in fine powder. ii) Concentration of Ore: Normally the ores are associated with the impurities or gangue like clay, silica, silicate etc. In the process of concentration of ore, the impurities are removed as much as possible and the relative % of the crucial metallic compound (from which the metal is to be extracted) is increased. (a) Gravity Separation : The Powdered ore is washed with the running stream of water. The lighter impurities washed away with water and heavier ore settle down. Thus the relative % of metal containg compound is increased. (b) Magnetic Separation: If any one of the impurities and the metallic ore are magnetic, then using a strong magneticfield, impurities can be separated from the metallic ore. Using this process, Cassiterite (SnO 2), ore of tin is separated from its magnetic impurity Wolframite (FeWO4). Again, Chromite (FeO.Cr2O3), ore of Cr is magnetic and is separated from its non- magnetic silicate impurities by magnetic separation. (c) Oil Floatation Process: In this process, the ground ore is taken in a tank and mixed with water, small amount of acid, pine or Flux: To remove the impurities (gangue) present in the ore, some reagents are added during smelting which by chemical reaction with the gangue forms a fusible mass called slag. These reagents are called flux. Slag: The fused mass produced by the chemical reaction between the flux and gangue during smelting is called slag, Flux + Gangue → Slag eucalyptus oil (froth forming agents), sodium xanthates. This mixture is stirred with the air. Froth is formed due to the mixing of water and oil. The ore becomes wet with oil and floats with foam on the surface of the mixture. On the other hand, the impurities become wet with water and settle down at the bottom of the tank.The concentrated ore is collected from the froth and is dried by hot air. This process is applied to concentrate sulphide ores like Copper Pyrites (CuFeS 2), Zinc Blende (ZnS), Galena (PbS) etc. Role of Xanthate: Xanthate is used as collector of the ore Role of Air: Air acts as stirrer. Role of pine or eucalyptus oil: It acts as froth forming agent. 27  EIJE _ Class Note _ Applied Chemistry_103(N)_AR (d) Leaching Process: In this chemical process, any suitable reagents like acid, base etc. are added to dissolve only the metal containing ore. Thus the ore is separated from the gangue which does not react with the added reagents. In this process, the impurities present in Bauxite are removed. In Bauxite (Al 2O3.2H2O) , SiO2, Fe2O3, TiO2 etc are present as the impurities. If the powdered bauxite is heated with 45% NaOH, Al2O3 and SiO2 reacting with NaOH forms NaAlO2 and Na2SiO3. Fe2O3 and TiO2 remain insoluble and separated by filtration. In the filtrate (NaAlO2 and Na2SiO3) little amount water is added followed by little amount of freshly precipitated Al(OH)3. Then Al(OH)3 is precipitated out from the solution and is separed by filtration from Na2SiO3. iii) Calcination: In this process, the crushed ore is heated at a temperature below its melting point in the presence or absence of water in such a way that all volatile impurities present in the ore are removed. Due to the calcinations, all volatile substances like water vapour,CO2, sulpher, arsenic, phosporous etc. are removed and the ore becomes hollow and porous and the reduction process in the next step becomes easier. Normally this process is applied for the carbonate ores and CO2 is removed from the ore. 4As + 3O2 → 2As2O3 ↑ S + O2 → SO2 ↑ Fe2O3.3H2O → Fe2O3 + 3H2O↑ ZnCO3 → ZnO + CO2↑ Sometimes, during calcinations the lower valent oxides get oxidized to the the higher valent oxides. FeCO3 → FeO + CO2↑ 4FeO+O2 →2Fe2O3↑ iv) Roasting: In the roasting process, the crushed ore is heated in the presence of air at a temperature, higher than the calcination temperature, but below the melting point of the ore. This process is applied for the sulphide ores. During roasting, all volatile substances like As, S, P, water vapour etc. present in the ore are removed. At the same time, sulphide ores are first partially converted to oxides. The produced oxides and the unchanged sulphides react to produce the metal. 2PbS + 3O2 → 2PbO + 2SO2 ↑ PbS + 2PbO → 3Pb + SO2 ↑ Similarities between Calcination and Roasting: Both calcinations and roasting processes are done at a lower temperature of the melting point. In both cases volatile substances like carbon dioxide, water vapour, S, As, P etc. are removed from the ore and the ore becomes hollow and porus. Both the processes are done at the penultimate step and are done in suitable furnace. Diffences between Calcination and Roasting: (a) Roasting is done at higher temperature than calcinations. (b) Roasting process is done at excess of air than calcinations. (c) Calcinations is done mainly for the carbonate ores whereas roasting is done mainly for sulphide ores v) Smelting: In the smelting process, roasted or calcinated ore, mixed with suitable reducing agents, is melted at a very high temperature in a furnace and the metal is produced by reducing the oxide ore. The impurities reacting with the flux form slag and. At the high temperature, the melted metals are present in the lower layer of the furnace and the slag floats on the upper layer. The metals are separated from the lower layer of the furnace. Smelting or reduction process can be done by following ways: (a) Carbon Reduction Process: In this process oxide ore or roasted or calcinated oxide ore is reduced at high temperature by coke (C) to produce metal in elemental state. Mainly the metals which are comparatively less electropositive such as Fe, Zn,Sn, Pb etc. (the standard reduction potential 0.00 volts to - 0.75 Volts) are extracted by this process. ZnO + C → Zn + CO, Fe2O3 + 3C → 2Fe + 3CO Produced CO also acts as reducing agents. ZnO + CO → Zn + CO2 ↑, Fe2O3 + 3CO → 2Fe + 3CO2 ↑ Carbonate and sulphide ores arefirst calcinated and roasted in air and oxidized to produce\oxide ore. These oxide ores are reduced by the cabon reduction process. Aluminium (Al) can not be extracted by carbon reduction method.--- Explain 28  EIJE _ Class Note _ Applied Chemistry_103(N)_AR Strongly electropositive metal Al has strong affinity for electronegative oxygen and form stable oxide, alumina (Al 2O3). It requires a very high temperature to reduce oxide compound to produce Al. Moreover, during reduction of Al 2O3 by carbon at such high temperature, aluminum carbide Al4C3 is produced. b) Self Reduction Process: This process is applied for sulphide ores like PbS, Cu2S etc. In the controlled flow of air, the ore is partially roasted into oxide and the produced oxide and unchanged sulphide reacts with each other to produce the metal. The unchanged sulphide itself acts as reducing agent and no external reducing agent is added externally. Hence this process is called self reduction process. During the extraction of Cu, copper pyrites (CuFeS2 or Cu2S. Fe2S3) or sulphide ore of Cu is partially roasted in the presence of air to form Cu2O. Newly formed Cu2O and the unchanged Cu2S react with each other to produce metallic Cu. Thus Cu is extracted in self-reduction process. Partial Oxidation: 2Cu2S + 3O2 → 2Cu2O + 2SO2 ↑ Self Reduction: Cu2S + 2Cu2O → 6Cu + SO2 ↑ In similar way, Pb can also be extracted from its sulphide ore, Galena. 2PbS + 3O2 → 2PbO + 2SO2 ↑, PbS + 2PbO → 3Pb + SO2 ↑ c) Electrolytic Reduction: In this process, the metal is extracted by the electrolysis of the fused or aqueous solution of metal salt and the metal is deposited at the cathode. Normally by this method the highly electropositive metals like Na, K, Ca, Mg, Al etc. metals are extracted by electrolysis of their halides or oxides in fused state. Flux are added to decrease the melting point and increase the electrical conductivity of the electrolyte. Metals which are above H in electrochemical series are more electropositive than hydrogen, hence the electrolysis of the aqueous solution of the metal salt releases H 2 at cathode instead of the metal. Hence the fused solution of metal salt is taken for metal extraction. On the other hand , the metals (Ag, Au, Cu etc.), below H in electrochemical series, are less electropositive than H and hence the aqueous solution of these metal salts releases the respective metals. For example, Al is extracted from the electrolysis of alumina (Al2O3). But alumina is not a good conductor of electricity and its melting point is quite high. Cryolite (AlF3.NaF) and fluorspar are added to decrease the melting point and to increase the electrical conductivity of the electrolyte mixture. Electrolysis: Al2O3 2Al3+ + 3O2- Cathode: 2Al 3+ + 6e → 2Al Anode: 3O2- -6e → 3[O], 3[O] + 3[O] → 3O2 d) Thermit Process. Al is more electropositive and hence more active than the metals below Al in the electrochemical series. Moreover highly electropositive Al has a strong affinity to oxygen at high temperature. Hence Al acts as a reducing agent to extract metals like Fe, Cr, Mn etc from their oxide ores. Fe2O3+ 2Al → Al2O3 + 2Fe, Cr2O3+ 2Al → Al2O3 + 2Cr, 3Mn3O4+ 8Al → 4Al2O3 + 9Mn. Thermit: Mixture of Fe2O3 and Al powder in 3:1 ratio is called Thermit. Molten iron is liberated by the reaction between Fe2O3 and Al and is used for welding and joining purposes. e) Reduction by Other Metal: ▪ More electronegative metal is used to liberate the less electronegative metal from the its aqueous salt solution. For example, if Fe chips are added to the CuSO4 solution, metallic Cu is deposited. ▪ Mg is more electropositive than Al, hence Mg can be used as reducing agent to extract metals from their oxides. f) Refining of Metals: Extracted crude metals are to be refined to remove the associated impurities by following ways a) Fractional Distillation: If the metal and impurities in it have different boiling points, then the more volatile one is separated by fractional distillation. For example, Zn has lower boiling point than its impurity Cd. Hence more volatile Zn is separated from Cd by distillation. b) By aerial oxidation the impurities present in metal are oxidized to the oxides. The volatile oxides evaporates and non- volatile oxides floates as slag on the fused metals.In purification of Pb and Cu metals this process is used. c) When impurity is less or more fusible than the metal itself, then the metal is purified by liquation process. The metals like Sn, Pb etc. Can be purified by this method. 29  EIJE _ Class Note _ Applied Chemistry_103(N)_AR d) Electrolytic Refining: In this process very pure metal is obtained. In electrolytic refining, impure metal rod is used as anode and pure metal rod is used as cathode. An acidified aqueous solution of suitable salt of the extracted metal is used as electrolyte. During electrolysis, the pure metal from anode comes to the solution and is deposited to cathode from the solution. Soluble impurities remains dissolved in the solution and the insoluble impurities are precipitated out below the anode and are known as anode-mud. In this process metals like Cu, Al etc. are refined. In the purification of Cu, pure Cu rod, impure piece of Cu and acidified CuSO 4 solution are used as cathode, anode and electrolyte, respectively. Anode: Cu - 2e → Cu2+ Cathode: Cu2+ + 2e → Cu Electrolysis: CuSO4 Cu2+ + SO42- In this process, 99.99% pure Cu is obtained. Aluminium (Al) Atomic Number : 13, Electronic Configuration: 1s22s22p63s23p1, Atomic Weight 26.98, Valency 3 Main Ores of Al: (i) Bauxite (Al2O3.2H2O) (ii) Gibbsite (Al2O3.3H2O) (iii) Diaspore (Al2O3.H2O) (iv) Cryolite (AlF3.3NaF) (v) Felspar (K2O.Al2O3.6SiO2) (vi) Mica (Al2O3.2SiO2.2H2O) (vii) Kaolin (Al2O3.SiO2.2H2O) (viii) Curundum (Al2O3) ▪ Al can not be extracted by the electrolysis of anhydrous AlCl3 as AlCl3 on heating undergoes sublimation at 185oC. ▪ Al can not be extracted by electrolysis of aqueous solution of any aluminium salt. Aqueous solution contains H+ and Al3+ ions. Because H being less electropositive than Al, H2 gas is liberated as cathode instead of Al. Extraction of Al from Bauxite: Bauxite contains 50-60% alumina (Al2O3). Ferric oxide (Fe2O3), silica (SiO2), titanium dioxide (TiO2) etc. are present as impuries in it. If the amount of Fe2O3 is present in excess then the bauxite is red in colour. This kind of bauxite is called red-bauxite. If silica is present in excess then the bauxite is called white-bauxite. Extraction of Al from Bauxite involes three steps: (i) Preparation of pure Al2O3 from Bauxite: Impurities present in Al produced by direct electrolsis of bauxite makes Al more brittle and susceptable to corrosion.Hence it is necessary to use pure bauxite for electrolysis. Impurities present in The purification of Al2O3 from bauxite depends on the nature of the impurities present in it. It can be done by different processes: a) Bayer’s Process: This process is applied for red bauxite which contains more Fe 2O3 and less Al2O3. The finely crushed bauxite is calcinated. If any FeO is present as impurity, then during calcination it is completely converted into Fe 2O3. Then 30  EIJE _ Class Note _ Applied Chemistry_103(N)_AR the calcinated ore is heated with 45% NaOH solution in autoclave under 80lbs pressure at 150oC for about 8 hours.The alumina and silica present in bauxite react with NaOH and dissolve in the solution as NaAlO 2 and Na2SiO3. Fe2O3 and TiO2 remain insoluble and separated by filtration. In the filtrate (NaAlO2 and Na2SiO3) little amount water is added to dilute the solution and it is cooled at 20oC-30oC. Then little amount of freshly precipitated Al(OH)3 is added and the mixture is then stirred for several hours. NaAlO2 then undergoes hydrolysis and Al(OH)3 is precipitated out from the solution while Na2SiO3 remains in the solution. The precipitate is separated by filtration and is washed with water. On heating Al(OH)3 dissociates at 1100oC to produce pure and anhydrous Al2O3. Al2O3 + NaOH → 2NaAlO2 + H2O (in autoclave under 80lbs pressure at 150oC for about 8 hours) SiO2 + NaOH → 2Na2SiO3 + H2O NaAlO2 + 2H2O → Al(OH)3↓ + NaOH Δ 2Al(OH)3 → Al2O3 + 3H2O↑ (1100oC) In case of white bauxite, Bayer process can not be applied as Al is wasted due to the formation of insoluble sodium aluminisilicate (NaAlSi3O8). b) Serpeck’s Process: This process is suitable for white bauxite which contains more silica and less Fe 2O3. Crushed bauxite is mixed coke and heated at 1800oC in a current of N2. As a result AlN is produced. At such high temperature SiO2 is reduced by C to form Si which leaves the system in the form of vapour. AlN is heated with NaOH solution and on hydrolysis Al(OH)3 and NH3 are produced. The precipitate is separated by filtration and is washed with water. On heating Al(OH)3 dissociates at 1100oC to produce pure and anhydrous Al2O3. Al2O3 + 3C + N2 → 2AlN +3CO ↑ (1800oC) SiO2+ 2C → Si ↑+ 2CO ↑ AlN+NaOH + H2O → NaAlO2 + NH3↑ AlN +3H2O → Al(OH)3↓ + NH3↑ NaAlO2 + 2H2O → Al(OH)3↓ + NaOH Δ 2Al(OH)3 → Al2O3 + 3H2O↑ ((1100oC)) In this process ammonia (NH3) is obtained as by-product. c) Hall’s Process: This process is applicable for both type of bauxite. Bauxite is fused with Na 2CO3 at 1000oC-1100oC in rotatory kiln and at such high temperature, alumina, silica and ferric oxide etc. are converted into NaAlO 2, Na2SiO3 and NaFeO2. This mixture is heated with water at 80oC. NaAlO2 is dissoved into the water. NaFeO2 undergoes hydrolysis and precipitated as Fe(OH)3. Na2SiO3 on hydrolysis precipitated as sodium aluminosilicate (NaAlSi3O8). Now the solution is filtered. In the filtrate containig NaAlO2, CO2 is passed and Al(OH)3 is precipitated. The precipitate is separated by filtration and is washed with water. On heating Al(OH)3 dissociates at 1100oC to produce pure and anhydrous Al2O3. Al2O3 + Na2CO3 → 2NaAlO2 + CO2 (1000oC-1100oC in rotatory kiln) SiO2+Na2CO3 → Na2SiO3 + CO2 Fe2O3 + Na2CO3 → 2NaFeO2 + CO2 NaFeO2 + 2H2O → Fe(OH)3 ↓+ NaOH 3Na2SiO3 + NaAlO2 + 3H2O → NaAlSi3O8 ↓ + 6NaOH NaAlO2 + 3H2O + CO2 → 2Al(OH)3↓ + Na2CO3 Δ 2Al(OH)3 → Al2O3 + 3H2O (ii) Electrolysis of Alumina --- Extraction of Al: fused mixture of alumina (Al2O3), cryolite (AlF3.3NaF) and fluorspar (CaF2) in 20:60:20 ratio is used as electrolyte. Coke dust is spread over the electrolyte. Thick coating of gas carbon on the inside wall of the iron made electrolytic tank acts as cathode. The graphite rods partially dipped into the elctrolyte hanging from Cu rod act as anode. Temperature of electrolysis: 950oC Electrolysis: Al2O3 2Al3+ + 3O2- 3+ Cathode: 2Al + 6e → 2Al Anode: 3O2- -6e → 3[O], 3[O]+3[O] → 3O2 At cathode, Al (m.p. 559oC) is liberated and remains at molten state at the temperature of electrolysis. Being heavier than the electrolytic mixture Al settles down at the bottom of the electrolytic tank and is collected from there. Al obtained in this process is 99% pure. Alumina is added time to time to continue the electrolysis process. An electric bulb is attached to electrolytic cell. This bulb by its bright light gives the signal of shortage of alumina. 31  EIJE _ Class Note _ Applied Chemistry_103(N)_AR Role of Coke Dust: Oxygen (O2), evolved at anode attacks the graphite anode at high temperature and CO and CO 2 are produced. To minimize the corrosion of expensive graphite electrode by evolved O2, coke dust is spread over electrolyte. The coke dust reacts with the liberated oxygen and thus corrosion of graphite electrode can be culminated. Role of Cryolite and Fluorspar a. The melting point of pure alumina is 2050oC. So a large amount of electricity is required to continue the electrolysis of pure alumina in molten condition. At such high temperature Al, liberated, is lost by vapourisation. When alumina is mixed with cryolite (AlF3.3NaF) and fluorspar (CaF2) in 20:60:20 ratio, the melting point of the electrolytic mixture decreases to 950oC. b. Alumina is a weak electrlolyte and bad conductor of electricity. Cryolite is added to increase the electrical conductivity of the fused electrolyte. c. Flourspar decreases the viscosity of the electrolyte mixture. (iii) Electrorefining of Al: Al obtained by electrolysis of molten mixture of alumina, cryolite and fluorspar is 99% pure. In it small amount of Fe, SiO2, Cu etc. are present as impurities.. Hoop’s Process: The inner wall of the electrolytic cell is coated with thick layer of carbon. The electrolytic cell contains three types of molten substances of different sepecific gravities. (a) Bottom layer contains molten impure alumunim which is obtained by electrolysis of alumina. This layer acts as anode (b) The middle layer consists of molten mixture of cryolite (AlF3.3NaF) and BaF2. This layer acts as eletrolyte. (c) The upper layer conists of pure fused Al which acts as cathode. The graphite rods hanging from the Cu rod are dipped into this layer. On the passage of electricity, Al3+ ions from the middle layer accepting electrons are deposited at the cathode (layer of pure aluminium). At the same time, equivalent amount of Al3+ ions from the layer of impure aluminium (anode) come to the middle layer to maintain the Al3+ ion concentration. Thus pure Al is deposited at the cathode and the impurities settle down at anode. Iron (Fe) Atomic Number 26, Atomic Weight: 55.85, Elctronic Configuration:1s 22s22p63s23p63d64s2, valency 2, 3 The Main Ores of Fe: (i) Brown Haematite (Fe2O3, 3H2O) (ii) Red Hameatite (Fe2O3) (iii) Magnetite (Fe3O4) (iv) Siderite (FeCO3) Why is Iron Pyrites not used as an ore of Fe? Iron Pyrites (FeS2) has high S %. Sulpher can not be removed completely from the extracted iron. Presence of sulpher makes the iron brittle. Hence Iron Pyrites is not regarded as an ore of Fe. Extracttion of Fe from Haematite ore: i) Concentration of Ore: The iron ore is crushed and ground and is washed in the flow of water (gravity separation) to remove impurities as much as possible. Then the ore is concentrated by magnetic separation. ii) Calcination of Ore: Concentrated ore is calcinated in the presence of controlled air flow. Most of moisture is removed as vapour. Δ Fe2O3.3H2O → Fe2O3 + 3H2O↑ S, As, P etc. impurities are removed in the form of volatile oxides. 4As + 3O2 → 2As2O3↑, 4P + 5O2 → 2P2O5↑, S + O2 → SO2↑ If any FeO is present then during calcinations it is converted to Fe 2O3. iii) Smelting: The calcinated ore is then mixed with coke dust (reducing agent) and limestone or CaCO 3 (flux) in 8:4:1 ratio. This mixture is called charge. This charge is taken in the blast furnaceand heated strongly in the presence of hot air. Reaction occuring in blast furnaces: (a) Combustion Zone at lower region: The coke burns in the presence of hot air and is oxidised to form CO2. This reaction is an exothermic reaction. As a result the temperature of this zone is in between 1300 oC -1500oC 1500ºC: C + O2 → CO2 ↑ + 97 kcal CO2 ,when came into contact of coke, is reduced to CO. This reaction is endothermic. Hence the temperature reduces to 1200º -1300ºC. CO is used as reducing agents. 32  EIJE _ Class Note _ Applied Chemistry_103(N)_AR 1200º -1300ºC : CO2 + C → 2CO ↑ (b) Reducing Zone (Upper Region): At the reducing zone at upper region of blast furnace, CO comes in contact with Fe 2O3. CO acts as reducing agent. 400ºC: 3Fe2O3 + CO 2Fe3O4 + CO2 ↑ 500º- 700ºC: Fe3O4 + CO 3FeO + CO2 ↑ 700º -900ºC: FeO + CO Fe + CO2↑ Overall Reaction: Fe2O3 + 3CO 2Fe + 3CO2 ↑ As the reactions are reversible, hence small amount of Fe 2O3 remains unreduced. Rest of Fe2O3 is reduced by C (produced by reduction of CO) in the middle region at temperature 900º -1100ºC. 900º -1100ºC: 2CO CO2↑ + C, Fe2O3 + 3C → 2Fe + 3CO2↑ (c) Slag Formation Region (Middle Region): At temperature 900º -1100ºC, CaCO3 dissociates into CaO and CO2. Silica (SiO2) present in the ore reacts with the produced CaO and is reduced as CaSiO 3 (slag). 900º -1100ºC: CaCO3 → CaO + CO2↑, CaO + SiO2 → CaSiO3 CaO acts as flux. (d) Fusion Zone (lower region): In this region at 1100º -1300ºC extracted Fe and slag remain at fused condition. At this temperature, the impurities such as phosphate compounds, silica, MnO2 etc. present in the ore are reduced by C to produce P, Mn, Si etc. Ca3(PO4)2 + SiO2 + 5C → 3 CaSiO3 + 5CO↑+ 2P MnO2+ 2C→ Mn+2CO↑, Mn2O3 + 3C → 2Mn + 3CO↑ SiO2 + 2C → Si+ 2CO ↑ This impurities (C, P, Si, Mn, S etc.) remains in the extracted Fe. Extracted Fe and the slag both in molten condition forms two different layers in the bottom of the furnace at 1500ºC. The Slag being lighter floats on the molten Fe and saves it from oxidation. Fe is collected from the lower layer of the blast furnace. The molten iron is called cast iron. The iron solidifies and cut in the shape of iron. It is called pig iron. Copper (Cu) Atomic Number : 29, Electronic Configuration: 1s22s22p63s23p63d104s1, Atomic Weight 63.54, Valency 1,2 Names of the Main Ores: (i) Copper Pyrites or Chalcopyrities (CuFeS2 or Cu2S.Fe2S3) (ii) Copper Glance or Chalcocite (Cu2S) (iii) Malachite CuCO3.Cu(OH)2 (iv) Azurite 2CuCO3.Cu(OH)2 (v) Cuprite Cu2O Extraction of Cu from Copper Pyrites (CuFeS2 or Cu2S.Fe2S3) Copper Pyrites contains 2-3% Cu. In this ore, iron sulphide, SiO2, silicate compounds, As, Sb, S etc. are present as impurities. The following steps are involved in the extraction of Cu from Copper Pyrites: (i) Concentration: Powdered ore of copper pyrites is concentrated by oil floatation method by removing the impurities present in it as much as possible. The powdered ore is taken in a tank and mixed with water, small amount of acid, pine oil (froth forming agents), sodium xanthates (collector). This air is bubbled through the solution and acts as stirrer. Froth is formed due to the mixing of water and oil. The SiO2 and silicates become wet with water and settle down at the bottom of the tank. On the other hand, the ore becomes wet with oil and floats with froth on the surface of the mixture. In the concentrated ore, % Cu increases to about 25-30%. (ii) Roasting: The dry and concentrated ore is heated strongly in the presence of air at a temperature below the melting point in Reverberatory furnace. The following changes take place during roasting: (a) During roasting, S, As, Sb etc. are removed in the form of volatile oxides. CO 2, water vapour etc. are also removed from the ore. 4As + 3O2 → 2As2O3↑, 4Sb + 3O2 → 2Sb2O3↑, S + O2 → SO2↑ (b) Fe2S3 present in the ore is converted to ferrus sulphide (FeS) and sulphur (S). S is oxidized to SO 2 and goes out of the furnace. Fe2S3 → 2FeS + S, S + O2 → SO2↑ 2CuFeS2 + O2 → 2FeS + Cu2S + SO2↑ 33  EIJE _ Class Note _ Applied Chemistry_103(N)_AR (c) Metal sulphides are partially converted to oxides. But the affinity of Fe towards oxygen is more than that of Cu. Cu 2O reacting with FeS is again converted in to Cu2O. 2Cu2S +3O2 → 2Cu2O + 2SO2↑, 2FeS +3O2 → 2FeO + 2SO2↑ Cu2O + FeS → 2Cu2S + FeO 2CuFeS2 + 4O2 → 2FeO + Cu2S + 3SO2↑ In roasted ore, Cu2S, FeS, FeO, small amount of Cu2O and small amount of impurities are present. (iii) Smelting: Roasted ore is mixed with coke, silica (SiO2) and small amount of CaO. The mixture is heated in hot and dry air in the Blast furnace. Coke acts as fuel and silica acts as flux. (a) Maximum FeS present in the roasted ore is converted into FeO. 2FeS +3O2 → 2FeO + 2SO2↑ (b) Small amount of Cu2O present in the ore reacting with FeS forms Cu2S. Cu2O + FeS → 2Cu2S + FeO (c) FeO produced in the furnace reacts with SiO2 at high temperature to form FeSiO3 as slag and is removed. FeO + SiO2 → FeSiO3 (Slag), CaO + SiO2 → CaSiO3 (d) Molten mixture of Cu2S and FeS is collected for the next step. This Molten mass of Cu2S and FeS is called matte and it contains about 45% Cu and about 30- 35% Fe. (iv) Self Reduction: Molten matte is directly transferred to the Bessemer converter and SiO 2 is mixed with it. Hot air is blown through it. FeS present in the matte is converted completely into FeO. Produced FeO reacts with SiO 2 and forms FeSiO3. Molten FeSiO3 being lighter floats on the molten matte and is removed. Cu2S is partially oxidized to Cu2O. Produced Cu2O and unchanged Cu2S reacts with each other to form Cu. The unchanged Cu2S itself acts as reducing agent and no external reducing agent is added externally. Hence this process is called self reduction process. 2FeS +3O2 → 2FeO + 2SO2↑, FeO + SiO2 → FeSiO3 (Slag) 2Cu2S +3O2 → 2Cu2O + 2SO2↑ (partial oxidation) Cu2S + 2Cu2O → 6Cu + SO2 ↑ (Self-Reduction) Molten Cu settles down at the bottom of the converter and hence does not get oxidized with the air. Then the molten Cu is poured off. On cooling, dissolved SO2 bubbled out of it and blisters are formed on the surface of solidified Cu. Hence this Cu is called Blister Copper. It contains about 98% of Cu. Fe, S, As, Pb and small amount of Ag, Au and Pt etc. are present as impurities in it. (v) Refining: (a) Thermal Refining: The blister Cu is then melted in the controlled flow of air in the silica coated reverberatory furnace. S, As etc. are removed in the form of volatile oxide. Fe is oxidized to FeO and removed as FeSiO 3 reacting with silica coating of furnace. Small amount of Cu is oxidized to Cu2O and dissolved in molten Cu. Presence of Cu2O makes Cu brittle. To reduce Cu2O, the molten Cu is mixed with coke and stirred with green wood. Produced H 2 and CO from green wood reduce Cu2O to metallic Cu. This process is known as poling. Cu obtained in this process is 99.5% pure. 4As + 3O2 → 2As2O3↑, S + O2 → SO2↑ FeO + SiO2 → FeSiO3 (Slag) Cu2O + CO → 2Cu + CO, Cu2O + H2 → 2Cu + H2O (b) Electrolytic Refining: In the thermally refined Cu, small amount As, Bi, Sn, Pb, Fe,Ni, Zn, Ag, Au etc. are present as impurities. In the purification of Cu, pure Cu rod, thermally refined Cu slab and CuSO 4 solution mixed with 5-10% H2SO4 are used as cathode, anode and electrolyte, respectively. Cu dissolves from the anode and deposited at cathode. The impurities like Zn, Ni, Fe etc. dissolves in the solution as sulphates and impurities like Au, Ag, Pt etc. precipitate out below the anode and are known as anode-mud. Anode: Cu - 2e → Cu2+ Cathode: Cu2+ + 2e → Cu Electrolysis: CuSO4 Cu2+ + SO42- In this process, 99.99% pure Cu is obtained. Alloy Alloy is homogeneous or inhomogeneous mixture of two or more metals at different ratios and acts as single metal. Purpose of Alloying: Alloy is much harder than the pure metal. In duralumin, an alloy of Al (95%) with Cu, Mg and Mn , is harder than the metallic aluminium. Chemical reactivity of metal is modified in alloy. 34  EIJE _ Class Note _ Applied Chemistry_103(N)_AR Alloy is more resistant towards corrosion. Due to the presence of 12-15% Cr, stainless steel is more resistant towards corrosion than iron. Ductility, malleability, melting point etc. properties are also modified in the alloy. Ferrous alloys - Alloys in which iron metal is the main constituent are called ferrous alloys e.g. Steel is an alloy of iron and carbon. Non-ferrous alloys - Alloys in which iron is not the main constituent are called non-ferrous alloys. E.g. Bronze (Cu, Sn) and Brass (Cu, Zn). Name of Alloy Composition Uses Brass Cu : 60-90%, Zn: 10-40% Used in making utensils, musical instruments, imitation jewellery, statues etc. Bronze Cu 80-90%, Sn 10-20% Used in Jewellery, statues, utensils etc. German Silver Cu 50%, Zn 30%, Ni 20% Used in making utensils, decorative items etc. Duralumin Al: 95%, Cu: 4%, Mg: 0.5%, Mn: Used in aircraft and automobile industry. 0.5% Nichrome Ni 60%, Cr 15%, Fe 25% Used in electrical coils in furnaces, heater, iron etc. Bell metal Cu 80%, Sn 20% Used in making bells, utensils etc. Gun Metal Cu 85%, Sn 10%, Zn 5% Used in foundry works, gears, bearings etc Monel Metal (Ni based Cu 28-30%, Fe & Mn 3% and rest Used in making automobile engine parts, alloy) Ni corrosion resistant tools etc. Alnico Fe: 63%, Ni: 20%, Al: 12%, Co: 5% Used in making electromagnets, permanent magnets. Dutch Metal Cu: 80%, Zn: 20% Used cheap jewellery, name plates etc. Babbit Metal (white Sn 80-90%, Pb 8-12%, Cu 3-10% Used for making engine bearings. metal, Sn based alloy) Stainless steel 12-15% Cr, 8-10% Ni, 0.05-1% C, Used in making household utensils, surgical rest Fe equipments etc. ILO: Discuss about the fundamental ideas about metallurgy with special emphasis on extraction of Al, Fe and Cu from their respective ore. Questions: 1. How is Fe extracted from Heamatite ore? (write chemical reactions) 2. Why Cu is extracted using self reduction method? Write relevant chemical reactions. 3. How will you purify Cu using electrolysis process? 4. How Al is extracted from alumina? 5. Why Al is not extracted using C reduction process? 6. Give 1 example of each of ferrous and non-ferrous alloy? Write their composition. 7. Write down the name and chemical formula of each of Al, Fe and Cu. Class 16 &17 & 18: Topic: Portland Cement, Refractories and Composite Materials, Polymer Portland Cement: Portland cement was first discovered by Joseph Aspidin. It is the most commonly used and most reliable cement in the construction works. Portland cement is chemically defined as the finely ground mixture of calcium silicates and calcium aluminates in various proportions. It is obtained by calcining limestone and clay at a very high temperature (~ 1550oC) at right proportion and then grinding the calcinated mass. When mixed with water, it sets to harden in a hard stone like mass which resembles in colour and hardness to the famous Portland rock in England. So it is called Portland cement. Raw Materials of Portland cement: 35  EIJE _ Class Note _ Applied Chemistry_103(N)_AR Manufacture of Portland cement requires following materials: i) Calcareous Materials (which supply CaO) such as lime, chalk, marl or marine shell etc. ii) Argillaceous Materials (which supply Al2O3, SiO2, Fe2O3 etc.) such as clay, shale, slate etc. iii) Gypsum (CaSO4. 2H2O) iv) Powdered coal or fuel oil Composition of Portland Cement: Portland cement is the mixture of mainly four components: i) Belite or Dicalcium Silicate 2CaO.SiO2 (~25%) ii) Alite or Tricalcium Silicate 3CaO.SiO2 (~45%) iii) Celite or Tricalcium Aluminate 3CaO.Al2O3 (~10%) iv) Brownmillerite or Tetracalcium Aluminoferrite 4CaO.Al2O3.Fe2O3 (~9%) Besides these, small amount of free CaO, MgO, Na2O, K2O, TiO2, Fe2O3 etc. is also present. To control the setting time of the cement, 2-3% of gypsum is added to it. Setting and Hardening of Cement: The Portland cement on mixing with water forms a pasty mass which gradually sets to harden and becomes resistant to pressure. The setting process indicates the stiffening of cement paste from a fluid state to solid state. This process involves hydration followed by metastable gel formation. It takes place in two stages — intial set and final set. Initial set indicates the beginning of setting process when the cement paste starts losing its plasticity and begins to stiffen considerably. A period of 30 minutes is the minimum initial setting time, specified by ISI for ordinary Portland cement. The final setting occurs when cement completely loses its plasticity and can resist certain definite pressure.The final setting time indicates the time elapsed between the moment water is added to the cement and the time when the cement completely loses its plasticity and can resist certain definite pressure. A period of 600 minutes is the maximum final setting time, specified by ISI for ordinary Portland cement. Setting time may be affected by water, cement ratio, temperature, fineness of cement and the chemical composition of cement. Hardening is a slow process in which strength of cement is gradually developed due to the crystallization. It first occurs within 24 hours and the subsequent hardening requires 15-28 days. The strength of cement depends on amount of gelation and extent of hydration. As we have discussed earlier, the setting and hardening of cement is due to hydration and hydrolysis of its constituents (calcium aluminates and calcium silicates) which are all anhydrous. When the cement is mixed with water, at first tricalcium aluminate (3CaO.Al2O3) gets hydrated followed by tricalcium silicate (3CaO. SiO2) and dicalcium silicate (2CaO.SiO2) and tetracalcium aluminoferrite (4CaO. Al2O3. Fe2O3). [ref:building materials, S. S. Bhavikatti, pg-70] 3CaO.Al2O3 + water → 3CaO.Al2O3 (aq) 3CaO.SiO2 + water → CaO.SiO2(aq) + 2Ca(OH)2 Gel Crystalline 2CaO.SiO2 + water → CaO.SiO2(aq) + Ca(OH)2 Gel Crystalline 4CaO. Al2O3. Fe2O3 + water → 3CaO. Al2O3 (aq) + CaO. Fe2O3 (aq) Gel Crystalline These gels and crystals have the ability to surround the sand crystals, stones and other inert materials etc. and bind them very strongly through interlocking. Anhydrous crystals absorb water from the hydrated gels resulting in over-drying of surface of the cement. Hence the insufficient water may cause the cracking on the over-dried surface. Function of Gypsum (CaSO4.2H2O) The purpose of addition of gypsum (2-3%) is to prevent the early setting of cement at a rapid rate. When cement is mixed with water, tricalcium aluminate combines with water very rapidly with the evolution of large amount of heat. 3CaO.Al2O3 + water → 3CaO.Al2O3 (aq) + heat But gypsum retards the rapid dissolution of tricalcium aluminate by interacting with it forming insoluble calcium sulpho aluminate (3CaO.Al2O3.3CaSO4.32H2O) which does not have quick hydrating property. 3CaO.Al2O3 + water + CaSO4.2H2O → 3CaO.Al2O3.3CaSO4.32H2O Refractories and Composite Materials Refractories are the materials those have high melting points and hence can withstand high temperatures without softening or deformation in shape. These materials are chemically and physically stable at high temperatures. A good refractory material must possess following characteristics: 1) It must be resistant against the high temperatures under particular service condition. 2) It must be chemically inert against the corrosive and abrading actions of working environment at high operating temperatures. (3) It must expand and contract uniformly with rise and fall of temperatures. (4) It should have high mechanical strength so that it can bear the heavy load of constructions at operating temperatures. 36  EIJE _ Class Note _ Applied Chemistry_103(N)_AR Boron Carbide, Carborundum (Silicon Carbide) etc. are the examples of good refractory meterials. Refractories are extensively used for the constructions of the lining of the furnaces, kilns, reactors, crucibles etc. in different industries like metallurgy, cement, glass, ceramics etc. Silicon Carbide (SiC): Silicon Carbide is also known as carborundum. It is prepared by heating pure silica sand (60%) with Carbon (40%) in the form of finely ground coke at high temperatures in an electric furnace. Silicon carbide crystallizes in more than 200 structures known as polymorphs. It may exist in zinc blende, wurzite and moissanite structures. Properties: 1) It is a black solid compound. Melting point: 2700oC. and relative density 3.217 g/cm3. 2) It is insoluble in water, but soluble in molten alkali. 3) It is very hard and can be used as an abrasive. 4) It is characterized by low thermal expansion and high thermal conductivity. 5) It has high refractoriness and superior chemical inertness. 6) It tends to oxidize to silica when heated with air at temperatures around 900oC to 1000oC. 7) It can also be classed as semiconductors.. Uses: It is an excellent abrasive and is used in grinding wheels, sand papers, cutting, polishing and drilling tools. Due to high refractoriness, it is used in manufacture of high-temperatures bricks, in refractory linings for industrial furnaces, atomic reactors etc. SiC is a promising substitute for traditional semiconductors in high temperatures. Boron Carbide (B4C) It is manufactured by the reduction of boric oxide with petroleum coke in an electric furnace. B2O3 +7C = B4C + 6CO It is one of the hardest substances known, with hardness between 9 and 10 on Mohs’ scale (being exceeded only by cubic boron nitride and diamond). Properties: 1) It is a black solid compound. Melting point: 2350oC, boiling point: 3500oC and relative density 2.52 g/cm3. 2) It is soluble only in fused alkali. 3) It is one of the hardest substances known, with hardness between 9 and 10 on Mohs’ scale (being exceeded only by cubic boron nitride and diamond). 4) B4C can be polished to a mirror finished and has a good resistance to acids. It is a refractory and chemically inert , but is less resistant to oxidation than silicon carbide. 5) It has low thermal conductivity. Uses: It is mostly used as an abrasive. It is a good neutron absorber and hence used in control rods for nuclear power generation. Because of its hardness, together with its low density, it can be employed as a reinforcing agent for aluminium in military armour and high-performance bicycles etc. It is also used as scratch and wear resistant coating. Asbestos Asbestos is any one of a group of fibrous amphibole (needle-like) minerals (amosite, crocidolite, tremolite, anthophyllite and actinolite) or the fibrous serpentine (curly) mineral (chrysolite). Generally, asbestos is a hydrous silicate mineral. Each of these six minerals has individual formula. Different types of Asbestos: Chrysolite/white asbestos: hydrous magnesium silicate Amosite/Brown asbestos: hydrous magnesium iron silicate Crocidolite/blue asbestos: hydrous sodium iron silicate Tremolite asbestos: hydrous calcium magnesium silicate Anthophyllite asbestos: hydrous magnesium iron silicate Properties 1) It is highly porus inorganic refractory materials 2) It has high electrical and heat resistance property. 3) it is chemically inert Uses: Asbestos has widespread commercial uses. The fibres may be spun and woven into fireproof cloth for use in protective clothing, curtains, brake linings, etc. It can be used as insulating materials. Warning: Short asbestos fibres have been recognized as a cause of asbesto