How the Water Treatment Process works

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UNIT - I WATER TREATMENT PROCESS “Nothing on earth can function without water”. -Thiruvalluvar. “Water is the driver of life on earth”. -Leo Nardo Davinci. Introduction  The Molecular Formula H O 2  The Molecular Mass 18  Structure O H H  Water is nature’s gift of life to us. It is the most widely distributed compound, which exists naturally as liquid, solid and gas.  It covers about 80% of the earth’s surface; about 70% of the human body is water.  The water content in the human body accounts for more than half of its total weight.  It is the most important compound for the existence of human beings, animals and plants. All plants, animals (Human being 70%) and fruits (Water melon 99%, tomato 95%) contain water.  In 1781, Cavendish prepared water by the combustion of hydrogen in air. Later Lavoisier proved that water is a compound of hydrogen and oxygen. Composition of Water  One molecule of water contains atleast two atoms of hydrogen and one atom of oxygen. There fore, the atomic ratio of hydrogen and oxygen is 2:1.  The atomic weights of hydrogen and oxygen are 1 and 16.  The gravimetric composition (by weight) of water is 1:8 and the volumetric composition of water is 2:1. Volumetric composition of water  The ratio by volume of hydrogen and oxygen present in water is called volumetric composition of water.  Its value is 2:1. It was established by Davy in 1800 AD. It is determined by using Hopeman’s voltameter.  Water has greater applications in industries such as textiles, chemicals, paper, pharmaceuticals, food processing, leather etc., www.annauniversityplus.com Water is mainly used in power generation.  It is also used as a coolant in atomic reactors, as well as in chemical plants.  It is also largely used in irrigation for agricultural purpose and fire fighting.  Now – a – days, the quantity of water is gradually deteriorating due to pollution. So, engineers need to have a wide knowledge about the quality of water, the problems posed by hard water in industries and its treatment processes. Sources of water Water is essential for the survival of all living organisms. About 80% of the earth’s surface is occupied by water. The main sources of water are, 1. Surface water. 2. Underground water. Surface water Surface runoff precipitation that does not infiltrate the ground or return to the atmosphere or return to the evaporation (including transpiration). This runoff flows into streams, lakes, wetlands, estuaries and reserviours. It can be further classified into four major sources. 1. Rain water 2. River water. 3. Lake water. 4. Sea water. Rain water It is the purest form of water. It is made impure by polluted atmosphere, like CO , SO , and NO etc., 2 2 2 River water River water starts from spring water and fed by rain water. Chlorides, sulphates, bicarbonates of Na, Ca, Mg and Fe are some of the major mineral salts present in river water. Lake water Lake water has constant chemical composition. It usually contains fewer amounts of dissolved minerals and a high quantity of organic matter. Sea water This is the most impure form of natural water. It contains NaCl (2.6%), sulphates of Na, bicarbonates of K, Mg and Ca and bromides of K, Mg and a number of other compounds. www.annauniversityplus.comUnderground water A part of rain water which falls on earth surface percolates into the earth and continues its journey till it meets a hard rock where it may be stored or come in the form of spring. Properties of water Physical properties (i) It exists in three states ie., solid, liquid and gas. The solid form of water is 0 known as ice. It exists only below 0 c. It exists as a liquid between 0 and 0 0 100 c and as gas (steam) above 100 c. Hence the boiling point of water is 0 0 100 c (373 K) and freezing point is 0 c (273 K). (ii) Pure water is a transparent, colourless, odourless and tasteless liquid. (iii) It is a good solvent for the ionic compounds and it dissolves almost all substances (solid, liquid or gas). Hence, it is known as universal solvent. (iv) Pure water is a bad conductor of electricity, but acidified water is an electrolyte. 0 -3. (v) The density of water is maximum at 4 c, which is equal to 1000 kg m Chemical properties (i) Heating Process (ii) Action with metals (iii) Action with non – metals (iv) Action with metallic oxides (v) Action with non – metallic oxides (vi) Action with carbides, phosphides and nitrides. (i). Heating Process At very high temperature, water is decomposed into hydrogen and oxygen. 0 2000 c 2H O ∆ 2H + O ↑ 2 2 2 Water Hydrogen + oxygen (ii). Action with metals (a). Cold water reacts with metals like sodium, potassium and calcium to form hydrogen and their respective hydroxides. 2Na + 2H O 2Na(OH) + H ↑ 2 2 Sodium Sodium hydroxide 2K + 2H O 2K(OH) + H ↑ 2 2 Potassium Potassium hydroxide Ca + 2H O Ca(OH) + H ↑ 2 2 2 Calcium Calcium hydroxide www.annauniversityplus.com(b). Metals like magnesium and zinc react with hot water or steam to form hydrogen and their respective oxides. Mg + H O MgO + H ↑ 2 2 Magnesium Magnesium oxide Zn + H O ZnO + H ↑ 2 2 Zinc Zinc oxide (iii). Action with non – metals (a). Carbon reacts with water at red hot condition to produce water gas (a mixture of carbon monoxide and hydrogen). 1273 K C + H O ∆ CO + H ↑ 2 2 Carbon Carbon monoxide (b). Chloride reacts with water in the presence of sun light to form oxygen and hydrochloric acid. 2Cl + 2H O 4HCl + O ↑ 2 2 2 Chlorine Hydrochloric acid (iv). Action with metallic oxides Water reacts with metallic oxides like sodium oxide and potassium oxide to form their respective hydroxides. Na O + H O 2Na(OH) 2 2 Sodium oxide Sodium hydroxide K O + H O 2K(OH) 2 2 Potassium oxide Potassium hydroxide (v). Action with non – metallic oxides Non – metallic oxides like carbon dioxide, sulphur dioxide and sulphur tri oxide react with water to form their respective acids. CO + H O H CO 2 2 2 3 Carbon dioxide Carbonic acid SO + H O H SO 2 2 2 3 Sulphur dioxide Sulphurous acid SO + H O H SO 3 2 2 4 Sulphur trioxide Sulphuric acid (vi). Action with carbides, phosphides and nitrides Water decomposes carbides, phosphides and nitrides to produce methane, phosphine and ammonia respectively. Al C + 12H O 4Al(OH) + 3CH 4 3 2 3 4 Aluminium carbide Aluminium hydroxide + Methane www.annauniversityplus.com Ca P + 6H O 3Ca(OH) + 2PH 3 2 2 2 3 Calcium phosphide Calcium hydroxide + Phosphine Ca N + 6H O 3Ca(OH) + 2NH 3 2 2 2 3 Calcium nitride Calcium hydroxide + Ammonia Types of impurities The impurities present in water may be broadly classified into four types. They are, 1. Dissolved impurities 2. Suspended impurities 3. Colloidal impurities and 4. Micro organisms Dissolved impurities The dissolved impurities are mainly the carbonates, bi-carbonates, Chlorides and sulphates of Ca, Mg, Fe, Na and K. The dissolved impurities also include dissolved gases like O , CO , etc., 2 2 The presence of these salts imports hardness to water. Suspended impurities There are two types of suspended impurities. They may either be; 1. Inorganic suspended impurities: Clay and sand 2. Organic suspended impurities: Oil globules, Vegetable and animal matter. The inorganic and organic suspended impurities impart turbidity, colour and odour to water. Colloidal impurities Finely divided silica and clay, organic waste products, complex protein amino acids, etc., Micro Organisms They include algae, fungi and bacteria. Water Treatment Among the sources of water (surface and underground water) are normally used for domestic and industrial purposes. Such water must be free from undesirable impurities. The process of removing all types of impurities from water and making it fit for domestic or industrial purposes is called water treatment or water technology. Hardness of water Define – Hardness Hardness is the property present in water which prevents lathering of soap. www.annauniversityplus.comClassification of water Water from different sources, differ in taste and odour. This difference is due to the presence of dissolved salts and minerals. Based on the quality, water can be classified into two types. They are, 1. Hard water. 2. Soft water. Define - Hard water “Water which does not give much foam lather with soap solution is called hard water”. On the other hand, it forms a white scum or precipitate. The hardness of water is due to the presence of soluble bicarbonates, chlorides and sulphates of ‘Ca’ and ‘Mg’. 2C H COONa + CaCl (C H COO) Ca ↓ + 2NaCl 17 35 2 17 35 2 Sodium stearate Calcium chloride Calcium stearate (Sodium soap) (Hardness causing salt in water) 2C H COONa + MgSO (C H COO) Mg ↓ + Na SO 17 35 4 17 35 2 2 4 Sodium stearate Magnesium sulphate Magnesium stearate (Sodium soap) (Hardness causing salt in water) Define - Soft water “Water which gives good foam lather with soap solution is called soft water”. This is due to the absence of ‘Ca’ and ‘Mg’ salts. Give the Reason for Hardness? Reason for Hardness - The hardness of water is due to the presence of bicarbonates (HCO ), 3 2- - 2- carbonates (CO ), chlorides (Cl ) and sulphates (SO ) of calcium or magnesium 3 4 or both. How to identify Hardness? Hardness of water can be identified into two ways. 1. Reaction with soap solution. 2. Reaction with EBT indicator. Reaction with soap solution When, the water is treated with soap solution, if it does not give much foam lather. This water is called hard water. 2C H COONa + CaCl (C H COO) Ca ↓ + 2NaCl 17 35 2 17 35 2 www.annauniversityplus.comSodium stearate Calcium chloride Calcium stearate (Sodium soap) (Hardness causing salt in water) 2C H COONa + MgSO (C H COO) Mg ↓ + Na SO 17 35 4 17 35 2 2 4 Sodium stearate Magnesium sulphate Magnesium stearate (Sodium soap) (Hardness causing salt in water) Reaction with EBT indicator When the water is added two to three drops of EBT indicator, if it gives wine red colour, the water is hard water. Classification of Hardness of water How is Hardness in water classified? Give Example? On the basis of dissolved ions, hardness of water can be classified into two types. They are, 1. Temporary hardness (or) Carbonate hardness (or) Alkaline hardness. 2. Permanent hardness (or) Non-carbonate hardness (or) Non-Alkaline hardness. Temporary hardness (or) Carbonate hardness (or) Alkaline hardness If bicarbonates of Ca and Mg are present in water, such hardness is called carbonate hardness or temporary hardness or alkaline hardness. It can be easily removed by boiling the water and Clark’s process. Removal of Temporary hardness Temporary hardness of water may be removed by the following methods. 1. Boiling 2. Clark’s process (i). Boiling When the temporary hard water is heated strongly, the following reactions take place. Ca(HCO ) CaCO ↓+ H O + CO 3 2 3 2 2 Calcium bicarbonate Calcium carbonate Mg(HCO ) MgCO ↓+ H O + CO 3 2 3 2 2 Magnesium bicarbonate Magnesium carbonate i.e, Calcium bicarbonate and magnesium bicarbonate are decomposed into calcium and magnesium carbonate. These salts are insoluble in water and settle at the bottom of the vessel. It can be removed by filtration. The filtrate obtained, is soft water. (ii). Clark’s process www.annauniversityplus.comIn this process a calculated quantity of slacked lime (calcium hydroxide) is added to temporary hardness of water. This converts the soluble bicarbonates into insoluble carbonates which are removed by filtration. Filtered water is thus free from calcium and magnesium bicarbonates and is soft. Ca(HCO ) + Ca(OH) CaCO ↓+ 2H O 3 2 2 3 2 Calcium bicarbonate Calcium hydroxide Calcium carbonate Mg(HCO ) + Ca(OH) MgCO ↓+ CaCO ↓+ 2H O 3 2 2 3 3 2 Magnesium bicarbonate Calcium hydroxide Magnesium carbonate Permanent hardness (or) Non - carbonate hardness (or) Non - alkaline hardness If chlorides and sulphates of Ca and Mg are present in water, such hardness is called permanent hardness or non – carbonate hardness or non – alkaline hardness. It cannot be removed by boiling the water, because permanent hardness producing salts do not decompose on heating. But it can be removed by the following methods. 1. Lime – soda (washing soda – sodium carbonate) process 2. Calgon process (Internal conditioning method) 3. Zeolite or Permutit (External conditioning method) process 4. Ion – exchange or Demineralisation or Deionisation process 5. Reverse osmosis method Removal of Permanent hardness Lime – soda process When washing soda is added to hard water, the chlorides and sulphates of calcium and magnesium are converted into their respective carbonates. CaCl + Na CO CaCO ↓+ 2NaCl 2 2 3 3 Calcium chloride Sod. Carbonate Calcium carbonate Calgon process ‘Calgon’ is the commercial name of sodium hexameta phosphate. It means ‘Calcium gone’. When Calgon is added to hard water, the magnesium and calcium salts present in it are converted into soluble complex salts and soft water is produced. As these salts are soluble in water, filtration is not required. 2CaSO + Na Na (PO ) Na Ca (PO ) + 2Na SO 4 2 4 3 6 2 2 3 6 2 4 Units of hardness The following four common units are used in hardness measurements. 1. Milligrams per litre (mg/l) 2. Parts per million (ppm) 0 3. Degree Clark’s ( Cl) 0 4. Degree French ( Fr) 5. Milliequivalent per litre (Meq/l) www.annauniversityplus.com Milligrams per litre It is defined as the number of milligrams of CaCO present in one litre of 3 water. Parts per million It is defined as the number of parts by weight of CaCO present in million 3 6 (10 ) partsof water. Degree Clark’s It is defined as the number of parts of CaCO equivalent hardness per 70,000 3 parts of water. Degree French It is defined as the number of parts of CaCO equivalent hardness per 3 5 1, 00,000 (10 ) parts of water. Milliequivalent per litre It is defined as the number of Milliequivalents of hardness present per litre. Relation between the hardness units 0 0 1 mg/lit = 1 ppm = 0.07 Cl = 0.1 F = 0.02 Meq/l Disadvantages of using hard water 1. Hard water when used for drinking affects the digestive system and leads to formation of kidney stones (Calcium oxalate). 2. Hard water is not suitable for laboratory analysis because the hardness 2+ 2+ producing ions (Ca and Mg ) interface in various reactions. 3. When hard water is used for cooking, more fuel and time consumption are required. Because of the presence of salts Ca and Mg, this increases the boiling point of water. 4. When hard water is used for steam production, the boiler affected by the problems like Scale – Sludge formation priming and foaming and caustic embrittlement. 5. When hard water is used for concrete making, the hydration of the cement and the strength of the concrete are affected. Expression of hardness in terms of equivalents of CaCO 3 The concentrations of hardness producing salts are usually expressed in terms of an equivalent amount of CaCO . 3 The reason for choosing CaCO as the standard for calculating hardness of 3 water is because,  Its molecular weight is exactly 100 and equivalent weight is 50, which makes the calculations easier.  It is the most insoluble salt, thus can be easily precipitated in water treatment processes. Amount equivalent to CaCO 3 = Weight of hardness producing salt or ions (cations) www.annauniversityplus.com _________________________________________ × Molecular Weight of CaCO 3 Molecular Weight of hardness producing salt or ions Molecular weight of some hardness producing salts Hardness producing Molecular Hardness producing Molecular salt & Cation salt & Cation weight weight Ca(HCO ) 162 Mg(HCO ) 146 3 2 3 2 CaCl 111 MgCl 95 2 2 CaSO 136 MgSO 120 4 4 CaCO 100 MgCO 84 3 3 2+ 2+ Ca 40 Mg 24 Ca(NO ) 164 Mg(NO ) 148 3 2 3 2 Problems based on hardness in terms of Calcium Carbonate equivalents Example: 1 A water sample contains 120 mgs of MgSO per litre. Calculate the hardness 4 in terms of CaCO equivalents. 3 Solution The amount of MgSO = 120 mgs/lit 4 Amount equivalent to CaCO 3 = Weight of hardness producing salt or ions (cations) _________________________________________ × Molecular Weight of CaCO 3 Molecular Weight of hardness producing salt or ions We know that, the molecular weight of MgSO = 120 4 Amount equivalent to CaCO = 120/120 X 100 3 = 100 mgs/lit Result Therefore amount equivalent to CaCO = 100 mgs/lit 3 Example: 2 A water sample contains 204 mgs of CaSO per litre. Calculate the hardness 4 in terms of CaCO equivalents. 3 Solution The amount of CaSO = 204 mgs/lit 4 www.annauniversityplus.comAmount equivalent to CaCO 3 = Weight of hardness producing salt or ions (cations) _________________________________________ × Molecular Weight of CaCO 3 Molecular Weight of hardness producing salt or ions We know that, the molecular weight of CaSO = 136 4 Amount equivalent to CaCO = 204/136 X 100 3 = 150 mgs/lit Result Therefore amount equivalent to CaCO = 150 mgs/lit 3 Example: 3 2+ If a sample of water contains 50 mgs of Ca ions per litre. Calculate its hardness in terms of CaCO equivalent. 3 Solution 2+ The amount of Ca = 50 mgs/lit We know that, the molecular weight of Calcium = 40 Amount equivalent to CaCO = 50/40 X 100 3 = 125 mgs/lit Result Therefore amount equivalent to CaCO = 125 mgs/lit 3 Example: 4 What is the hardness of a solution containing 0.585 grams of NaCl and 0.6 grams of MgSO per litre? 4 Solution 1. The amount of NaCl = 0.585 grms/lit = 0.585 X 1000 = 585 mgs/lit 2. The amount of MgSO = 0.6 grms/lit 4 = 0.6 X 1000 = 600 mgs/lit NaCl does not contribute to hardness. So it is ignored. We know that, the molecular weight of MgSO = 120 4 Amount equivalent to CaCO = 600/120 X 100 3 = 500 mgs/lit Result www.annauniversityplus.comTherefore amount equivalent to CaCO = 500 mgs/lit 3 Example: 5 A sample of water is found to contain the following analytical data in mgs/lit. (i) Mg(HCO ) = 14.6 mgs/lit 3 2 (ii) MgCl = 9.5 mgs/lit 2 (iii) MgSO = 6.0 mgs/lit and 4 (iv) Ca(HCO ) = 16.2 mgs/lit 3 2 Calculate the temporary and permanent hardness of the sample of water (Atomic weight of Ca, Mg, Cl, C, S, O and H are 40, 24, 35.5,12, 32, 16, and 1 respectively). Solution Name of the hardness Amount Molecular Amount equivalent producing salt in mgs/lit weight to CaCO 3 Mg(HCO ) 14.6 146 = 14.6/146 X100 3 2 = 10 mgs/lit MgCl 9.5 95 = 9.5/95 X 100 2 = 10 mgs/lit MgSO 6.0 120 = 6.0/120 X 100 4 = 5 mgs/lit Ca(HCO ) 16.2 162 = 16.2/162 X 100 3 2 = 10 mgs/lit Temporary hardness producing salts = Mg(HCO ) + Ca(HCO ) 3 2 3 2 = 10 + 10 = 20 mgs/lit Permanent hardness producing salts = MgCl + MgSO 2 4 = 10 + 5 = 15 mgs/lit Result Temporary hardness = 20 mgs/lit Permanent hardness = 15 mgs/lit Estimation of Hardness The estimation of hardness of water is very essential for its use in boilers for steam generation, as well as for industries uses. Methods of estimation of hard water www.annauniversityplus.comThere are mainly three basic methods of estimation of hard water. They are, 1. EDTA Method 2. Alkalinity Method 3. O.Hehner’s Method (i) Determination of Temporary hardness (ii) Determination of permanent hardness EDTA Method (or) Complexometric Method EDTA Ethylene Diamine Tetra Acetic acid Structure of EDTA HOOCH C CH COOH 2 2 N ─ CH ─ CH ─ N 2 2 HOOCH C CH COOH 2 2 Structure of disodium salt of EDTA HOOCH C CH COONa 2 2 N ─ CH ─ CH ─ N 2 2 NaOOCH C CH COOH 2 2 ‘EDTA’ is insoluble in water; its disodium salt is used as a complexing agent. This method is also known as “ Versenate” Method. Principle 2+ 2+ The amount of hardness causing ion (Ca and Mg ) can be estimated by titrating the water sample against EDTA using ‘Eriochrome Black –T’ (EBT) indicator at a pH range of 8 -10. Before starting the titration to the hard water, ammonia buffer and EBT are added which 2+ Ca pH 8 -10 Ca + EBT EBT 2+ Mg Mg Unstable complex with wine red coloured When this solution is titrated against EDTA, it replaces the indicator from an unstable complex and form stable EDTA complex. After the titration, all the hardness causing ions are complexed by EDTA, the indicator is set free. www.annauniversityplus.comTherefore the colour of the free indicator is steel blue. Thus the end point is the change of colour from wine red to steel blue. Ca Ca EBT + EDTA EDTA + EBT Mg Mg Unstable complex with Stable complex with steel blue coloured Wine red coloured (alkaline medium) Preparation of solutions 1. Preparation of EDTA solution Dissolve ‘4g’ of pure disodium salt of EDTA crystals in 1 litre of distilled water. Preparation of standard hard water Dissolve 1g of pure CaCO3 + dil HCl Residue (small quantity) upto dryness Dissolve the residue in one litre of distilled water. “1 ml” of this solution contains “1 mg” of CaCO . 3 2. Preparation of Buffer solution Add 67.5g of NH4CL to 570ml of concentrated NH solution and dilute he 3 solution with one litre of distilled water. 3. Preparation of EBT Indicator Dissolve 0.5g of EBT in 100ml alcohol. “Structure of EBT” Procedure In this method, three titrations are carried out to estimate the total, permanent and temporary hardness. Standardization of EDTA solution The burette is filled with EDTA solution. 50ml of standard hard water is pipetted out into a clean conical flask. Add 10-15ml of buffer solution and few www.annauniversityplus.comdrops of EBT indicator. The wine red solution present in the conical flask is titrated against the burette EDTA solution till the wine red colour changes to steel blue colour. Let the volume of EDTA consumed be V ml. 1 Estimation of total hardness Pipette out 50ml of sample hard water into the conical flask add the ammonia buffer (NH OH + NH Cl) and EBT indicator and titrate it against the 4 4 same EDTA burette solution to get the end point. Let the volume of EDTA consumed be V ml. 2 Estimation of permanent hardness The water sample of 250ml is taken in a beaker and evaporates nearly to 50ml. the temporary hard salts settle down. Filter and wash thoroughly and make up the solution again to 250ml. Pipette out 50ml of the made-up solution in to a clean conical flask and titrate it against the (EDTA) burette solution to get the end point. Let the volume of EDTA consumed be V ml. 3 Calculations A.Total hardness (i) V ml of EDTA is consumed by 50ml of standard hard water. 1 V ml of EDTA = 50 mg of CaCO 1 3 Therefore 1ml of EDTA = 50/V mg of CaCO 1 3 (ii) V ml of EDTA is consumed by 50ml of sample hard water. 2 1ml of EDTA = 50/V mg of CaCO 1 3 V ml of EDTA = 50/V X V mg of CaCO 2 1 2 3 Therefore 50ml sample hardwater contain = 50/V X V mg of CaCO 1 2 3 Therefore 1000 ml sample hardwater = 50/V X V /50X1000mg/l 1 2 = V /V X1000mg/l 2 1 Therefore total hardness = V / V X 1000mg/l of CaCO (ppm) 2 1 3 B.Permanent hardness 50ml of sample hard water after removing temporary hardness consumes V 3 ml EDTA. 1ml of EDTA = 50/V mg of CaCO 1 3 Therefore V ml of EDTA = 50/V X V mg of CaCO 3 1 3 3 50ml of sample hard water (after boiling) contain = 50/V X V mg of CaCO 1 3 3 Therefore 1000ml of sample hard water = 50/V X V /50 X 1000mg/l 1 3 www.annauniversityplus.com = V /V X 1000mg/l 3 1 Therefore permanent hardness = V /V X 1000mg/l of CaCO (ppm) 3 1 3 C.Temporary Hardness Temporary Hardness = Total hardness – permanent hardness = (V /V X 1000) – (V /V X 1000) 2 1 3 1 = 1000 (V /V ) – (V /V ) 2 1 3 1 = 1000 X (V –V /V ) ppm 2 3 1 Problems based on EDTA method Example: 1 50 ml of a standard hardwater containing 1 mg of pure CaCO per ml 3 consumed 24 ml of EDTA. 50ml of sample of hard water consumed 16 ml of EDTA. Calculate the total hardness in ppm. Solution We know that, 1ml of std. hard water = 1mg of CaCO 3 24ml of EDTA = 50 ml of std. hard water Therefore 1ml of EDTA = 50/24 ml of std. hard water = 50/24 mg of CaCO equivalent 3 Therefore 1ml of EDTA = 2.0833 mg of CaCO equivalent 3 16 ml of EDTA = 2.0833 X 16 = 33.3328 mg of CaCO equivalent 3 Therefore 50 ml of hard water contains =33.3328 mg of CaCO 3 Therefore 1000ml of hard water contains = 33.3328/50 X1000 Total hardness = 666.656 ppm Result Therefore total hardness = 666.656 ppm. Example: 2 (i) 50 ml of standard hard water containing 1mg of pure CaCO per 3 ml consumed 20 ml EDTA. (ii) 50 ml of sample consumed 25 ml of EDTA solution. (iii) 50 ml of water sample after boiling and filtering consumed 18 ml of EDTA. Calculate the temporary, permanent and total hardness. Solution A.Total hardness We know that, 1ml of std. hard water = 1mg of CaCO 3 20ml of EDTA = 50 ml of std. hard water Therefore 1ml of EDTA = 50/20 ml of std. hard water www.annauniversityplus.com = 50/20 mg of CaCO equivalent 3 Therefore 1ml of EDTA = 2.5 mg of CaCO equivalent 3 25 ml of EDTA = 2.5 X 25 = 62.5 mg of CaCO equivalent 3 Therefore 50 ml of hard water contains = 62.5 mg of CaCO eq. 3 Therefore 1000ml of hard water contains = 62.5/50 X1000 Total hardness = 1250 ppm B.Permanent hardness After boiling EDTA consumed = 18 ml 1 ml of EDTA = 2.5 mg of CaCO equivalent 3 18 ml of EDTA = 2.5 X 18 = 45 mg of CaCO equivalent 3 50 ml of sample hard water = 45 mg of CaCO equivalent 3 (after boiling) contains 1000 ml of sample hard water = 45/50 X1000 (after boiling) contains Permanent hardness = 900 ppm C.Temporary hardness Temporary hardness = Total hardness – Permanent hardness = 1250 – 900 = 350 ppm Result Temporary hardness = 350 ppm Permanent hardness = 900 ppm Total hardness = 1250 ppm Example: 3 (i) 25ml of standard hard water consumes 12 ml of standard EDTA solution. (ii) 25 ml of sample hard water consumes 8 ml of standard EDTAsolution. (iii) After boiling the sample 25ml of the boiled and cooled hard water consumes 6 ml of standard EDTA solution. Calculate the total, temporary and permanent hardness. Solution A.Total hardness We know that, 1ml of std. hard water = 1mg of CaCO 3 12ml of EDTA = 25 ml of std. hard water Therefore 1ml of EDTA = 25/12 ml of std. hard water www.annauniversityplus.com = 25/12 mg of CaCO equivalent 3 Therefore 1ml of EDTA = 2.0833 mg of CaCO equivalent 3 8 ml of EDTA = 2.0833 X 8 = 16.6664 mg of CaCO equivalent 3 Therefore 25 ml of hard water contains = 16.6664 mg of CaCO 3 Therefore 1000ml of hard water contains = 16.6664/25 X1000 Total hardness = 666.656 ppm B.Permanent hardness After boiling EDTA consumed = 6 ml 1 ml of EDTA = 2.0833 mg of CaCO equivalent 3 6 ml of EDTA = 2.0833 X 6 = 12.4998 mg of CaCO equivalent 3 25 ml of sample hard water = 12.4998 mg of CaCO equivalent 3 (after boiling) contains 1000 ml of sample hard water = 12.4998/25 X1000 (after boiling) contains Permanent hardness = 500 ppm C.Temporary hardness Temporary hardness = Total hardness – Permanent hardness = 666.656 – 500 = 166.656 ppm Result Temporary hardness = 166.656 ppm Permanent hardness = 500 ppm Total hardness = 666.656 ppm Example: 4 50 ml of a standard hardwater containing 1 mg of pure CaCO per ml 3 consumed 17 ml of EDTA. 50ml of sample of hard water consumed 12 ml of EDTA. Calculate the total hardness in ppm. Solution We know that, 1ml of std. hard water = 1mg of CaCO 3 17 ml of EDTA = 50 ml of std. hard water Therefore 1ml of EDTA = 50/17 ml of std. hard water = 50/17 mg of CaCO equivalent 3 www.annauniversityplus.com Therefore 1ml of EDTA = 2.9412 mg of CaCO equivalent 3 12 ml of EDTA = 2.9412 X 12 = 35.2944 mg of CaCO equivalent 3 Therefore 50 ml of hard water contains =35.2944 mg of CaCO 3 Therefore 1000ml of hard water contains = 35.2944/50 X1000 Total hardness = 705.888 ppm Result Therefore total hardness = 705.888 ppm. Example: 5 100 ml of a standard hardwater containing 1 mg of pure CaCO per ml 3 consumed 22 ml of EDTA. 100ml of sample of hard water consumed 18 ml of EDTA. Calculate the total hardness in ppm. Solution We know that, 1ml of std. hard water = 1mg of CaCO 3 22 ml of EDTA = 100 ml of std. hard water Therefore 1ml of EDTA = 100/22 ml of std. hard water = 100/22 mg of CaCO equivalent 3 Therefore 1ml of EDTA = 4.5454 mg of CaCO equivalent 3 18 ml of EDTA = 4.5454 X 18 = 81.8172 mg of CaCO equivalent 3 Therefore 100 ml of hard water contains =81.8172 mg of CaCO 3 Therefore 1000ml of hard water contains = 81.8172/100 X1000 Total hardness = 818.172 ppm Result Therefore total hardness = 818.172 ppm. Uses of EDTA 1. It is used to measure the total hardness of water. 2+ 2+ 2. EDTA is used in volumetric and gravimetric analysis of metal (Ca , Mg ) ions. 3. The formation of scale (CaSO ) in the boilers can be prevented by EDTA 4 solution. 4. Fruits, fruit juices and food stuffs are preserved by the addition of EDTA. BOILER FEED WATER www.annauniversityplus.com In industry, one of the main uses of water is generation of steam by boilers.  Water used in boilers for steam production is known as boiler feed water.  Boiler feed water should be free from dissolved salts (MgCl , CaCl ), gases 2 2 (O , CO ), suspended impurities (silica and oil) etc., 2 2 Essential requirements of boiler feed water Boiler feed water should be free from, 2+ 2+  Hardness producing ions like Ca and Mg to avoid scale and sludge formation.  Turbidity, oil and non – scaling dissolved salts to produce priming and foaming.  Caustic alkali (NaoH) to remove caustic embrittlement and  Dissolved oxygen and CO in order to prevent corrosion in the boiler. 2 Disadvantages of using hard water in boilers (or) Boiler Troubles Hard water when used in boiler, it leads to the following troubles. 1. Sludge and scale formation 2. Priming and foaming 3. Caustic embrittlement 4. Boiler corrosion Sludges and scale formation: Due to continuous evaporation of water in boilers, the concentration of dissolved salts increases gradually and get deposited as precipitates on the inner walls and bottom of the boiler. This precipitate is known as sludge or scale. Sludge If the precipitate is a soft, loose and slimy it is called sludge. Scale If the precipitate forms hard and adherent coating on the inner walls of the boiler, is called scale. www.annauniversityplus.com

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