Chem Unit 1 PDF

# 1 - Water ## Introduction Water is one of the most abundant commodities in nature but also the most misused one. Water covers nearly three-fourth of the earth’s surface as liquid or solid ice. Water was believed to be an element until Cavendish prepared water in 1781 by igniting a mixture of two...

# 1 - Water ## Introduction Water is one of the most abundant commodities in nature but also the most misused one. Water covers nearly three-fourth of the earth’s surface as liquid or solid ice. Water was believed to be an element until Cavendish prepared water in 1781 by igniting a mixture of two volumes of hydrogen and one volume of oxygen. Due to this experiment, it was proved that water is a chemical compound of hydrogen and oxygen. Water is not only essential for the lives of animals and plants but also occupies a unique position in industries. Probably its most important use as an engineering material is in the steam generation. Water is also used as a coolant in power and chemical plants. In addition to it, water is widely used in other fields such as production of steel, rayon, paper, atomic energy, textiles, chemicals, ice and for air conditioning, drinking, bathing, sanitary, washing, irrigation, fire-fighting etc. ## Sources Of Water The main sources are 1. Rain water 2. Surface water * River water * Sea water * Lake water 3. Underground water * Well water * Spring water ## Classification Of Impurities From Water 1. **Suspended Impurities** * e.g.: - Particles of soil, sand, organic matter, industrial waste material etc. * Particle size greater than 1000A° & these are visible. They can be removed by filtration or sedimentation. 2. **Colloidal Impurities -** * Particle size < 1000A°. which make the water turbid. e.g.: - organic and inorganic matter. These impurities do not settle at the bottom of tank nor we can filter. To overcome this problem add small amount of coagulant. * Due to the addition of coagulant, colloidal impurities swells and ppt settle at the bottom of tank or you can filter. * Prevention:- They are separated by coagulation followed by sedimentation or filtration 3. **Dissolved Impurities** * Divided into two types. 1. Dissolved gases. 2. Dissolved Solid. * **Dissolved gases** :- like CO2, Cl2, H2S, SO2, NO2 etc soluble in water. * Prevention :- 1. By thermal process. 2. By mechanical aeration method * **Dissolved Solid :-** like bicarbonates, Sulphates, Chlorides, nitrates of Ca, Mg, Fe, Al, Na, K etc. * Prevention - These are removed by some chemical treatment like zeolite, process, ion exchange process etc. 4. **Biological Impurities** * These impurities, in water includes bacteria, algae and other small aquatic animals. These are present on the surface of water. * These are removed by first filtration and then sterilization. ## Causes Of Hardness of Water Rain water dissolved no. of gases from atmosphere during its journey through air and then falls on the ground. When water flows over the rocks, various minerals from it slowly get dissolved in water 1. **Dissolution :-** Some minerals from rocks are readily soluble in water such as NaCl, CaSO4 etc. 2. **Hydration** * CaSO4 + 2H2O Hydration) (CaSO4 2H2O gypsum 3. **Oxidation** * Dissolved oxygen present in water bring oxidation of rock minerals, which then undergo hydration and get dissolved in water. * 2 FeS2 + 7O2 + 2H2O oxidation) Hydration 2 FeSO4 + 2 H2SO4 ## 4. Action of dissolved carbon dioxide * CaCO3 + CO2 + H2O → Ca(HCO3)2 * MgCO3 + CO2 + H2O → Mg(HCO3)2 It converts insoluble carbonates of Ca, Mg and Fe into soluble bicarbonates ## Alkalinity Of Water * The alkalinity in water is due to far hydroxyl icon (OH), carbonate (CO32-) and bicarbonate (HCO3-). * Hydroxide and carbonates are stronger bases than bicarbonates. * The knowledge of alkalinity of water is necessary for controlling the corrosion in boiler, for calculating the amount of lime and soda needed for water softening etc. * Alkalinity is imp. to aquatic organisms because it protects them against rapid changes in pH. The knowledge of exact amount of alkalinity is imp: to the boiler feed water. * Strength of the bases * OH > CO32- > HCO3- ## Procedure:- 1. **Alkaline water sample (pH=12)** * Add 1-2 drops of Phenolphthalein indicator. * Turn pink - colourless (pH=8.2) * Add 2-3 drops Methyl orange * Turn Yellow - orange (pH=4.2) 2. **Reactions:** * OH + H+ → H2O * CO32- + H+ → HCO3 * HCO3 + H+ → H2O + CO2 ## 7 * The alkalinities due to all the three types of ions in water can be easily determined by neutralization titration against a std. strong acid titrant. * Take 10 ml of the alkaline water sample in a conical flask and add 1-2 drops of Phenolphthalein indicator in it. Titrate this sample against acid from burette, till pink colour changes to colourless. Note the burette reading as V1 ml. * Then add 1-2 drops of Methyl orange indicator into the same titrating mixture and continue the titration till yellow colour changes to orange. Note the burette reading as V2 ml (from intial) #### P = Phenolphthalein alkalinity * V1 x 2 x 50 x 1000 / vol. of H2O PPM #### M = Methyl orange alkalinity - Total Alkalinity * V2 x 2 x 50 x 1000 / vol.of H2O ### Types of alkalinities or Possible combinations 1. Only OH- 2. Only CO32- 3. Only HCO3 4. OH & CO32- together 5. CO32- & HCO3 together ## 8 The amount of alkalinities due to the OH, CO32-, HCO3 types are calculated from table | Alkalinity | Quantity of OH | Quantity of CO32- | Quantity of HCO3 | |---|---|---|---| | P = 0 | 0 | 0 | M | | P = M | 0 | 2P | 0 | | P = ½ M | P | 0 | 0 | | P < ½ M | 0 | 2P | M-2P | | P > ½ M | 2P-M | 2(M-P) | 0 | ### Numerical 1. 50 ml of water requires 3.7 ml of .025 N H2SO4 upto Phenolphthalein end point and further 4.8 ml upto the methyl orange end point. What are the types and amounts of alkalinities present in the water sample. * V1 = 3.7 ml * V2 = 3.7 + 4.8 = 8.5 ml #### P = V1 x 2 x 50 x 1000 / vol. of H2O PPM * 3.7 x 0.025 x 50 x 1000 = 92.5 PPM #### M = V2 x 2 x 50 x 1000 / vol. of H2O PPM * 8.5 x 0.025 x 50 x 1000 / 50 = 212.5 PPM We observe that P < ½ M Hence the alkaline water sample contains CO32- and HCO3- alkalinities. * CO32- = 2P = 2 x 92.5 = 185 PPM * HCO3 = M-2P = 212.5 - 185 = 27.5 PPM ## 10 ### Solve the Numericals 1. 50 ml of al alkaline water sample requires 5.2 ml of N/50 HCl up to phenolphthalein end point and 15.4 ml for complete neutralization. Find the type and amount of alkalinity in the water sample. (May 2012, 4 Marks) 2. 100 ml of water sample requires 4.3 ml of 0.02 N HCl upto phenslphthalium end point and total 11.9 ml upto methyl orange end point. Calculate the type and amount of alkalinity present. (May 2011, 3 Marks) 3. A water sample is not alkaline to phenolphthalein. However, 100 ml of the sample an titration with N/50 HCL required 16.9 ml to obtain end point using methyl orange as indicator. What are the types and amount of alkalinity present in the sample ?(May2014 & Dec 10 3 marks) 4. 50 ml of an alkaline water sample requires 9.2 ml of N/50 HCl upto phenolphthalein end point and total 13.1 ml of the acid for complete neutralization. Find the types and amounts of alkalinities in the water sample. (May 2009, 4 Marks) 5. 100 ml water sample requires 4 ml N/50 H2SO4 for neutralization upto phenolphthalein end point. Another 16 ml of the same acid was needed for further titration to methyl arrange end point. Determine the type and amount of alkalinities. (May 2010, 4 Marks) ## 11 ### Determination of Hardness by EDTA Method Here the analysis is done by complexometric titration using EDTA and EBT. * EDTA (Ethylene diamine tetra Acetic acid) the titrant, complexes with Mg and Ca ions removing them from association with the indicator. When all the Mg and Ca are complexes with EDTA, the indicator will turn blue. This is the end point of the titration. * HOOCCH2N-CH2-CH2-NCH2COOH HOOCCH2 EDTA CH2COONa This compound does not dissolved in water. If we take disodium salt of EDTA then it is easily dissolved in water. * NaOOCCH2N-CH2-CH2-NCH2COONa HOOCCH2 Na2-EDTA CH2COONa ### Principle of EDTA * Total hardness can be determined by titrating Ca2+ & Mg2+ in water with against Na2 EDTA soin. * NH4Cl - NH4OH buffer solution Add in conical before the titration. * EBT indicator form wine red colour (unstable Complex with Ca2+/Mg2+ ions in Hard water) * This complex is broken by EDTA soln during titration & give stable complex with ions. Here color changes from wine red to blue. ### PH - 10 * Ca2+/Mg2+ + EBT NH4Cl-NH4OH Mg2+ EBT Ca2+ * unstable complex (wine red) Titrated against EDTA soln. * Mg2+ EDTA + EBT Ca2+ blue colour Stable soluble complex <div align=center> <img src="https://i.imgur.com/y2B9x61.jpg" width = 250> </div> ## 13 * When all the Mg & Ca are complexed with EDTA the indicator will turn blue. This is the end point of the titration. * Note down the reading when colour changes from wine red to blue. * In this way, the hardness of water can be determined by this method. ### Numerical 2 Types 1. When conc. of EDTA is absent then use 3 step Procedure * std. standardization of EDTA * Determination of Total Hardness * Determination of Permanent Hardness 2. If conc. of EDTA is given then use #### Hardness = Y x Z x 100 x 1000 / V * Y = Burette reading * V = vol. of water sample * Z = Conc. of EDTA ## 14 ### Numericals 1. 60 ml of SHW containing 1 mg of pure CaCO3 per ml consumed 22 ml of EDTA. 40 ml of water sample consumed 20ml EDTA solution using EBT indicator. 40ml of water sample after boiling consumed 15ml of EDTA using EBT indicator. Calculate temporary Hardness. * EDTA = N1V1 = SHW = N2V2 * N1 x 22 = 60 = 1 x 60 * N1 = 60 / 22 = 2.72 mg of CaCO3 2. Calculate Total Hardness * 40 ml water sample consume = 20 ml EDTA = 20 x 2.72 = 54.4 mg of CaCO3 * Total Hardness = 54.4 x 1000 / 40 = 1363.6 PPM 3. Calculate Permanent Hardness * 40 ml boiled H2O consume = 15 ml EDTA = 15 x 2.72 = 40.8 mg of CaCO3 * Total Permanent H. = 40.8 x 1000 / 40 = 1020 PPM ## 15 ### Solve the Numericals 1. 50 ml of standard hard water containing 1 mg of pure CaCO3 per ml consumed 20 ml of EDTA. 50 ml of water sample consumed 25 ml of the same EDTA solution using EBT indicator. Calculate total hardness of water. 2. 60 ml of std hard water containing 1 mg of pure CaCO3 per ml consumed 22 ml of EDTA. 40 ml of water sample consumed 20 ml EDTA solution using EBT indicator. 40 ml of water sample after boiling consumed 15 ml of EDTA using EBT indicator. calculate temporary Hardness. 3. 50 ml of water sample requires 12.2 ml of 0.02 M EDTA during the titration. 50 ml of boiled water sample requires 8.2 ml of EDTA. Calculate temporary, permanent and total hardness. 4. 50 ml of a water sample requires 12.7 of 0.02 M EDTA during titration. Calculate total hardness of the water. ## 16 ### External treatment Methods OR Softening Methods. ### Zeolite OR Permutit Process This is the modern process used for removing both temporary and permanent hardness of water. Zeolite means hydrated alumino silicate. The chemical formula of sodium zeolite is Na2O. Al2O3. *X*SiO2. *Y*H2O Where *X* = 2-10, *Y* = 2-6 These silicates hold sodium ion loosely hence these are called as sodium permutit (Na2P) or sodium zeolite (Na2Z) Where Permutit and zeolite stand for Al2Si2O8. 6H2O The remarkable property of sodium permutit (Na2P) is that when it comes in contact with hard water, it exchanges its Na+ ions with Ca2+ and Mg2+ ions to form insoluble calcium permutit (CaP) and magnesium permutit (MgP). ### Process Sodium permutit is placed in a suitable container and hard water is allowed to pass through it. The Ca2+ and Mg2+ salts react with it forming insoluble Ca2+ and Mg2+ permutits. These salts are retained in the filter bed and water <start_of_image> Schematic Rep of Zeolite Process: <div align=center> <img src="https://i.imgur.com/h259E0Z.jpg" width = 250> </div> After reaction free from Ca2+ and Mg2+. Only the harmless sodium salts are left in the water. The following reaction takes place in the process. ### Reactions * Ca(HCO3)2 + Na2P → CaP↓ + 2NaHCO3 * Mg(HCO3)2 + Na2P → MgP↓ + 2NaHCO3 * CaSO4 + Na2P → CaP↓ + Na2SO4 * MgSO4 + Na2P → MgP↓ + Na2SO4 When the process is continued for about 12 hours, all the Na+ ions from the permutit are replaced by Ca2+ and Mg2+ ions and it is found that the permutit stops working and water is no more softened. ## 18 When the permutit is completely converted into CaP and MgP, it is regenerated by treating with 10%. NaCl soln (brine) for a few minutes. Sodium permutit (Na2P) is formed and can be used for softening of more hard water. * CaP↓ + 2NaCl → Na2P + CaCl2 * MgP↓ + 2NaCl → Na2P + MgCl2 ### Limitations * If hard water contains large quantities of coloured ions like manganese ions (Mn2+) or ferrous ions (Fe2+) then these ions form manganese permutit and ferrous permutit. * If hard water is turbid then turbidity blocks the pores of permutit. And it restricts the flow of water * If mineral acids are present in water, destroy the zeolite bed and therefore they must be neutralized with soda before water being entered into zeolite plant. ### Advantages * Soft water obtained has hardness 5-10 PPM * Equipment used is compact and occupies less space. * It is a quite clean process and requires less time for softening. * It requires less skill in maintenance as well as operation. ## 19 ### Numericals on Zeolite 1. An exhausted zeolite reformer was regenerated by passing 150 liters of sodium chloride solution having strength 150 gm/liters of NaCl. How many liters of hard water sample having hardness 400 PPM can be softened by using this softener? * Step 1: * 1 lit of liq. NaCl = 150 gm of NaCl * 150 lit of liq. NaCl = ? * Total amount of NaCl required for regeneration = 150 x 150 = 22,500 gm = 22,500 x 1000 mg = 225 x 105 mg * Step 2: Conversion into CaCO3 eq. * 225 x 105 x 100 / 117 mg of CaCO3 = 192.31 x 105 mg of CaCO3 * Step 3: * Vol. of H2O = CaCO3 eq / Hardness = 192.31 x 105 / 400 = 48.077. 5 lit ## 20 ### 2. A zeolite softener gets exhausted on softening 4000 lits of hard water, calculate hardness of water, if the exhausted zeolite requires 10 liters of 10% Nach solution for regeneration. (DLC 2013, 3 Marks) ### Solve the Numericals 1. A zeolite bed exhausted by softening 4000 litres of water requires 10 liters of 15% NaCl solution for regeneration. Calculate the hardness of water sample.(May 2012, 4 Marks) 2. A zeolite softener was completely exhausted and was regenerated by passing 100 litre of NaCl containing 120 gm/lit of NaCl. How many litre of a sample of water of hardness 500 ppm can be soften by this softener? (Dec. 2009, 4 Marks) 3. Find the hardness of water sample from the given data. A zeolite bed gets exhausted on softening 2400 litre of water and requires 10 litre of 8% NaCl for regeneration.(May 2011, 3 Marks) ## 21 ### Ion-Exchange or Demineralisation or Deionization Methods This is the modern development in water softening method. Certain organic compounds possess a property like zeolite i.e. they are capable of exchanging ions. Such organic synthetic compounds are known as resins. ### Principle * When hard water is fed to ion exchanger, the cation exchanger resin removes all cation from water in exchange of its H+ ions. * The anion exchanger resin removes all anions from water in exchange of its OH- ions. Finally, the water becomes soft. ### There are 02 Types Resin 1. **Cation exchange resin (RH2)** <div align=center> <img src="https://i.imgur.com/K1tR5O6.jpg" width = 250> </div> * SO3H+ band is loose. * Sulphonation on styrene (Polymer) 2. **Anion Exchange resin [R'(OH)2]** <div align=center> <img src="https://i.imgur.com/mG1h63r.jpg" width = 250> </div> * N(CH3)3 On * Amination on styrene (Polymer) These resins are capable of exchanging anion by OH- ions. * R'(OH)2 + SO42- → R'SO4 + 2OH- From the above, it is clear that, if hard water is passed first through cation exchanger and then through anion exchanger, the resulting water will be free from both cations and anions and water is said to be deionized. ## Process It consists of two cylindrical towers, out of which the first tower consists of cation exchanger (RH2) and another one consists of anion exchanger R'(OH)2. Hard water is first allowed to pass through a tower containing cation exchanger, which removes all the cations like <div align=center> <img src="https://i.imgur.com/15m8H2y.jpg" width = 250> </div> * Ca2+, Mg2+, Fe2+ etc and releases H+ ions. * RH2 + Ca2+ → RCa↓ + 2H+ * RH2 + Mg2+ → RMg↓ + 2H+ Thus the anions like chlorides, sulphates and bi-carbonates are converted into corresponding acids HCl, H2SO4. * SO42- + 2H+ → H2SO4 * Cl- + H+ → HCl In other words, water from cation exchanger is free from all cations but it is acidic. This acidic water is then passed through another tower containing an anion exchanger, where acids are converted into water. * R'(OH)2 + H2SO4 → R'SO4 + 2H2O * R'(OH)2 + 2HCl → R'Cl2 + 2H2O Thus water is free from all ions (cation and anions) but dissolved gases present in it. The water is finally free from dissolved gases like CO2, by passing it through a degasifier. Degasifier, which is a tower whose sides are heated and which is connected to vacuum pump. High temp and low pressure reduces the quantity of dissolved CO2 and O2 in water. In this way, we will get soft H2O that is free from dissolved gases. ## Regeneration When the resins get exhausted (i.e. when their capacity to exchange H+ and OH- ions are lost) they are regenerated. The exhausted cation exchanger is then regenerated by passing a dilute solution of acid (e.g. dil HCl or H2SO4). * RCa↓ + 2HCl → RH2 + CaCl2 * RMg↓ + 2HCl → RH2 + MgCl2 Similarly exhausted anion exchanger is regenerated by passing dil NaOH. * R'SO4 + 2NaOH → R'(OH)2 + Na2SO4 * R'Cl2 + 2NaOH → R'(OH)2 + 2NaCl ## 25 ### Ill effects of Hard water The most important use of water in industry is for steam generation using boilers. So the boiler feed water should be free from impurities in order to avoid trouble inside the boiler. When hard water is used in the boiler following are the ill effects of Hard water. 1. **Priming and Foaming** * **a) Priming :-** In boiler during the production of steam some water droplets are carried. This phenomenon is called Priming. **Causes -** * Large amounts of dissolved salts are present i.e. high TDS * Improper boiler design. * Sudden increase in steaming rate. **Prevention -** * Use soft water. * Maintain low water levels. * Avoid rapid change in steaming rate. * Use Mechanical steam purifiers. * **b) Foaming :-** In boiler, during the production of steam continuous bubble or foam formation takes place and which do not break easily. This phenomenon is called Foaming. **Causes -** * Presence of oil and soap * Bubbles ↓ Foam * Prevention - * Foam → Add antifoaming agent like castor oil * Oil → Oil impurities remove by sodium aluminate. 2. **Sludge and Scale :-** * **a) Sludge -** In the boiler, on continuous conversion of the water into steam, boiler feed water get’s conc. and form loose and slimy mass of precipitate is called sludge. * Sludge generally formed at cooler part of the boiler and they deposited on edges, bends, valves etc. * They have greater solubility in hot water than in cold water. e.g.:- MgCl2, MgSO4, CaCl2 etc. * Disadvantages * Reduces the flow of steam. * Causing choking of pipes. * Poor conductor of heat + wastage of energy. * Decreases the efficiency of the boiler. * Prevention - * Use soft water * Blow down operation (remove conc. water & replacing it by fresh water or remove last portion of water containing high amount of salt) * Removal of sludge by brush, scrappers etc. <div align=center> <img src="https://i.imgur.com/x98c2wQ.jpg" width = 250> </div> * Fig:- Top view of Boiler -> Boiler Metal -> loose ppt * b) **Scale** - Hard and strong coating formed inside the boiler which can not be removed easily is called Scale. * The Scale is formed when boiler feed water contains dissolved salts such as bicarbonates, CaSO4, Mg salts, Silica etc. <div align=center> <img src="https://i.imgur.com/r4Y8t44.jpg" width = 250> </div> * Fig:- Top view of Boiler -> Hard Scale * Causes of Scale * **Presence of Bicarbonates -** * If water contains Ca(HCO3)2 and Mg(HCO3)2, decomposition of bicarbonates will take place. * i) Ca(HCO3)2 Δ> CaCO3↓ + CO2↑ + H2O * In low pressure boiler scale is soft but in high pressure boiler scale is hard. * CaCO3 + H2O → Ca(OH)2↓ + CO2↑ Sticky Ppt * ii) Mg(HCO3)2 Δ> Mg(OH)2 + 2CO2↑ Sticky Ppt * These sticky products get deposited at the inner surface of boiler as a scale. * Solubility of CaSO4- * CaSO4 has lesser solubility at high temp. so at high temp. CaSO4 present in boiler feed water will precipitate as hard scale forming material. * Hydrolysis of Mg Salt - * At high temp. Mg salt undergo hydrolysis to form sticky ppt and converted into scale. * MgCl2 + 2H2O Δ> Mg(OH)2 ↓ + 2HCl↑ Sticky Ppt * Presence of Silica - * If sand filter is used the sand (SiO2) will be converted into silicate. * SiO2 → (CaSiO3 or MgSiO3) (Sticky and very firm) * CaSiO3 and MgSiO3 are sticky and firm strongly adhered coating at the inner wall of boiler, that is very difficult to remove. * Prevention of Scale * Use soft water * Use frequent blow down operation * Boiler feed water should be free from suspended, colloidal & dissolved impurities. * To overcome this problem with internal treatments like Phosphate, colloidal, EDTA, Sodium Aluminate and Calgon Conditioning. ## 29 ### Difference Between Sludge and Scale: | Sludge | Scale | |-------------------------------|------------------------------------------------------------------------| | These are loose and slimy ppt | These are hard and adherent ppt. | | These are formed at cooler parts of the boiler | These are formed at hot parts or regions of the boiler. | | Sludges are formed by the salts which are less soluble in cold water but highly soluble in hot water | Scales are formed by the salts which are less soluble in hot water but highly soluble in cold water eg: - CaSO4 | | Sludges are formed due to increase in conc. of the salt eg:- MgCO3, MgCl2, CaCl2 etc. | Scales are formed due to decomposition of Mg-salts, less solubility in hot water (CaSO4) and Silica. | | They can be removed by blow-down operation. | They can be removed by EDTA Treatment. | ## 30 ### 3) Boiler Corrosion Boiler Corrosion is the most serious problem, caused by using hard H2O. Corrosion is the destruction of metal by surrounding environment. Corrosion in boiler is due to I. Dissolved gases II. Acids from dissolved salts **I. Dissolved gases** * A) Due to dissolved **gases oxygen**- * Most commonly used boiler metal is iron or alloy of iron. * If boiler feed water contains oxygen, then causes the corrosion of iron. ## 31 * i) 2 Fe + (H2O + O2) → 2Fe(OH)2↓ Ferrous hydroxide * ii) 4 Fe(OH)2↓ + O2 → 2 [Fe2O3.2H2O]↓ Rust Ferric oxide Hence boiler metal gets corroded. ### Prevention - 1. with the help of chemicals * a) Hydrazine * N2H4 + O2 → N2↑ + 2H2O Hydrazine * b) Sodium Sulphite * 2 Na2SO3 + O2 → 2 Na2SO4 * c) Sodium Sulphide * Na2S + 2O2 → Na2SO4 * j) with the help of Mechanical deareation method. 2. Due to dissolved **CO2 -** * CO2 + H2O → H2CO3 Carbonic acid * ↓ * It is a weak acid & has a slow corrosive material. ## 32 ### Prevention 1. Addition of NH3 (ammonia) * CO2 + 2NH3 + 2H2O → (NH4)2CO3 + H2O Ammonium Carbonate 2. By Mechanical deareation method. 3. Acids from dissolved salt OR Hydrolysis of Mg salt * i) MgCl2 + 2H2O → Mg(OH)2+ + 2HCl↑ Librate acid Mg hydroxide on hydrolysis * ii) Fe + 2HCl (g) → FeCl2 + H2↑ Ferrous Chloride * iii) FeCl2 + 2H2O→ Fe(OH)2 ↓ + 2HCl ### Prevention - 1. By various internal methods such as Phosphate conditioning, colloidal conditioning etc. 2. Blow down operation. 3. By using Zealite Process, ion exchange process etc. 4. Caustic Embrittlement ## 33 ### Caustic Embrittlement For softening of hard water, Lime soda can be used. Soda (Na2CO3) will remove the hardness causing impurities. But the unreacted or excess soda will make the water alkaline at high temp. * Na2CO3 + H2O Δ> 2NaOH + CO2↑ The hydroxide will react with the boiler metal through cracks. * 2 Fe + 2NaOH + O2 → 2 NaFeO2 th2 Sodium Ferrote Sodium Ferroate is brittle in nature. Thus boiler becomes brittle due to alkalinity (caustic) of water. This phenomena is called Caustic embrittlement. ### Prevention * Use of Phosphate instead of sadalime for softening of water. * The cracks in the boiler metal can be covered with tannin or lignin. ## 34 ### Desalination of Brackish Water The process of removing common salt (sodium chloride) from the water, is known as desalination. Brackish water is water that has more salinity than fresh water, but not as much as seawater. Brackish water is totally unfit for drinking purpose. Commonly used methods for the desalination of brackish water are ### 1. Electrodialysis :- Electrodialysis is a method in which the ions (of the salts present) are pulled out of the salt water by passing direct current, using electrodes and thin rigid plastic membrane pair (natural or synthetic). When direct current is passed through saline water, the sodium ions (Na+) start moving towards negative pole (cathode) while the chloride ions (Cl-) start moving towards the positive pole (anode), through the membrane. As a result, the conc. of brine decreases in the central compartment; while it increases towards side compartments. Pure water (desalinated brine) is removed from the central compartment from time to time, while conc. brine (side compartments) is replaced by fresh brine or sea water. <div align=center> <img src="https://i.imgur.com/0m9R9dC.jpg" width = 250> </div> * Fig 1.8: Electrodialysis cell For more efficient separation, usually ion selective membranes are employed. An ion selective membrane has permeability for only one kind of ions with specific charge. For eg. a cation selective membrane is permeable to cations only, because of the presence of charged fixed (inside the membrane) functional groups (such as RSO3 or RCO2), reject anions having the same charge as that of fixed functional groups) Similarly anion selective membrane has positively charged fined functional groups such as R4N+ Cl-. An electrodialysis cell consists of a large number of paired sets of rigid plastic membranes. Saline water is passed under a pressure (of about 5-6 kg mm2) between membrane Pairs and an electric field is applied perpendicular to the direction of water flow. ## 36 Just as magnets of like charges repel each other, the fixed positive charges inside the membrane repel positively charged ions (Na+), yet permit negatively charged ions (Cl-) to pass through. Similarly, the fixed negative charges inside the other type of membrane repel negatively charged ions (Cl-), yet permit positively charged ions (Na+) to pass through. Therefore water in one compartment of the cell is deprived of its salts, while the salt conc. in adjacent compartments is increased. Thus we get alternate streams of pure water and conc. brine. ### Advantages 1. It is the most compact unit. 2. The cost of installation of the