Water Technology PDF - Unit 4

WATER TECHNOLOGY 24-12-2024 -  Water Analysis Water is the nature’s most abundant and essential compound useful for lives of animals and plants. It is used for both domestic and industrial purposes. The surface sources of water contain severa...

WATER TECHNOLOGY 24-12-2024 -  Water Analysis Water is the nature’s most abundant and essential compound useful for lives of animals and plants. It is used for both domestic and industrial purposes. The surface sources of water contain several components and they affect our health, agriculture and industry. Some of the components in water to be studied in this chapter with regard to their sources and effects, experimental determination and calculations include: 1. Hardness 2. Nitrate content 3. Suphate content 4. Chloride content 5. Dissolved Oxygen content HARDNESS OF WATER Hard water: “A sample of water containing soluble salts of Calcium, Magnesium, Iron, Manganese etc., which consumes a lot of soap and does not produce free lather with it is called hard water”. These salts in water are derived from soils and rocks. They form insoluble white scum with sodium soap. For example: 2 C17H35COONa + Ca2 → [C17 H 35COO]2 Ca ) + 2Na+. A Sodium soap (in water) Cal. soap ppt(scum) Thus hard water consumes large amount of soap. In addition, hard water clogs pipe and damage boilers by forming boiler scales due to the deposition of the salts on the inner wall of the boilers. Hardness of water: Hardness of water is a measure of its capacity to precipitate soap. It gives the magnitude of the presence of soluble salts responsible for making the water hard. Types of hardness: There are two types of hardness- 1) Temporary Hardness (Carbonte hardness): This type of hardness is due to dissolved bicarbonates of Ca and Mg. It can be removed by boiling. For example: Ca(HCO3)2 (in water) (boil for 20 -30 minutes) ----> CaCO3 (ppt) +H2O + CO2. 2) Permanent Hardness (Non- carbonate hardness): This type of hardness is due to soluble non carbonate salts of Ca, Mg etc, namely, chlorides, nitrates, sulphates, hydroxides etc. They cannot be removed by heating. They need special chemical treatments to remove them. A water sample may contain both types of hardness and hence, Total Hardness = Temporary hardness + Permanent hardness Engineering Chemistry Notes, 2024-2025 Syllabus Page 1 Units of Hardness of water: Hardness of water is expressed in several ways 1. Parts per million (ppm) : ie., parts of CaCO3 equivalents per million parts of water. For example, 1 ppm is 1 part of CaCO3 equivalent hardness in 1 million parts of water. 2. Milligram per litre (Mg/L): It is the number of milligrams of CaCO3 equivalent hardness in 1 litre of water. 3. Degree of General Hardness (dGH or German degree): It is defined as 10 mg/L CaO or 17.848 ppm.  Determination of Hardness of Water (EDTA Method) Principle: To a sample of hard water containing Ca or Mg salts, when the blue coloured Eriochrome Black-T (EBT) indicator is added at pH 10, it forms a pink complex with Ca or Mg ions. However, when it is titrated against a solution of disodium salt of Ethylene diamine tetra acetic acid (EDTA), Ca and Mg ions from the EBT complex react with EDTA releasing free EBT which is blue in colour. Hence by knowing the quantity of EDTA required it is possible to find the hardness of water Ca2+ + EBT indicator → Ca-EBT complex (in hard water) (wine red) Ca-EBT complex(pink) + EDTA → Ca-EDTA + EBT indicator complex (colourless) (Blue) Procedure: a) Preparation of Std. EDTA solution: Dissolve a known wt. of soluble salt of EDTA in water in a 250 ml std. flask and make up to the mark with water and mixed well. Molarity of EDTA = Wt of EDTA salt x 4/Mol wt. of EDTA salt) Let its molarity be ‘x’ M. b) Determination of Total hardness: Pipette out 25 ml of the water sample in to a clean conical flask. Add 2 ml of NH3- NH 4Cl buffer and a pinch of EBT indicator. Titrate this against std. EDTA solution, until the colour changes from pink to just blue. Note the volume of EDTA consumed, (say V1 ml). Calculations: 1000ml of 1 M EDTA= 100 g. of CaCO3 Hence, V1 ml of ‘x’ M EDTA = (V1 X ‘ x’ X100/1000X1M) = ‘A’ g. of CaCO 3 in 25 ml water sample. Hence, Total hardness = [A/25] 106 ppm = -------- ppm Engineering Chemistry Notes, 2024-2025 Syllabus Page 2  c) Determination of Permanent Hardness: Pipette out 25 ml of the same water sample in to a beaker, boil gently for 20-30 minutes, cool and filter in to a conical flask. To the filtrate add 2 ml NH 3- NH 4 Cl buffer and a pinch of EBT indicator. Titrate this against the same std. EDTA until the colour changes from pink to just blue. Note the volume of EDTA consumed, say V2 ml. Calculations: 1000 ml 1 M EDTA = 100 g. of CaCO 3 Hence, V2 ml of ‘x’ M EDTA =(V2 X ‘x’ X 100/1000X 1M) = ‘B’ g. of CaCO 3 in 25 ml water boiled. Hence, Permanent hardness = (B/25)106 ppm =T2 ppm d) Calculation of Temporary Hardness: Total Hardness = Temporary hardness + Permanent Hardness Hence, Temporary hardness = Total hardness - Permanent hardness = T1-T2 = T3 ppm = ……. mg dm-3  Dissolved Oxygen Dissolution of oxygen from atmosphere gives dissolved oxygen content in water. Algae and rooted aquatic plants also give out oxygen in to water through photosynthesis. Dissolved oxygen is required for biological oxidation of pollutants. However, oxygen being an oxidizing agent it causes corrosion in industry. Determination of Dissolved Oxygen content in water (Winkler’s Method) Dissolved oxygen cannot be directly estimated by titration, but however oxygen being an oxidizing agent it can be estimated in an indirect way. Principle: When a strong alkaline solution of MnSO4 (divalent) is added to a water sample containing dissolved oxygen, it picks up oxygen to form basic MnO(OH)2 ( tetravalent) which is used to oxidise KI to I2. The liberated Iodine is found by titrating against a standard Na2S2O3 thereby oxygen content is estimated. MnSO4 + 2 KOH → Mn(OH)2 + K2SO4 2Mn (OH)2 + O2 (Dissolved Oxygen) → 2MnO(OH)2 MnO (OH)2 + H2SO4 → MnSO4 + 2H2O + [O] 2 KI + H2 SO4 + [O] → K2SO4 + H2O + I2 I2 + 2 Na2S2O3 → 2 NaI + Na2S2O4 Engineering Chemistry Notes, 2024-2025 Syllabus Page 3 Procedure: In a stoppered bottle containing 300 ml of water sample add 3 ml MnSO4 and 3 ml alkaline KI. Stopper the bottle and shake well for 10 to 15 minutes. Allow to stand for 5 minutes. Then add 1 ml conc. H2 SO4 and mix well to dissolve MnO2 ppt and to liberate I2.. Pipette out 100 ml of the above solution in to a conical flask and titrate against standard Sodium thiosulphate solution using 2 ml of starch solution as indicator near the end point until the blue colour is just discharged. Note the volume of Na2S2O3 required (say V1 ml). Calculations: Volume of water sample taken = 100 ml Volume of Na2S2O3 consumed = V1 ml Normality of Na2S2O3 used = N1 So, Normality of D.O in water = ( N1 x V1) /100 = N2 Hence, Dissolved Oxygen content = N2 x 8 x 1000 = ……… mg/dm3.  Biological Oxygen Demand (B.O.D) Sewage water contains a lot of organic pollutants which undergo anaerobic oxidation to give products with stinking smell. Such waste water requires oxygen for the oxidation. Definition: “The amount of oxygen required in milligrams for the biological oxidation of organic matters under aerobic condition, present in one litre of waste water over an incubation period of five days at 20o C, is called biological oxygen demand”. Oxygen present in water is capable of oxidizing the biologically oxidizable pollutants, with the help of bacteria and in this way the oxygen in water is consumed. Determination of B.O.D: Principle: When a sample of water containing dissolved oxygen is treated with MnSO4 and alkaline KI solutions, dissolved oxygen oxidizes Mn2+ to Mn4+. On acidification Mn4+ liberates I2 from KI which is estimated by standard NasS2O3. By determining the dissolved O2 content in water before and after incubation period it is possible to find B.O.D Procedure: a) Preparation of two separate diluted waste water sample: Pipette out a known volume of sewage water sample (‘A’ ml) and dilute it to a known volume (‘B’ ml) and equally divided and fill in two BOD bottles (Bottle -1 and Botttle -2). Engineering Chemistry Notes, 2024-2025 Syllabus Page 4  b) Blank titration: In Bottle -1 find D.O immediately as follows: Add 2 ml MnSO4 + 3 ml alkaline KI. Stopper the bottle and mix well for 10-15 minute. Wait for 5 minutes and then add 1 ml conc. H2SO4 to dissolve MnO(OH)2 and to liberate I2. A known volume of this solution (‘V’ ml) is titrated against std. Na2S2O3 using 2 ml of starch as indicator near the end point. Note the volume of Na2 S2O3 required ( say V1 ml). c) Back titration: Bottle -2 is incubated for 5 days at 20o C. After 5 days, add 2 ml MnSO4 + 3 ml alkaline KI. Stopper the bottle and mix well for 10-15 minutes. Wait for 5 minutes and then add 1 ml of conc. H2 SO4 to dissolve MnO2 ppt and to liberate I2. Pipette out same known volume of this solution (‘V’ ml) and titrate against the same std. Na2S2O3 as above. Note the volume Na2S2O3 required (say V2 ml ). Calculations: D.O from Blank titration: Volume of water sample taken = V ml, Volume of Na2S2O3= V1 ml Normality of Na2S2O3 = N1. Normality of D.O in water = N = (N1 X V1)/ V Hence, Amount of D.O in mg = N2 X Eq. wt of O2 (8) x1000 = D1 mg/ dm-3 D.O from back titration (After Incubation for 5 days): Volume of water sample = V ml, Volume of Na2S2O3 = V2 ml. Normality of Na2S2O3 = N1 Normality of D.O in water= N = ( N1 V2 )/V Hence, Amount of D.O in mg = N3 X Eq. Wt of O2 x 1000 = D2. mg/dm3 Therefore, B.O.D of the sample = (D1- D2) (B/A) =………mg dm-3  Chemical Oxygen Demand (C.O.D) Effluent coming out of domestic and industrial areas contains both biologically oxidisable organic pollutant and chemically oxidisable organic and inorganic pollutants. For oxidation of such pollutants a strong oxidizing agent such as K2Cr2O7 which can release a large amount of oxygen is required Definition: “The amount of oxygen required in milligrams for the chemical and biological oxidation of organic matters present in one litre of waste water using acidified K2Cr2O7 as an oxidizing agent is called chemical oxygen demand”. Engineering Chemistry Notes, 2024-2025 Syllabus Page 5 Determination of chemical oxygen demand (COD) of the given industrial waste water sample Principle: The Chemical Oxygen Demand (COD) test is extensively employed as a means of measuring the pollutional strength of industrial wastes. Chemical oxygen demand ‘is a measure of the total quantity of oxygen required per liter of water for oxidation of organic and inorganic wastes to their corresponding oxides by a strong oxidising agent’. This parameter is particularly valuable in surveys designed to determine the losses to sewer systems. Results may be obtained within a relatively short time and measures taken to correct errors on the day they occur. Waste water contains organic impurities which include straight chain aliphatic compounds, aromatic hydrocarbons, straight chain alcohols, acids and other oxidisable materials. Straight chain compounds, acetic acid, etc. are oxidised more effectively when silver sulphate is added as catalyst. But silver sulphate reacts with chlorides in the waste water to form precipitates which are oxidised only partially by this procedure. This difficulty is overcome by adding mercuric sulphate to the sample. Procedure: a) Preparation of standard Mohr’s salt solution (FAS solution) Weigh about 4.0 g of Mohr’s salt accurately into a 250cm3 volumetric flask. To prevent the hydrolysis of ferrous salt, add two test tubes of dilute sulphuric acid and dissolve the crystals. Dilute the solution with water upto the mark and shake well. b) Back Titration: Pipette out 25 cm3 of the waste water sample into a conical flask, add 1 g of AgSO4 and 1 g of HgSO4 followed by 10 cm3 of standard potassium dichromate solution and half test tube of 1:1 dilute sulphuric acid. The solution is refluxed for 2 hours. Cool, add 2 to 3 drops of ferroin indicator and titrate against standard Mohr’s salt solution until the solution turns from blue green to reddish brown. Repeat for agreeing value. The role of AgSO4 is, it acts as a catalyst, oxidizes straight chain organic compounds, aromatics and pyridine. HgSO4 avoids the interference of Cl- ions by forming complex with them. c) Blank Titration: Pipette out 10cm3 of standard potassium dichromate solution. Add half test tube of 1:1 dilute sulphuric acid followed by 2-3 drops of ferroin indicator. Titrate against standard Mohr’s salt solution until the colour turns from blue green to reddish brown. Engineering Chemistry Notes, 2024-2025 Syllabus Page 6 CALCULATION: 1 ml of 1N FAS solution ≡ 8 mg of oxygen ≡ 1milli equivalent of oxygen (c – b) cm3 of a N FAS solution ≡ 8 × (c – b) × a = _____________ (d) mg of oxygen i.e., 25 cm3 of waste sample required =................ d mg of oxygen 1000cm3 of waste water sample required = mg of oxygen COD of waste water sample = ____________ mg of oxygen/dm3  Nitrate content The sources of nitrate in water are due to the use of synthetic fertilizers mostly containing nitrates, to the soil, decay of animal and vegetable products in the soil where in the nitrogenous compounds get oxidized to nitrates, and due to fixation of nitrogen in nature. Water with more than 45 mg of nitrate per litre is bitter in taste and causes many physiological disorders like gastrointestinal cancer, alzeimer, reduces oxygen carrying capacity of blood in children causing a condition called blue-baby syndrome. Determination of Nitrate in Water Principle: Nitrate reacts with Phenol Di-sulphonic Acid (PDA) to give a nitro derivative in alkaline solution, which is yellow in colour. The colour produced is proportional to the concentration of nitrate content according to Beer’s law. Hence by preparing a calibration curve using standard nitrate solutions the nitrate content in water sample can be found using a colorimeter or Spectrophotometer. Procedure: a) Preparation of PDA solution: Dissolve 12.5 g of Phenol in 75 ml of conc.H2SO4. Add further 80 ml of the acid and heat for 2 hours on a water bath and cool. b) Preparation of Calibration curve: Take 2, 4, 6, 8, 10 ml of standard KNO3 solution (Prepared by dissolving a known weight of about 0.367 g of KNO3,in 1000 ml of water) in 5 separate 50 ml beakers and evaporate to dryness. Add 5 ml PDA solution to each one and dissolve the residue. Transfer each quantitatively in to separate 50 ml standard volumetric flasks. Add 10 ml conc. NH3 and dilute up to the mark with water and mix well. Measure the absorbance or optical density using 410 nm wave length filter in a colorimeter. Plot the graph to get the calibration curve. Engineering Chemistry Notes, 2024-2025 Syllabus Page 7 c) Determination Nitrate content in the given test sample: The given test solution is in to a 50 ml beaker and evaporate to dryness. Add 5 ml of PDA and dissolve the residue. Transfer it quantitatively in to a 50 ml standard flask. Add 10 ml of conc. NH3 and dilute to the mark with water and mix well. Measure the absorbance or optical density for this sample also and find the concentration of nitrate from the graph. Flask No. Volume Volume OD KNO3 taken, ml PDA added, ml 0 0.0 5 (Blank solution) 0.0 1 2.0 5 2 4.0 5 3 6.0 5 4 8.0 5 5 10.0 5 6 ‘c’ 5 b Unknown water sample Calculations: Wt. of KNO3 in 1000 ml (Given) = ‘0.367 g. 101 g of KNO3 contain 62 g of NO3- (Mol. wt of KNO3 = 101, Mol. wt of NO3- = 62) ‘w’ g KNO3 contain = 0.367 x 62/ 101 = ‘0.225’ g of NO3- in 1 lit Or Weight of NO3- 1.0 ml = ‘0.225’/1000 = ‘0.00024’ gm of NO3- O.D value of unknown sample is ‘b’, and the corresponding volume is ‘c’ (from the calibration curve), Hence, NO3- content in water = ‘0.00024 x c g/dm3 Or Weight of NO3- = 0.00024 x c x 10000 = ………. mg/ dm3 (or ppm) of NO3- Engineering Chemistry Notes, 2024-2025 Syllabus Page 8  ↑ Optical Density Volume of KNO3 →  Potable Water Definition: “Water that is fit for human consumption and meets all the safe microbiological and chemical standards of quality is called potable water.” Water that is fit for drinking should be i) reasonably soft ii) have high degree of clarity iii) free from smell and colour iv) should have agreeable taste v) free from pathogenic bacteria vi) pH should be around 7 to 7.2 vii) should not contain minerals like lead, arsenic, chromium etc., Purification of water Membrane based Technology for water purification: Reverse Osmosis Process Reverse Osmosis (RO) is the most economical and highly effective method of removing several impurities from water such as total dissolved solids, turbidity, toxic metals, organic compounds, pesticides etc. Principle: When pure water and a solution of a salt are separated by a semi permeable membrane, which can permit only water molecules to pass through, pure water will flow in to the salt solution by a natural tendency called osmosis. This natural tendency can be reversed by applying an opposite pressure more than the osmotic pressure, to the salt solution so that water from the salt solution tends to flow out as pure water. This is the basis of reverse osmosis. Engineering Chemistry Notes, 2024-2025 Syllabus Page 9  Fig. Reverse Osmosis Process Method: It consists of several horizontal steel compartments separated by semi permeable membrane made out of Cellulose acetate or poly methyl methacrylate. A pressure of 15 to 40 kg cm-2 is applied to the brackish water or sea water to force its pure water to pass out through the semi permeable membrane in to the neighboring compartments leaving behind concentrated sea water. The pure water passed in to the different compartments are collected while the concentrated sea water left out is discarded. Advantages of the method: 1) This method can remove ionic, non-ionic, colloidal and high molecular weight impurities. 2) The life time of the membrane is very high- about 2 years 3) It has low capital cost 4) Membrane can be replaced easily 5) Maintenance is low. Engineering Chemistry Notes, 2024-2025 Syllabus Page 10