Water and its Technology PDF

Water and its Technology  Water – a unique resource Liquid over wide temperature – 0 – 100oC High specific heat – cools down and heat up slowly High latent heat of vaporization- cooling effect during evaporation Excellent solvent – carrier of nutrients and pathogens as w...

Water and its Technology  Water – a unique resource Liquid over wide temperature – 0 – 100oC High specific heat – cools down and heat up slowly High latent heat of vaporization- cooling effect during evaporation Excellent solvent – carrier of nutrients and pathogens as well High surface tension – rise through heights Anomalous expansion behavior – expands when cooling – ice floats  Sources of water o Rains, Rivers, Seas, glaciers, Springs, Lakes etc. o Rain water is the purest form of water. o Water has different physical, chemical and biological impurities which can cause problems in both domestic and industrial areas  Impurities in water o Physical (insoluble) - Inorganic such as clay, sand - Organic such as oil globules, vegetable/animal matter - Colloidal such as Fe(OH)3, Complex proteins, amines o Chemical (soluble) - Anions such as Cl-, SO42-, CO32-,HCO3-, NO3- - Cations such as Ca2+, Mg2+, Na+, K+, Fe3+, Al3+ - Dissolved gases such as O2, N2, CO2, H2S, NH3 o Biological - Microorganisms such as algae, fungi, bacteria (Pathogenic causing Malaria, diarrhea, typhoid etc.)  Hardness of water o Hardness of water is due to dissolved salts of mainly calcium and magnesium as well as iron and other heavy metals. o Hardness is two types: a) Temporary : - Due to dissolved bicarbonates of calcium and magnesium and carbonates of iron and other heavy metals. - Can be easily removed by boiling where CO2 gas gets expelled removing the hardness. b) Permanent: - Due to dissolved chlorides and sulphates of calcium and magnesium. - Can be removed through zeolite, Lime-soda, ion-exchange processes.  Effect of hardness of water o Water hardness can be identified when soap does not form lather. o Problems of using hard water: a) Domestic problems: - wastage of fuel & time - improper cleaning (wastage of soap) - health related issues ( Urinary infections, kidney stones, dryness and hardening of skin) b) Industrial problems: - Boiler troubles (scale, sludge, caustic embrittlement, priming and foaming) - wastage of fuel - process related problems - problems in textile, sugar, paper, laundry, pharma industries  Unit of hardness o Hardness of water is measured in parts per millions (ppm) as calcium carbonate equivalents. o Reasons for expressing hardness in CaCO3 equivalents: - its molecular weight is 100 ; equivalent weight is 50 - it is the most insoluble impurity found in water commonly. o Units of hardness: - parts per million in CaCO3 equivalents (1 mg/L is 1ppm) - if 146 mg/L of MgSO4 is present in water, the hardness of water is 146 ppm. as MgSO4 o When expressed in CaCO3 equivalents, the formula for conversion is: Mass of hardness causing substance (mg/L) X 50 (Eq. wt of CaCO3) Chemical Eq of hardness causing substance Expressing water hardness in different units 1. ppm: 1 part of CaCO3 equiv. in 106 parts of water (1 mg/L). 2. Clark’s degree: 1 part of CaCO3 equiv. in 70,000 parts of water. 3. French degree: 1 part of CaCO3 equiv. in 105 parts of water. * Conversion: 1ppm = 1 mg/L = 0.1of = 0.07oCl  Example of hardness calculation Numerical: A sample hard water drawn from a jute mill near Kolkata contains, 8.1 mg/L Ca(HCO3)2 ; 7.5 mg/L Mg(HCO3)2; 13.6 mg/L CaSO4; 12.0 mg/L MgSO4 , 2.0 mg/L MgCl2 and 110 ppm NaCl. Calculate temp., perm. and total hardness in Clo and oFr. Constituent Multiplication CaCO3 equivalents factor Ca(HCO3)2= 8.1 mg/L 100/162 8.1 x 100/162 = 5.0 mg/L Mg(HCO3)2 = 7.5 mg/L 100/146 7.5 x 100/146 = 5.14 mg/L CaSO4= 13.6 mg/L 100/136 13.6 x 100/136 = 5.0 mg/L MgSO4= 12.0 mg/L 100/120 12.0 x 100/120 =10.0 mg/L MgCl2 = 2.0 mg/L 100/95 2.0 x 100/95 = 2.11 mg/L  How to measure hardness? Determination of Ca2+ and Mg2+ in water - Complexometric titration using EDTA and Eriochrome Black T (EBT) – common method - Estimation of Ca and Mg using flame photometry  EDTA EDTA metal complex EBT (Chelate complex) Water sample will be titrated with standard EDTA solution. The volume of EDTA required for complexing the free metal ions in the solution will be found out. The Volume of EDTA will be normalized to CaCO3 equivalent by titrating it with std. CaCO3 (CaCl2) solution. So from the titre value, hardness can be calculated. The indicator EBT-Ca2+ complex is wine red colour and it change to steel blue at the end point. EDTA is stronger chelating agent than EBT so it snatches ca ions from it and free EBT is liberated so steel blue colour appears. Procedure:  First EDTA Solution is standardized using standard hard water (1 mg/ml of CaCO3 equivalents is prepared as standard hard water).  For this, first known aliquot of Standard hard water is taken and 10-15 mL of ammonia buffer is added to bring the pH between 9-10.  Then a few drops of EBT solution is added to form the unstable complex giving wine red colour.  This solution is titrated with the EDTA solution till the solution turns to steel blue indicating the formation of stable EDTA- Metal ion complex.  This volume of EDTA is noted as V1.  The above procedure is repeated with sample hard water of unknown hardness.  Volume of EDTA is noted as V2. o Then sample hard water of 100 mL is taken and boiled for the temporary hard salts to settle down. o The solution is filtered and washed thoroughly and made up again to 250mL. o From this solution, 20 mL is pipetted out and titrated in similar manner as done with standard hard water. o Volume of EDTA is noted as V3. Calculations: a) Total hardness: V1mL of EDTA is consumed by 20 mL of std. hard water V1mL of EDTA = 20 mg of CaCO3 1 mL of EDTA = 20/V1 mg of CaCO3 EDTA consumed by sample hard water = V2 mL So, V2 mL of EDTA = 20/V1 x V2 mg of CaCO3 Hence, 20 mL of sample hard water contains 20/V1 x V2 mg of CaCO3 Therefore, 1000mL of sample hard water = 20/V1 x V2/20 x 1000 mg/L i.e. Total hardness of sample hard water = V2/V1 x 1000 mg of CaCO3 (ppm.) Permanent hardness: 20 mL of sample hard water after removing temporary hardness consumed V3 mL of EDTA. 1 mL of EDTA = 20/V1 mg of CaCO3 Therefore, V3 mL of EDTA = 20/V1 x V3 mg of CaCO3 20 mL of sample hardwater after boiling contained 20/V1 x V3 mg of CaCO3 20 V3 Therefore, 1000 mL of sample hard water contains x X 1000mg/L V1 20 V3 Permanent hardness = X 1000 mg/L of CaCO3 (ppm.) V1 Temporary hardness: Temporary hardness = Total hardness – permanent hardness { V2 V1 X 1000 { { V3 V1 X 1000 { = 1000 X { V2 V1 [[[[ { { V2 V1 [[[[ { ppm. { V2 – V3 { = 1000 X ppm. V1  Numerical Problem on EDTA method 0.5 g (or 500 mg) of CaCO3 was dissolved in HCl and the solution was made up to 500 ml with distilled water to make a std. hard water. 50 ml of the solution required 48 ml of EDTA for standardization titration. 50 ml of hard water sample required 15 ml of EDTA and after boiling and filtering, required 10 ml of EDTA solution. Calculate the hardness.  Dissolved Oxygen (DO) Dissolved oxygen is nothing but oxygen dissolved in water It is 48.89 mL/litre at 32oF (0oC) and atmospheric pressure. The amount of oxygen dissolved in water depends on physical, chemical and biological activities taking place in water According to Henry’s Law, the solubility of a gas is directly proportional to the absolute pressure. It means, if pressure is increased, there is increase in dissolved oxygen at given temperature DO level decreases when Temp. increases Non-bonded oxygen molecules in water  DO Vs altitude and temp. At 25oC and 1 atm pressure DO will be  8.2 ppm  Importance of DO It is needed for living organisms to maintain their biological activities D.O. is also important in precipitation and dissolution of inorganic and organic substances in water It is necessary for degradation of pollutants It causes corrosion in boilers  DO and waste water treatment Sewage is the liquid waste which includes domestic wastes, industrial wastes, street washings, storm waters and other wastes It consists of about 99.9%water and 0.1% of organic and inorganic matters in dissolved, suspended and colloidal states. Sewage contains both aerobic and anaerobic bacteria from natural sources or synthetic which is the result of waste water discharge. DO is necessary for aerobic bacteria to bring about oxidation of organic compounds present in it.  Determination of DO Dissolved oxygen is usually determined by Winkler’s method It is based on the fact that dissolved oxygen oxidized potassium iodide (KI) to iodine The liberated iodine is titrated against standard sodium thiosulphate solution using starch indicator Since dissolved oxygen in water is in molecular state. It as such cannot oxidize KI. Hence Manganese Hydroxide is used as an oxygen carrier to bring about the reaction between KI and Oxygen. Manganese hydroxide, in turn, is obtained by the action of NaOH on MnSO4 Winkler’s method  Biochemical Oxygen Demand (BOD) BOD is a quantitative measure of oxygen required for the biological oxidation of the organic matter in a polluted water sample under aerobic conditions during five days incubation period at 20oC BOD value indicates the amount of decomposable organic matter in the sewage Higher BOD will consume more DO Source of effluent BOD (ppm) Domestic sewage 320 Cow shed sewage 3010 Paper mills 8190 Tannery effluents 12360 BOD is an important parameter to design effluent treatment  How BOD is measured? This analysis is performed using 300 ml incubation bottles (amber bottles) in which buffered dilution water is dosed with seed microorganisms and stored for 5 days in the dark room at 20 °C to prevent DO production via photosynthesis The initial and final DO levels are measured to find out BOD What is the desired BOD value for potable water?  Limitations of BOD measurement BOD values of effluents of rayon, paper and pulp industries and chemical industries are much less They contain enough organic matter in the form of cellulose and other non-degradable organic matter which limit BOD measurements COD values should be considered in these cases  Chemical Oxygen Demand (COD) Measurement of the oxygen equivalent of any oxidisable impurity present in a water sample that is susceptible to oxidation by strong chemical oxidants COD value of a sample is generally greater than its BOD value (normally 1.5 to 2.5 times) COD determination takes about 3 hours compared to 5 days for BOD determination  Importance of COD In such industries where effluents are cellulose or other non-degradable organic matter, COD value reveals the real pollution potential. It is not affected by the presence of toxins and other unfavorable conditions for the growth of microorganisms. It is very important parameter in designing and management of the treatment plants. COD test does not differentiate between bio inert and bio degradable materials.  COD measurement An excess amount of potassium dichromate (or any oxidizing agent) is added to water sample along with catalyst (Ag+- AgSO4) and chlorine scavenger (mercuric sulphate) The contents are boiled to facilitate oxidation and then the samples are cooled and the oxidizing agent consumed is back titrated generally using ferrous ammonium sulphate solution with Ferroin indicator (Titre value V2) A blank (distilled water ) is also titrated in the same way (Titre value V1). The difference in titre value is used for calculations. (V1 – V2) X N1 X 8 X 1000 COD = ------------------------------ ppm Volume of water sample N1- normality of FAS  pH pH of water sample depends on the number of free hydrogen ions. pH = - log [H+] A solution is more acidic when it contains more hydrogen ions. The level of acidity is important to the plant and animal life. Most animals are adapted to live in neutral conditions pH of water in the range 8.5 is considered as harmful Highly acidic water causes corrosion to pipes and highly basic water causes staining pH is measured using a pH meter  Total Dissolved Solids (TDS) The source of dissolved solids is various kinds of minerals present in water. Simply it can be determined taking a known amount (say 100 mL) of water and by evaporating the contents carefully to dryness. The residue (W gm) left after evaporation of the filtered sample shows the total dissolved solids present in that particular water sample. T.D.S. must be less than 500 ppm in potable water Higher TDS also indicates possibility of higher hardness TDS = (W/100) x 106 mg/L or ppm  TDS measurement TDS is commonly determined by gravimetry, chemical analysis, or conductivity. The gravimetric protocol is given in the previous slide The chemical analysis protocol requires that the sample be measured for major ions (such as sodium, potassium, calcium, magnesium, chloride, sulfate, phosphate, and fluoride) and other parameters, such as nitrate and alkalinity. The results are used to calculate TDS. The conductivity protocol requires only a conductivity measurement to be made. This measurement is multiplied by a factor (previously determined) to yield an estimate of TDS Measure the conductivity of a representative number of your typical sample. Then perform the gravimetric test on the same samples. Calculate the TDS factor as: the TDS in mg/L (gravimetric results) divided by the conductivity in uS/cm. TDS measurement -conductivity A TDS Meter indicates the Total Dissolved Solids (TDS) of a solution, i.e. the concentration of dissolved solids present in it. Since dissolved and ionized solids such as salts as well as minerals increase the conductivity of a solids, measurement of the conductivity of the solution gives a direct indication of TDS present. A TDS meter typically displays the TDS in parts per million (ppm). For example, a TDS reading of 1 ppm would indicate presence of 1 milligram of dissolved solids in each litre of water  Alkalinity Alkalinity of water is due do dissolved anions like HCO3-, CO32- and OH- Alkalinity can be determined by titrating water with standard solution of HCl or H2SO4 OH- and HCO3- ions cannot coexist because they react immediately to form CO32- and water To estimate the amount, nature of ions and types of alkalinity present in a given sample of water Alkalinity of water is attributed to the presence of the (i) Caustic alkalinity (due to OH- and CO32-) and (ii) Temporary hardness (due to HCO32-) Alkalinity can be estimated by titrating the given sample of water against standard acid (H2SO4) using phenolphthalein, P (pH = 10-8.3, end point is the just disappearance of pink color) and methyl orange, M (pH = 3.1 - 4.4, end point is the color change from pale yellow to orange) indicators. The determination of alkalinity is based on the following reactions. (i) [OH-] + [H+] ----> H2O P (ii) [CO32-] + [H+] ----> [HCO3-] M (iii) [HCO3-] + [H+] ----> H2O + CO2 Titration – I: Standardization of sulphuric acid: 0.1 N Na 2CO3 Vs 0.1 N H2SO4 Titration – II: Estimation of Alkalinity: Phenolphthalein alkalinity, P (mg/L) as CaCO3 equivalents 20 ml of given water sample = [p] ml of H2SO4 (V1) x Strength of H2SO4 (N2) x 50 1000 ml of given water sample = V1 x N2 x 50 x 1000/20 mg/L = ……………. [P] ppm Methyl orange alkalinity, M (mg/L) as CaCO3 equivalents 20 ml of given water sample = [m] ml of H2SO4 (V2) x Strength of H2SO4 (N2) x 50 1000 ml of given water sample = V2 x N2 x 50 x 1000/20 mg/L = ……………. [M] ppm Alkalinity OH- (ppm) CO32- (ppm) HCO3- (ppm) p=0 0 0 [M] p=m [M] 0 0 Alkalinity Table p = ½m 0 2[P] 0 p > ½m 2[P]-[M] 2[M] - 2[P] 0 p < ½m 0 2[P] [M] – 2[P]  Alkalinity –numerical problem 100 ml of water sample require 5 ml of N/50 HCl for neutralization in the presence of phenolphthalein. Another 17 ml of the same acid was needed for further titration to methyl orange end-point. Determine the type and amount of alkalinity.