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So, 1 part per million ( I ppm) solution would be 1 kg of, solute per I million kilograms of solution., And these masses are just too large to be useful in, Chemistry laboratory., But we can divide the masses of solute and solution by J, million to arrive at more useful units:, 1 ppm, , =, , =, , =, , I kg solute, 1 000 000 kg solution, , I kg solute/I 000 000, I 000 000 kg solution/I 000 ()(X), , 1o-6 kg solute, I kg solution, , Recall that there are I 000 grams in a kilogram, so now we, can write, , ***, , 139

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Mathematically,, A{a.ss of sol.ute, , Mass per cent = - - - - - x 100, Mass of solution, , PPM:This is a way of expressing very dilute concentrations, of substances. Just as per cent means out of a hundred, so, parts per million or ppm means out of a million. Usually, describes the concentration of something in water or soil., One ppm is equivalent to I milligram of something per liter, of water (mg/1) or 1 milligram of something per kilogram, , soil (mg/kg) parts per million is abbreviated as ppm For, very dilute solutions, weight/weight (w/w) and, weight/volume (w/v) concentrations are sometimes, expressed in parts per million., NOTE: weight/weight (w/w) may also be referred to as, mass/mass (m/m), NOTE: weight/volume (w/v) may also be referred to as, mass/volume (m/v), I ppm is one part by weight, or volume, of solute in l, mi Ilion parts by weight, or volume, of solution., , In weight of solute per volume of solution (w/v or rn/v), terms,, , l ppm = lg m- 3 = l mg L-1 = 1 µg mL- 1, In weight of solute per weight of solution (w/w or m/m), , terms,, , 1 ppm= 1 mg kg-I = 1 µg g-1, in SI units, w/w concentration (m/m concentration) would, be given in kilograms of solute per kilograms of solution., , 138

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Molality: -, , " Molality is defined as the "total moles of a solute, contained in a kilogram of a solvent.", Molality is also known as molal concentration. It is a, measure of solute concentration in a solution. The solution, is composed of two components; solute and solvent. There, are many different ways to express the concentration of, solutions like molarity, molality, normality, formality,, volume percentage, weight percentage and part per million., The term needs to calculate the mass of the solvent and, moles of solute., Molality Formula, , M olalit y (M), , =, , M olalit y (M), , =, , Numbe r of mole.a of MJlute, M a&a of advent in kga, gx lOOO, , W xm, , Molarity: -, , It is one of the most widely used unit of concentration, and is denoted by M. It is defined as no. of moles of solute, present in 1 liter of solution. Thus,, , M l, O, , ., , =, No. of moles of solute, arity · Volum e of sol.utiun(in Litres ), , Mass Per Cent or weight percent (w/w %), , It is the ratio of the mass of solute to the mass of, solution multiplied by 100 to calculate mass percent. It is, also known as weight percent and is represented by (w/w, %). You may have seen this symbol on the back of, medic ines and tablets. It is one of the most comm only used, , units of representing concentration., , 137

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A secondary standard solution is a solution that is, made specifically for a certain analysis. A secondary, standard is a substance whose active agent contents have, been found by comparison against a primary standard. This, means it is usually standardized against a primary standard., Secondary standard solutions are used to calibrate, analytical equipment and analytical techniques., However, the purity of these solutions is less, compared to primary standards and the reactivity is high., Due to this high reactivity, these solutions get contaminated, easily. Some common examples are anhydrous sodium, hydroxide and potassium permanganate. These compounds, are hygroscopic., , Units: In analytical chemistry, a standard solution is a, , solution containing a precisely known concentration of an, , element or a substance. A known mass of solute is, dissolved to make a specific volume. It is prepared using a, standard substance, such as a primary standard. Standard, solutions are used to determine the concentrations of other, substances, such as solutions in titration., The concentrations of standard solutions are nonnally, expressed in units of moles per litre (mol/L, often, abbreviated to M for molarity), moles per cubic decimetre, (mol/dm3), kilomoles per cubic metre (kmol/m3) or in, terms related to those used in particular titrations (such as, titres A simple standard is obtained by the dilution of a, single element or a substance in a soluble solvent with, which it reacts., , 136

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compound to act like (proton) acceptor. The basicity of a, chemical compound is expressed by the acidity of the, , conjugate acid, Primary standard and secondary solution: Primary Standard Solution, Primary standard solutions the solutions made fonn, primary standard substances. These substances have a high, purity which nearly equals 99.9% purity. We can dissolve, this substance in a known volume of solvent in order to, obtain the primary standard solution. These solutions can, involve chemical reactions. Therefore, we can use this, , reagent to determine the unknown concentration of a, solution that undergoes a particular chemical reaction., These solutions have specific chemical and physical, properties. For example, these solutions have a high purity, and are highly stable. In titrations, we should standardize, all the solutions we use for the titration before doing the, experiment. This is because, even though we take the exact, amounts of substances to make those solutions, they may, not have the exact concentration that we expect because, , those substances are not that much pure. Some examples of, primary standards include potassium bromate (KBr03),, sodium chloride, zinc powder, etc., , Secondary Standard Solution?, Secondary standards solutions are the solutions made, from secondary standard substances. We prepare these, solutions for a specific analytical experiment. We should, determine the concentration of these solutions using, primary standards. Most of the times, these solutions are, , useful for the calibration of analytical instruments., , 135

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Where,, Ca is the analyte concentration, typically in molarity., Ct is the titrant concentration, typically in molarity., V is the volume of the titrant which is used, typically in, liters., , M is the mole ratio of the analyte and reactant from the, balanced equation., V is the volume of the analyte, typically in liters., , Many non-acid-base titrations are needed a constant pH, throughout the reaction. Therefore, a buffer solution can be, added to the titration chamber to maintain the pH value., , Requirements of titration:•, •, •, •, •, •, •, , Burette, white tile (used to see a colour change in the solution), pipette, Indicator (the type depends on the reactants), Erlenmeyer or conical flask, titrant (a standard solution of known concentration; a, common example is aqueous sodium carbonate), analyte, or titrant (the solution of unknown, concentration), , Definition of Equivalent weight :Equivalent weight (also known as gram equivalent) is, the mass of one equivalent, that is the mass of a given, substance which will combine with or displace a fixed, quantity of another substance., The equivalent weight of an element is the mass, which combines with or displaces 1.008 gram of hydrogen, or 8.0 grams of oxygen or 35.5 grams of chlorine. These, , 133

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8. If the reagent or reactant that we use is to be made into, a standard solution then we can weigh and dissolve the, reagent into a solution, so that it is in a definitive, volume within a volumetric flask., Basic principles of volumetric analysis, , The basic principle of the titration is the following: A, solution - a so called titrant or standard solution - is added, to sample to be analyzed. The titrant contains a known, concentration of a chemical which reacts with the substance, to be determined. The titrant is added by means of a, bu.rette., 1. The solution to be analyzed contains an unknown, , 2., , 3., 4., , 5., , amount of chemicals., The reagent of unknown concentration reacts with a, chemical of an unknown amount in the presence of an, indicator (mostly phenolphthalein) to show the endpoint. It's the point indicating the completion of the, reaction., The volumes are measured by titration which completes, the reaction between the solution and reagent., The volume and concentration of reagent which are, used in the titration show the amount of reagent and, solution., The amount of unknown chemical in the specific, volume of solution is determined by the mole fraction, of the equation., , When the endpoint of the reaction is reached, the volume of, reactant consumed is measured and applied to carry, volumetric analysis calculations of the analyte by the, following formula,, Ca= Ct Vt M I Va, 132

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In other words, measuring the volume of a second, substance that combines with the first in known proportions, is known as Volumetric analysis or titration. It is this, method of quantitative analysis that allows us to determine, the concentration of the analyte., , Procedure for Volumetric Analysis, I. A typical titration starts with a beaker or flask, containing a precise volume of the analyte and small, amount of indicator placed underneath a calibrated, burette or pipette containing the titrant., 2. The solution that needs to be analysed needs to have an, accurate weighed in the sample of +/- 0.0001g of the, material to be analysed., 3. Choosing the right kind of material to be analysed is, also very important, as choosing the wrong type of, titrant will give us the wrong results. A substance that, reacts rapidly and completely to produce a complete, solution is chosen., 4. Small quantities of titrant are added to the analyte and, indicator till the indicator changes colour in reaction to, the titrant saturation threshold reflects the arrival at the, endpoint of the titration., 5. The titration has to be continued up until the reaction is, complete and the amount of reactant added is exactly, the amount that is needed to complete the reaction., 6. Another important step is in measuring the right volume, of the standard solution since molarity is a standard, metric to calculate the number of moles present in a, solution., 7. Based on the desired endpoint, single drops or less than, a drop of the titrant makes a difference between a, permanent and temporary change in the indicator., 131

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Volume of FAS for back titration= 10.2 cm3 =A, Volume of Sample V = 50 cm3, , N = 0.1 N, , (B-A) = 31.1-10.2 = 20.9 cm3, COD = (B-A) X N X8000, , v., 20.9 X .01 X 8000/ 50 = 324.4 mgldm3, , 5. Make use of the following data, in a COD analysis an, effluent sample consumed 35ml and 19.5ml of Std., FAS, normality of FAS is 0.05N, the amount of sample, used for the titration is 25ml. Calculate the COD of the, effluent sample., Solution:, Given, B = 35, A= 19.5, N = 0.05 N, V = 25, COD= 8000(b-A) x N, , = 8000(35 - 19.5) x 0.05, , 25, , V, , COD =, , 248 mg of oxygen / dm3, , 6. In a COD test 28.1ml and 14ml of 0.05N FAS solution, were required for blank and back titration respectively., The volume of test sample used was 25ml, calculate the, COD of the sample solution., Given, B = 28.1ml, A= 14ml, z = 0.05 N, v = 25, COD= 8000(B-A) x z, = 8000(28.1 - 14) x 0.05, , 25, , V, , COD=, , 3, , 225.6 mg of oxygen / dm, , Volumetric Analysis, , Volumetric analysis 1s a quantitative analytical, method which is used widely. As the name suggests, this, method involves measurement of the volume of a solution, whose concentration is known and applied to determine the, concentration of the analyte., 130

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COD = 1760 mg of 02/I, , COD = 1760 mg of 02/1, 2. Calculate the COD of the effluent sample when 25cm3, of effluent sample requires 8.5 cm3 of 0.001N, K2Cr2O7 solution for complete oxidation, Solution: B-A = 8.5cm3, V = 25cm3, , COD of the sample=, , N{B-A) x 8000, V, , = 0.00 I X 8.5X8000 /25 = 2 .72 mg of Oz/I, , 3. Calculate COD of 25cc of an effluent sample that, requires 8.3cc of 0.00 I M K2CnO1 for the complete, oxjdation., Solution: B-A = 8.3cm3, V = 25cm3, COD of the sample =, , N{B-A) x 8000, , V, 0.001 X 8.3 X 8000/25 = 2.65 mg of Oi/1, , 4. 50cm3 of sewage water was refluxed with 20cm3 of, 0.1 N acidified K2Cr2O1, the unreacted acidified, K2Cr2O7 consumed 10.2 cm3 of0.1N FAS. 20cm 3 of, O. lN acidified K2Cr2O1 when treated under identical, conditions consumed 31.1 cm3 of 0.1 N FAS, Calculate, COD of waste water., Solution:, Given: Volume of FAS for blank titration = 31.1 cm3 =B, , 129

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Temporary Hardness = Total hardness - permanent, hardness = 1200 - 480, , = 720ppm, Prewt, , '"'"'"'", '"', tTJ 1T, I I I I I ~ · - - • I I 11 I, I I I I I, I I I I I I, , Biological Oxygen Demand, "BOD is defined as the amount of oxygen required for the, biological oxidation of organic matter under aerobic, conditions at 20oC for the period of five days"., , BOD= DOI - DOS, DO- Dissolved Oxygen, , Chemical Oxygen Demand, "COD is defined as the amount of oxygen required in, mg to oxidize both organic & inorganic wastes present in 1, litre of waste water using strong oxidizing agents such as, K2Cr2O7". This is expressed in terms of milligrams of, oxygen, Experiment: Back titration: Pipette out 25 ml (V) of, water sample and 25cm3 K2CriO7 solutions into 250 cm 3, conical flask. Add 20cm3 of I: I H2S04 containing silver, sulphate with constant stirring. Boil the mixture for half an, hour and cool. Add 2 drops of ferroin indicator and titrate, against std. Ferrous ammonium sulphate solution till the, solution turns from blue green to reddish brown. Note, , 127