1. 30. Water and Sewage treatment

2. Hardness in water Def: Water is said to be hard when it is difficult to form a lather with soap. It is caused by the presence of Ca +2 and Mg +2 ions being dissolved in water There are two different types of hardness in water: 1. Temporary hardness 2. Permanent hardness

3. 1. Temporary hardness Def: Temporary hardness is hardness in water that can be removed by boiling. Found in parts of the country where limestone (calcium carbonate) is found in the earth.

4. What causes temporary hardness Water (rain), carbon dioxide( in the air) and Calcium carbonate (limestone) all react together: Calcium hydrogen carbonate Ca(HCO 3 ) 2 is a made and dissolves in the water.

5. Removing temporary hardness from water 1. It can be removed by boiling! This reaction occurs: Ca(HCO 3 ) 2 + heat CaCO 3 + H 2 O + CO 2

6. Limescale inside a kettle

7. Removing temporary hardness from water 2. It can be removed using an ion exchanger - the hardness causing calcium and magnesium ions are swapped with sodium ions, which do not cause hardness.

9. Permanent hardness Permanent hardness is hardness in water that cannot be removed by boiling. It is caused by the presence of calcium sulfate CaSO 4 and magnesium sulfate MgSO 4 in water.

10. It can be removed by: using an ion exchanger Removing permanent hardness from water

11. Deionisation –

12. Deionisation In a deioniser any dissolved salts present ae replaced by water molecules.

13. Relative purity of deionised and distilled water.

14. Experiment – estimation of the total hardness of a water sample

15. 1.Find the concentration of the Ca +2 ions in the water using results from the titration: M1 X V1 = M2 X V2 N1 N2 2. Find concentration of CaCO 3 in the water in g/L and ppm Doing the calculations

16. Questions relating to the experiment 1.Why is it important that the reaction between the edta and the metal ions in solution (i) is rapid and (ii) goes to completion? 2.The water sample could contain metal ions other than Ca2+ and Mg2+. How would the reliability of the result be affected if this were the case? Suggest two other metal ions that could be present in the water. 3.This reagent cannot distinguish between temporary and permanent hardness. List the compounds of calcium and magnesium that cause hardness, and indicate those which cause temporary hardness. 4.Suggest a method of establishing the amount of permanent hardness in a water sample. 5.What is the function of the buffer solution?

17. Q243

18. Water treatment

19. Involves the stages: 1. screening 2.sedimentation, flocculation, 3.filtration, 4.chlorination, 5. fluoridation and 6.pH adjustment

21. 1. Screening Water is passes through screens to remove large floating solids.

22. 1. Flocculation Aluminium sulfate (a flocculating agent) is added to water to help suspended solids to clump together.

23. 2. Sedimentation The clumped suspended solids are allowed to settle to the bottom of tanks called sedimentation tanks.

25. Filtration The water is passed through beds of sand and gravel to remove any remaining suspended solids.

26. Chlorination Chlorine compounds are added to the water to destroy bacteria.

27. Fluoridation Fluorine compounds are added to the water supply to help prevent tooth decay

28. pH adjustment Bases(lime) or acids ( sulfuric acid) are added so that the pH is adjusted to an acceptable level – range of 4 - 6.

29. Suspended solids Insoluble particles that are very light and so float in the water. They make the water cloudy and interfere with plant photosynthesis Usually measured in ppm ( = mg in a litre)

30. Determination of total suspended solids (expressed as p.p.m.) in a water sample by filtration 1. Find the mass of a dry filter paper. 2. Filter 1 litre of the water sample. 3. Dry the filter paper at about 105 oC overnight. 4. Find the mass of the dried filter paper. 5. Calculate the mass of total suspended solids in mg/l (p.p.m.).

31. Mass of dry filter paper before use = Mass of the dried filter paper after filtration = Mass of the suspended solids = Volume of water sample = Total suspended solids in mg/l (p.p.m.) =

32. Determination of total dissolved solids (expressed as p.p.m.) in a water sample by evaporation 1.Find the mass of a clean dry 250cm3 beaker. 2.Add 100cm3 of a filtered water sample to the beaker. 3.Evaporate to dryness 4.Find the mass of the cool dry beaker. 5.Calculate the total dissolved solids in the water sample in mg/l (p.p.m.). =

34. pH meter Used to measure the pH of any solution. Common uses to measure the pH of rivers, drinking water supplies and swimming pool water

35. Calculations involving suspended solids The total suspended solids in 500cm 3 of water is 0.23g. Express the concentration in ppm. Total suspended solids = 0.23g in 500cm 3 = (0.23g X 2) IN 1000cm 3 = 0.46g in 1000cm 3 To convert grams to milligrams multiply by 1000! = 460mg in 1000cm 3 = 460ppm

36. Calculations involving suspended solids The total suspended solids in 100cm 3 of water is 0.05g. Express the concentration in ppm. Total suspended solids = 0.05g in 100cm 3 = (0.05g X10) IN 1000cm 3 = 0.50g in 1000cm 3 To convert grams to milligrams multiply by 1000! = 500mg in 1000cm 3 = 500ppm

37. Calculations involving suspended solids The total suspended solids in 200cm 3 of water is 0.07g. Express the concentration in ppm. Total suspended solids = 0.07g in 200cm 3 = (0.07g X5) IN 1000cm 3 = 0.35g in 1000cm 3 To convert grams to milligrams multiply by 1000! = 350mg in 1000cm 3 = 350ppm

38. Sewage treatment

39. 1. Primary treatment Sewage : involves screening and settlement Screening: sewage is passed through grids to remove solid material. Settlement: remaining sewage is then transferred to tanks where suspended solids are allowed to settle to the bottom.

40. 2. Secondary treatment The nutrients in the sewage is broken down through biological digestion by microorganisms. Microorganisms feed and reproduce in the waste and become known as activated sludge. This activated sludge will be removed or recycled into the first stage Solid settled waste will be compressed and can be spread out on land or dumped at sea

41. Tertiary treatment.It has a high cost so is not always used. Sewage is passed through sand filters and then is discharge to sea. Removes phosphates and nitrates from the sewage so helps to prevent eutrophication.

42. Eutrophication Caused when water becomes over enriched by nutrients such as phosphates and nitrates. Algae reproduce in large numbers Causes the death of many forms of animal life due to the lack of oxygen in the water. Can be prevented by the tertiary treatment of sewage. EU limits for species in water!!

43. Pollution by heavy metal ions in water especially Pb 2+, Hg 2+ and Cd 2+. Cause death to aquatic life, and are passed along the food chain. Known to cause brain damage in humans. Sources – car batteries, pesticides dumping EU limits for species in water!!

44. Removal by precipitation Lime Ca (OH) 2 is added to the water. The heavy metal ions form metal hydroxide which precipitates out the water and settles to the bottom. Levels can be checked using atomic absorption spectroscopy

45. Atomic Absorption Spectroscopy Every element shows a characteristic emission of a certain wavelength of light.. Due to different electronic configurations Emission of light can be measured with an atomic absorption spectroscope and can identify the presence of an element in a water sample examples Cadmium. Lead and mercury

46. Organic mattere.g. sewage, industrial waste, silage, milk. discharged into a water acts as a food source for the bacteria present there. The bacteria will multiply and use up the available dissolved oxygen This may cause fish kills. bacteria will produce hydrogen sulphide and ammonia( anaerobic conditions) Biochemical Oxygen Demand

47.  The level of dissolved oxygen in a water sample is an indicator of the quality of the sample. Biochemical oxygen demand test  The amount of dissolved oxygen used up by biochemical action when a sample of water is kept in the dark at 20 o C for 5 days. Note – Compare before and after readings! Sample is kept in the dark to prevent new oxygen being produced by photosynthesis

48. It is not possible to directly measure the amount of dissolved oxygen in a water sample directly. The dissolved oxygen does not directly react with another suitable reagent, an indirect procedure was developed by Winkler. An iodine/thiosulfate titration can be used to measure the dissolved oxygen present in a water sample. The Winkler method

49. Manganese sulfate in alkaline conditions 1. Mn 2+ (aq) + 2OH - (aq)  Mn(OH) 2(s) ( white precipitate) Mix with sample under water - This reacts with the dissolved oxygen to produce a brown precipitate. 2. 4Mn(OH) 2(s) + H 2 0 + O 2 (aq)  4Mn(OH) 3(s) Adding concentrated H 2 SO 4 - enables the Mn(IV) compound to release free iodine from KI. 3. 2Mn(OH) 3(s) + 6H + (aq) + 2I - (aq)  2Mn 2+ (aq) + I 2(aq) + 6H 2 O (l) Titration : The free iodine is then titrated with standard sodium thiosulfate 4. I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

50. Standard solution = Sodium thiosulfate Iodine solution = unknown concentration Colour change = Red/ Brown – Pale yellow but hard to see the end point Add starch indicator near the end point to see a clear colour change. Blue /Black ( iodine present) – Colourless ( iodine absent

51. 1.Find the concentration of the iodine solution using results from the titration: M1 X V1 = M2 X V2 N1 N2 2. Using the known ratios of how the dissolved oxygen reacted to make the iodine solution you can work out the molarity of the oxygen in the solution 3. Use the molarity of the oxygen in the solution to find g/L and ppm Doing the calculations

52. 1.Why is the reagent MnSO 4 used? 2.Why is concentrated H 2 SO 4 used? 3.Why must the bottles be shaken vigorously after adding the Manganese sulfate and alkaline potassium iodide? 4.Why are the bottles completely filled with water? 5.If the white precipitate remains on addition of manganese(II) sulfate solution and alkaline potassium iodide solution, what does this indicate about the water sample? 6.State and explain what the letters B.O.D. mean. 7.Why are the bottles used during B.O.D. measurements stored in the dark? Questions relating to the experiment

53. Question 233g V 1 X M 1 = V 2 x M 2 n 1 n 2 (50)X (M1) = (6) x (0.01) 1 2 M 1 = (6) x (0.01) x (1) (2) x (50) M 1 = 0.0006 The concentration of the iodine solution is 0.006M (moles per litre) I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

54. Ratios of reactions making I 2 1. Mn 2+ (aq) + 2OH - (aq)  Mn(OH) 2(s) ( white precipitate) 2. 4Mn(OH) 2(s) + H 2 0 + O 2 (aq)  4Mn(OH) 3(s). 3. 2Mn(OH) 3(s) + 6H + (aq) + 2I - (aq)  2Mn 2+ (aq) + I 2(aq) + 6H 2 O (l) Overall ratio is 1O 2 2I 2 M 1 = 0.0006 M 1 = 0.0006 /2=.0003

55. Ppm = parts per million ( mg in 1000cm 3 ) f) i)Concentration of O 2 in g/L in water = 0.0003 X 32 = 0.0096g/L ii) Concentation of of O 2 in mg/L in water (ppm).0096 x 1000 = 9.6ppm e)f) Finding the concentration of oxygen in ppm

56. Question 235 – Sample 1 V 1 X M 1 = V 2 x M 2 n 1 n 2 (200)X (M1) = (27) x (0.01) 1 2 M 1 = (27) x (0.01) x (1) (2) x (200) M 1 = 0.000675 The concentration of the iodine solution is 0.000675M (moles per litre) I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

57. Ratios of reactions making I 2 1. Mn 2+ (aq) + 2OH - (aq)  Mn(OH) 2(s) ( white precipitate) 2. 4Mn(OH) 2(s) + H 2 0 + O 2 (aq)  4Mn(OH) 3(s). 3. 2Mn(OH) 3(s) + 6H + (aq) + 2I - (aq)  2Mn 2+ (aq) + I 2(aq) + 6H 2 O (l) Overall ratio is 1O 2 2I 2 M 1 = 0.000675 M 1 = 0.000675 /2=.0003375

58. Ppm = parts per million ( mg in 1000cm 3 ) f) i)Concentration of O 2 in g/L in water tested = 0.0003375 X 32 = 0.0108g/L ii) Concentation of of O 2 in mg/L in water tested(ppm).0108 x 1000 = 10.8ppm Water tested was diluted down ten times so original sample had 10.8 x 10 = 108ppm e)f) Finding the concentration of oxygen in ppm

59. Question 235 – Sample 2 V 1 X M 1 = V 2 x M 2 n 1 n 2 (200)X (M1) = (4.8) x (0.01) 1 2 M 1 = (4.8) x (0.01) x (1) (2) x (200) M 1 = 0.00012 The concentration of the iodine solution is 0.00012M (moles per litre) I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

60. Ratios of reactions making I 2 1. Mn 2+ (aq) + 2OH - (aq)  Mn(OH) 2(s) ( white precipitate) 2. 4Mn(OH) 2(s) + H 2 0 + O 2 (aq)  4Mn(OH) 3(s). 3. 2Mn(OH) 3(s) + 6H + (aq) + 2I - (aq)  2Mn 2+ (aq) + I 2(aq) + 6H 2 O (l) Overall ratio is 1O 2 2I 2 M 1 = 0.00012 M 1 = 0.00012 /2=.00006

61. Ppm = parts per million ( mg in 1000cm 3 ) f) i)Concentration of O 2 in g/L in water tested = 0.00006 X 32 = 0.00192g/L ii) Concentation of of O 2 in mg/L in water tested(ppm).00192x 1000 = 1.92ppm Water tested was diluted down ten times so original sample had 1.92 x 10 = 19.2 ppms e)f) Finding the concentration of oxygen in ppm

62. BOD of the water 1 st reading – 2 nd reading 108ppm – 19.2ppm = 88.8ppm

63. Water analysis

64. As the concentration of a coloured solute in a solution increases so does the intensity of the colour! We say they are directly proportional to each other, (as one goes up so does the other!)

65. A colorimeter measures the amount of light that can pass through a solution, and uses this to calculate the concentration of the solution. Is used by scientists to measure the amount of lead in water and fertilisers

66. Mandatory experiment Determination of the amount of chlorine in swimming pool water using a colorimeter

67. Free chlorine in a solution In water chlorine comes from two sources Chloric acid HOCl and Chlorate ion – OCl The chlorine present in water does not give a strong enough colour intensity to be useful for analysing its concentration

68. Adding colour to the solution Cl 2 + 2KI I 2 + 2KCl Iodine produced has a strong colour and its colour can be used to determine its concentration Once the concentration of iodine is known, the concentration of Chlorine can be determined too!!! Acidic conditions are required for this reaction!!!

69. Using a colorimeter Standard solutions must be prepared and a calorimeter is used to measure their absorbance of light The absorbance of an unknown sample unknown concentration of chlorine can then be measured using a calorimeter and its concentration can be worked out.

70. Specimen results and calculations Absorbance of solution in flask A = 0.00 Absorbance of solution in flask B = 0.06 Absorbance of solution in flask C = 0.12 Absorbance of solution in flask D = 0.24 Absorbance of solution in flask E = 0.43

72. Anions in Water Chemists do tests for the presence of anions found in water such as: chloride, phosphate, nitrate, sulfate and hydrogen carbonate Sources of these anions come from sewage, industrial and farm effluent