Water Quality Parameters - PDF

2.0 WATER QUALITY PARAMETERS To illustrate the quality of a tested water QUALITATIVELY and QUANTITATIVELY. Can be divided into three types: Physical Chemical Biological 2  Water Quali...

2.0 WATER QUALITY PARAMETERS To illustrate the quality of a tested water QUALITATIVELY and QUANTITATIVELY. Can be divided into three types: Physical Chemical Biological 2  Water Quality Parameters Physical Chemical Biological Suspended Odour Total Alkalinity Pathogen Indicator Solid & Taste Dissolved Solid Organisms Organisms Temperature Turbidity Hardness Organic Compounds Inorganic Nutrients Compounds Nitrogen Phosphorus 3 2.1 Physical Parameters a. Turbidity - measured in NTU/FTU Source(s): i.Inorganic compounds such as clay, sand ii.Organic compounds such as plant fibre, human waste Effect(s): i. Aesthetic ii. Adsorption point/centre for chemicals and micro-organisms iii. Health aspect b. Odour and Taste Source(s): i. Inorganic compounds such as minerals, metals, salts (all of them give taste to water but no odour) ii. Organic compounds from petroleum and/or degradation of organic matters. (odour and taste) Effect(s): i. Aesthetic ii. Health problems [reaction from sources and other chemicals such as chlorine (Cl2)] c. Temperature - measured in oC or oF Source(s): i. Effect from ambience ii. Industrial activities - cooling system Effect(s): i. Disturb biological activities such as micro-organism and aquatic life ii. Chemical properties such as the degree of gas solubility, density and viscosity 4  d. Suspended solid - measured in mg/L Source(s): i. Same as in (a) Effect(s): i. Same as in (a) 2.2 Chemical Parameters a. Total dissolved solid (TDS) - Solid left in water after the water is filtered and dried. Source(s): i. Inorganic compounds - minerals, metals & gases ii. Organic compounds – product from degradation of organic matters, organic gas Effect(s): i. Cause taste, colour and odour problems ii. Health aspect iii. Small amount of TDS – water becomes corrosive to attain equilibrium Measured in either mg/L (organic or inorganic) or mS/m (millisiemens per meter) – measuring unit for conductivity (the potential of water allowing electric current to flow in it) b. Organic compounds Definition All organic compounds contain carbon in combination with one or more elements. Source(s): i. Nature: fibres, vegetable oils, animal oils and fats, cellulose, starch, sugar. 5  ii. Synthesis: a wide variety of compounds and materials prepared by manufacturing processes. E.g. DDT, polyvinylchloride. iii. Fermentation: Alcohols, acetone, glycerol, antibiotics, acids. Effect(s): i. Depletion of the dissolved oxygen in the water - Destroying aquatic life - Damaging the ecosystem ii. Some organics can caused cancer - Trihalomethane (THM-carcinogenic compound) are produced in water and wastewater treatment plants when natural organic compounds combine with chlorine added for disinfection purposes. c. Inorganic compounds Definition When placed in water, inorganic compounds dissociate into electrically charged atoms referred to as ions. All atoms linked in ionic bond. Can be classified into two: i. Metal – Non toxic and toxic Non-toxic – Ca2+, Mn2+, Na+, Fe2+, Mg2+, Al3+, Cu2+, Zn2+ dangerous for health if the concentration is high Source(s): i. Mineral, readily available from nature Effect(s): i. Colour, odour, taste and turbidity ii. Deteriorate health (at high concentration) Toxic – As2+, Ba2+, Cd2+, Cr2+, Pb2+, Hg2+ Stored up in food chain 6  Source(s): i. Human activities such as mining and industries Effect(s): Dangerous diseases such as cancer, abortion and deformation in new born baby ii. Non-metal – e.g. Si4+, Cl-, NO3- Source(s): i. Mineral Effect(s): i. Diseases - heavy metal, NO2- → “blue baby syndrome” ii. Aesthetic - Si4+ → turbidity iii. Fluoride (F-) a. Not good for health if it is taken in high concentration b. Concentration of 1 mg/L is good for the growth of children teeth c. Excessive concentration – colour on teeth and problem in bone growth d. Nutrients - Crucial elements needed by animals and plants to live - Important elements – C, N, P - C – easily obtained from CO2, degradation of organic compounds - N, P – limiting factors Nitrogen (N) Source(s): i. Element for protein, chlorophyll and biological compounds ii. Decomposition to a simple compound Protein → Amino Acid → NH3 ⎯+⎯→ ⎯ O2 NO2- ⎯+⎯→⎯ O2 NO3- iii. Animals and human wastes; chemicals (fertilisers) 7  Effect(s): i. NO3- poisoning in human and animal babies (human babies below than 6 month old) “blue baby syndrome” NO3- NO2- (in acidic condition) – will substitute O2 in blood vessel Babies will breathe less oxygen and eventually die ii. Excessive algae breeding and aquatic plants Phosphorus (P) Exists in a form of “orthophosphate”, “condensed phosphate”, and “organic phosphate”. Source(s): i. Readily present in soil ii. Fertilisers iii. Human wastes (“organic phosphate”) iv.Domestic wastes (element in detergent) Effect(s): i.Algae breeding and aquatic plants ii.>0.2 mg/L – disturb coagulation process in water treatment plant e. Alkalinity Definition The quantity of ions in water to neutralise acid or a measure of water strength to neutralise acid. Main constituents are bicarbonate (HCO3-), carbonate (CO32-), and hydroxide (OH-) ions. Source(s): i. Mineral dissolved in water and air. ii. Human activities such as detergent (in wastewater), fertilisers, pesticide etc. Effect(s): i. Non pleasant taste ii. Reaction between alkaline constituent and cation (positive ion) produces precipitation in pipe. 8 f. Hardness Definition A measure of “multivalent” cations in water such as Ca 2+, Mg2+, Fe2+, Mn3+. Ca2+ and Mg2+ are very important Source(s): i. Natural mineral on earth Effect(s): i. Excessive soap usage (a waste!!) Na2CO2C17H33 Cation2+(CO2C17H33) + Cation2+ 2Na+ ii. Precipitate form on hardware iii. Precipitate in pipe - temperature and pH increased Two kinds of hardness: (1) Carbonate hardness (2) Non-carbonate hardness Carbonate Hardness Non-Carbonate Hardness 1. Temporary in character 1. Permanent in character e.g. Ca(HCO3)2, Mg(HCO3)2 e.g. CaCl2, CaSO4, MgCl2, MgSO4 2. Precipitated through boiling 2. Eliminated through chemical softness method and/or ion exchange 9 Measurement of alkalinity and hardness A. mg/L Unit mg  substance L  medium mg In water, = ppm L 1 part 1 mg s ubs tance ppm = = 10 6 part 10 6 mg water   1000 kg 103 g We know that water density is m3 = L = 10 6 mg L ¨10 6 mg of w ater = 1 L of w ater 1 mg substanc e 1 ppm = 1L w ater 10 B. meq/L Unit For measurement that involve combination of several different elements 𝒎𝒈 𝟏 = 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 ( ) × 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑴𝒂𝒔𝒔 𝑳 where 𝑎𝑡𝑜𝑚𝑖𝑐 𝑜𝑟 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 Equivalent Mass = 𝑣𝑎𝑙𝑒𝑛𝑐𝑦 Example 1 : Equivalent Mass for Ca2+ = ???? Atomic mass for Ca2+ = 40 g Valency = 2 40  Equivalent Mass for Ca2+ = = 20 2 C. Hardness & Alkalinity Measurement Unit (mg/L as CaCO3) Equivalent mass for CaCO3= Equivalent Mass of Ca2+ + Equivalent Mass of CO32- 60 Equivalent mass of CO32- = = 30 2 Equivalent mass for CaCO3 = 20 + 30 = 50 11 Example 2: Results from water analysis is as shown below: Calcium = 29.0 mg/L; Magnesium = 16.4 mg/L; Sodium = 23.0 mg/L; Potassium = 17.5 mg/L; Bicarbonate (as HCO3-) = 171.0 mg/L; Sulphate = 36.0 mg/L; Chloride = 24.0 mg/L. (1) Convert these concentrations from mg/L to meq/L, (2) List down the hypothetical combination, (3) Calculate the water hardness in term of mg/L as CaCO3. Solution: (1) Component mg/L Equivalent Mass meq/L Ca2+ 29.0 20.0 1.45 Mg2+ 16.4 12.2 1.34 Na+ 23.0 23.0 1.00 K+ 17.5 39.1 0.45 Total cations 4.24 HCO3- 171.0 61.0 2.81 SO42- 36.0 48.0 0.75 Cl- 24.0 35.5 0.68 Total anions 4.24 12 (2) Hypothetical Combination: Ca(HCO3)2 - 1.45 meq/L Mg(HCO3)2 - 1.34 meq/L NaHCO3 - 0.02 meq/L Na2SO4 - 0.75 meq/L NaCl - 0.23 meq/L KCl - 0.45 meq/L (3) Hardness (Ca2+ + Mg2+) = 2.79 meq/L Equivalent Mass for CaCO3 = 50.0  Hardness = 2.79 x 50.0 = 140 mg/L CaCO3 13 The Periodic Table of the Elements 14 2.3 Biological Parameters Micro-organisms that bring diseases are called “PATHOGEN”. Their quantities are very small compared to other micro-organisms. The experiment to determine the presence of all pathogens takes a long time and very expensive. It is only carried out for very specific cases. The presence of pathogenic micro-organisms is shown by indicator micro- organisms. Pathogen Organisms o Live and breed in host and disseminated through faeces. o Small in quantity o Their presence is hard to detect Indicator Organisms Their presence shows that pollution has occurred and suggests the TYPE and LEVEL of pollution. Indicator organisms properties: o Can be used for all type of waters o Always present when pathogen is present o Always absent when pathogen is absent o Easily experimented and give reliable results o Not pathogen micro-organisms Typical indicators used are coliform groups Coliform groups: - o Fecal coliforms e.g. E.Coli o Total coliforms e.g. Fecal Coliforms, Soil Coliforms & any others Determination experimental Methods: 1. Membrane Filtration Method 2. Most Probable Number (MPN) 15 Membrane Filtration Method - to determine the number of coliform organisms that are present in water - advantage: faster than MPN procedure and gives a direct count of the number - can be determined by passing a known volume of water sample through a membrane filter that has a very small pore size. The bacteria are retained on the filter then contacted with an agar that contains nutrients necessary for the growth of the bacteria. After incubation, the coliform colonies can be counted and the concentration in the original water sample determined. - General formula colonies 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑙𝑜𝑛𝑖𝑒𝑠 𝑐𝑜𝑢𝑛𝑡𝑒𝑑 ×100 = 100mL 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 Example 1 Assume that filtration of volumes 75, 25, 10, 3 and 1 mL produced FC colony counts of 210, 89, 35, 11 and 5, respectively. What is the FC density for the sample? Solution Select the membrane filter (MF) with the number of colonies in the acceptable range. The acceptable range for FC is 20-60. The MF with 35 FC colonies is selected. Thus, the density is 35×100 FC/100 mL = = 350 10 16 Most Probable Number (MPN) - The equation proposed by Thomas can be used to estimate the MPN. The Thomas equation: Number of positive tubes x 100 √[(mL of sample in negative tubes) x (mL of samples in all tubes)] - in practice, many analytical laboratories are using the MPN tables in Standard Methods. Example 2 A bacterial analysis for a surface water yielded the following results for the standard confirmed test for total coliform. Determine the coliform density (MPN) using the MPN tables and the Thomas equation. Size of portion, Number Number mL positive Negative 10.0 4 1 1.0 4 1 0.1 2 3 Solution: From Table F-1, Selecting the combination of positive as 4, 4, 2, the MPN/100 mL is 47 Using Thomas equation, Number of positive tubes = 4 + 4 + 2 = 10 mL sample in negative tubes = (1x10) + (1x1.0) + (3x0.1) = 11.3 mL sample in all tubes = (5x10) + (5x1.0) + (5x0.1) = 55.5 10×100 MPN/100 mL = = 40/100 mL √11.3×55.5 17 18 Example 3 Calculate the MPN value by the Thomas equation. Sample Positive Five tubes, mL used Combination 1 0.1 0.01 0.001 Of positive A-raw 5/5 5/5 3/5 1/5 5-3-1 B x 10-3 5/5 5/5 3/5 1/5 5-3-1 C 5/5 3/5 2/5 0/5 5-3-2 D 4/5 3/5 1/5 1/5 4-3-1 E 0/5 1/5 0/5 0/5 0-1-0 Solution Sample A, Step 1 Number of positive tubes =3+1=4 mL sample in negative tubes = 2 x 0.01 + 4 x 0.001 = 0.024 mL sample in all tubes = 5 x 0.01 + 5 x 0.001 = 0.055 The Thomas equation: Number of positive tubes x 100 √[(mL of sample in negative tubes) x (mL of samples in all tubes)] 4×100 = √0.024×0.055 = 11 000 Sample B, As step 1, in addition multiply by 103 MPN/100 mL = 11 000 x 103 Sample C, (3+2)×100 MPN/100 mL = = 1406 √0.23×0.55 Sample D, 19  (4+3+1)×100 MPN/100 mL = = 305 √1.24×5.55 Sample E 1×100 MPN/100 mL = = 18 √5.45×5.55 20