Unit-V Water Technology

8/2/2019 Unit-V Water Technology

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Unit-V

WATER TECHNOLOGY

Natural sources of water:

(a) Surface water:1. Rain water is probably the purest form of natural

water, since it is obtained as a result of evaporation from thesurface water. However, during the journey downwards throughthe atmosphere, it dissolves a considerable amount of industrialgases (like CO2, SO2, NO2 etc.) and suspended solid particles, bothof organic and inorganic origin.

2. River water: Rivers are fed by rain and spring waters.Water from these sources flow over the surface of land, dissolves

the soluble minerals of the soil and finally falls in rivers. Riverwater thus contains dissolved minerals of the soil such aschlorides, sulphates, bicarbonates of sodium, calcium, magnesiumand iron. River water also contains the organic matter, derivedfrom decomposition of plants, and small particles of sand and rockin suspension. Thus, river water contains considerable amounts ofdissolved as well as suspended impurities.

3. Lake water has a more constant chemical composition.It usually, contains much lesser amounts of dissolved mineralsthan even well water, but quantity of organic matter in it is quitehigh.

4. Sea water is the most impure form of natural water.Rivers join sea and throw in the impurities carried by them.Moreover, continuous evaporation of water from the surface ofsea makes sea water continuously richer in dissolved impurities.Sea water contains, on an average, about 3.5% of dissolved salts,out of which 2.6% sodium chloride. Other salts present aresulphate of sodium; bicarbonates of potassium, magnesium andcalcium; bromides of potassium and magnesium and a number of

other compounds.Surface water, generally contains suspended matter, whichoften contains the disease-producing (or pathogenic) bacterias.Hence, such waters as such are not considered to be safe forhuman activities.

(b) Underground water: A part of the rain water, which reachesthe surface of the earth, percolates into the earth. As this water

journeys downwards, it comes in contact with a number of mineralsalts present in the soil and dissolves some of them. Water


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continues its downward journey, till it meets a hard rock, when itretards upwards and it may even come in the form of spring.

Spring and well water (or underground water), in general iscleaner in appearance due the filtering action of soil, but containsmore of the dissolved salts. Thus water from these sourcescontains more hardness. Usually, underground water is of highpurity.

Impurities in water:The water found in nature is never pure and contains a large

number of impurities in varying amounts. The major types ofimpurities found in water are of the following type:

i). Dissolved gases: Most water contains dissolved gasessuch as oxygen, carbon dioxide, sulphur dioxide, ammonia and

oxides of nitrogen all of which are derived from atmosphere.ii). Dissolved solids: Dissolved solids mainly consist ofbicarbonates, chlorides and sulphates of calcium, magnesium andsodium. In addition, small amounts of nitrates, nitrites, silicates,ammonia and ferrous salts are also present.

iii). Suspended impurities:The suspended matter may beinorganic or organic nature. The inorganic materials includeparticles such as sand, clay, silica, hydroxides of iron andaluminium etc. derived from erosion of soil. Some of theseparticles have large size and therefore settle down readily. Others

are fine particles and colloidal in nature. Such particles do notsettle down easily. The organic suspensions are decayingvegetable matter and due to microorganisms. These are alsocolloidal form. The presence of suspended matter, particularly thecolloidal particles impart turbidity to water.

iv) Microscopic matter: Many pathogenic bacteria andmicroorganisms are also present in water. They are main causesfor the water borne diseases.

Water analysis:

Hardness of water: Hardness in water is thatcharacteristic, which prevents the lathering of soap. This is dueto presence in water certain salts of calcium, magnesium andother heavy metals dissolved in it.

A sample of hard water, when treated with soap (sodium orpotassium salt of higher fatty acid like oleic, palmitic or stearic)does not produce lather, but on the other hand forms a whitescum or precipitate. This precipitate is formed, due to theformation of insoluble soaps of calcium and magnesium.

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2C17H35COONa + CaCl2 (C17H35COO)2Ca + 2NaCl

Sodium stearate (Hardness)

(Sodium soap)

Calcium stearate

(Insoluble)

2C17H35COONa + MgSO4 (C17H35COO)2Mg + Na2SO4Magnesium stearate

(Insoluble)

Thus, water which does not produce lather with soap solution readily, but

forms a white curd, is called hard water. On the other hand, water which lathers

easily on shaking with soap solution is called soft water. Such water, consequently,

does not contain dissolved calcium and magnesium.

Hardness of can be classified as temporary and permanent hardness.(1) Temporary or carbonate hardness is caused by the presence of dissolved

bicarbonates of calcium, magnesium and other heavy metals and carbonate of iron.Temporary hardness is mostly destroyed by mere boiling of water, when

bicarbonates are decomposed, yielding insoluble carbonates or hydroxides, which

are deposited as a crust at the bottom of vessel. Thus:

Ca(HCO3)2 CaCO

3+ H

2O + CO

2

Heat

Calcium bicarbonate (Insoluble)

Mg(HCO3)2 Mg(OH)2 + 2CO2

Heat

Magnesiumbicarbonate Magnesiumhydroxide

(2) Permanent or non-carbonate hardness is due to the presence ofchlorides and

sulphates of calcium, magnesium, iron and other heavy metals. Unlike permanent

hardness is not destroyed on boiling.

The sum of temporary and permanent hardness is referred to as total

hardness of water.

Equivalents of calcium carbonates:

The concentration of hardness as well as non-hardness constituting ions are,

usually, expressed in terms of equivalent amount of CaCO3, since this modepermits the addition and substraction of concentration, when required. The choice

of CaCO3 in particular is due to its molecular weight is 100 (equivalent weight =

50) and moreover, it is the most insoluble salt that can be precipitated in water

treatment. The equivalents of CaCO3 equal to

Mass of hardness producing substance x Chemical equivalent of CaCO3

Chemical equivalent of harness-producing substance

Mass of harness producing substance X 50Chemical equivalent of hardness-producing substance

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Units of hardness:

(i) Parts per million (ppm) is the parts of calcium carbonate equivalent hardness

per 106 parts of water, i.e., 1 ppm = 1 part of CaCO3 eq hardness in 106 parts of

water.

(ii) Milligrams per litre (mg/L) is the number of milligrams of CaCO3 equivalent

hardness present per litre of of water. Thus:

1mg/L = 1 mg of CaCO3 eq. hardness of 1 L of water

But 1L of water weighs = 1kg = 1000g = 1000 1000 mg = 10 6 mg.

Therefore 1 mg/L = 1 mg of CaCO3 eq per 106 mg of water.

= 1 part of CaCO3 eq per 106 parts of water = 1ppm.

Determination of Hardness using EDTA:

Hardness of water is determined by EDTA method, which involves

complexometric titration.

Priciple: The total hardness of a water sample could be determined by titrating

against ethylenediaminetetraacetic acid (EDTA). EDTA is a hexadentate ligand and

forms complexes with Ca2+ and Mg2+ ions. EDTA, commonly represented as H4Y

has two replaceable hydrogen ions in its molecule. It has the structure:

N CH2 CH2 N

CH2 COOH

CH2 COOH

HOOC H2C

HOOC H2C

The ionization in solution is represented as

H4Y H2Y2- + 2H+

The anion formed in the ionization forms complexes with metal ions, M2+

which can be given as

M2+ + H2Y2- MY2- + 2H+

where M2+ is a Ca2+ or Mg2+.The total hardness of water can be determined by titrating a

known volume of water against standard EDTA solution at pH of10 using Eriochrome black-T indicator. The colour of the freeindicator at pH 10 is blue. Eriochrome black-T forms a wine redcomplex with M2+ ions. On titration, EDTA first gets complexedwith all the free M2+ ions and then with M2+ ions of M2+-indicatorcomplex. Thus the indicator gets freed and consequently gives acolour change from wine red to blue at the equivalence point.Such indicators are referred to as metal-ion indicators. Since thereaction involves the release of H+ ions, a buffer mixture (NH4OH-NH4Cl) is used to main a pH of 10.

The total hardness is determined by titrating a known

volume of water sample against EDTA. To determine thetemporary hardness, another sample of the same volume of water

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is boiled to convert the bicarbonates to carbonates andprecipitated calcium carbonate is filtered off. The filtrate, aftercooling, is titrated against EDTA in the same away. This givespermanent hardness.

The difference between the total hardness and thepermanent hardness gives the temporary hardness.

Procedure: (i) Total hardness: 25 ml of given hard water ispipetted out into a clean conical flask. 5 ml of NH 4OH-NH4Cl bufferand 3 drops of Eriochrome black-T indicator are added. Theresulting wine red coloured solution is titrated against EDTA(0.01M) until clear blue without any reddish tinge is obtained. Letthe volume of EDTA required be V1 ml.(ii) Permanent hardness: 25 ml of given hard water is pipetted

out into a clean 500 ml beaker and boiled for 20-30 minutes. It iscooled and filtered directly into a 250 ml conical flask. 5 ml ofNH4OH-NH4Cl buffer and 3 drops of Eriochrome black-T indicatorare added. The resulting wine red coloured solution is titratedagainst EDTA (0.01M) until clear blue without any reddish tinge isobtained. Let the volume of EDTA required be V2 ml.Calculations:

1000 ml of 1M EDTA 100 g CaCO3 (Mol. Wt. of CaCO3 =100)

1ml of 1M EDTA (1000

100) g of CaCO3

V1 ml of 0.01M EDTA 1000

10001.01 Vg of CaCO3

25 ml of sample water contains1000

10001.01 Vg of CaCO3

106 (1 million) ml of water sample 251000

10001.01

V106 g of CaCO3

= 40 V1 g of CaCO3Total hardness of water sample = 40 V1 ppm of CaCO3 equivalent

Similarly, Permanent hardness =251000

10001.02

V106 g of CaCO3

= 40 V2 g of CaCO3Temporary hardness = 40 (V1- V2) ppm of CaCO3 equivalent.

Alkalinity:

Alkalinity in water arises due to the substances that can cause the formation

of hydroxyl (OH-) ions and in turn can react with strong acids. Alkalinity of a water

sample is a measure of its capacity to neutralize acids. Substances that cause the

alkalinity in water are of three types.

(i) Hydroxides eg., NaOH, Ca(OH)2, Mg(OH)2(ii) Carbonates eg., Na2CO3, CaCO3, MgCO3(iii) Bicarbonates eg., NaHCO3, Ca(HCO3)2, Mg(HCO3)2

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Total alkalinity, At is the sum of the alkalinity due to hydroxyls, Ah, alkalinity due

to carbonates, Ac and alkalinity due to bicarbonates, Ab.

Total alkalinity, At = Ah + Ac +Ab.

When a sample of alkaline water is treated with a strong acid such as HCl or

H2SO4, the following reactions occur.

NaOH + HCl NaCl + H2O (for hydroxyls) .....(

Na2CO

3+ HCl NaHCO

3+ NaCl (for carbonates) ....

NaHCO3

+ HCl NaCl + H2O + CO

2(for bicarbonates) (3

Determination of Alkalinity:

Alkalinity is determined by titrating a known volume of water sample

against acid using suitable indicator. Based on the indicator used-methyl orange or

phenolphthalein-two types of alkalinity can be evaluated.(i) Alkalinity when methyl orange is used Amo: When methyl orange is used as

indicator, the colour change at the end point is obtained only after all the threereactions given by the equations (1), (2) and (3) are complete. Hence methyl orange

gives total alkalinity. At = Amo = Ah + Ac + Ab (4)

(ii) Alkalinity when phenolphthalein is used Aph: When phenolphthalein is used

as indicator, the colour change at the end point is obtained after reactions (1) and

(2) are complete and before (3) occurs. Thus the alkalinity due to phenolphthalein

is attributed to hydroxides and half the carbonate as the colour change occurs

before the bicarbonates reacts [see reactions (2) and (3)].

Aph = Ah + Ac (5)

Procedure: (i) 100 ml of water sample is pipetted out into a clean conical flask.Two drops of methyl orange indicator is added and titrated against 0.02N HCl till

the colour changes sharply from yellow to orange. Let the volume of HCl

consumed be X ml.

(ii) To another sample of 100 ml water 2 drops of phenolphthalein indicator is

added and titrated against 0.02N HCl till the colour changes sharply from pink to

colourless. Let the volume the HCl consumed be Y ml.

Calculations: (i) Alkalinity due to methyl orange:1000 ml of 1N HCl 50 g of CaCO3 (equivalent wt. of CaCO3 =50)

1ml of 1N HCl 1000

50g of CaCO3

X ml of 0.02 N HCl 1000

02.050 Xg of CaCO3

100 ml of water sample contains1000

02.050 X g of CaCO3

106 (1 million) ml of water sample contains1001000

02.050

X 106

= 10 X g of CaCO3

That is, Amo = 10 X g of CaCO3

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(ii) Alkalinity due to phenolphthalein: The alkalinity due to phenolphthalein iscalculated in a similar way.

Aph =1001000

02.050

Y 106

= 10 Y g of CaCO3

The possible combination of alkalinity causing substances in water could be given

as follows:(i) Only hydroxyls (OH-)

(ii) Only carbonates (CO32-)

(iii) Only bicarbonates (HCO3-)

(iv) Hydroxyls + carbonates (OH- + CO32-)

(v) Carbonates + bicarbonates (CO32- + HCO3

-)

(vi) Hydroxyls + bicarbonates (OH- + HCO3-)

(vii) Hydroxyls + carbonates + bicarbonates (OH- + CO32- + HCO3

-).

Combination of (vi) and (vii) are not stable because hydroxyls and bicarbonates

mutually neutralize in water following reaction

NaOH + NaHCO3 Na2CO3 + H2O

The alkalinity in the combinations mentioned can be calculated based on the titre

values and relations,

Aph = Ah + 0.5 Ac and At = Amo = Ah + Ac + Ab

(i) If water contains only hydroxidesAph = Amo

since Ac and Ab are equal to zero.

(ii) If water contains only carbonates

From Eq. (5) Aph = Ah + 0.5Ac,

Since Ah = 0, Aph = 0.5Ac orAc = 2Aph

From eq. (4) Amo = Ah + Ac + Ab

Since hydroxyls and bicarbonates are absent, Ah = 0 and Ab = 0

Therefore, Amo = Ac = 2Aph.

(iii) If water contains only bicarbonates;

Ab = 0 and Ac = 0,

From eqs. (4) and (5), Aph = 0 and Ab = Amo.

(iv) If only hydroxides and carbonates are present

Ab = 0,

From Eq. (5), Aph = Ah + 0.5 Ac or r Ac = 2Aph 2Ah .(6)

From eq. (4), Amo = Ah + Ac

Ac = Amo Ah ..(7)Therefore 2Aph 2Ah = Amo Ah

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i.e.,Ah = 2Aph Amoand

From equation (7) Ac = Amo Ah

= Amo [2Aph Amo]

i.e., Ac = 2 (Amo Aph)

(v) If only carbonates and bicarbonates are present

Ah = 0,

From Eq. (5), Aph = 0.5 Aci.e., Ac = 2Aph

From Eq.(4), Amo = Ac + AbSubstituting for Ac,Ab = Amo - 2Aph

The relation between Aph and Amo for different combinations is summarized in

Table-1

Table-1

Alkalinity Relation between Aph and Amo1. Only Ah Aph = Amo2. Only Ac Ac = Amo = 2Aph3. Only Ab Aph = 0, Ab = Amo4. Ah + Ac Ac = 2(Amo Aph), Ah = 2Aph - Amo5. Ac + Ab Ac = 2Aph, Ab = Amo - 2Aph

Determination of Chloride (Cl-) by Mohrs (Argentometric)method:

Chloride ions present in water are due to chlorides of

calcium, magnesium, sodium etc.Principle: Chloride content in water is determined by titrating aknown volume of the water sample with standard silver nitratesolution using potassium chromate as indicator. Silver nitrateprecipitates chloride ions as silver chloride.

AgNO3 + Cl- AgCl + NO 3

-

When all the chloride ions are precipitated as silver chloride, the excess drop of

silver nitrate solution reacts with chromate ion to form a red colored precipitate.

2AgNO3 + CrO42-

Ag2CrO4 + 2NO3-

Therefore the end point of the titration is indicated by the appearance of reddish

tinge color in the solution.

Procedure: 100 ml of the water is pipetted out into a clean conical flask. 3-4 drops

of potassium chromate indicator is added and titrated against 0.02N silver nitrate

solution till the colour changes from yellow to brick red tinge. Let the volume of

AgNO3 consumed be a ml.

Blank titration: If the water sample is acidic it has to be neutralized by adding

about 1 g of CaCO3 powder before the titration is carried out. In such cases, a blank

titration is carried out. For this 100 ml of distilled water is pipetted out into a

conical flask. One gram of CaCO3 and 3-4 drops of potassium chromate indicator is

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added. It is titrated against 0.02N silver nitrate solution till the colour changes from

yellow to brick red tinge. Let the volume of AgNO3 consumed be b ml.

Calculations:

Volume of AgNO3 required for Cl- estimation = (a-b) ml = V mL.

1000 ml of 1N AgNO3 35.45 g of Cl-

1 ml of 1N AgNO31000

45.35g of Cl-

V ml of 0.02N AgNO3 1000

45.3502.0 Vof Cl-

i.e., 100 ml of water sample contains1000

45.3502.0 Vg of Cl-

Therefore Cl- content in the 106 ml of water sample =1001000

45.3502.0

V106 mg

= 7.09V ppm

Determination of fluoride by SPADNS method:Principle: Under the acidic conditions, fluorides react with zirconium SPADNS

solution and colour of SPADNS reagent [Sodium 2-(p-sulphophenylazo)-1,8-

dihydroxy-3,6-naphthalene disulphonate] gets bleached. Bleaching is a function of

fluoride ions and is directly proportional to the concentration of fluoride ions.

Procedure:

(i) Preparation of the reagent: 958 mg of SPADNS is dissolved in distilledwater and diluted to 500 ml. 133 mg of zirconyl chloride octahydrate

(ZrCl2.8H2O) is dissolved in 25 ml distilled water and 350 ml of

conc.HCl is added and diluted to 500 ml with distilled water. Equalvolumes of SPADNS solution and zirconyl acid solutions are mixed.

(ii) Preparation of calibration curve: 0.221 g of anhydrous sodium fluoride

is dissolved in water and diluted upto one liter. The stock solution is

further diluted to get standard solution having 10 mg per liter of fluoride.

(iii) 1,2,3,4,5 and 6 mls of this solution is pipetted out into 50 ml standard

flasks. 10 ml of zirconyl-SPADNS reagent and one drop of NaAsO2 (to

remove residual chlorine) are added to each of the solutions. It is diluted

up to the mark and mixed well. The absorbance of the solutions are

measured at 570 nm against a reagent blank and a calibration plot is

constructed by plotting absorbance against concentrations usingcolorimeter.

(iv) Suitable aliquot of water sample is taken and repeated the step 3.

(v) Using the calibration curve, the concentration of F-/L is calculated.

Determination of nitrate by phenoldisulphonic method:

Principle: Nitrate reacts with phenoldisulphonic acid (PDA) to produce a nitro

derivative, which in alkaline solution develops a yellow colour. The development

of yellow colour is attributed to rearrangement in the structure of the nitro

derivative. The colour produced follows Beers law and is proportional to the

concentration of NO3- present in the sample. The concentration of NO3- isdetermined using a colorimeter or spectrophotometer.

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Procedure:

(i) A calibration curve is prepared by using suitable aliquots of standard nitrate

(KNO3) solution in 5-500mg NO3- range.

(ii) 5,10,15, 20, 25 mls of standard KNO3 solutions are taken in 5 separate

beakers and evaporated dryness on a hot plate. To each of them 2 ml of PDA

is added and residue is dissolved. 10 ml of conc. NH3 (to develop colour) is

added and diluted to 100 ml standard volumetric flasks. Contents were

mixed well and the solutions from each of the standard flasks are transferred

to a cuvette. The absorbance is measured at 410 nm wavelength filter using

colorimeter.

(iii) 25 ml of the given water sample is taken in a beaker and evaporated

to dryness on a hot plate. 2 ml of phenoldisulphonic acid is added into the

beaker and the residue is dissolved. 10 ml of conc. NH3 (to develop colour)

is added diluted to 100 ml standard volumetric flask. Contents were mixed

well and the solution from the standard flask is transferred to a cuvette. The

absorbance is measured at 410 nm wavelength filter using colorimeter.

(iv)A blank solution is prepared by excluding the water sample.

(v) A calibration curve is drawn by plotting absorbance against the

concentration of NO3-

(vi)Using the calibration curve, the concentration of NO3- in the water sample is

determined.

Determination of Sulphate (SO42-

) by Gravimetric method:Principle: It gives the most accurate results and is the recommended procedure for

sulphate concentrations above 10 mg/mL. The sulphate ions in the sample are

precipitated by the addition of barium chloride solution to water sample acidified

with hydrochloric acid and kept near the boiling point.

SO42- + Ba2+ BaSO4.

In highly alkaline water maintained near the boiling temperature, BaCO3 may get

precipitated and eliminate this, the sample is acidified. To precipitate sulphate ions

as completely as possible, excess barium chloride is used. The precipitate of BaSO4is highly insoluble and hence there is considerable tendency for most of the

precipitate to form in colloidal condition which cannot be removed by ordinaryfiltration procedures. To facilitate the conversion of colloidal form to crystalline

form, the samples at temperatures near the boiling point for a few hours are

digested.

Procedure: About 200 mL of water sample is transferred into a beaker. Two drops

of methyl red indicator is added to it. Conc.HCl is added to it drop by drop till the

colour changes to pink. Two drops of conc.HCl are added in excess. It is heated to

nearly boiling and reduce the volume to 50mL. Hot barium chloride solution is

added to it, with stirring until the formation of white precipitate is complete. Two

drops of hot BaCl2 are added in excess. The precipitate (BaSO4) is digested forabout 2 hours (or until the precipitate becomes settles down). It is filtered using

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Whatman filter paper No. 42 (ash less filter paper), quantitatively. The precipitate

is washed several times with distilled water until the washings are free from

chloride ions. After it completely drains out, the filter paper along with the

precipitate is transferred carefully into previously weighed Gooch crucible and

ignited at 800-9000C till all traces of filter paper are burnt. The crucible is cooled in

a desiccator and finally weighed. Let the amount of BaSO4precipitated be W g.

Calculations: We know that, 233.4 g of BaSO4 contains 96.06 g of sulphate.

Therefore, W g of BaSO4 contains4.233

06.96 Wg of sulphate.

That is, 200 mL of water sample contains4.233

06.96 Wg of sulphate.

Therefore 1000 mL of water sample contains2004.233

100006.96

Wg of sulphate.

Determination of Dissolved Oxygen by Winklers method:Principle: The principle involved in the determination of dissolved oxygen is that

the divalent manganese solution along with a strong alkali is added to water

sample. The DO present in water sample oxidizes divalent manganese to tetravalent

manganese. The basic manganese oxide formed acts as oxygen carrier to enable the

dissolved oxygen in molecular form to take part in the reaction. Upon acidification,

tetravalent manganese reverts to divalent state with the liberation of nascent

oxygen, which oxidizes KI to I2. The liberated iodine is titrated against sodium

thiosulphate solution using starch indicator.

MnSO4 + 2KOH Mn(OH)2 +K2SO4

2Mn(OH)2 + O2 2MnO(OH)2

MnO(OH)2 + H2SO4 MnSO4 + 2H2O +[O]

2KI + H2SO4 + [O] K2SO4 + H2O + I2

I2 + 2Na2S2O3 Na2S4O6 + 2NaI

KI is added as alkaline KI which consist of a solution of sodium azide, KI and

NaOH in water. Sodium azide destroys the nitrites in water and thereby reduces the

error due to nitrites.

NaN3 + H+

HN3 + Na+

HN3 + NO2+ + H+ N2(gas) + N2O + H2O

Procedure: 250 cm3 of water sample is collected in a BOD bottle (avoiding the

contact with air, as far as possible). Immediately 2 cm3 of manganese sulphate is

added by means of pipette (dipping the end well below the water level). Similarly,

2 cm3 of alkaline iodide-azide mixture is added carefully. The bottle is stoppered

and shaken thouroughly. 2 cm3 of conc. H2SO4 is added slowly into it. The bottle is

stoppered and shaken thouroughly.

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100 cm3 of the solution pipetted into a clean conical flask from the bottle and

titrated against Na2S2O3 solution using 1 cm3 starch as indicator near the end point.

End point is colour change from blue to colourless.

Calculations:

100.0 cm3 of water sample (iodine) V ml of 0.01 N Na2S2O3 solution.

1000 cm3 of 1N Na2S2O3 solution 8 g of oxygen

Therefore V cm3 of 0.01N Na2S2O3 solution 1000

801.0 Vg of oxygen

The amount of oxygen present in 106 cm3 of water 1001000

801.0

V106

The dissolved oxygen present in the water = 0.8 V ppm

POTABLE WATER:

Water that is fit for human consumption and meets the stringent

microbiological and chemical standards of quality to prevent waterborne diseases

and health risks from toxic chemicals is called potable water.

Desalination: The process of partial or complete demineralization of highly

saline water such as the sea water is referred to desalination. In partial

demineralization, the amount of dissolved salts is reduced to such a level, that

water is rendered potable. Several methods such as flash evaporation, reverse

osmosis and electrodialysis are available for desalination and are described below.

i) Flash Evaporation:

Principle: The process takes place the advantage that water is volatile

compound whereas salts are non-volatile. Saline water is introduced into a flash

chamber in which the pressure maintained less than vapor pressure of saline water.

This results in rapid evaporation or flash distillation of water. The steam on

condensation produces fresh water. When this process is repeated several times, it

is multi stage flash distillation. Latent heat of steam is used for the preheating of the

saline water.

Method: A simplified multistage flash distillation unit is shown in Fig. 1

The preheated brine is heated further by steam in pre-heater. The hot brine (1000C)

now passes into the first flash chamber. Since the pressure is lower in this flash

chamber, a portion of the saline water flashes off (rapid evaporation) to form

water vapor. The flashed water vapor condenses on the condenser coils, thus

releasing its latent heat of condensation. The heat so released is utilized in heating

the influent brine passing through the coils. The brine that remains after some of

the water has been evaporated of flashed off is cooler. It then passes into the second

flash chamber in which the pressure is little lower than in the first. A bit more of

water evaporates and brine is cooled still further. In each successive stage, the brine

becomes progressively more concentrated and is discharged from the last flash

chamber. The fresh water from each chamber is collected and pumped off.

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Fig.1. Multi stage flash distillation

ii) Reverse osmosis:

Principle: When two solutions of unequal concentrations are separated by a

semi-permeable membrane (which selectively does not permit the passage of

dissolved solute particles, i.e., molecules, ions, etc.), flow of solvent takes place

from dilute to concentrated sides, due to osmosis. If, however, a hydrostatic

pressure in excess of osmatic pressure is applied on the concentrated side, thesolvent flow reverses, i.e., solvent is forced to move from concentrated side to

dilute across the membrane. This is the principle of reverse osmosis. Thus, in

reverse osmosis (R.O) methods, pure solvent (water) is separated from its

contaminates rather than removing contaminants from the water. This membrane

filtration is some times also called super-filtration or hyper-filtration.

Method: In this process, pressure (of the order 15 to 50 kg/cm2) is applied to

the sea-water/impure water (to be treated) to force its pure water out through the

semi-permeable membrane; leaving behind the dissolved solids (both ionic as well

as non-ionic). The principle of reverse osmosis, as applied for treating saline/sea

water, is illustrated in Fig.2. The membrane consists of very thin films of celluloseacetate polymethacrylate or polyamide, affixed to either side of a perforated tube.

Fig.2.Reverse osmosis cell

iii) Electro-dialysis:

Principle: Electro-dialysis is a method in which the ions (of the salts

present) are pulled out of the salt water by passing direct current, using electrodes

and thin rigid plastic membrane pair (natural or synthetic). Fig.3a illustrates the

method desalination by electro-dialysis. When direct electric current is passed

through the saline water the sodium ions (Na+) start miving towards the negative

pole (cathode): while the chloride ions (Cl-) start moving towards the positive pole

(anode), through the membrane, as a result, the concentration of brine decreases in

the central compartment; while it increases in two side compartments. Desalinated

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brine (or pure water) is removed from the central compartment from time to time;

while concentrated brine (in the side compartments) is replaced by fresh brine/sea

water.

For more efficient separation, usually, ions selective membranes are

employed. An ion-selective membrane has permeability for only one kind of ions

with specific charge, a cation-selective membrane is permeable to cations only,

because of the presence of charged fixed functional groups (such as RSO3- or

RCOO-), which rejects anions. Similarly, anion-selective membrane has positively

charged fixed functional groups such as R4N+Cl-.

Fig.3a. Line diagram of electro-dialysis

An electro-dialysis cell (Fig. 3b) consists of a large number of paired sets of

rigid plastic membranes. Saline water is passed under pressure (of about 5-6 kg/m2)between membrane pairs and an electric field is applied perpendicular to the direct

of water flow. Just as magnets of like charges repel each other, the fixed positivecharges inside the membrane repel positively charged ions (Na+), yet permit

negatively charged ions (Cl-) to pass through. Similarly, the fixed negative charges

inside the other type of membrane repel negatively charged ions (Cl-), yet permit

positively charged ions (Na+) to pass through. Therefore, water in one compartment

of the cell is deprived of its salts: while the salt concentration in adjacent

compartment is increased. Thus, we get alternate streams of pure water and

concentrated brine.

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Fig.3b. Electrodialysis cell

Water pollution:

Water pollution is defined as any any alternation in the physical, chemical

and biological properties of water as well as contamination with any foreign

substances which would constitute a health hazard or otherwise decrease the utility

of water.

Sources of water pollution: The following sources are mainly responsible for

water pollution.

i). Excreta and organic industrial wastes.ii). Pathogenic bacteria

iii). Plant nutrients

iv). Organic pesticides.

v). Heavy metals and Chemicals.

vi). Radioactive wastes

vii). Heat

Water containing any one or a combination of the above inputs is referred to

as sewage or effluent.

Hazardous Chemicals and their ill effects:Chemicals which are combustible, oxidizers, explosives flammable,

pyrophoric, unstable (reactive), water reactive, carcinogenus, toxic agents,

reproductive toxins, irritants, corrosives, hepatotoxins, nephrotoxins the release of

which may substantially endanger to public health, public welfare or the

environment are called hazardous chemicals.

Some hazardous chemicals with their sources and ill effects are listed in the

table below.

Hazardous

chemical

Source Ill effects

Cadmium(Cd)

Mining wastes, effluentsfrom plating industries.

Maladies including renal failure and agenerative bone disease called itai itai, High

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blood pressure, kidney damage, destruction

of red blood corpuscles, affect stereostruture

of enzymes in the body imparting their

biological activities.

Lead (Pb) Discharges from mining,

metallurgical operations,

plumbing, lead acid

batteries.

Severe dysfunction of the kidneys,

reproductive systems and lever, impairment

of central and peripheral nervous systems.

Mercury

(Hg)

Mineral processing

operations, as an

electrode in the

electrolytic generation of

chlorine, organomercury

fungicides, discharged

batteries, amalgams,tooth fillings.

Neurological damage including paralysis,

depression and irritability, blindness,

insanity, chromosome breakage and birth

defects.

Arsenic(As)

Erosion of naturaldeposits, runoff from

otchardes, runoff from

glass and electronics

production of wastes.

Skin damage, problems with circulatorysystems, increased risk of getting cancer.

Sewage: Sewage may be broadly classified into domestic sewage and industrial

effluents. Domestic sewage is the liquid wastes conducted away from residences,

institutions and buildings. They essentially contain organic wastes, pathogenicbacteria, plant materials, pesticides, detergents and other waste materials. Liquid

wastes conducted away from industrial establishment such as chemical plants,fertilizer industries, leather tanneries, sugar and paper industries, breweries, textile

mills, oil refineries, pharmaceutical industries, metal plating units and so on

constitute industrial effluent. Industrial effluents generally contain oils, vegetable

and animal matter and offensive and complex chemicals like acids, alkalis,

detergents, phenols, toxic metals, pesticides and related materials. In recent years,

however, indiscriminate use of various chemicals and unlawful discharges of

industrial effluents into rivers without treatment have posed several environmental

problems. The sewage can be rendered harmless by suitable treatment. The organicwastes present in sewage undergo degradation that is, complex molecules are

broken down into simple ones by bacteria. Bacterial degradation of organic matter

is of two types.

a) Aerobic degradation brought about by bacteria thriving on oxygen

dissolved in water producing harmless products such as carbon dioxide and

water; and

b) Anaerobic degradation brought about by bacteria in the absence of air

producing harmful end products such as methane, hydrogen sulphide and

ammonia. Anaerobically treated sewage is sometimes referred to as septic

sewage.

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Biological oxygen demand (BOD):

Natural water contains dissolved oxygen (8.7 ppm) and dissolved oxygen

(DO) is capable of oxidizing many of these pollutants particularly the organic

wastes such as dead plant matter and animal wastes. In this way the dissolved

oxygen is consumed. However, in running water such as streams and rivers there is

a continuous replenishment of oxygen maintaining the DO level and hence the

degradation is aerobic. The degradation products are CO2 and water which are

harmless. On the other hand, in stagnant waters such as in lake and well waters,

their is a gradual decrease in the DO level ultimately causing anaerobic (absence of

air) degradation of organic wastes releasing obnoxious gases such as H2S, CH4 and

NH3.

The amount of organic matter present in a sample of water is measured in

terms of the amount of dissolved oxygen required by microorganisms to oxidize the

organic matter. This is called the biological (or biochemical) oxygen demand.

Definition: BOD is defined as the amount of oxygen required by

microorganisms to oxidize the organic wastes present in one liter of water under

aerobic conditions at 200C and for a period of five days.

BOD determination: The test is based upon determination of dissolved

oxygen prior to and following a 5 days period at 200C. A known volume of sample

of sewage is diluted with known volume of dilution water (water containing

nutrients for a bacterial growth), whose dissolved oxygen is pre-determined. The

whole solution is incubated in a closed bottle at 200C for a period of 5 days. After

this, unused oxygen is determined. The difference between the original oxygen

content in the diluted water and unused oxygen of solution after 5 days gives BOD.

It is expressed in mg/dm

3

.Importance: BOD is the most important in sewage treatment, as it indicates

the amount of decomposable organic matter in the sewage. Larger the concentration

of decomposable organic matter, greater the BOD and consequently, more the

strength or nuisance potential. BOD has a special significance in pollution control,

as it enables us to determine the degree of pollution at any time in the sewage

stream.

Chemical oxygen demand (COD):

Definition: It defined as the amount of oxygen consumed in the chemical

oxidation (using oxidizing agent such as acidified K2Cr2O7) of organic andinorganic wastes present in one litre of waste water. COD values are expressed in

mgdm-3.

The principle of the method is the oxidation of organic matter using

chemical oxidizing agents such as acidified potassium dichromate in the presence

of a catalyst such as silver sulphate (which catalyzes the oxidation of organic

matter) and mercuric sulphate (which forms a complex with chloride ions present in

water thus preventing its interference). A typical reaction representing the oxidation

of organic matter is given below.

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3CH2O + 16H+ + 2Cr

2O

7

2- 4Cr3+ + 3CO2

+ 11H2O

Ag2SO

4

HgSO4

COD determination: A known volume (25 ml) of the waste water is pipetted out

into a round bottomed flask. A known excess of K2Cr2O7 (25 ml) is pipetted intothe same flask along with one test tube full of H2SO4 containing HgSO4 andAg2SO4. The flask is refluxed for 2 hours. The contents are cooled and transferred

to a conical flask. 5 drops of ferroin indicator is added to it and titrated against FAS

taken in the burette till the red colour changes from blue green to reddish brown.

Let the volume of titrant required be a ml. A blank titration is performed taking

the same amount of water in place of waste water. Let the volume required be b

ml.

Calculations: Volume of K2Cr2O7 required for the sample be b ml.

COD of the sample = N (b-a) 8000 / V mgdm-3

Where N= Normality of FASV=Volume of waste water sample.

Difference between BOD and COD: COD is a measure of oxidisable impurities

present in the sewage. Where as, the BOD measures the oxygen consumed by

living organisms while assimilating organic matter present in the water; the COD is

a measure of both the biologically oxidisable and biologically inert organic matter

such as cellulose. Consequently, COD values are, generally, higher than BOD

values. The COD value can be employed to estimate appropriate BOD values. The

main advantage the COD is that its determination takes about 3 hours, compared to

less than 5 days for the BOD determination.

Treatment of domestic sewage:

The treatment of sewage is carried out in three stages.(a) Primary (Physical and chemical) treatment:

The removal of coarse solids in the sewage water is effected by means of

racks, screens, grid chambers and skimming tanks. Then water is passed into a

sedimentation tank where it is allowed to settle. The non-settleable solids are

removed by coagulation by treatment with coagulating agents like alum, ferric

chloride or lime.

(b) Secondary (Biological) treatment:Activated sludge process: The waste water after the primary treatment is

allowed to flow into large tanks (Fig.4) where biological treatment is carried out.

Activated sludge containing microorganisms (from a previous operation) is sprayed

over the water. The microorganisms present in the sludge form a thin layer and

thrive on the organic wastes in the sewage. Air is passed vigorously from the centre

of the tank in order to bring good contact between the organic wastes and bacteria

in presence of air and sunlight. Under these conditions, aerobic oxidation of organic

matter occurs. The sludge formed is removed by settling or filtration. A part of the

sludge is reused and rest is used as fertilizer. The residual water is chlorinated to

remove bacteria and finally discharged into running water or used for watering

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plants. The activated sludge process operates at 90-95% efficiency of BOD

treatment.

Fig.4. Activated sludge process

(c) Tertiary treatment:

If the treated water contains a high concentration of phosphates, heavy metal

ions, colloidal impurities and non-degradable organic compounds, the water is

subjected to tertiary treatment. The process includes

(i) Treatment with lime for the removal of phosphates as insoluble calcium

phosphates.

(ii) Treatment with S2- ions for the removal of heavy metal ions as insoluble

sulphides.

(iii). Treatment with activated charcoal to absorb remaining organic

compounds.

(iv). Treatment with alum to remove the colloidal impurities not removed in

the previous treatments to further reduce BOD level.

******************************

Problems on Hardness:

1). 100 ml of a sample of water required 18 ml of 0.01M EDTA for titration using

Eriochrome Black-T indicator. In another experiment, 100 ml of the same sample

of water was gently boiled and the precipitate was removed by filtration. The

filtrate required 9.0 ml of 0.01M EDTA using Eriochrome black-T indicator.

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Calculate a) the total hardness of the water sample b) permanent of the water

sample and carbonate hardness.

a)Total hardness

1000 ml of 1M EDTA = 100 g CaCO3

1 ml of 1M EDTA = 100/1000 g of EDTA

18 ml of 0.01M EDTA = (18 0.01100 / 1000) g of CaCO3

= 0.018 g of CaCO3

100 ml of sample water contains 0.018 g of CaCO3

106 (1 million) ml of water sample contains 180 ppm of CaCO3

Therefore total hardness of the water sample is 180 ppm of CaCO3

b) Permanent hardness:1000 ml of 1M EDTA = 100 g CaCO3

1 ml of 1M EDTA = 100/1000 g of EDTA

9.0 ml of 0.01M EDTA = (9 0.01100 / 1000) g of CaCO3

= 0.009 g of CaCO3

100 ml of sample water contains 0.009 g of CaCO3

106 (1 million) ml of water sample contains = 0.009 106 ppm

of CaCO3

= 90 ppm of

CaCO3

Permanent hardness of the water sample = 90 ppm of CaCO3

temporary hardness of the water sample = 180-90 = 90 ppm of

CaCO3

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Problems on Alkalinity:

2). The carbonate alkalinity of a water sample was found to be 60

ppm CaCO3 equiv. Its total alkalinity increased to 350 ppm CaCO3

equiv after lime treatment. Find out the excessive amount of

Ca(OH)2 present in the water lime treatment. Express your answer

in mg of Ca(OH)2 = 74 g, CaCO3 = 100 g).

Total alkalinity, At = Ah + Ac or,

Ah = At - Ac

= 350 60 = 290 ppm CaCO3 equiv

Excessive amount of Ca(OH)2 = 290 ppm CaCO3 equiv

= 290 mg per litre of CaCO3

= 290 74 / 100 mg per litre

Ca(OH)2

= 214.6 mg per litre Ca(OH)2

3). Find the alkalinity of a water due to hydroxides and carbonates

if Aph = 90 ppm CaCO3 equiv and Am.o. = 155 ppm CaCO3 equiv.

For water containing hydroxides and carbonates,Ah = 2Aph Amo Ah = 2 (Amo Aph)

= 2 90 155 = 2 (155 35)

= 35 ppm. = 240 ppm.

4). How many times is it necessary to evaporate a boiler water sample whose

alkalinity due to NaOH is 5 mg-equiv so that the concentration of NaOH increases

to 10 g per litre (atomic weight: Na = 23 g; O = 16 g and H = 1 g)

Molecular weight of NaOH = 23 + 16 +1 = 40 g.

5 mg equiv per litre of NaOH = 5 40 = 200 mg per litre of NaOH

Number of times to be evaporated = 10000 mg per litre/ 200 mg per litre

= 50 times to be evaporated.

Problems on BOD:

1). Calculate BOD of a water sample having organic compoundwith formula CH2O containing 9.6 mg/dm3.

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Solution:

CH2O + O2 CO2 + H2O

Molecular weight 30 32

Mol. weight of CH2O is = 30.

According to the equation 30 g of CH2O requires for complete oxidation, 32 g of

oxygen.

Therefore 9.6 mg of CH2O requires 9.6 mg 32 103 mg / 30 103 mg of

oxygen. i.e., equal to 10.24 mg of oxygen.

Therefore BOD = 10.24 mg/dm3.

2). Claculate BOD of an effluent sample containing 8.2 mg/dm3 of organic matter

represented by the formula CH2O (Ans- 8.6 mg/dm3).

3). What would be 5 day BOD value for sample containing 200 mg/dm3 of glucose

assuming that it was completely oxidized in the BOD test. (At. Wt. of C = 12, H =

1, O = 16).Solution:

C6H12O6 + 6O2 6CO2 + 6H2O

Molecular weights: 180 6 32 6 40 6 18

180 g of glucose requires 192 g of oxygen for complete oxidation.

Therefore 200 mg of glucose requires 200 mg 192 103 mg / 180 103 mg of

oxygen for oxydation = 213.33 mg of oxygen

Therefore BOD of the waste water sample = 213.33 mg/dm3

Problems on COD:

1). 25ml of a sewage water sample was refluxed with 10 ml of 0.25N K2Cr2O7solution in presence of dil.H2SO4. The unreacted dichromate required 5.5 ml of

0.1N FAS solution. 10 ml of the same K2Cr2O7 solution and 25 ml distilled water

under the same conditions as the sample required 26.0 ml of FAS of 0,1N.

Calculate the COD of the sample.

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COD = NFAS (b-a) 8000 / Vwaste water= 0.1(26 5.5) 8000 / 25

= 656mg/dm3

2). In a COD experiment, 29.5 cm3 and 20 cm3 of 0.025N FAS solution were

required for blank and sample titration respectively. The volume of test sampleused is 25 cm3. Calculate the COD of the sample solution. (Ans-76 mg/dm3).

3). In a COD experiment, 28.1 cm3 and 14 cm3 of 0.05N FAS solution were

required for blank and sample titration respectively. The volume of test sample

used is 25 cm3. Calculate the COD of the sample solution. (Ans-225.6 mg/dm3).

4). 20 ml of sewage for COD is reacted with 25 ml of K2Cr2O7 solution and

unreacted K2Cr2O7 requires 9.0 ml of n/4 FAS solution. Under similar conditions, in

blank titration 15.0 ml of FAS is used up. Calculate the COD of the sample. (Ans-

600 mg/dm3).

5). In a COD experiment, 30 cm3 of an effluent sample required 9.8 cm3 of 0.001M

K2Cr2O7 solution for oxidation. Calculate the COD of the sample.

Solution: (i) Evaluate the oxygen equivalent

1000 cm3 of 1M solution of K2Cr2O7equivalent to 3 equivalent of oxygen.

= 3 16 g of oxygen.

Therefore 9.8 cm3 of 0.001M K2Cr2O7equivalent to

9.8 cm3 0.001M 3 16 g / 1000 cm3 1

= 0.0004704 g of oxygen.

= 0.4704 mg of oxygen.

(ii) To evaluate COD value

30 cm3 of effluent is equivalent to 0.4704 mg of oxygen.

Therefore 1000 cm3 of effluent is equivalent to

1000cm3 0.4704 mg / 30 cm3

= 15.68 mg of oxygen.

Therefore COD of effluent = 15.68 mg/dm3.

6). Calculate the COD of the effluent sample when 25 cm3 of an effluent sample

required 8.3 cm3 of 0.001M K2Cr2O7 solution for complete oxidation (Ans- 15.93

mg/dm3) .

7). Calculate the COD of the effluent sample when 25 cm3 of an effluent sample

required 10.5 cm

3

of 0.005M K2Cr2O7 solution for complete oxidation.

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8). 25 cm3 of an industrial effluent requires 12.5 cm3 0.5N K2Cr2O7 for complete

oxidation. Calculate COD of the sample. Assuming that the effluent contains only

oxalic acid, calculate the amount of oxalic acid present in 1 dm3 (given equivalent

weight of oxalic acid as 45).

Solution: 1 equivalent of K2Cr2O7equivalentto 1 equivalent of oxalic acid.

i.e., 1000 cm3 of 1N K2Cr2O7equivalent45 g of oxalic acid.

Therefore 12.5 cm3 of 0.5N K2Cr2O7equivalentto 45 12.5 0.5 /1000 1

= 0.28125 g or 281.25 mg of oxalic acid.

Amount of oxalic acid present in one litre of the effluent = 0.28125 g 1000 /25

= 11.25 g.

C2H2O4 + O 2CO2 + H2O

For complete oxidation,

1 equivalent of oxalic acid equivalent to 1 equivalent of oxygen

i.e., 45 g of oxalic acid equivalent to 8 g oxygen.

Therefore, 11.25 g of oxalic acid equivalent to 11.25 8 / 45

= 2 g or 2000 mg of oxygen.

Therefore COD of effluent sample = 2000 mg/dm3.

9). 25 cm3 of an industrial effluent when subjected to COD test required 22.5 cm3

of 0.5N K2Cr2O7 solution for complete oxidation. Calculate the COD of the sample.Solution:

1000 cm3 of 1N K2Cr2O7 solution contains 49 g of K2Cr2O7

22.5 cm3 of 1N K2Cr2O7 solution contains 22.5 cm3 0.5N 49 g /1000 1N

= 0.55125 g or 551.25 mg of K2Cr2O7

K2Cr2O7 + 4H2SO4 K2SO4 + Cr2(SO4)3 + 4H2O + 3(O).

294 g of K2Cr2O7equivalent to 48 g of oxygen.

551.25 mg of K2Cr2O7 equivalent to 551.25 mg 48 g / 294 g.= 90 mg of oxygen.

Therefore COD of effluent sample = 90 mg 1000 cm3 / 25 cm3

= 3600 mg of oxygen/dm3

Alternate method:

1000 cm3 of 1N K2Cr2O7 solution equivalent to 8 g or 8000 mg of oxygen.

22.5cm3 of 0.5N K2Cr2O7 equivalent 22.5 cm3 0.5N 8000 mg / 1000 cm3 1N

= 90 mg of oxygen.

Therefore COD of effluent sample = 90 mg 1000 cm3 / 25 cm3

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= 3600 mg of oxygen/dm3

10). 25 cm3 of an industrial effluent consumed 5 cm3 of 0.5N K2Cr2O7 solution for

oxidation. Calculate the COD of the sample (800 mg/dm3).

Unit-V Water Technology

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Transcript of Unit-V Water Technology