env- i pract

8/7/2019 Env- I pract

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EXPERIMENT 2

ACIDITY

OBJECT

To determine the acidity of the given sample of water.

APPARATUS

Burette, Pipette, Conical flask and Glazed tile.

REAGENTS

0.02N NaOH solution, Methyl orange solution and Phenolphthalein solution.

THEORY

The acidity of water may be used by presence in uncombined carbon-dioxide, mineral acids and salts of strong acid

and weak bases. It is defined as the capacity of a solution to neutralize a standard alkali.

It can be noted that for carbonic acid the starchiometric end point is not reached unti1 the pH has been raised to

about 8.5 which indicates that all waters, having a pH lower that 8.5 contains acidity. Usually, the phenolphthalein

end point at pH 8.2 to 8.4 is taken as the reference point. Inspection of curve further indicates that at pH 7.0

considerable carbon-dioxide remains is to be neutralized but alone CO2 will not depress the pH below a value of

about 4.5.

Considering the nature of the curve which is for a strong acid, it may be concluded that neutralization of acid is

essentially complete at pH 4.5. It is incidentally, the methyl orange end point. Thus, it becomes obvious that the

acidity of neural water is caused by carbon dioxide or by strong mineral acids. Titration to methyl orange end point

(pH 4.5) is defined as the acidity which gives a measures of relatively strong acids such as mineral acids and

titration to the phenolphthalein end point (pH 8.3) is defined as totally acidity and it includes also the weak acids,acids salts and some acidity due to hydrolysis.

1 2 3 4 5 6 7 8 9 10 11

Practical range of mineral Range of CO2 acidity

PROCEDURE

(a) Total Acidity

Its determination should be made on spot on a fresh sample collected in a bottle and stoppered immediately to

prevent escape of carbon dioxide. Take 50 ml to 100 ml of the sample in an Erlenmeyer flask, add 3 drops of

phenolphthalein indicator and titrate over a white surface with 0.02 N NaOH until faint colour appears.

(b) Mineral Acid Acidity

Take 50 mL or l00 ml of the sample in an Erlenmeyer flask, add 2 drops of Methyl orange indicator, and titrate over

a white surface with 0.02 N NaOH until colour changes to faint orange. Brome-Cresol green indicator can also be

used in place of methyl orange; it gives a sharp end point.


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OBSERVATIONS

S. No. Volume of sample Initial burette

reading

Final burette

reading

Volume of NaOH

required

(A) Total Acidity Indicator phenolphthalein

1

2

3

4

5

(B) Mineral Acidity - Indicator methyl orange


reading

Final burette

reading

Volume of NaOH

required

1

2

3

4

5

CALCULATION

(a) Total Acidity

Mg / lt. total acidity as CaCO3 =

= ml. 0.02 N NaOH x 1000 x 50 = ml. of Sample

(b) Mineral acid acidity

Mg / lt. total mineral acid acidity =

= ml. 0.02 NNaOHx 1000 x 50 = ml. of Sample

RESULT


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The total acidity of sample B with initial pH is found to be mg/lt. and mineral acidity is .mg /lt.

The acidity of sample A (Tap water) is ..mg/lt. The total acidity is due to mineral acidity & weak aids. For

pH range more than 8.5 the acidity due to OH- ions

SIGNIFICANCE

Acidity is of little concern from a statutory or public health view point carbon dioxide is present in malt andcarbonate beverages in concentration greatly in excess of any concentration known in natural water and no

deleterious effects due to the carbon dioxide have been recognized. Water that contains mineral acidity is usually so

unpalatable that problems related to human consumption arc non-existent.

Acid water is of concern to Sanitary engineer because of their corrosive characteristics and the expenses involved in

removing or controlling the corrosion producing substances. The corrosive factor in most water is carbon dioxide,

but in many industrial wastes it is mineral acidity. Carbon dioxide must be reckoned with water softening problems

where the lime or lime soda ash method is employed.


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EXPERIMENT NO. 3

ALKALINITY

OBJECT: To determine alkalinity of given sample of water.

APPARATUS: Burette, pipette, conical flask and glazed tile

REAGENTS: 0.02 NH2SO4 solution, phenolphthaleinindicator and Methyl orange indicator.

THEORY: Alkalinity is measure of the basic constituents of water and is defined as the capacity of a solution toneutralize a standard acid. In natural water it is usually present as the carbonate andbicarbonate salts of calcium,

magnesium, sodium and potassium.

Bicarbonates represent the major form of alkalinity since they are formed in considerable amounts from the action of

carbon dioxide upon basic materials in the soil. Under certain conditions natural water may contain appreciable

amount of carbon and hydroxide alkalinity. Chemically treated water may contain appreciable amounts of carbonate

and hydroxide alkalinity. Chemically treated water, particularly those produced in lime or lime soda ash softening ofwater, contain carbonates and excess hydroxide.

Thus it is obvious that alkalinity is caused by three major classes of materials may he ranked in order of their effect

on pH as hydroxides, carbonates, bicarbonates and other salt of weak acids.

Alkalinity is determined by titration with a standard solution of a strong acid to certain end point as given by

indicator solutions. Phenolphthalein is satisfactory indicator for the first end point (pH approx 8.3) contributed by

hydroxide and carbonate. Methyl orange is used for the second end point (pH approx 4.5) contributed by

bicarbonates. The phenolphthalein end point of titration is defined as P alkalinity and the end point observed bycontinuing the titration with same solution using methyl orange indicator is knownas total or T-alkalinity.Following table can be used for working out OH, CO3 and HCO3 alkalinity individually after completing titration.

Table

Result of Titration Value of radical expressed in terms of Calcium Carbonate

OH- CO32- HCO3

-

P=0 0 0 T

P(T/2) 2P-T 2(T-P) 0

P=T T 0 0

PROCEDURE:

Phenolphthalein alkalinity

The 50 or 100 ml of sample in an Erlenmeyer flask, add two drops phenolphthalein indicator and titrate over a white

tile with 0.02 N l-{2S04 until the pink colour just disappears.


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Total or methyl orange alkalinity

Add two drops methyl orange indicator to the same sample in which phenolphthalein alkalinity has been determined

previously and titrated with 0.02 N H2S04 until the colour changes from yellow to faint orange.

OBSERVATIONS:

(a) Sample ..

(b) Initial pH of given sample is ..

Table for phenolphthalein alkalinity


reading

Final burette reading Volume of H2S04

1

2

3

Table for methyl orange alkalinity


reading

Final burette reading Volume of H2S04

1

2

3

CALCULATIONS:

Initial pH of the sample is .

Mg/lt. phenolphthalein alkalinity as CaCO3 =

(ml. of 0.02N H2S04 x 10 0 0 x 5 0) = ml of sample

Mg/lt, of total or methyl orange alkalinity as CaCO3 =

Total ml. of 0.02 N H2S04 x 1000 x 50 =. ml. of sample.

RESULT:

Methyl orange alkalinity as CaCO3 is mg/lt. and Phenolphthaleine alkalinity is

..mg/lt.

Total alkalinity due to bicarbonate is got by using methyl orange indicator, it comes mg/It.

CONCLUSION:


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Since alkalinity of tap water is ..mg/It, which is very large/moderate/low. Thus it can be used/not used as

drinking water because according to IS-10500:1991 range of alkalinity for drinking water is 200-600 mg/It. OH- ion

is mainly responsible for alkalinity. Due to only OH- ion alkalinity has range of pH 8.3 to 14 and practical range of

alkalinity comes pink to colorless solution of alkalinity above 600 mg/lt, is not good for human point of view.

SIGNIFICANCE: Within regional limit alkalinity has sanitary significance, but it is very important in connection

with coagulation, softening and corrosion preservation, Alum used in coagulation is an acid salt which when added

in small quantity to natural water, reacts with alkalinity present to form flocs. If insufficient alkalinity is present to

react with all the alum, coagulation will be incomplete and soluble alum will be left in the water. It may thereafter be

necessary to add alkalinity in the form of soda ash or lime to complete the coagulation or to maintain sufficient

alkalinity to prevent the coagulated water for being corrosive. Ordinarily the total alkalinity determined with methyl

orange indicator; gives sufficient information for the control of coagulation and corrosion prevention when pH is

also determined.

Many regulatory agencies prohibit the discharge of waste containing caustic alkalinity to receiving water. Municipalauthorities usually prohibit discharge of waste containing caustic alkalinity to sewers. Alkalinity as well as pH is an

important factor in determining the amenability of waste water to biological treatment.

Lastly from public health point of view, alkaline water is usually unpalatable and consumer tends to seek other

supplies. Chemically treated water some time has rather high pH values, which have met with some objections on

the part of consumers. For these reasons, standards are sometimes established on chemically treated water. Where

biological processes of treatment are used the pH must ordinarily be maintained whim the range of 6 to 9.5. This

criterion often requires adjustment of pH to favorable levels and calculations of the amount of chemical needed is

based upon acidity values in most cases.


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EXPERIMENT 5

HARDNESS TEST

OBJECT: To determine the total hardness and calcium hardness of a given sample of water.

APPARATUS: Burette, Pipette, Conical flask, etc.

REAGENTS: Standard EDTA solution (N/50), Ammonia buffer solution and NaOH solution, Eriochrome black T

indicator and Murex indicator (dry power), inhibitor.

THEORY

Water that consumes considerable quantity of soap to produce lather and or produces scale in hot water pipes,

heater, boilers and utensils used for cooking is called hard water.Hardness is caused by divalent metallic anions that are capable of reacting with soap to form precipitates with

cations present in water to form scale. Principal actions causing hardness and the major anions associated with them

are as listed below:CATIONS ANIONS

Ca++ HCO3-

Mg++ S04--

Sr++ Cl-Fe++ N03-

Mn++ Si03--

Calcium and magnesium are primarily the constituents of chalk and limestone. When rain falls it takes up carbon

dioxide from the atmosphere and forms a weak acid and this percolates underground, it then dissolves calcium and

magnesium forming hard water. In general hard water originates in the areas where the topsoil is thick and limestone

formation are present. Soft water originates in areas where the topsoil is thin and limestone is either sparse or absent.

0

50

100

150

Over

50 ppm

100 ppm

150 ppm

250 ppm

250 ppm

Soft

Moderately soft

Slightly hard

Moderately hard

Hard

Hardness may be classified as:

(a) Carbonate and non carbonate hardness

(b) Calcium and magnesium hardness, and

(c) Temporary and permanent hardness.


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PRINCIPLE

In alkaline condition EDTA (Ethylene-diamine tetra acetic acid) or its sodium salt forms a soluble chelated complex,

which is stable with Ca and Mg. Also Ca and Mg form a weak complex with the indicator Eriochrome black T,

which has wine red color. During titration when all free hardness ions are complexed by Eriochrom black Tindicator end point. The pH has to maintain at 100.1.

At higher pH i.e. about 12.0 mg ion precipitates and only Ca++ ions remain in solution. At this pH murex indicator

from a pink colour with Ca++, gets complexed resulting in a change from pink to purple, which indicates and point

of the reaction.

INTERFERENCE

Metal ions do interfere but can be overcome by addition of inhibitors.

PROCEDURE

A. TOTAL HARDNESS

1. Rinse burette, pipette, and flask, etc.

2. Take 25 or 50 ml of well-mixed sample in a flask.

3. Add 1-2 ml buffer solution followed by 1 ml inhibitor.

4. Add a pinch of Eriochrome black T and titrate with standard EDTA solution till wine colour changes to blue.

Note down the volume of EDTA required.

B. CALCIUM HARDNESS

I. Take 5 ml of sample in a flask.

2. Add 3 drops of NaOH (N/10) to raise pH to 12 and a pinch of indicator. Note initial burette readings.

3. Titrate with EDTA till pink colour changes to purple Note the final burette readings.

4. Repeat the procedure for other sample s till concurrent readings are obtained.

C. MAGNESIUM HARDNESS

1. Take l00 ml of sample ,add 1.5 ml of the buffer solution and 2.3 ml of a saturated solution of ammonia oxalate.

2. Mix the solution and allow it to stand for two hours or overnight if possible.

3. Filter using a No. 42 Watman filter paper.

4. Pipette cut 25 ml from the filtered solution and add Eriochrome black T indicator (1-2 drops) and titrate with

EDFA solution till the colour changes from wine red to blue.


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5. Take two concurrent readings.

OBSERVATIONS FOR TOTAL HARDNESS

S. no. Volume of

sample

Initial reading Final reading Net volume of

EDTA (ml)

Total hardness

mg/lt as CaCO3

12

3

4

OBSERVATIONS FOR CALCIUM HARDNESS

S. no. Volume of

sample

Initial reading Final reading Net volume of

EDTA (ml)

Calcium

hardness mg/lt

1

2

3

4

SAMPLE CALCULATIONS

(a) Total hardness:

Total hardness (mg/litre) =ml. of EDTA x 1 x 1000

ml. of sample

(b) Calcium hardness:

Calcium hardness (mg/litre) = ml. of EDTA x 1 x 1000

ml. of sample

(c) Magnesium hardness:

Magnesium hardness (mg/litre)= Total hardness - Ca

RESULT

For the given tap water sample the hardness is found to be mg/lt., Calcium hardness is mg/lt. and Magnesium

hardness is mg/lt.

CONCLUSION

As the total hardness and calcium hardness arc below/above the maximum tolerable value i.e. 600 and 200respectively. The water can be used/not used for domestic purposes.

SIGNFICANCE


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The determination of hardness is helpful in deciding the suitability of water for domestic and industrial purpose. The

design of softening process depends upon the relative amounts of carbonate and non-carbonate hardness present in

water. The amount of calcium and magnesium hardness decides the suitability of water for boiler use.

1.

calcium hardness

mg/It.

EXPERIMENT 6

DISSOLVED OXYGEN

OBJECT

To determine the dissolved oxygen in a given sample of water.

APPARATUS

500 ml ground glass, stopper bottles, nesslers tubes, pipettes, DO bottles, etc.

REAGENTS

MnSO4 solution concentrated H2S04, Alkali iodide azide solution, N/40, Na2S2O3 solution.

THEORY

Adequate dissolved oxygen is necessary for the use of fish and aquatic life .The dissolved oxygen concentration may

also he associated with corrositivity of water and its septicity. The D.O. test is used in B.O.D. determinations.

Determination of dissolved oxygen has no importance from the point of view of potability, but the dissolved oxygen

will indicate the power of self purification of water .Amount of dissolved oxygen in water varies with temperature

and also with dissolved salts present in water. Greater the temperature, lesser the dissolved oxygen and greater the

percentage of dissolved salts, lesser the dissolved oxygen.

There are two methods of determination of dissolved oxygen in a given sample of water.

1. winkler method

2. poplarograplic method

In determining the dissolved oxygen, the various ions and compounds present in water cause interference in desired

reactions and hence a specific modification is desired to be adopted in every case depending upon the type of ions

present in water sample. The most common interference is caused by nitrites ferrous and ferric ions in the surface

water and incubated waste water sample for B.O.D. hence the modification known as Winklers Azide modification

for nitrite is used.


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In all the modification the principle is dependent on the reaction which releases free iodine from potassium iodine

and the amount of iodine released is directly proportional to the amount of oxygen originally presented. In this

method Fe++ and NO2 will oxide iodide to free iodine. Mn++ (manganese) salts are oxidized to Mn+++ salts in alkaline

solution. Mn+++ is capable of oxidized from iodide to iodine in acidic conditions. The reducing agent used in this

case is sodium thiosulphate, with starch as indicator. As stated above to avoid interferences of NO2 and Fe+++, S, SO3

etc. The alkali azide reagent is added, the titration is then proceeded with.

CHEMICAL REACTIONS

Winklers method

1. MnSO4 + 2KOH = Mn(OH)2 + K2SO4(white ppt)

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

3. MnO(OH)2 +2H2SO 4 = MN(S0)2 + 3H2O

4. Mn(SO4) + 2KI = MNSO4 + I2 + K2SO4

5. 2Na2S2O3 + I2 = 2NaI + Na2S4O6

Winklers method with Azide modification

1. 2NaN3 + 2H2S04 = Na2SO4 + 2NH3

2. 2NaNO2 + H2S04 = 2HNO2 + Na2SO4

3. HN3 + HNO2 = N20 + N2 + H2O

PROCEDURE

1. Sample is collected in D.O. bottle from source and air bubbles removed.

2. D.O. is fixed by adding 2 ml of each aikali-azide and manganese sulphates solution to the sample filled in bottle

water cause interference in desired.

3. These bottle samples are brought to the laboratory.

4. Sufficient quantity of H2SO4 acid is added to D.O. bottle sample.

5. Iodine is liberated from KI. The amount of I2 liberated is proportional to amount of D.O.

6. 100 ml of solution thus obtained is taken and two drops of starch is added as an indicator.

7. This is treated against sodium thiosulphate and volume of it is noted.

OBSERVATION

S. No. Sample Volume of Burette reading Volume of Dissolved

(white ppt)

(brown ppt)(D.O.)


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sample sodium

thiosulphate

required

oxygen (D.O.)Initial

reading

Final

reading

1

2

3

CALCULATION

The ml of N/40 Na2S2O3 (sodium thiosulphate ) used for 200 ml of sample original is equivalent to D.O. in mg/l as

oxygen.

1)D.O. = ..mg/l

N1V1 = N2V2

N/40X x = N2 X 200

N2 = x/40 X 200

D.O. in mg/l as O2 = N x equivalent weight of oxygen x 1000

= x X 8 X 1000/40 X 200

RESULTS

The dissolved oxygen content in given sample of water is found to be mg/l.

COMMENT

In the experiment when 2 ml of MnSO4 followed by 2 ml of NaOH + KI + NaN3 is added to the sample 4 ml of

original sample is lost. Thus 203 ml taken for titration corresponding to 200 ml of original sample

200 X 300 / (300 - 4) = 203 ml

If initially 300 ml of sample is taken.

SIGNIFICANCE

Living organisms require oxygen to maintain their metabolic process. Dissolved oxygen is very important in

precipitating and dissolution of organic substances in water depends upon its temperature. Analysis of dissolved

oxygen is the main key test in sanitary engineering.

Surface water contains dissolved oxygen from atmosphere. This oxygen is used by the aquatic animals for their life.

Also this dissolved oxygen gives freshness and sparkling to water. Thus the determination of D.O. is necessary. It

helps in determination of B.O.D. greater D.O. may use corrosively to pipe materials. D.O. decreases with rise in

temperature. Water used for domestic purposes DO. should be 5 - 6 ppm.

PRECAUTIONS


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1. The initial pH should be taken.

2. During Ca hardness pH should be maintain sufficiently high.

3. Other actions should be present in small quantity.


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PROCEDURE

Total solids

Evaporate 100ml of the sample in an ignited and tarred dish, on anwater bath, dry to constant weight at 103 C to 105 C cool indesiccators and weigh. Drying for 1 hour at 103 C to 105 C is usuallysufficient. Record the loss in weight of dish as total solids.

Total volatile and fixed solids

Ignite the residue remaining in the dish at a low red heat over burner,or in the muffel furnace at 500 to constant weight, cool in desiccatorsand weigh. Record the loss in weight on ignition on total volatile solids.

Suspended solids

Prepare an asbestos fiber mat in a 30ml gooch crucible by addingsufficient homogeneous suspension of the asbestos cream, to producea mat 2.3mm thick. Dry in an oven at 103 to 105 C for 1 hour, ignitecool and weight. Filter 50ml to 100ml of the well mixed sample throughthe prepared gooch crucible under suction. Wash with distilled water,

dry at 103 to 105 C for 1 hour, cool in desiccators and weigh. Theincrease in weight equals the suspension solids.

Volatile and fixed suspended solids


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Ignite the residue remaining in the crucible after the determiningsuspended solids in a muffle furnace at 500 C for one hour, cool in adesiccators and weigh. Record the loss in weight on ignition as volatilesuspended solid. The residue represents fixed suspended solids.

Dissolved solids

Dissolved solids may be obtained by difference between total solidsand suspended solids. Dissolved solids may also be determined byevaporating a filtered sample.

CALCULATION

Mg/litre = (mg.residue x 1000) / ml of sample

Suspended solids in mg/lit = ______ x 1000 /100

= ______ mg/lit

Settable solids

Settable solids may be determined and reported on a volume or weightbasis.

(a)By volume: Fill in Imhoff cone to the liter mark with a thoroughlymixed sample. Allow to settle for 45 minutes, gently stir the sides ofthe cone with a rod or by spinning, and settle for a further period of 15minutes. Record the volume in ml of the settable matter in the cone.


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(b)By weight: This technique defines settable solids as that matter insewage which will not stay in suspension during the settling periodeither by virtue of settling to the bottom or floating to the top.

Determine the suspended solids in the sample under investigation.

Form a well mixed sample into a glass vessel not less than 9cm indiameter, using a quantity of samples not less than 1 liter andsufficient to ensure a depth of 20cm. A glass vessel of greaterdiameter and larger volume may be used.

Allow the sample top stand quiescent for 1 hour and siphon 250mlfrom the center of container at a point half way between the surface ofthe settled sludge and the liquid surface with out disturbing the settled

material or that which may be floating. Determined the suspendedmatter in mg/lit in all or in a adequate proportion of this supernatantliquor. This value mg/lit is equivalent to the non settling solids.

Mg/liter settable solids = Mg/liter suspended solids Mg/liter

non-settling solids

OBSERVATION TABLE

Sample = artificial

Weight of flask = _____________ gm

Weight of flask + residue = _____________ gm

Total residue = _____________ gm

CALCULATION


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Total solids in mg/lit =(__________) x 1000 / ml ofsample

= _____________ gm x 1000 / 10

= _____________ mg/lit

Total dissolved solids = total solids suspended solids

= _____________

= _____________ mg/lit

Results the amount of solids impurities present is given in the watersample is found to be :

Total solids = _____________ mg/lit

Total suspended solids = _____________ mg/lit

Total dissolved solids = _____________ mg/lit

RESULT

CONCLUSION

SIGNIFICANCE


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EXPERIMENT 10

DETERMINATION OF AVAILABLE CHLORINE FROM BLEACHING POWDER CaOCl2 AND

RESIDUAL CHLORINE IN THE GIVEN SAMPLE OF WATER

OBJECT

1. To determine the available chlorine in bleaching powder.

2. To determine the residual chlorine in given sample of water.

APPARATUS

Burette, pipette, conical flask, stirrer.

REAGEN TS

KI solution, Glacial acetic acid, Distilled water, Starch, Sodium thiosulphate solution.

THEORY

In small water works, chlorine required for disinfections is usually obtained from bleaching powder. For thispurpose iodometric method i.e. oxidation-reduction method is employed. The iodometric method is more precise

when the residual chlorine concentration is greater than one ppm. Chlorine will liberate free iodine from KI solution

when its pH is 8 or less. The liberated iodine is titrated against standard solution of sodium thiosulphate using starchas indicator. When blue colour disappears then all the liberate I iodine will have reacted. This indicates the end

point.

PROCEDURE

1.5 gram bleaching powder is taken and dissolved in 1 liter of distilled water. The solution thus prepared is to be

tested for available chlorine. 20 ml. of 10% KI is taken in i clean, dry conical flask, 2 ml of glacial acetic acid is

added into the flask to reduce 1 3 to 4.12. 12 5 ml of bleaching powder solution is then pipetted out and is added into

the flask. The colour of the solution will be brown. Titrate this solution against N/40 sodium thiosulphate solution,

till pale or straw yellow colour is developed. At this stage, add 2 drops of freshly prepared starch solution, which

results in appearance of blue colour. The titration against sodium thiosulphate solution is continued till the blue

colour disappear. This indicates the end point. Initial reading and final reading of sodium thiosulphate solution in the

burette is noted, the difference is then found out. The whole experiment is repeated and the mean difference is taken.

OBSERVATIONS

S. No. Burette reading Mean difference

Initial Final Difference

1.

2.


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3.

CALCULATIONS

Quantity of chlorine in = Number of ml. of thiosulphate solution of

mg/lt n the water sample N/40 normality remove the blue colour

Chlorine of given sample = N (Na2S2O3) X Volume of Na2S2O3

= mg/lt. as chlorine

Quantity of chlorine in mg/lt. in the sample = ..(ml.of sodium thiosulphate as required)

CHLORINATION

Chlorine is widely used for disinfection of water for removing odour since it is a powerful oxidizing agent and is

cheaply available. It can be used in molecular from or in hypochlorite form. For effective disinfection, dose of

chlorine, optimum contact period and residual chlorine are required to be found out.

PRINCIPLE

Chlorine combines with water to form hypochlorous and hydrochloric acid. Hypochlorous acid dissociates to giveOCl- ion. Quantities of OCl- and HOCl- depend on pH of solution. Hypochlorides also gives the OCl- ions, HOCl-

rupture the cell membrane of microbes, the disease producing organisms. These also reacts with the impurities like

ammonia, oxidisable inorganic matter like ferrous ion, nitrites etc. to from chloramines and stable ions of the latter

respectively.

INTERFERENCE

Oxidisable organic and inorganic matter.

REAGENTS

1. Bleaching powder

2. Concentrated acetic acid

3. Potassium iodide crystals

4. Standard sodium thiosulphate 0.1 N-- Dissolve 25g Na2S2O3.5 H20 and dilute to 1000 ml in freshly boiled and

cooled distilled water. Add about 5ml chloroform as preservative.


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5. Starch indicator: Prepare slurry by adding small quantity of water to 1 g starch powder. Add it to 100 ml boiling

water and continue boiling for few minutes then cool and uses.

6. Standard chlorine solution: Procedure outlined under analysis of bleaching

REACTION

When chlorine is added to water it forms hypochlorous acid or hypochlorine ions.

Cl2 + H2O pH>5 HOCl + HCl

Hypochlorous acid is unstable and may break into hydrogen ions and hypochlorite ions.

HOCl pH > 8 H+ + OCl-

Moreover the chlorine will immediately react with ammonia present in water to form various chloramines

NH3 + HOCl NH2Cl2 + H2O

PROCEDURE

1. Take l000ml sample in 12 stoppered bottles.

2. Add standardized chlorine solution in ascending order. If chlorine demand of treated water is being estimated,

doses from 0 to 300mg Cl2/1. I will be found useful. However, if the sample is polluted, doses from 0.1 mg to 3 mg

C12/l may be required as in case of treated effluents etc.

3. A low a contact period of 30 minutes of probable water and suitably higher for poliute:1 ater, or secondary

effluents.

4. Estimate residual chlorine iodometrically as described under analysis of bleaching powder.

5. Plot residual chlorine versus chlorine added. In case of organically polluted samples, a distinct break point can be

obtained. But in case of treated water sample, it is possible at only a straight line is obtained in absence of any

ammonium. A residual 0.2 mg Cl2, it. after the break point is recommended.

CONCLUSION

We have found the result in the above test is .. mg/lt. chlorine that chlorine is satisfied/not satisfied as

compared to permissible limit. The permissible limit is 0.2 according to IS: 10500, hence given sample of water

does not fit for drinking purpose The amount of available chlorine in a sample indicates that bacteria are reduced up

to safer limits but when it increases above the permissible limit. It may lead to the water born disease. In this water

sample the chlorine is below / average / above/ as per permissible limit. So it is fit / not fit for drinking and can be

accepted/ rejected.

RESULT

The amount of available chlorine in given sample of water is mg/lt, as chlorine

pH < 7


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SIGNIFICANCE

If the amount of available chlorine present in water is more than the permissible limit then the water should have

unpleasant taste characteristics. But if it is present within limit it forms hydrochlorous acid and kills the bacteriapresent in the water as it has been described earlier.

EXPERIMENT 11

BREAK POINT CHLORINATION

OBJECT: To determine the break point chloride demand of given sample of water.

APPARATUS: A number of bottles with stoppers, pipette, O.T. test comparator, stock solution of Cl2 of known

strength.

THEORY: The determination of break point chlorine demand of water is in effect the extension of experiment

already performed for determination of chlorine demand of water. As discussed already in the said experiment the

residual chlorine appears only after the demand has been met after adding a particular dose of chlorine, at definite

period of contact.

In the absence of ammonia or its derivatives in water, the residual chlorine is the free available chlorine in the form

of H0CL and/or OOL which are oxidants and react with orthotolidine to show residual chlorine. Thus once theresidual appears, it will go on increasing with increase in applied dose.

However, when ammonia is present, the hypochlorus acid i.e. HOCL reacts with it form chlorines first mono and

then dichlormine if excess of HOCL is available

HOCl+NH3 = NH2Cl+H20

and HOCl+NH3Cl = NHCl2+H20

Both dichloromine and monochloromine are oxidizing agents as if HOCl and react with orthotolidine to show

residual chlorine. Hence, in the presence of ammonia in water, the residual chlorine is the sum total of the action of

HOCl and chloromines is called the Combined available chlorine and lie total residual of chloride is both due to

free and combined available chlorine. An interesting stage comes in when all the ammonia present has been


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converted into monochloromine with addition of Cl2. At this stage further addition of chlorine will not results in

increase of residual because of the following reaction:

NHCl2+HOCl =NCl3+H20

The conversion of NHCI2 to NCl3 rather results in drop of residual chlorine from the level already attained. This is

due to the fact that NCl3 is non-oxidizing and does not react with O.T. and thus the drop in NHCl2 results in the dropof residual chlorine with added dose of chlorine. Thus, once again the chlorine demand goes on increasing because

of consumption of chlorine in producing NCl3

, which does not show any residue. A stage reaches when all theNCHCl2 present is converted into NCl3 and there is no combined available chloride at all, the chlorine demand is

maximum at this product. This is called the break point in chlorination. Further addition of Cl2 will again show

residue but this will be in the form of free chlorine i.e. HOCl and OCl as all the NH3 has been converted into NCl3 or

oxidized to free nitrogen or other gases at the break point.

SIGNIFICANCE OF THE BREAK POINT

The form of chlorine available as residue after break point is entirely free available chlorine and under controlled

conditions of pH it may be entirely available as HOCL (and not as OCl because at lower pH values, the ionization of

HOCl will be minimum).Which is the most efficient as bactericide, the chloramines and OCl being poor in action as

disinfectants. The break point chlorination therefore achieves:

a) Complete oxidation of ammonia and other compound

b) Destruction of tastes and odors of biological and chemical nature

C) Removal of color due to organic matter.

d) Removal of manganese

e) Improved bacteriological quality of water and destruction of virus infection.

PROCEDURE

A number of sample bottles are taken and to each of them water sample (containing NH3 and its derivatives) in equal

quantity, say 100 ml is put. to those sample are added at 5 minutes interval one by one gradual increasing does of Clsolution from a stock solution of chlorine of known strength. Residual chlorine is tested in each bottle after a

contact time of 10 minutes by means of O.T. reagent and comparators.

OBSERVATIONS

Strength of chlorine stock solution= ppm. Therefore 0.1 ml

contains =. ppm ml added to 1 .00 ml contributes ppm chlorine.

Sample No. ml of Cl2 Solution added Dose of Cl2 in ppm Residual chlorine

after 10 minutes1 2 3 4 5


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CALCULATIONS

CONCLUSION AND RESULT

SIGNIFICANCE

EXPERIMENT No. 18

BIOCHAMICAL OXYGEN DEMAND

OBJECT

To determine the biochemical oxygen demand of the given

sample of sewage.

APPARATUS


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B.O.D. incubator, Burette, Pipette and flasks.

REAGENTS

Manganese sulphate solution, alkaline azide iodide solution,concentrated sulphuric acid, N/40 sodium thiouslphate, starchsolution, FeCl3 MgSO4 CaCl2 NaCO3 and phosphate buffer.

THEORY

The biochemical oxygen demand may be defined as the amountof oxygen required by bacteria to stabilize organic matter underaerobic condition. The B.O.D. test is widely used to determine thepollution strength of the sewage, industrial waste etc. it is a testof prime importance in the evaluation of the purifying capacity ofthe receiving bodies of water . it is a device test. It involves the

measurement of the dissolved oxygen contents of a samplebefore and after a bioassay process in which living oraganismserve as a medium for the oxidation B.O.D. in terms of oraganicmatter as well as the amount of oxygen used during its oxidation,under atmospheric condition. nitrogen nearly twice as soluble inthe in water as in oxyfen. Most of the critical condition related todissolved oxygen deficiency in sanitry engineering pratice occurin prioud of high temperature

The kinetics of B.O.D, reaction indicate that they are first orderreaction, in which the rate of the reaction is proportion to theamount if oxidisable organic matter at any time as modified bypopulation of active organisms. This can be express in the from of


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an equation , Y=L(1-10-kt) , where Y=B.O.D. at any time and Y=(I-

10-kt).

Experiment have shown that a reasonably large %(68-70) of theB.O.D is exerted in the first five day incubation. The test is ,therefore, carried out with incubation for five days.

For samples having a very high B.O.D. value, dilution tooppropriate value is essential. So that oxygen depletion of at leastmg/L is the end of incubation period. The maximum D.O. in thesample bottle of the end of encubation period is 0.5mg/L.

PROCEDURE

Take well aerated distilled water and to it add nutrients such asFeCL3,MgSO4,CaCL2,NaCO3 and phosphate buffer in approximatedquantity to serve as food for growth bacteria and for buffer

action..

Take three portions of sample of known volume and dilute inthree B.O.D. bottles to obtain 0.5%, 1% and 2% of the sewagewith aerated water containing nutrients also take one blankB.O.D. bottle containing aerated water, keep all the four bottles inincubation at 200C for 5 days and determine the dissolved oxygen

of sample and blank.

OBSERVATION (A) For zero day


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S.No Sample N/40 Sodium

thiosulphate

Initial Final

reading

reading

Volume D.O. in ppm=

volume

cosumed

O2 consumed =

D.O. of Blank D.O.

of diluted sample

1.

2.

3.

4.

OBSERVATION TABLE (A) For 5 day

S.No Sampl

e

N/40 Sodium

thiosulphate

Initial Final

readingreading

Volume D.O. in ppm=

volume

cosumed

O2 consumed =

D.O. of Blank D.O.

of diluted sample

1.

2.


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3.

4.

CALCULATION

BOD At 20 C in mg/lt. =O2 consumed/ dilution factor

O2 consumed in 5th day in

(1) 1% sample=. =mg/lt as O2

(2) 2% sample= 4.4/6/300 =mg/lt as O2

(3) 3% sample= 6.2/9/300 =mg/lt as O2

RESULTS

BOD for the sample in 5 day at 20C is as following-

(1) 1% dilution = ..mg/lt as O2



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COMMENTS

B.O.D. (biochemical oxygen demand) is the amount of oxygen

required by microorganism while stabilizing decomposableorganic matter in a waste under aerobic condition . the BOD isalways less than C.O.D. the interferences in the test are lack ofacclimated seed organism, presence of heavy metals RCl2. thestandard condition is Ph near to 7 nutrients supply absence ofmicrobial growth inhibiting substance.

REACATION

Organic matter +O2 micro organism

..> new cells +CO2+H2O+stable products

(if O2 is not present )

Dry matter micro organism

> new cells+CO2+H2O+H2S+NH3


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Anaerobic (unstable product whichproduced bed smell)

SIGNIFICANCE

BOD test is used to determine:

(1) pollution load of waste water.

(2) Degree of pollution in lakes and stream at any andthree self purification capacity.

(3) Efficiency of waste water treatment methods.

LIST OF CHEMICAL


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1. Ferric chloride FeCl3

2. Mangnesium sulphate MgSO4

3. Calcium chloride CaCl2

4. Sodium carbonate Na2CO3

5. Phosphate Buffer.

env- i pract

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