Experiment No

Aim : To determine the sulphate content of the given water sample

Principle : Sulphate ions ( SO4-2

) is precipitate in an acelic acid medium with barium

chloride ( BaCl2) so as to form barium sulphate (BaSO4) crystals of uniform size. Light

absorption of the BaSO4 suspension is measured by photometer and the sulphate

concentration is determined by comparison of reading with standard curve.

Apparatus: 1) Magnetic stirrer

2) Klett summerson colorimeter or spectro photometer

3) Measuring spoon

Reagent:

a) Conditioning reagent – Mix 50 ml glycerol with a solution containing 30 ml

conc. HCL, 300 ml distilled water, 100 ml 95% ethyl alcohol and 75 gm

sodium chloride.

b) Barium chloride crystals ‘AR grade’.

c) Standard Sulphate solution – Prepare by diluting 10.41 ml of the standard 0.02

NH2SO4 to 100 ml with distilled water.

Procedure:

a) Formation of barium sulphate turbidity – Measure 100ml sample or a suitable

aliquot made up to 100ml into a 250ml Erlenmeyer flask. Add exactly 5 ml

conditioning reagent and mix in the stirring apparatus. While the solution is

being stirred add a spoon full of barium chloride crystals stir

b) Measurements of barium sulphate turbidity – Immediately after the stirring period

is over, pour some of the solution into the absorption cell of the photometer and

measure the absorption at fifth minute. Maximum turbidity is usually achieved

within 2 min. and the reading remains constant there after for 3-10 min.

c) Read mg SO4 present in the sample on the calibration curve prepared by standard

solutions,

Calculation:

mg / l SO42-

= mg SO42-

x 1000 / ml sample

Conditioning

Reaction - SO42-

+ BaCl2 BaSO4

SO4Solution

Result :

Experiment No

Aim : To determine the chloride content (Avgentometric method) in the given

water samples.

Theory : : Chloride in the form of chloride Cl – ions, is one of the major inorganic

anions in water and waste water. The chloride concentration is higher in

the waste water than the raw water because of sodium chloride ( NaCl) is

a common ingredient of diet and passes unchanged through the

digestive system. At the sea coast, chloride may be present in high

concentration because of leakage of salt water into the sewage system. It

may be also increased by industrial process. High chloride content may

harm metal pipes and structures as well as growing plants.

Principle : In neutral or slightly alkaline solution K2CrO4 indicates the end of silver

chloride gets quantitatively precipitated before red silver chromate is

formed.

Reagents : i) Chloride free water: Demonized water

ii) K2CrO4 indicator: Dissolve 50 gm in distilled water. Add AgNO3

solution till a definite red precipitation is formed. Allow to stand for 12

hours filter. Dilute the filtrate to 1 liter D.W.

iii) Standard AgNO3 (0.0141 N): Dissolve 2.395 gm in distilled water and

dilute to 1 Liter. 1ml of standard silver nitrate (0.0141 N) is equivalent to

500µg Cl-

iv) Standard NaCl (0.0141 N): Dissolve 824.1 mg water and dilute to 1 lit.

1ml of this solution is equivalent to 500µg Cl-.

Procedure: : i) Use 100ml or a suitable aliquot diluted to 100ml with distilled water.

ii) If the sample is colored, add 3ml Al(OH)3 wash combines the filtrate

and washing.

iii) Check the pH and adjust it to near neutrality.

iv) If sulphide, sulphite or thiosulphate is present, make the water alkaline

to Phenolphthalein with NaOH. Add 1ml H2O2, stir, neutralize with

H2SO4.

v) Titrate the sample with AgNO3 after adding 1ml K2CrO4 to the sample

till orange red color appears.

vi) Run the blank taking distilled water as the sample.

Calculation :

1000ml of 1N AgNO3 = 35.45gm Cl-

... 1 ml of 0.0141 AgNO3 = 35.45 x 0.0141mg Cl

-

= 0.499mg Cl- or say 500µg Cl

- or 0.5mg Cl

-

Formula :

(A-B) x 0.5 x1000

Mg/l Cl- =

ml of sample

Where A = ml titrant of sample

B = ml titrant for blankV

Reaction: Cl- + AgNO3 --� AgCl+ NO3

–

K2CrO4 + 2AgNO3 --� Ag2CrO4 + 2KNO3

If pH> 8.3 - Ag (OH) 2, is precipitated,

If pH < 7 – Cr2 O7, is precipitated.

Ksp AgCl = 3 x 10-10

, Ksp Ag2CrO4 = 5x 10-12

Observation Table :

S.N Type of sample (A-B) x 0.5 x1000

Mg/l Cl- =

ml of sample

Result:

Conclusion :

Experiment No

Aim : To determine Biochemical oxygen demand ( B.O.D.)of given

water samples

Theory : BOD is the amount of O2 required by microorganism for stabilizing

biologically decomposable organic matter in water sample under

aerobic condition. It is mainly used to determine pollution load of

the waste water, degree of pollution in lake or stream and the

efficiency of waste water treatment system.

Principle : : The method consist of airtight bottle of specified size overflowing

with sample and incubating at specified temperature for five days.

Dissolved oxygen ( D.O.)of blank and sample is measured and BOD is

calculated using the formula. It is necessary to provide standard

condition such as nutrients, pH, temp and mixed group of organism is

seed for determination of BOD. Temperature is controlled at 20O C. The

test is conducted for 5 days as 70 to 80 % is stabilized during the period.

Apparatus : Specially prepared BOD glass bottles provided with exactly fitting

ground glass stoppers and surrounding well to accommodate 5 ml of

water so as to exclude exchange of gases. BOD incubator working at

200C.

Reagents :

i) Distilled water of highest purity and thoroughly aerated so as to

saturate with DO at a lowered temp. of 200C.

ii) Phosphate buffer solution – 8.5 gm KH2PO4 , 21.75 gm

K2 HPO4, 33.4 gm Na2 HPO4.

iii) Magnesium sulphate solution–22.5gm MgSO4H2O dissolved in

1-1 of distilled water.

iv) Calcium chloride solution – 27.5 gm anhydrous CaCl2 dissolved

in 1-1 distilled water.

v) Ferric chloride solution – 0.25 gm FeCl3, 6 H2O in 1-1 distilled

water, All other regents are similar to those in Do measurement.

Preparation of dilution water: Place required amount of aerated distilled water at 200C. Add 1

ml each of phosphate buffer, magnesium sulphate, calcium

chloride and ferric chloride per liter of water, Seed the dilution

water, if necessary by adding 1 to 10 ml of settled sewage (24

to 36 hours old) per liter (seed should not exert more then 0.5

mg/l of depletion of DO in the blank). Seeded dilution water

should be used the day it is prepared..

Dilution Of sample : When the BOD value is expected to be more than 5.0 mg/l,

dilution of the sample is necessary neutralized at pH 7.0. the

sample should be free from residual chlorine. If it contain

chlorine then sodium sulphade should be added . Make several

dilutions of the prepared in DO. Depletion in DO should not be

less than than 2 mg/ L and dissolved oxygen should not be less

than 1 mg/L after 5 days. Generally following dilutions are

suggested

0.1 to 1 % for strong trade waste

1 to 5 % for raw and settle waste / sewage

5 to 25 % for oxidized effluent

25 to 100 % for polluted river water

Procedure for BOD set up : Select a definite volume of sample (less than 300 ml), add to

BOD bottle and fill completely with dilution water. All

concentrations should be in duplicate. Keep one bottle of each

concentrate in the BOD incubator for 5 Days at 200C and subject

the duplicate of that concentration to do determination on the

same day. That will be O – Day do. After 5 days do. Similarly put

one or two bottles for finding out the depletion of DO in blank

(Seeded dilution water only). Find out the difference in

between O day Do and 5 day DO values.

Calculations : 5- BOD mg/l = Initial DO (mg/l) – 5 day DO (mg/l)

Reaction : CnHaObNc + (n+a/4 – b/2 – 3/4C)O2 - CO2 + (a/2 – 3/2C) H2O +

NH3

Result :

Conclusion :

Experiment No

Aim : To determine the Chemical oxygen demand (COD ) of given sample.

This is determined by refluxing the sample with an excess of potassium

dichromate in a highly acid conditions and estimating by titration the amount of

dichromate used. With a reducing agent like ferrous ammonium sulphate.

Theory : The COD is used to measure of oxygen equivalent of organic matter contain

of sample i.e. susceptible to oxidation by a strong chemical oxidant. For ex.

Specific source. COD can be related empirically to BOD organic carbon or

organic matter. The test is useful for monitoring and controlled after

correlation has been established. The dichromate reflux method is prefer to

procedure using other oxidants because of superior oxidizing ability,

applicability to wide varieties of samples and each of the manipulation

oxidation of most organic compound is 95 to 100 % of the theoretical value.

Principle : Most types of organic matters are oxidized by the biolling mixtures and

chromic and sulphuric acid . A sample is reflux in strong acid solution with

the known excess amount of potassium dichromate. After digestion the

reaming unreduced K2Cr2O7 is treated with ferrous ammonium sulphate

(FAS )to determine the amount of K2Cr2O7 consumed and oxidasible organic

matter is calculated in term of oxygen equivalent . Mercuric sulphate is

added to remove inference of chlorides. Silver sulphate is added as a catalyst

as it catalyst oxidation of long chain alphatic compounds.

Interference: Chlorides – 1mg/l Cl- exerts 0.23 mg/l of COD. There fore correction as mg/l

Cl- x 0.23 should be applied the COD of Cl

- from the total COD. Nitrites

exert COD of 1.1 mg/mgN.

Limitations: Amino nitrogen gets converted to ammonia nitrogen. All organic compounds

with few exceptions (e.g. aromatic hydrocarbons, straight aliphatic

compounds and pyridine) are oxidized by this procedure.

Reagents:

i) Standard potassium dichromate 0.25 N

ii) Conc. H2SO4 (A.R. Grade)

iii) Ferroin Indicator – Dissolve 1.485gm 1-10 phenanthroline monohydrate together

with0.695 gm ferrous sulphate (FeSO4, 7H2O) in distilled water and dilute to 100 ml.

iv) Catalyst – Silver Sulphate ( for 8 straight chain sulphatic compds) mercuric sulphate

(for Cl-).

v) Sulphomic acid – Required only if the interference of NO2 is to be eliminated. Add 10

mg sulphamic acid/mg NO2 – N if present, in the refluxing flask. (Do not forget to add

in blank also in this case)

Procedure: Place 50 ml sample or aliquot diluted to 50 ml with distilled water in a 300 ml

round bottom refluxing flask with ground glass joint. Add 25 ml K2Cr2O7 and 75 Conc.

H2SO4 gently shake. Attach refluxing condenser and reflux the mixture for 2 hr. After

refluxing wash the condenser with distilled water. Cool the mixture. Dilute the mixture

with distilled water. Titrate with ferrous ammonium sulphate (0.25 N) with ferroin

indicator till the red colour apprers after the intermediary green colour

Note: -

1. While refluxing if the colour changes to green discard the mixture as

potassium Dichromate is not sufficient to oxidize the solution.

2. For small volumes i.e. 10, 20, 30, 40 ml of samples, proportionate

reduction of potassium Dichromate & sulphuric acid may done)

Calculation:

COD mg/l = (A-B) N x 8000 / V

Where A = Volume in ml. Ferrous ammonium sulphate for blank

B = Volume in ml. Ferrous ammonium sulphate for Sample

V = Volume of sample

N = Normality of ferrous ammonium sulphate

Experiment No

Aim To prepare filter sand from the stock sand and determine effective size and

uniformity coefficient

Theory Natural run –of – bank sand may be too coarse or too fine for a projected filter.

Within economical limits, specified sizing and uniforming can be obtained by

screening out coarse grains and washing out fines. To get the purified water , the

sand shoud have specific properties i.e. D10 , D60.

Formula :

Pusable = 2 ( P60.- P10 )

Ptoo fine = {P10 – 0.2 (P60.- P10 ) }

Ptoo coarse = {P10 + 1.8 (P60.- P10 )}

Procedure Wash the sand and then filter the stock sand by arreging the various screens. Shake

the screens for the 20-30 minutes and the measure the wt of sand retained on each

sieve. From the data calculate the % retained on each sieve . Plot the graph by taking

% finer on ‘Y’ axis and diameter of grain size on ‘ X’ axis. From the graph find out

the value of D60 and D10. Assume the value of coefficient of sand ( CU) in the range

of 1.3-1.8. calculate the Pusable, Ptoo fine , Ptoo coarse .

Experiment No

Aim : To determine the Available Chlorine and Residual Chlorine in a Given Water

Sample

Theory :

Chlorine determination includes:

a) Available chlorine in case of chlorine solution, bleaching powder or chlorine tablets.

b) Chlorine demand (assessment of the requirement of the quantity of chlorine to be added).

c) Residual Chlorine- i) Free residual during Break Point chlorination, ii) Total combined during

Break point Chlorination.

Apparatus: Chlorine Determination Kit (Colour Comapatometer)

a) Available Chlorine (Bleaching Powder):

Reagents:

ii) Bleaching powder solution: Make a paste of 1 g. bleaching powder (CaCl, OCL, H2O) in

minimum water and dilute the paste with distilled water to a volume of 100 ml. Take care

to see that the paste is transferred tot eh volumetric flask quantitatively.

iii) 0.025N sodium Thisulphate: 6.25 g. Na2S2O3, 5H2O is dissolved in 1 liter of distilled water.

iv) Glacial acetic acid.

v) Potassium iodide crystals.

vi) 0.1 N potassium iodate solution: 812 mg dissolved in 250 ml. distilled water.

vii) Starch indicator: 5 g soluble starch mixed with little water and ground in a pestle and

mortar so as to prepare a paste. Paste is than transferred quantitatively to 1 liter of boiling

water. Mixture is then allowed to settle overnight and the supernatant is used.

Procedure:

i) Take 10 ml of bleaching powder solution in a conical flask and add to it KI crystals.

Sufficient distilled water and approximately 2 ml of glacial acetic acid mix.

ii) Titrate the sample till dark amber colored solution turns to pale straw color.

iii) Add starch indicator and Mix.

iv) Titrate till the blue colored starch iodide complex becomes colorless.

v) Prepare reagent blank using distilled water. Note the volume of thiosulphate required.

Calculations:

1 ml of 1 N Thio = 35.45 mg Cl2

1 ml of 0.025 N Thio = 35.45 x 0.025

= 0.88625 or Say 0.89

ml of thio x 0.89 x 100

∴ Percent available= ---------------------------- x (100 in bleaching chlorine ml of B.

P. Solution Powder)

Interference:

Mn, Fe, and NO2 interfere in Iodometric titration. To overcome this interference acetic acid is

used in place of H2SO4 for acidification of the sample.

Reactions:

Ca.OCl.Cl. H2O + 2CH5 COOH � (CH3 COO) 2 Ca + 2H2O+ Cl2

Cl2 + 2 KI �2 KCI + I2

I2 + Starch�Blue colored Starch-iodide complex (Qualitative Test)

I2 + 2Na2 S2 O2� 2Nal + Na2 S4 O6 colorless (Qualitative Test)

Chlorine Demand:

General:

Disinfection is the unit operation which cannot be missed in the treatment plants from the point

of view of supply of safe water. Chlorine either in the form of gas or bleaching powder, is a

universally accepted disinfectant. In addition to destruction of pathogons, chlorination is also used to

achieve oxidation of Fe2+

, NH3 Mn +, H2S removal of taste and odour problems and oxidation of

organic compounds in water or waste waters by forming their chloro-derivatives.

In order to have an effective disinfection by chlorine in water treatment, following points need

scientific thinking viz., determination of chlorine dose so as the have residual chlorine of 0.2 mg/l after

a contact time of 30 min. These points are very well considered in this concept of breakpoint

chlorination.

Principal:

Bleaching powder or chlorine gas when added to the water reacts with organic matter, if

present, and kills pathogens resulting in the formation of chloroderivatives and free residual chlorine.

The dose which achieves this, leaving behind, o.2 to 0.4 mg/l free residual chlorine can be taken as

chlorine demand of that water for the particular contact period, temp. and PH.

Interference:

Unsaturated organic compounds, Fe2+

Mn2+, NO2, organic nitrogen and NH3 are the main

interfering substances.

These can be eliminated by changing the sequence of regent additions using break point

chlorination.

Apparatus:

Chlorine determination kit (color comparator) to measure residual chlorine.

Reagents:

1. Orthotolidine: Dissolve 1.35 g orthotolidine dihydrochloride in 500 ml distilled water.

Add it to a mixture of 350 ml distilled water + 150 ml Conc. HCl. Store in a colored

bottle.

2. Sod. Arsenite: Dissolve 5 g NaAsO2 and dilute to 1000 ml.

3. Standard chlorine solution: Prepare chlorine solution from bleaching powder and

determine its strength as described in the test for available chlorine. This solution has to

be prepared freshly.

Procedure:

1. Take identical aliquots of 100ml well water in 12 conical flasks or bottles (stoppered).

2. Add chlorine solution in increasing quantity to the bottles or flasks to result in a concentration

in the range of 0.1 to 3.00 gm/l. Mix well.

3. Allow to stand for predetermined contact period usually 30 to 60 min.

4. Determine the residual chlorine present in each bottle by OT test as described.

5. Plot chlorine residual Vs chlorine added and determines chlorine required at breakpoint. This

will give the chlorine demand of that specific water.

6. This can also be determined by measuring free residual chlorine left after contact period. The

dose where 0.2 mg/l free residual chlorine is left out will give the amount of chlorine required

to disinfect the water.

Zone 1 : Major reaction in between Cl2 & NH4+

Zone 2 : Formation of dichloramine & the further

oxidation of it

Zone 3 : Free residual Cl2

7.55

CHLORINE DOSE

321

ZONEZONEZONE

RESIDUAL CHLORINE

INCRESING CHLORINE DEMAND

FREE RESIDUAL CHLORINE

CURVE EXPECTED IN ABSENCE OF

CHLORINE DEMANDING MATTER

OXID

ATION

OF CHLORAMINES

BREAK POINT CHLORINATION

Reaction : Cl OH

| |

i) –C = C – + HOCL � - C – C – (unsaturated organic compounds)

| | | |

H H H H

ii) Fe2+

Mn2+

/NO2- + HOCL � Fe

3+ /Mn

3+ /NO3

- (Reducing

substances)

iii) H2S + 4Cl2 + 4H2O � H2SO4 + 8HCl (Reducing substances)

iv) NH3 + HOCl � NH2Cl (Monochloramine) + H2O

NH3 + 2HOCl � NHCl2 (Dichloramine) + 2H2O

NH2Cl + HOCl � NHCl2 + H2O

NH3 + 3HOCl � NCl3* (Trichloramine) + 3H2O

NH2Cl + HOCl � NCl3* + H2O

NH2 + NHCl2 + HOCl � N2O ↑ + 4H2O

NH2Cl + NHCl2 � N2* + 3HCl

Formation of chloramines with respect to pH values:

If pH > 8.4, Monochloramines.

If pH is in the range of 4.4 – 5.5, Dichloramines.

If pH < 4.4, Trichloramines

Complete oxidation state of ammonia or the ‘Break Point.’ After this point no more chlorine demand

will be exerted.

At pH in between 5.5 = 8.4 Mono and Dichloramine exist together Relation of these two species of

chloramines is fixed by the pH Value. At pH 7, Monochloramine: Dichloramino = 50: 50.

Residual Chlorine:

Reagents:

i) Orthotolidine: Weigh 1.35 g. Orthololidine dihydrochloride in 50 ml distilled water. Add to

this mixture 150 ml conc. HCl and make this Vol. to 1 liter. Store the solution in brown

bottle or in the dark.

ii) Sodium arserite solution: Dissolved 0.5 g. Na As O2 (Sodium meta arsenate) in 1000 ml of

distilled water. Take care to avoid ingestion since it is toxic.

iii) Copper sulphate solution: Dissolve 1.5 g. copper sulphate in 50 to 60 ml distilled water,

add 1 ml conc-sulphuric acid and make up to 100 ml.

Potassium dichromate solution: Dissolve 0.25 g of potassium dichromate in distilled water,

add 1 ml of conc. Sulfuric acid and make up to 1 liter.

Standard color solution: Measure in the 100 ml. Nessler cylinder the volumes of copper

sulphate solution and potassium dichromate solution so as given in Table and dilute to 100

ml with distilled water. In column 2 of the table are given amounts of chlorine to which the

color solutions are equivalent.

Procedure:

Use three Nessler cylinders and designate then as cylinders A, B and C. In cylinder a add 1 ml

of O-tolidine regent, 100 ml of the sample, mix and add immediately 2 ml of sod. Arsenite solution,

mix

and after 5 seconds, compare the color with standard color solution.

This reaction (FR) represents the total of free residual chlorine and of interfering substances.

In cylinder B, add 2 ml of sodium arsenite solution and 100 ml of the sample mix and add

immediately 1 ml of orihotolidine reagent. Mix and match the color with standard solutions. This

reading (B1) is the blank for interfering substances after 2 min standing. Also allow it to stand for 5

minutes and record the result. This reading (B2) is the blank for interfering substance after 5 minutes

standing.

In cylinder ‘C’ add 1 ml of O-tolidine reagent and 100 ml of the sample, mix and after 5

minutes match the color. This reading (TR) gives the total residual chlorine plus interfering

substances.

NOTE: Use 0.5 ml Orhotolidine for 10 ml. sample

0.75 ml Orhotolidine for 15 ml. sample

TABLE : Standing color solutions for residual chlorine determination

Sr. No. Chlorine Copper Sulphate Solution Potassium Dichromate Solution

Mg/l ml Ml

I 0.01 0 0.8

II 0.02 0 2.1

III 0.03 0 3.2

IV 0.04 0 4.3

V 0.05 0.4 5.5

VI 0.06 0.8 6.6

VII 0.07 1.2 7.5

VIII 0.08 1.5 8.2

IX 0.09 1.7 9.0

X 0.10 1.8 10.0

XI 0.15 1.8 15.0

XII 0.2 1.9 20.0

XIII 0.25 1.9 20.0

XIV 0.30 1.9 30.0

XV 0.35 1.9 34.0

XVI 0.40 2.0 38.0

XVII 0.50 2.0 45.0

XVIII 0.60 2.0 51.0

XIX 0.70 2.0 58.0

XX 0.80 2.0 63.0

XXI 0.90 2.0 67.0

XXII 1.00 3.0 72.0

Calculation:

a) Free residual chlorine (as Cl) mg/l = FR-B1

b) Total residual chlorine (as Cl) mg/l = TR-B2

c) Combined residual chlorine (as Cl) mg/l = (TR-B2) - (FR-B1)

Range:

The method is applicable up to 5 mg/l of chlorine.

setting time Settled solids t Dissolved Solid +

(Total solids) -----------� Colloidal or suspended solids

Volatile: organic fraction

Dissolved Solids (filtrate) --

Fixed: Residue after burning

At 6000C for 1 hr.

Filter Settled Water -�

Volatile: organic fraction

Suspended Solids --

(Retained on filter paper) Fixed: Residue after burning

At 6000C for 1 hr.

Procedure:

Total Solids:

Pipette out 100 ml of well mixed sample in to the clean dry weighing dish and evaporate to

dryness at 1050C. Cool the weighing dish at R.T. and weigh.

mg of residue x 1000

Total solids mg/1 =

Ml sample.

Fixed residue:

Ignite the residue obtained in total solids determination or volatile solids determination at 550C

to 6000C for one hour in the platinum crucible in the muffle furnace. Cool the crucible and weigh.

Find out the amount of fixed residue in the and contains dissolved

nonvolatile as well as suspended nonvolatile residue. This is known as ignited residue also.

mg of residue x 1000

Ignited residue, mg/1 = ------------------------------------

Ml sample.

Or fixed residue

Volatile residue = Total solids - Ignited residue

Dissolved Solids:

Pipette out 100 ml of the well mixed sample and filter through a filter paper. Collect the filtrate

in a clear and previously weighed evaporating dish at room temperature and weigh,

mg of residue x 1000

Total dissolved solids mg/1 =

Ml sample.

Suspended matter – It is determined generally indirectly as follows-

Suspended matter mg/ 1 = Total solids mg/1 – Dissolved solids mg/1

If the Glass Fiber filter paper GFC grade (What man make) is available this determination can

be done directly. This paper keeps the consistency not only at 1050C but at 600

0C also. While using

this filter paper a separate filter assembly known as Hartley’s Filter assembly is made use of.