PRINCIPAL HOD LAB INCHARG

Dr.S. D. Deshmukh Dr.V.M.Arole S.N.Deshmukh

PRINCIPAL HOD LAB INCHARG

Dr.S. D. Deshmukh Dr.V.M.Arole S.N.Deshmukh

2013

MGM’SMGM’SMGM’SMGM’S Jawaharlal Nehru Engineering College,

N-6, CIDCO, Aurangabad.

LAB MANUALLAB MANUALLAB MANUALLAB MANUAL

FIRST YEAR (EC & ES)

PRINCIPAL, HOD Dr. S. D. Deshmukh Dr. V. M. Arole

Lab In-charge: - Prof. S. N. Deshmukh

MGM’s

Jawaharlal Nehru Engineering College,

N-6, CIDCO, Aurangabad.

Lab Manual

Engineering Chemistry and

Environmental Chemistry

Academic Year: - 2013-14

Principal HOD Lab In-charge

List of Experiments

Sr. No. Name of Experiment

1 To prepare phenol-formaldehyde resin.

2 To determine the acid value of given plastic material.

3 To determine the acid value of given oil sample.

4 To determine the normality of acids by conductometric titration.

5 To determine the cell constant of conductivity cell by using KCl solution.

6 To determine the calcium and magnesium hardness of water by EDTA method.

7 To determine the pH of unknown solution by pH meter and pH paper.

8 To study the factor’s influcing on rate of electrochemical corrosion.

9 To determine the relative viscosity of different liquid with respective water at different

temperatures.

10 To determine the chlorine content from given water sample.

11 To determine the Dissolve Oxygen from given water sample.

Experiment No: - 01

Aim: - Preparation of Phenol-Formaldehyde resin (Bakelite).

Apparatus: - Beaker, Conical flask, glass rod, measuring cylinder, fractional weight box etc.

Chemicals: - Glacial acetic acid, 40 % formaldehyde solution, phenol, conc. HCl, Dist.water etc.

Theory: - Phenol resins are condensation polymerization product of phenolic derivative with

aldehyde (like formaldehyde, furfural) It is prepared by condensing phenol with formaldehyde in

presence of acid or alkaline catalyst.

Reaction:-

OH

C

O

H H

OH

H

OHH2C

OH

OH

H2C OH

Ortho Para

Hydroxry,Benzyl Alcohols

Step I:- Formation of ortho and para hydroxy benzyl alcohols from phenol and formaldehyde:-

Step II:- Formation of Novalac from ortho-hydroxy,benzyl alcohol

OHH2C OH

n

n

OHH2C OH Heat/ -H2O

OH

H2C

OH

Novalac

Step III:- Formation of Bakelite from Novalac:-

OH

H2C

OH

OH

H2C

OH

H

H H

HC O

H

H

C O

H

H

Hexamethylene Tetraamine

H2C

OH

CH2

OH

*

CH2

H2C

OH

*

OH

NovalacBakelite

Phenol Formaldehyde

Ortho-Hydroxy Benzyl Alcohol

Procedure:-

1) Place 5 ml of glacial acetic acid and 2.5 ml of formaldehyde solution in 500 ml beaker.

2) Add 2 gr of phenol and 1 ml of conc. HCl solution in it.

3) Heat the solution slowly with constant stirring for 5 minutes.

4) A large mass of pink Plastic is formed.

5) The residue obtained is washed several times with distilled water.

6) Dry the product and calculate the yield accurately.

Result: - The weight of obtained Bakelite is ………….. grams.

Question-

1. Define the term resin.

2. Explain condensation polymerization.

3. Which types of monomers are used for making Bakelite?

4. What is the role of catalyst in the reaction?

5. Explain the properties of Bakelite.

6. Uses of Bakelite.

Experiment No: - 02

Aim: - To determine the acid value of given plastic material.

Apparatus: - Burette, pipette, conical flask, glass rod etc.

Chemicals: - Ethyl alcohol, Std. KOH solution, Plastic, Phenolphthalein, Acetone, Benzene etc.

Theory: - The acid value is number of milligrams of KOH which are needed to neutralize the acidity

present in 1 gr of material. For its determination material is dissolved on suitable solvent usually

neutral ethyl alcohol. It is then titrated against standard alkali solution (0.1 N KOH) using

phenolphthalein as an indicator. Acid value is calculated by formula,

Acid Value = Volume of 0.1 N KOH used in ml × 5.6

Weight of sample taken

Procedure:-

1) Weight exactly around 5 gr of the sample on watch glass and dissolved it into 50 ml of neutral

ethanol.

2) Heat the solution for one minute in water bath then cool it.

3) Add 2 to 3 drops of phenolphthalein indicator in it.

4) Titrate the solution with 0.1 N KOH till the color of solution is faint pink color.

Observation Table:-

Sr. No Weight of Sample (in gm) Burette reading (in ml) Mean

1

2

3

Calculations: - Acid Value = Volume of 0.1 N KOH used in ml × 5.6

Weight of sample taken

Result: - Acid value of given plastic material is …………… mg of KOH.

Questions-

1. What is the acid value?

2. How acid value of Plastic sample is determined?

3. What happen? When acid value of sample is more than permissible value.

Experiment No: - 03

Aim: - To determine neutralization number of Oil.

Apparatus: - Burette, pipette, conical flask, glass rod etc.

Chemicals: - Ethyl alcohol, Std. KOH solution, Oil, Phenolphthalein, Acetone, Benzene etc.

Theory: - The acid value of lubricating oil is defined as number of milligrams of KOH required to

neutralize the free acid present in one gram of oil sample. In good lubricating oil acid value should

minimum increase in acid value should be taken as an indicator of oxidation of oil which may lead to

gum and sludge formation decides corrosion. A known weight of oil sample is dissolved in a suitable

solvent and titrates with a standard alcoholic KOH solution.

Reaction:-

H+ + OH

- H2O --------- (1)

RCOOH + KOH RCOOK + H2O -------- (2)

Procedure:-

1) Weight exactly around 5 gr of the sample on watch glass and dissolved it

In to 50 ml of neutral ethanol.

2) Heat the solution for 30 minute in water bath cool it and add 2 to 3 drops of

Phenolphthalein indicator.

3) Titrate the solution with 0.1 N KOH till the color of solution is faint pink color.

Observation Table:-

Sr. No Weight of Sample ( in gm) Burette reading (in ml) Mean

1

2

3

Formula:-

Neutralization number = Volume of 0.1 N KOH used in ml × 5.6

Weight of sample taken

Calculations:-

Result: - Neutralization number of given Oil sample is …………… mg of KOH.

Question:-

1. Explain the properties of good lubricant.

2. Define term Neutralization number.

3. What happen? When acid value of lubricant is more than permissible value.

4. Why the acid value of lubricant is determined before using in lubrication process?

Experiment No: - 04

Aim: - To find the normality of hydrochloric acid and acetic acid by titrating the mix of HCl &

CH3COOH against sodium hydroxide solution conductometrically.

Apparatus: - Conductometer, Magnetic stirrer, Beaker, Pipette, Burette, etc

Chemicals: - Hydrochloric acid, Acetic acid, Sodium hydroxide, distilled water etc.

Theory: - Conductometry can be used to determine the end point of a titration. This method is based

on the measurement of conductance during the course of titration. The conductance varies

differently before and after the equivalence point. This is due to the reason that electrical

conductance of a solution depends upon the number of ions present and their ionic mobility’s i.e.

speeds. When conductance values plotted against volume of titrants added, U shape is obtain, the

point of intersection of lines gives the end point.

Reaction:-

HCl + NaOH NaCl + H20

CH3COOH + NaOH CH3COONa + H20

Procedure:-

1. Prepare the 10 ml mixture of 5ml hydrochloric acid solution and 5ml of acetic acid solution in a

beaker. Add few ml of distilled water in it.

2. Immerse the conductivity cell in the solution so that the electrodes completely dip in solution.

3. Note down the conductance of solution.

4. From the burette add 0.3N NaOH solution in 0.5 ml lots with continuous stirring and measure

the conductance of solution at each 0.5ml addition of NaOH.

5. Plot a graph between observed conductance values along y-axis against the volume of alkali

added along x-axis. The end point of intersection gives the amount of alkali required for

neutralization of acid.

Observation Table:-

Sr.No. Volume of NaOH added (in ml) Observed conductance (in mhos)

1 0 ml

2 0.5 ml

3 1 ml

19 9 ml

20 9.5 ml

Calculations: -

1) N1V1 = N2V2 2) N3V3 = N4V4

HCl = NaOH CH3COOH = NaOH

N1 × 5 = 0.3 × V2 N3 × 5 = 0.3 × V4

3) Strength of HCl = N1 × Eq. Wt 4) Strength of CH3COOH = N3 × Eq. Wt

Result: -

1. Normality of given hydrochloric acid is = …………… N

2. Normality of given acetic acid is = ………………….. N

3. Strength of hydrochloric acid in given solution is = ………….. gr/lit

4. Strength of acetic acid in given solution is = …………… gr/lit

Question -

1. What is mean by conductance?

2. How you can calculate the normality of acid by conductometrically.

3. Which acid first neutralize &why.

4. By plotting graph how you calculate normality of acids.

5. After neutralization of both acids excess of NaOH added what is the effect on conductance

Experiment No: - 05

Aim: - Determination of cell constant of a conductivity cell.

Apparatus: - Conductometer, Conductivity cell, beaker, Standard flask etc.

Chemicals: - Potassium chloride, distilled water etc.

Theory: - Conductance is the reciprocal of resistance. It is depend upon three factors, number of ions,

and nature of ions and mobility of ions towards their respective electrodes. The specific conductance

of electrolyte is decreased by increasing its dilution and equivalent conductance is increased by

increasing its dilution. The observed conductivity of an electrolyte will be equal to its specific

conductivity if cell constant is one.

Procedure: - Connect conductivity cell to Conductivity Bridge and keep the cell deepen in distilled

water.

Prepare the solutions of 0.1, 0.01, 0.02, 0.005 and 0.001 N of KCl.

PART I – Determination of cell constant.

Note down the conductance of 0.1 N and 0.02 N KCl solutions by using conductometer and

determine the cell constant.

PART II –

Determine the conductance of all prepared KCl solutions and calculate the specific and equivalent

conductance of each solution.

Plot a graph Equivalent conductance against concentration of solution.

Formula:-

Specific conductance

Cell Constant = Observed conductance

Specific Conductance = Cell constant × Observed conductance.

K × 1000

Equivalent Conductance = C

Observation Table:-

PART I –

Sr

No

Conc. OF KCl

solution

Observed

Conductance

Specific

(k)Conductance

Cell

Constant

Mean

1 0.1 N 0.01313

2 0.02 N 0.02819

PART II –

Sr

No.

Conc. of KCl solution Observed

Conductance

Specific

Conductance

(k)

Equivalent

conductance

(λ)

√C

1 0.02 N

2 0.01 N

3 0.005 N

4 0.001 N

5 0.1 N

Calculations:-

1) Specific conductance:-

2) Equivalent Conductance:-

Result: - The cell constant of the given conductivity cell is = ………………… /cm

Question-

1. Define the term Specific & equivalent conductance?

2. What is effect of dilution on Specific &equivalent conductance?

3. Why we can use KCl solution for determination of cell constant?

4. How you calculate cell constant?

5. Difference between Specific &equivalent conductance?

Experiment No: - 06

Aim: - Determination of total hardness of water by EDTA method.

Apparatus: - Beaker, conical flask, burette, pipette, standard flask etc.

Chemicals: - Calcium carbonate, EDTA, Buffer solution, EBT dye etc.

Theory: - In hard water sample the total hardness can be determined by titrating the Ca2+ and Mg2+

present in an aliquot of the water sample at pH10 with EDTA using EBT indicator. Permanent

hardness can be determined by precipitating the temporary hardness by prolonged boiling followed

by titration with EDTA solution. The difference in the titre values corresponds to the temporary

hardness of the water sample.

When Eriochrome Black-T dye is added to the hard water at pH around 10 it gives wine red colored

unstable complex with Ca2+ and Mg2+ ions of the sample water. Now when this wine red-colored

complex is titrated against EDTA solution (of known strength) the color of the complex changes from

wine red to original blue color showing the end point.

Reaction:-

H2C

H2C NN

H2C

H2C

H2C

CH2

C

C

O

O

OH

OH

C

C

O

O

HO

HO

EDTA

Ca2+

Mg2+

+ Erinchrome Black T

Ca2+

Mg2+

Erinchrome Black T Complex

of water Unstable complex (wine red)

Ca2+

Mg2+

Erinchrome Black T Complex

Unstable complex (wine red)

Ca2+

Mg2+

EDTA Complex + Erinchrome Black T

Ethylene diamine tetra-acetic acid

Blue (Stable complex)

Procedure:-

Standardization of EDTA solution with standard Hard water:

1. Take 25ml of standard hard water in a conical flask.

2. Add 5 ml of buffer solution and 2 to 3 drops of EBT indicator, the color of solution turns wine

red. Titrate the flask solution against standard EDTA solution from the burette until the color changes

from wine red to blue at the end point. Take at least two concordant readings. Let the volume of

EDTA solution used = V1 ml

3. Titrate similarly unknown hard water and find out volume of EDTA solution used. Let the volume

of EDTA used with unknown hard water = V2 ml

4. Take 250 ml of the hard water sample in 500 ml beaker & boil gently for about one hour, cool,

filter into a 250 ml measuring flask & make the volume up to the mark with distilled water. Take 25

ml of this solution and proceed in the same way as in step (2). The volume of EDTA used (V3)

corresponds to permanent hardness of the water sample. Temporary hardness is calculated by

subtracting permanent hardness from total hardness. Let volume used = V3 ml

Observation Table:-

1. For standard Hard water: -

Sr.No.

Volume of Standard Hard

Water taken in conical Flask A

(ml)

Volume of EDTA

solution used up (ml)

Mean (V1 ml)

1 10 ml

2 10 ml

3 10 ml

2. For unknown Hard Water: -

Sr.No. Volume of Sample Hard Water

taken in conical Flask A (ml)

Volume of EDTA

solution used up (ml)

Mean (V2 ml)

1 10 ml

2 10 ml

3 10 ml

3. For Boiled Water: -

Sr.No. Volume of Boiled Water taken

in conical Flask A (ml)

Volume of EDTA

solution used up (ml)

Mean (V3 ml)

1 10 ml

2 10 ml

3 10 ml

Calculations:-

1) Standardization of EDTA solution:- 2) Total Hardness:-

1 ml EDTA = 1 mg Ca ++

1ml of 1M EDTA=100mg of CaCO3

1ml of 1M EDTA=100mg of CaCO3 V1 ml of 0.01M of EDTA=------of CaCO3

=V1 × 100×0.01

1

10ml of EDTA=------mg of CaCO3

1000ml of EDTA=----- mg of CaCO3

3) Permanent Hardness = 4) Temporary hardness= Total hard. - Permanent hard.

1ml of 1M EDTA=100mg of CaCO3

V2 ml of 0.01M of EDTA=------of CaCO3

=V2 × 100×0.01

1

10ml of EDTA=------mg of CaCO3

1000ml of EDTA=----- mg of CaCO3

Result: - The given water sample contains,

• Temporary hardness = ………………………. ppm.

• Permanent Hardness = ……………………… ppm.

• Total Hardness = ………………………………… ppm.

Question:-

1. What is the other name for temporary and permanent hardness?

2. What is mean by Hardness of water?

3. Which factor responsible for hardness.

4. What is unit of hardness?

5. Complete the reaction.

Mg (HCO3)2 Boil ?

Experiment No: - 07

Aim: - To determine pH value of solution by Indicator pH meter and pH paper.

Apparatus: - Standard flask Beaker PH Meter PH Paper Glass Electrode etc.

Chemicals: - Buffer solution, hydrochloric acid, sodium hydroxide etc.

Theory: - The pH value of solution is defined as negative logarithm of hydrogen ion concentration. It is

expressed in gram/ion lit.

pH = -Log [H+] = Log10 1/[H

+]

AT 250 the ionic product of water

Kw = [H+] [OH

-] =10

-14

Also in pure water [H+] = [OH

-]

In neutral solution [H+] = [OH

-]

[H+] [OH

-] = (1×10

-7) (1×10

-7)

pH of neutral solution = -Log10 10-7

= 7

Consequently, PH of neutral solution is 7; in acidic solution PH<7, since [H+]>10

-7 on the other hand PH

of the alkaline sol.n is >7, Since [H

+] <10

-7

pH value of distilled water = 7

pH of the sol.n is tasted by using pH paper. This gives different colour with different values.

Procedure:

1) Take the standard solution in 100mlBeaker

2) Deep the pH electrode in the Beaker & note down the pH.

3) Then take the another solution in a beaker deep the electrode & note pH of solution also

deep the pH paper & observe the colour Change& note down the nature of solution.

4) Take the four different reading of four different solutions & Write the result either the solution

are acidic, basic or neutral.

Observation Table:-

Sr.No Solution pH paper pH of

solution

Change in

Colour

Type of

solution

1 Unknown Solution-1 Neutral pH

paper

2 Unknown Solution-2 Neutral pH

paper

3 Unknown Solution-3 Neutral pH

paper

Result: - 1) pH of first unknown solution is ------ hence the nature of solution is --------

2) pH of second unknown solution is ------ hence the nature of solution is --------

3) pH of third unknown solution is ------ hence the nature of solution is --------

Question:-

1. Define term pH value.

2. What is the pH of water?

3. Which different methods are used for determination of pH.

4. Explain the construction and working of ph electrode.

Experiment No: - 08

Aim: - To Study the factors influencing on rate of Electrochemical Corrosion.

Apparatus: - Daniel Cell etc.

Chemicals: - Sulphuric acid etc.

Theory: - Corrosion is defined as disintegration of deterioration of metals by chemical or

electrochemical reaction with its environment. The electrochemical corrosion occurs when a

conducting liquid is in contact with metal or when two dissimilar metals or alloys either immersed of

disposed partially in a solution.

This corrosion occurs due to existence of separate anodic and catholic part between which current

flows to the conducting solution. When there is acidic environment evolution of hydrogen gas takes

place. Thus this type of corrosion causes displacement of hydrogen ions from all acidic solution by

metal ions.

All metals above hydrogen in electrochemical cells have a tendency to get dissolved in acidic

solution with simultaneous evolution of hydrogen gas.

Daniel cell consist of Zinc electrode dipped in Zinc Sulphate solution (Oxidation takes place) and

Copper electrode dipped in Copper Sulphate solution (Where reduction takes place).

Daniel Cell can be represented as,

Zn / ZnSO 4 // CuSO4 / Cu

Zinc is present at higher stage in electrochemical series hence oxidation occurred at Zinc

electrode and Zinc get corrode.

Diagram:-

Galvanic Cell

Conclusion:-

Question-

1. What is mean by electrochemical corrosion?

2. Which factor affecting on electrochemical corrosion?

3. What is mean by Oxidation-reduction?

4. What is the role of electrolyte in electrochemical corrosion?

5. Difference between Dry corrosion and Wet corrosion?

Experiment No: - 09

Aim: - To determine the relative viscosity of given liquid with respect to water at room temperature by

Ostwald’s viscometer.

Apparatus: - Viscometer, stopwatch, beaker, water bath, thermometer etc.

Chemicals: - Dist. Water, ethanol acetone etc.

Theory: - The property of resistance to flow when a stress is applied to a liquid is called “Viscosity”. In the

process of flow the molecule comprising the fluid move fast one another and viscosity arises from what can

the termed the frictional effect of relative motion. When the liquid is flowing to a circular tube the flow

pattern is called streamlines or viscous or laminar. The viscosity is increased by increasing molecular weight

and decrease by increasing the temperature.

Formula: - n = ℓ2 / ℓ 1 × t2/t1 × n1

ℓ 1 = Density of H2O at 25oc

ℓ2 = Density of given liquid at toc

n1 = Viscosity of water at toc

n2 = Viscosity of given liquid at toc

t1 = Time flow of given liquid at toc

Procedure:-

1) Clean the viscometer with chromic acid and then wash thoroughly with distilled water. It & finally washed

with acetone and dried.

2) A sufficient volume of distilled water is introduced by pipette in bulb B so that the bent Portion of tube

and half or a little more than a half of bulb B is filled up.

3) Clamp the viscometer in quite vertical position.

4) Trough the rubber tube attach to upper arm of bulb A, suck up water until it rises above the upper mark

C and allow it to flow under its own weight.

5) The time of flow of water from C to D is continued by starting the stop watch as the Meniscus just

reaches upper mark C and stopping the watch as the meniscus just passes the lower mark D.

6) Take at least three reading of water at different temperature. Also take the similar reading of ethanol at

various temperatures.

Observation Table:-

Sr. No Different liquids Temperature Flow of time in

sec.

Density Viscosity

1 Ethanol 25oc 0.998

2 Ethanol 50oc 0.985

3 Water 25oc 0.786 0.8004

4 Water 50oc 0.773 0.5083

Calculations:- ƞ25oc= ℓ2/ℓ1 × t2/t1 × n1 w

ƞ50oc= ℓ2/ℓ1 × t2/t1 × n1 w

Result: - The relative viscosity of the liquid with respective with water at 250C temperature

is …………… poise and at 500C temperature is …………… poise

Question:-

1. Define term viscosity.

2. What is effect of molecular weight on viscosity

3. Explain the another method used for determination of viscosity of liquid,

Experiment No: - 10

Aim: - Preparation of Urea-Formaldehyde resin.

Apparatus: - Beaker, glass rod, water bath, funnel etc.

Chemicals: - Formaldehyde (40%), Urea, Conc. Sulphuric acid, dist. Water etc.

Theory: - Urea resins are condensation polymerization product of urea with formaldehyde. It is

prepared by condensing urea with formaldehyde in presence of acid or alkaline catalyst.

Reaction:-

O

C

H H

O

C

H2N NH2

O C

HN

NH2

H2C OH

O

C

H Hi)

iii)n n

O

C

H2N NH2

O

C

H H

Urea FormaldehydeMonomethylol Urea

O C

HN

HN

H2C

H2C

OH

OH

Dimethylol Urea

Heat

Heat

UreaFormaldehyde

O C

N

N

H2C

H2C

N

N

CO

H2C N

H2C N

C

FormaldehydeUrea

O

Procedure:-

1) Place 5 ml of 40% formaldehyde solution in 100 ml beaker.

2) Add about 2.5gr of urea while stirring until a saturated solution is obtain.

3) Add a few drops of concentrated sulphuric acid with stirring continuously during addition.

4) All of a sudden a voluminous white solid mass is appear in a beaker.

5) When the reaction is complete wash the residue with water and dry the product and calculate the

yield of the product formed.

Result: - The yield of obtained urea-formaldehyde resin is = ………………… gram.

Experiment No: -11

Aim: - To estimate the amount of dissolved oxygen in given water sample.

Apparatus: - Burette, pipette, iodine flask.

Chemicals:- Conc. HCl, Sodium hydroxide, potassium iodide, MnSO4, Sodium Azide, 0.1 N Sodium

thiosulphate , water sample.

Theory: - Oxygen dissolved in water to the extent of 7-9 mgs/lit at a temperature range of 250 -35

0 C.

The estimation of dissolved oxygen in water is useful in studying corrosion effect of boiler feed water

and in studying water pollution. The amount of dissolved oxygen in water is estimated using winkler’s

reagent (potassium bromide + Potassium bromated).

Water sample is collected carefully aeration/deaeration in ground stopperd flask. Initially manganese

sulphate and alkaline-iodide reagents are added and the reaction occur as follows

2KOH + MnSO4 → Mn (OH)2 + K2SO4

Mn2+

+2OH- →Mn (OH) 2↓ (white)

2 Mn (OH)2 +

+ O2 →2 MnO (OH) 2 ↓ (yellow brown)

The precipitate dissolves in concentrated sulphuric acid liberating iodine and the liberated iodine is

titrated against Na2S2O3.

MnO (OH)2 + 2H2SO4→Mn(SO4)2 + 3H2O

Mn (SO4)2 + 2KI → MnSO4 + K2SO4 + I2

Procedure:-

1. Take 50 ml water sample in a iodine flask.

2. Add to it 1ml of MnSO4solution and 1ml of basic KI solution.

3. Shake it, the precipitate is allowed to settle down half a way and shake again. Repeat the

procedure for three times.

4. Then add 1ml of Conc. H2SO4 to dissolve the precipitate.

5. Then add 2-3 drops of starch indicator.

6. Titrate the solution against the thiosulphate solution till colour changes from blue to colourless.

Observations:-

In conical flask: - 50ml water + 1ml of MnSO4 solution + 1ml of basic KI solution

In burette: - 0.1 N Sodium thiosulphate

Indicator: - Starch

End Point: - Blue to Colorless

Observation table:-

Sr.No

Volume of water sample

taken

Burette Reading (in ml)

MBR

Initial Final Difference

1. 50 ml

2. 50 ml

3. 50 ml

Calculations:-

Volume of sodium thiosulphate V1 = …………………. ml

Strength of sodium thiosulphate N1 = ………………… N

Volume of water sample V2 = 50ml

Strength of dissolved water sample N2 =?

V1N1=V2N2

N2= V1N1/V2

=……………ml × ………..N/50

Amount of dissolved oxygen in 1000 ml of H2O = normality × eq.wt.of O2×1000

One litre of tap water = ……………….N × 8 × 1000

= ………………..mgs/lit

Result: - Amount of O2 present in given water sample is __________ ppm

Question:-

1 What is the role of buffer solution?

2 Why estimation of dissolved oxygen is of great significance?

3 What is the role of KI used in estimation?

4 Complete the reaction Mn (SO4)2 + 2KI

4 What is the amount of dissolved oxygen present in pure water?