Applied Chemistry I Lab Manual

TSSM’s

Bhivarabai Sawant College of Engineering & Research,

Narhe, Pune - 41.

Department of Applied Chemistry

List of Experiments

Semester – I

1. Preparation of phenol formaldehyde and its characterization.

2. To determine Molecular weight of a polymer using Ostwald

Viscometer.

3. To determine Chloride content in given sample of water by

using Volhard’s method.

4. To Standardize KMnO4 solution by preparing standard oxalic

acid and to estimate ferrous ions.

Laboratory Incharge Principal

Experiment No: 1

Aim: Preparation of phenol formaldehyde and its characterization.

Apparatus: Beakers, Glass rod, measuring cylinder etc.

Chemicals: Glacial acetic acid, 40% formaldehyde solution, Conc. HCL, Phenol

etc.

Theory: Phenoplasts are the condensation polymerization products a phenol with

an aldehyde.The phenolic substance can be phenol or resorcinol and the aldehyde

can be formaldehyde or furfural. They undergo condensation polymersation

reaction to give a network structure (three dimensional).Phenol formaldehyde is

thermosetting type of polymer. There are three stages in the preparation of the

thermosetting Phenol formaldehyde plastic. The reaction can be given as

Stage I: Phenol is heated with formaldehyde in the presence of catalyst acid to get

methylol phenols.

OH OH OH

H+ CH2OH

+ HCHO OR

CH2OH

Phenol Formaldehyde Monomethylol phenol

Stage II: The methylol phenols can undergo polymerization by condensation to

form low molecular weight linear polymer, called as novolac.

The Novolac resin is a linear polymer, thermosofting and soluble in few aromatic

solvents.

OH OH OH

CH2 CH2

Novolac

Stage III:

Further heating of novolac above 150°C in the presence of hexamethylene

teteramine, in a mould, produces highly cross linked structure called as Bakelite,

which is hard and rigid. The formaldehyde so formed act as cross linking agent for

novolac, to convert in to bakelite.The CH2O removes H-atoms from Para-positions

of rings on novolac chain and connect them through –CH2-linkages.

OH OH

CH2

CH2 CH2

CH2

OH OH

Bakelite

Procedure:

1. Take 5ml of glacial acetic acid in a 100 ml beaker.

2. And add 2.5 ml 40% formaldehyde solution by measuring Cylinder in a

beaker and 2gm of phenol.

3. Add few drop of Conc.HCl into the mixture.

4. Stir the mixture continuously (since the reaction is vigorous addition

should be done carefully preferably in a fuming chamber) Phenol

formaldehyde (a pink colour resin) is produced within five minutes.

Result: A pink colored phenol formaldehyde resin is formed.

Experiment No: 2

Aim: To determine molecular weight of a high polymer using its solution with

different concentrations.

Apparatus: Ostwald’s viscometer, stop watch, pipettes.

Chemicals: Pure solvent, 0.1%, 0.2%, 0.3%, 0.4%, 0.5%, solutions of unknown

polymer.

Theory: Viscosity measures the resistance of flow offered by fluid layers to each

other during the flow. The coefficient of viscosity (η). Is the force required per unit

area of the layer to maintain a unit difference of velocity between two parallel

layers of the liquid one centimeter apart at the given temperature. It is measured in

‘poise’ in CGS unit.

Viscosity determinations are very important in the study of high

polymers. Using simple viscosity measurements an average molecular weight of

the polymer may be determined.

According to Posieuilles equations,

OR …………… I

Where

(η)= coefficient of viscosity (poise)

P = difference in pressures at the two end of the tube to maintain uniform

rate of flow.

r = radius of the capillary tube (cm)

t = time (seconds) required to flow ‘V’ volume.

V = volume (ml) of the liquid flowing out of the capillary in time‘t’ seconds.

For two liquid of the same volume and with the use of same viscometer the

equation can be written as

η1 and

= =

As pressure is directly proportional to the density of the liquid (P

By comparing the viscosities of pure solvent and the solution of polymer in the

same solvent, the molecular weight of high polymer is given by following

expression-

…………… 3

Where M = Molecular weight of high polymer

K = Constant for a given polymer for the particular solvent and

Temperature

α = Function of the geometry of the molecules.

C = Concentration of polymer in g 100 ml-1

solution.

(η) = Viscosity of solution of high polymer.

(η0)= Viscosity of pure solvent

As we know

[η1/η0-1] = ηsp = specific viscosity

This equation is valid only for very dilute solution (less than 1%)

The graph of ηsp/C against 'C’ is extrapolated to zero concentration. The

extrapolated value is known as the intrinsic viscosity (η).

The intrinsic viscosity (η).can also be determined from the plot of 2.303/C

log η/η0 against C. The plot is straight line giving intercept on y-axis. The intercept

value describes intrinsic viscosity (η). The molecular weight of polymer is given

by following equation known as Mark- Houwink expression.

(η) = KMα

----------------- 4

The values of k and α, listed below, have been determined.

Part I: To determine the time of flow for a given pure solvent.

1. Take a clean and dry viscometer and clamp it in a perfectly vertical

position to an iron stand.

2. Keeps the viscometer in a thermostat or beaker containing water in order

to retain constant temperature.

3. Introduce exactly 20 ml of pure solvent (water) to the viscometer from

clamp side.

4. Suck the liquid by means of rubber tube, above the mark ‘X’.

5. Release the liquid and start the stopwatch when liquid leaves upper mark

‘X’.

6. Stop the stopwatch when liquid reaches lower mark ‘Y’.

7. Take three time of flow (in seconds) for a liquid leaves upper mark ‘X’.

8. Record three readings; Remove the pure solvent from the viscometer.

9. Then wash the viscometer carefully with acetone. Dry it by passing

current of hot air (hair dryer can be used).

Part II: To determine the time of flow for a given solution of polymer.

1. Introduce exactly 20 ml of the 0.1% solution of unknown polymer in to

the viscometer from clamp side.

2. Record the procedure from (4) to (6) as given in part I.

3. Record the time of flow (in seconds) for 0.1% solution of polymer.

4. Similarly repeat the procedure for 0.2%, 0.3%, 0.4%, 0.5% s solution of

polymer and the time of flow.

Figure:

Ostwald’s Viscometer

Observation Table (Part I) :

Solvent Time of flow (seconds) Mean Time t0 (seconds)

1 2 3

Observation table (Part II) :

Conc.

of

polymer

Time ‘t’ (sec) t/to=ŋ/ ŋo

ŋ/ŋ0-1= ŋsp

ŋsp/C 2.303/C. log ŋ/ ŋo

1 2 Mean

0.1%

0.2%

0.3%

0.4%

0.5%

Graph:

1. Plot the graph of ŋsp/C against % concentration of polymer. Find out

Intrinsic viscosity (ŋ) by extrapolation to zero concentration.

2. Also plot the graph of 2.303/c. log ŋ/ ŋ0 against % concentration of

Polymer.

Find out intrinsic viscosity (ŋ) by extrapolation to zero concentration

Ask for values of k & α

Thus knowing the values of (ŋ) intrinsic viscosity of given polymer from

the graph and the constant k & α (supplied), the molecular weight of

high polymer can be calculated by using the formula.

(ŋ) = KMα i.e. M

α = (ŋ)/k

α log M = log

Log M = log

Knowing (ŋ), K, and α, find log M and then M

2.303/C. log ŋ/ ŋo

ŋsp/C

(0,0) % concentration (0,0) % concentration

Result Table:

Molecular Weight

From graph (I)

From graph (II)

Experiment No: 3

Aim: To determine calcium from the given sample of cement by

Volumetric method.

Apparatus: Burette, Pipette (25ml), conical flask, Beakers, Test tube,

Volumetric flask (250ml) etc.

Chemicals: Cement sample,0.01 M EDTA, Conc. HNO3, NH4Cl, 1:1NH3,

EBT indicator.

Theory: Cement is essential building material used for construction, plaster

& concrete making. It is the mixture of calcium silicate & calcium aluminates,

which has the property of setting & hardening in presence of water cement is

chemically analyzed in terms of % of oxides of calcium, silicon, aluminium,iron,

Magnesium as CaO, SiO2, Al2O3, Fe2O3, and MgO respectively. But the calcium

is the major constituent of cement.

Procedure : Perform the experiment in two parts

Part I : Dissolution of cement sample

1. Weight about 1gm of cement sample and take it in to the

250 ml beaker.

2. Add 10 ml of conc. HCl and 2 ml distilled water.

3. Heat it on low flame on wire gauze with stirring till it dissolve.

4. Cool it, filter it, and collect the filtrate in 250 ml volumetric flask.

5. Dilute the solution up to mark with distilled water and use it as a stock solution.

6. Pipette out 50ml of this stock solution in beaker.

7. Add 1 ml conc. HNO3 and boil the solution, then add 1gm NH4Cl and

One test tube 1:1 NH3 solution with constant stirring. So Al & Fe will

Precipitate in the form of hydroxide. Filtrate solution to remove precipitate.

8. Collect the filtrate in 250 ml volumetric flask and dilute up to mark.

Part II : Estimation of calcium volumetrically.

1. Pipette out 10 ml of above sample solution in conical flask.

2. Add 5 ml of buffer solution (PH=10) and 5 drop of EBT indicator.

3. Fill the burette with 0.01 M EDTA solution from burette till wine

red colour changes to blue, call this burette reading as ‘X’ ml

4. Titrate the sample solution against 0.01 M EDTA solution from

burette till wine red colour changes to blue, call this burette reading as ‘X1’ ml.

5. Repeat the procedure to get at least three reading and Note down C.B.R as

‘X’ ml.

6. Reading corresponds to Calcium and Magnesium, but Mg is very negligible, so

take it for Calcium.

Observations :

Part II : Estimation of Calcium Volumetrically

Burette: 0.01 EDTA

Pipette: Cement sample solution

Indicator: EBT

End Point: Wine red to blue

Reactions:

Ca2+

+ EBT Ca-EBT

Blue wine red colour complex

Ca2+

+ Na2EDTA Ca-EDTA + 2 Na+

Colorless

Ca-EBT +Na2EDTA Ca-EDTA + 2 Na

+ + EBT

Wine red colour complex Colorless Blue

Observation Table :

Pilot Reading

Burette Reading (ml) CBR ‘X’ ml

X1 X2 X3

Calculations :

Step I: Quantity of Ca2+

in 10 ml of cement sample solution :

1000 ml of 1M EDTA = 40 gm of Ca2+

1000 ml of 0.01 M EDTA = 0.4 gm of Ca

2+

‘X’ ml 0.01 M EDTA = Y gm of Ca2+

=………… gm Ca2+

Step II: Since 10 ml of cement sample solution contains Y gm =………gm of Ca2+

250 ml of cement sample solution contains=Z gm of Ca2+

=……gm of Ca2+

But 250 ml of cement sample solution is prepared by dilution of 50 ml

Stock solution

As 50 ml of stock solution contains Z gm =………….. gm of Ca2+

250 ml of stock solution contains A gm

=………….. gm of Ca2+

Step III: % of Ca2+

in cement:

As 1 gm of cement contains A gm =……………gm of Ca2+

100 gm of cement contains B gm =……………gm of Ca2+

therefore % of Ca2+

in cement

=……………%

Result Table:

Sr.No Description Result

1.

Amount of calcium in given cement sample

=…………………….%

Experiment No : 4

Aim : To standardize KMnO4 solution by preparing standard oxalic acid

and to estimate ferrous ions.

Apparatus : Beakers, Pipette, Measuring cylinder, Stand, Conical flask, etc.

Chemicals : KMnO4(0.1N approx), A.R.Grade oxalic acid, 2N H2SO4.

Theory : The amount of iron in the given solution can be determined by titrating it

against standard KMnO4.This titration is the example of redox titrations in which

reduction – oxidation occurs, oxidizing reagent is the one which forces other

molecules or atom or ion to liberate electrons, is the oxidizing agent. Here KMnO4

act as oxidizing agent and oxalic acid act as reducing agent, in this reaction there is

transfer of electron from oxalic acid to KMnO4 solution is always filled in burette.

This titration is carried out in acidic and hot conditions. The oxalic acid or sodium

oxalate solution pipetted,is heated along with about 20ml dil H2SO4 up to 60-70 0C

and the hot mixture is titrated against KMnO4 to get the end point colorless to

faint pink. The normality of either oxalic acid or KMnO4 is the end point colorless

to faint pink. The normality of either oxalic acid or KMnO4 is known in the

beginning and normality of other can be calculated by N1V1=N2V2 formula.

Oxalic acid dihydrate H2C2O4.2 H2O being a primary standard is used

for standardizing the KMnO4 solution.

Procedure : Perform the experiment in two parts

Part I : Standardization of KMnO4 solution.

1. Fill the burette-1 by 0.1 N KMnO4 solution.

2. Take by burette-2, 9 ml 0.1 N standard solution of oxalic acid in a 100 ml

conical flask and add to it 10ml 2N H2SO4 solution.

3. Heat the solution on wire gauze to 70oC and titrate this solution with

KMnO4 solution added from burette-1.

4. End point of titration is noted when faint pink colour appears to the

Solution. Call this burette reading as ‘X1’ ml.

5. To the same solution, add 1 ml of oxalic acid solution by burette-2 an

heat the flask to 70oC the solution becomes colorless.

6. To this hot solution, add KMnO4 solution from burette-1 till faint pink

colour appears to the solution. Call this burette reading as ‘X2’ml.

7. To the same solution, add 1 ml of oxalic acid solution by burette- 1

and heat the flask to 70oC the solution becomes colourless.

8. To this hot solution, add KMnO4 solution from burette-1 till faint pink

colour appears to the solution. Call this burette reading as ‘X3’ ml.

9. From these burette reading, find out normality of KMnO4 solution.

Part II : Estimation of ferrous (Fe2+

) ions :

1. Pipette out 10ml of the given iron solution in conical flask and add nearly

10ml 2N H2SO4 solution in it.

2. Titrate it against standardized KMnO4 solution, taken in burette,

till light colour appears.

3. Note the end point of titration.

4. Repeat the procedure to get at least three concordant readings.

Observation Table :

Part I : Standardization of KMnO4 solution :

Given : KMnO4 (0.1 N approx.)

To find : Extract normality of KMnO4 solution

Burette 1: KMnO4 solution

Burette 2: oxalic acid (0.1 N exact)

Indicator : KMnO4 itself

End point : Colourless to faint pink

Reaction :

2 KMnO4 + 3 H2SO4 + 5 H2C2O4 K2SO4 + 2 MnSO4 + 10 CO2 + 8 H2O

Burette-1

KMnO4

‘X1’ = ‘X2’= ‘X3’= X=

Burette-2

Oxalic acid 9 ml 10 ml 11 ml

Calculations:

To calculate exact normality of KMnO4 :

KMnO4 = H2C2O4

N1V1 = N2V2

N1 x X1 = 0.1 x 9

N2 x X2 = 0.1 x 10

N3 x X3 = 0.1 x 11

Exact normality of KMnO4 = N4 = N1+N2 N3 =……………………N

3

Part II: Estimation of ferrous (Fe 2+

) ions:

Given: FeSO4 (NH4)2SO4.6H2O solution.

To find: Estimation of ferrous (Fe 2+

) ions

Burette 1: KMnO4 solution

Flask : Ferrous (Fe 2+

) solution

Indicator: KMnO4 itself

End point: Colorless to faint pink

Reaction:

2 KMnO4 + 8 H2SO4 + 10 FeSO4 K2SO4 + 10 MnSO4 + 5 Fe(SO4)3 + 8 H2O

Burette-1 KMnO4

‘Y1’=

‘Y2’=

‘Y3’=

Y=

Burette-2

Fe 2+

solution

10 ml

10 ml

10 ml

Calculations:

i) Volume of KMnO4 used for titration = Y ml

ii) Volume of sample solution taken = 10 ml

iii) Normality of KMnO4 = N4

iv) Normality of ferrous ion =

Fe2+

= KMnO4

N1V1 = N2V2

N1 ( Normality w.r.t. Fe2+

) = N2V2

V1

N1 =

N1 =

Strength of Fe2+

ions = Equivalent weight x Normality

= 55.5 x Normality

=

=…………….gm/liter

Result:

Sr. No.

Description

Value

1.

Exact normality of KMnO4(N4)

2.

Strength of Fe2+

ions in gm/liter

Experiment No. 3

Aim: To determine chloride content in the given sample of water by Volhard’s

method.

Apparatus: Burette, Pipette (10 ml), Conical Flask, Burette stand, Beakers,

Volumetric flask.

Chemicals: Chlorides ions solution, 0.1 N Ammonium Thiocynate Solution,

0.1N AgNO3 Solution, 50%HNO3 Solution and Ferric ion solution.

Theory: - In Volhard’s method Conc. of Halide ion can be determine with

thiocyanate solution performed in acidic medium using ferric ion as an indicator.

In this method excess amount of silver nitrate is added in solution to precipitate all

halide ion as AgX and remaining amount of excess Ag+ ions i.e. silver ions is

found by titration with standard potassium thiocyanate solution.

Reaction: Ag + X AgX

(Silver halide)

Ag + SCN AgSCN

(Silver thiocyanate)

The indicator used is Fe3+

which forms soluble red colour at the end point.

Fe3+

+ SCN Fe [SCN] +2

(Blood Red colour complex/Precipitate)

Procedure:

A) Blank Titration:-

Fill the burette with 0.1N NH4SCN.Take 10ml AgNO3 (0.1N) in the conical flask.

Add 5ml 50% HNO3 and 1ml Ferric ion solution (Ferric sulphate/Ferric nitrate) as

indicator. Titrate this mixture against NH4SCN Solution, till the formation of

brick-red precipitate, and note the end point as X ml. Take three readings

Observations:-

Burette : 0.1N Ammonium Thiocynate solution

Pipette out : 10ml 0.1N AgNO3

End Point : Colourless to Brick-red

Indicator : Ferric ion solution

Observation Table:-

Sr.No. Volume of

AgNO3 solution

pipette out in ml.

Burette Reading C.B.R

X ml Final Initial Difference

1 10ml 0.0ml

2 10ml 0.0ml

3 10ml 0.0ml

B) Back Titration: - Fill the burette with 0.1N NH4SCN. Take the given 10ml

Halide solution (Chloride ion soln) in the conical flask and add 10ml 0.1N AgNO3

to it and white ppt. is obtained. Filter and wash the ppt. of (AgX) Silver halide with

distilled water and use filtrate for titration. Add 5ml of 50% HNO3 & 1ml ferric ion

solution as indicator. Titrate against 0.1N NH4SCN solution till the formation of

brick-red precipitate and note the end point as Y ml. Take three readings.

Observations:-

Burette : 0.1N Ammonium thiocynate solution

Pipette out : 10ml Halide solution+10ml. 0.1N AgNO3 solution

End Point : Brick -red colour

Indicator : Ferric ion solution

Observation Table:-

Sr.No. Volume of

AgNO3 solution

pipette out in ml.


Y ml Final Initial Difference

1 10ml 0.0ml

2 10ml 0.0ml

3 10ml 0.0ml

Calculations:-

1) 1000ml of 1N NH4SCN = 1000ml of 1N metal chloride

= 35.5gm of chloride ion

2) Chloride ion solution = Blank titration reading – Back titration reading

= (X – Y) ml

3) 1000ml of 1N NH4SCN = 35.5gm of chloride ion

1ml of 0.1N NH4SCN = 35.5 x 0.1 gm of Chloride ion

1000

= 35.5 x 0.1x 1000 mg of Chloride ion

1000

= 3.55mg of Chloride ion

1ml of 0.1N NH4SCN = 3.55 gm of Chloride ion

(X – Y) ml of 0.1N NH4SCN = (X – Y) x 0.1 x 3.55

1 x 0.1

=(X – Y) x 3.55mg of chloride ion

4) But, 10ml of water sample contains = (X – Y) x 3.55mg of chloride ion

1000ml of water sample contains =1000 x (X – Y) x 3.55

10

1000ml of water sample contains = (X – Y) x 3.55 x 100

= _____ ppm or mg of Chloride ions per lit.

Just like hardness, Chloride content is also measured in ppm.

1ppm = 1mg/Lit.

Result Table:-

Sr.No. Description Value

1.

Total Chloride content of water sample

_______ ppm.

Conclusion:-The given sample of water can be used for ______________

Tolerance limit of Cl- in ppm for different industries and domestic purpose are as

follows.

Washing Drinking Irrigation Textile Steam

roller

Sugar

500 250 100 100 100 20

TSSM’s

Bhivarabai Sawant College of Engineering & Research,

Narhe, Pune - 41.

Department of Applied Chemistry

List of Experiments

Semester – II

1) To determine the Moisture, Volatile matter and ash content of a

given sample of coal.

2) To determine Chloride content in the given sample of water by

using Mohr’s method.

3) To determine Temporary and Permanent hardness of water by

EDTA method

4) To determine Total Alkalinity of water sample.

Laboratory Incharge Principal

Experiment No: 1

Aim : To determine moisture ,volatile matter and ash contents in a given

coal sample by proximate analysis.

Theory : Coal is an important fossil fuel. It occurs in layers in the crust.

It is formed by the partial decay of plants buried under earth millions of

years ago. It is highly carbonaceous matter and process of conversion of

vegetable matter into coal can be represented in the following sequence;

Plant -- Peat—lignite (brown coal) -- bituminous coal—Anthracite --

Graphite.

The composition of coal varies depending upon its type, source and

age .The carbon content lowest in peat and highest in anthracite. The

composition of coal further depends upon the combine effect of heat and

pressure to each it has been put. It is necessary to analysis of coal sample

to determine its quality and thus the industrial use to which can be put.

The quality of coal is ascertained by the following two types of analysis.

1) Proximate analysis: It is the simplest types of coal and it tells us

about the partial utilization of coal. It includes the determination of (a)

moisture content (b) ash content (c) fixed carbon as percentages of the

original weight of the coal sample.

2) Ultimate analysis: It includes the estimation of carbon, hydrogen,

sulphur, oxygen and as content. The analysis is useful in calculation the

calorific value of the coal sample.

A) Moisture: Coal contents two types of moisture

Free or surface moisture: The moisture which is lost on air drying is

called surface or free moisture. It depends on the factor such as size of

coal and its rank, treatment given to the coal, the degree of drying and

nature of this surface.

Moisture reduces the calorific value. A considerable amount of heat is

wasted in evaporating in moisture during combustion. evaporating in

moisture during combustion.

Further the cost is added up as water is brought and transported at

the fuel price. However the presence of moisture in coals by 5 to 10 %

forms a uniform bed of coal and also reduces the formation of fly

ash and the yield and quality of metallurgical coke is improved.

Internal moisture: Internal moisture is that types of moisture which

is retained by air dried coal. It is rough indicator of the coal maturity.

B)Volatile Matter: The Matter which evaporates during combustion

is called Volatile Matter . A considerable amount of heat is wasted in

evaporating in volatile Matter. Volatile matter reduces the calorific

value.

C) Ash content: The non-combustible content left after burning of the

organic matter form coal is termed as ash. The ash which is intimately

interspersed within the mass of coal is called fixed ash or inherent ash,

whereas the which occurs in different layers of coal is known as free ash

or extraneous moisture ash.0nly the extraneous or free ash can be

removed by washing. Ash reduces the heating value of the coal. Ash

usually consist of silica, alumina, iron oxide and small quantities of

lime, magnesia etc. It composition is of considerable important in

metallurgical operation as it effect the slag and metal composition and

consequently is a prime consideration in selecting the flux. When coal is

used in a boiler, the fusion temperature of ash is particular significance

as it is dependent on its composition. Fusion temperature of ash is

generally between 10000C to 1700

0C. Ash is fusion temperature below

12000C is called fusible ash and that about 1430

0C is called refractory

ash. If the ash fuses at the working temperature when coal is burnt on

grates it leads to the formation of clinkers and reduces the primary air

supply and efficiency of production and distribution of heat are

adversely affected.

Fixed Carbon: The amount of fixed carbon increases from low ranking

coals, i.e. peat to Anthracite side more the amount of fixed carbon in

coal more will be its calorific value. The volume of furnace required for

the combustion of such a coal sample will be low. Fixed carbon amount

in a coal varies inversely with volatile matter.

Apparatus: Powdered coal, Analytical balance, Electric oven

Desiccator, Silica crucible with lid, long legged tong, Muffle furnace,

given coal sample

Procedure:

Part A: Moisture Content:

Take some known quantity of powered coal sample in previously

weighed silica crucible. Heat the sample in an electric oven at about

100-1100C about an hour. Take out the crucible, cool it in desiccators

and weigh it. Repeat the procedure of heating, cooling and weighing the

crucible till constant weigh is obtained. Note the constant weight.

Part B: Volatile matter

Take some known quantity of powered dry coal sample in previously

weighed silica crucible and place a ventilated lid over it. Heat the sample

in Muffle furnace at about 925-9500C,Stir the residue and ignite it for 7

min at same temperature.Take out the crucible, cool it in desiccators and

weigh it. Repeat the procedure of heating, cooling and weighing the

crucible till constant weigh is obtained. Note the constant weight.

Part C:Ash Content:

Take some known quantity of powdered dry coal sample in previously

weighed silica crucible. Heat the sample in Muffle furnace at about

700-8000C, Stir the residue and ignite it at same temperature for half an

hour. Take out the crucible, cool it in desiccators and weigh it. Repeat

the procedure of heating, cooling and weighing the crucible till constant

weigh is obtained. Note the constant weight.

Observation table and calculation:

Observations:

Sr. No. Description Value

1. Weight of empty crucible W1g

2. Weight of crucible and coal before heating W2g

3. Weight of coal before heating (W2-W1)g

4. Weight of crucible and coal after heating W3g

5. Weight of coal after heating (W3-W1)g

Calculation:

Loss in weight due to heating = Moisture content of sample

= (W2-W1) - (W3-W1) g

= (W2-W3) g

Loss in weight

% of moisture = * 100

Weight of coal sample

(W2-W3)

% of moisture = * 100

(W2-W1)

% of moisture =

Part B: Volatile Matter:

Observations:


1. Weight of empty crucible X1g

2. Weight of crucible and coal before heating X2g

3. Weight of coal before heating (X2-X1)g

4. Weight of crucible and coal after heating X3g

5. Weight of coal after heating (X3-X1)g

Calculation:

Loss in weight due to heating = Moisture content of sample

= (X2-X1) - (X3-X1) g

= (X2-X3) g

Loss in weight

% of Volatile Matter = * 100

Weight of coal sample

(X2-X3)

% of Volatile Matter = * 100

(X2-X1)

% of Volatile Matter =

Observation table and calculation:

Part C: Ash content:

Observations:


1. Weight of empty crucible Y1g

2. Weight of crucible and dry coal before heating

Y2g

3. Weight of dry coal (Y2-Y1)g

4. Weight of crucible and Ash after heating Y3g

5. Weight of Ash after heating (Y3-Y1)g

Calculation:

Loss in weight due to heating = Ash content of sample

= (Y2-Y1) - (Y3-Y1) g

= (Y2-Y3) g

Weight of Ash

% of Ash = * 100

Weight of dry coal sample

(Y3-X1)

% of Ash = * 100

(Y2-Y1)

% of Ash =

Result Table:


1

Percentage of Moisture

=----------------

2

Percentage of Volatile matter

=----------------

3

Percentage of Ash

=----------------

Experiment No: 2

Aim: To study the effect of pH medium on Corrosion.

Theory: Destruction of metals (and their alloys) by medium (atmospheric

or aqueous conducting) through chemicals or electrochemical actions

starting at metal surface is called corrosion. The corrosion is of two types

depending on nature of corroding medium. If corrosion take place, when

the metals are exposed to atmosphere then it is called dry corrosion. In this

type of corrosion metals are attacked by the atmospheric gaseous such as

O2, CO2, SO2,and H2S etc. Dry corrosion is slow process. Rate of dry

corrosion may increases due to the factors such as temperature, moisture,

nature of metal undergoing corrosion and the nature of film formed on the

surface of metal.

If metal undergo corrosion when they are attacked by

aqueous conducting medium, by formation of anodic and cathodic areas,

such corrosion is known as wet corrosion.

Nature of metal (in other words, position metal in

electrochemical and galvanic series) also plays an important role in

corrosion. In galvanic series (Which is formetals, alloys and metal in

combined state) metals are arranged in their increasing standard reduction

potential. Metals that are placed at higher position will act as anode and

hence undergo oxidation easily and are more prone to corrosion and

metals whose position is lower in galvanic series will act as cathode and

will undergo reduction and are less prone to corrosion.

The aim of the present experiment is to highlight effect of pH of

corroding medium on the rate of corrosion. When the metals are exposed

to corroding medium, it can have any pH.

Effect’s of acidic medium:

If corroding medium is acidic, then corrosion takes place by

mechanism of hydrogen (H2) gas. Following figures illustrates the

mechanisms.

H+ medium

Iron copper Iron

Every metal has tendency to go in to its solution i.e. It has natural

tendency to undergo oxidation. During corrosion a part of metal where

oxidation is taking place forms anodic area. Then reaction taking place

at anode is

At Anode:

Fe Fe2+

+ 2e-

H2 H2 H2

e- e

-

These liberated electrons will be present on metal surface and they will

be accepted by acidic medium. The area on metal plate where this

reaction taking place acts as cathodic area. Then reaction at cathode is

At cathode: 2H+ + 2e

- H2

At anode: Fe Fe2+

+ 2e-

---------------------------------------------------

Net reaction Fe + 2H+

Fe2+

+ H2

Effects of basic and neutral medium:

If corroding medium is slightly basic or neutral, then corrosion takes

place by absorption of oxygen gas from the medium. Following fig.

illustrate the mechanism. Let us consider an iron plate on which there is

thin layer of tin and it is ruptured.

-------------------------------------------------Corrosion product

------------------------------------------------- Alkaline/ neutral medium

Oxide film

Iron plate

Reaction at anode:

Fe Fe

2+ + 2e

-

Reaction at cathode:

1/2O2 + H2O + 2e- 2 OH

-

Net Reaction: Fe + 1/2O2 + H2O Fe2+

+ 2 OH-

Apparatus: Beaker, Stand, Metal Plate, Distilled Water etc.

Chemicals: 1N HCL, 1N NaOH, Distilled Water etc.

Procedure:

1) Take three metal plates and remove the corrosion from its surface

if present.

2) Weigh accurately the metal plates.

3) Tie a thread to the hole of metal plates. Prepare 1N HCl and 1N

NaOH

4) Dip three plates respectively in 50ml solution of 1N HCl, 1N

NaOH and Distilled water.

5) Take care, that the thread is not dipped in the solution.

6) Remove the plates after 1hr and measure the area of portion of

plate that is dipped in solution.

7) Dry the plates and remove the corrosion formed due to dipping in

the solution.

8) Record the weight of the plates after removing the corrosion.

Result:

Sr.No. Medium Weight loss g/h.cm2

1) Acidic

2) Basic

3) Neutral

Experiment No: 3

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

Theory: The property of water which restricts or checks the lather

formation with soap is called hardness. The soap’s which are sodium

salts of higher fatty acids will not be forming lather with water till all the

hardness causing cation are lost from the water in the form of calcium

and magnesium state. Palmitate or oleates. Hardness of water is of two

types:

1) Temporary or carbonate hardness: It is caused by the presence of

carbonates and bicarbonates of calcium and magnesium and is removed

on boiling of water and sludge formation takes place.

2) Permanent or non carbonate hardness: It is caused by chlorides

and sulphates of calcium and magnesium and cannot be removed by

boiling. For its removal water has to be treated chemically.

In a hard water sample the total hardness can be determined by titrating

the Ca2+

and Mg2+

present in an aliquot of the sample at pH 10 with

EDTA using Eriochrome Black –T as indicator. Permanent hardness can

be determined by precipitating the temporary hardness by prolonged

boiling followed by titration with EDTA solution. The difference in 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 now

when this wine red colored complex is titrated against EDTA solution

(of known strength) the color of complex with Ca2+

and Mg2+

of the

metal ions of the water. Changes from wine red to original blue color

showing the end point. Eriochrome black –T is the metal ion indicator

used in the determination of hardness by comlexometric titration with

EDTA.

Apparatus: Conical flask ; Burette; Pipette; Beaker; Measuring flask;

0.1M EDTA solution; Buffer pH 10±0.1 solution Eriochrome Black T

indicator; Standard hard water; Given water sample.

Procedure:

Part A: Standardization of EDTA solution with ZnSO4.

1. Take 10 ml of 0.1 M ZnSO4 in 100 ml conical flask.

2. Add 5 ml of buffer solution having pH =10 and add 2-3 drops of

indicator (Eriochrome Black T) the colour of the solution turns

wine red.

3. Titrate the flask solution against standard EDTA solution from the

burette until the colour change from wine red to clear blue at the

end point.

4. Take at least three concordant reading. Let volume of EDTA

solution used = V1 ml.

Part B: Determination of the total hardness:

1. Take 25ml of given sample of water in 100ml conical flask.

2. Add 5ml of buffer solution having pH = 10 add 2-3 drops of


wine red.



end point.


solution used = V2 ml.

Part C: Determination of the Permanent hardness:

1. Take 25 ml of boiled water in 100 ml conical flask.

2. Add 5ml of buffer solution having pH = 10 add 2-3 drops of


wine red.



end point.


solution used = V3 ml, corresponds to permanent hardness of the

water sample. Temporary hardness is calculated by subtracting

permanent hardness from total hardness.

Reactions:

1) EBT + Mn+

M-EBT + 2H+

(Blue) (Wine red)

2) Mn+

+ Na2EDTA M-EDTA + 2Na+

(Colorless)

2) M-EBT + Na2EDTA M-EDTA + EBT

(Wine red) (Colorless) (Blue)

A) Observation Table for Standardization of EDTA:

Readings I(ml) II(ml) III(ml) CBR

Initial V1=-----ml Final

Difference

Calculations:

EDTA = ZnSO4

M1V1 = M2V2

M1 * CBR = 0.01* 10

M1=----------------M

B) Observation Table for Determination of the total hardness:



Difference

Calculations:

H2O = EDTA

M1V1 = M2V2

M1*25 = M1 of part A *CBR

M1 =--------------- M

M1 =---------------- M

Strength or CaCO3 equivalent = M1 * Molecular weight of CaCO3

= * 100

= g/l

Total hardness = ppm

C) Observation Table for Determination of the Permanent

hardness:



Difference

Calculation:

H2O = EDTA

M1V1 =M2V2

M1*25 = M1 of part A *CBR

M1 =--------------- M

M1 =---------------- M

Strength or CaCO3 equivalent = M1 * Molecular weight of CaCO3

= * 100

= g/l

Permanent hardness = ppm

Temporary Hardness = Total hardness – Permanent hardness

=

Temporary Hardness = ------------- ppm

Result Table:


1. Temporary Hardness ppm

2. Permanent Hardness ppm

3. Total Hardness ppm

Experiment No: 4

Aim: To determine types and amount of alkalinity of given water

sample.

Theory:

The knowledge of alkalinity of water is necessary for controlling the

corrosion, in conditioning the boiler feed water (internally), for

calculating the amount of lime and soda needed for water softening and

also in neutralizing the acidic solution produced by hydrolysis of salts.

In boilers for steam generation, high alkalinity of water may not only

lead to caustic embrittlement but also to the precipitation of sludges and

deposition of scales.

The alkalinity of water is due to the presence of hydroxide ion

(OH-), carbonate ion (CO3

2-) and bicarbonate ion (HCO3) present in the

given sample of water. These can be estimated separately by titration

against standard acid, using phenolphthalein and methyl orange

indicator.

The chemical reactions involved can be shown by equations

below:

1) OH- + H

+ H2O

2) CO32-

+ H+

HCO3-

3) HCO3

2- + H

+ H2O + CO2

The titration of the water sample against a standard acid up to

phenolphthalein end point shows the completion the reactions 1 & 2

only. This amount of acid used thus corresponds to hydroxide plus one

half of the normal carbonate present. The titration of the water sample

against a standard acid to methyl orange end point marks the completion

of the reaction 1, 2, 3. Hence the amount of acid used after the

phenolphthalein end point corresponds to one half of the normal

carbonate plus all the bicarbonates: while the total amount of the acid

used represent the total alkalinity (due to hydroxide, bicarbonate and

carbonate ions).

The possible combinations of ions causing alkalinity in water are:

1) OH- only HCO3

- only, CO3

2- only

2) OH- & CO3

2- together, CO3

2- & HCO3

- together.

3) The possibility of OH- & HCO3- together is not possible since they

combine together to from CO32-

OH- + HCO3

- CO3

2- + H2O

Apparatus: Burette, Pipette, Conical flask (250 ml capacity), Measuring

flask, 2N HCl, phenolphthalein indicator, methyl orange indicator.

Procedure:

1. Pipette out 25 ml of the water sample into a conical flask.

2. Add 1-2 drops of phenolphthalein indicator and titrate this sample

against 2N HCl from burette until the pink colour caused by

phenolphthalein just disappears

3. Note down this reading as phenolphthalein as end point V1 ml.

4. Add 1 drops of methyl orange indicator in solution. It will give light

yellow colour; continue the titration against 2N HCl until the light

yellow colour changes to red now. Note the total volume of acid as V2

ml. This is methyl orange end point. The same experiment is repeated

till the concordant reading is obtained.

5. Calculate the concentration of OH- , HCO3

- , CO3

- , from

phenolphthalein & methyl orange alkalinities.

The amount of alkalinities due to OH- ,CO3

2- ,HCO3

- , types are

calculated from following table

Alkalinity Quantity of OH-

Quantity of

CO32-

Quantity of

HCO3-

P = 0 0 0 M

P = ½ M 0 2P 0

P = M P 0 0

P <1/2 M 0 2P M-2P

P >1/2 M (2P-M) 2(M-P) 0

A) Observation Table for phenolphthalein indicator:


Initial

V1=-----ml Final

Difference

Calculations:

A) Phenolphthalein alkalinity:

H2O = HCl sample

N1V1 = N2V2

N1 * 25 = 2N * CBR

N1 = -------- N

Strength or CaCO3 equivalent = N2 * Equivalent weight of CaCO3

= * 50

Phenolphthalein alkalinity or P = ----- ppm.

B) Observation Table for Methyl orange indicator:


Initial

V2=-----ml Final

Difference

A. Methyl orange alkalinity:

H2O = HCl sample

N1V1 = N2V2

N1 * 25 = 2N * CBR

N1 = -------- N

Strength or CaCO3 equivalent = N2 * Equivalent weight of CaCO3

= 50

= g/l

= ppm

Methyl orange alkalinity or M = -------- ppm.

½ M = -------- ppm.

Result:

Sr.No. Types of alkalinity Amount of alkalinity

1)

2)

Experiment No:-2

Aim:-To determine Chloride content in the given sample of water by Mohr’s

method.

Apparatus: - Pipette (10ml), Burette (25ml), Conical flask (100ml), Burette stand,

Beakers (200ml) etc.

Chemical: - 0.1N AgNO3 solution, 0.1N KCl solution, 5% K2CrO4 indicator,

Water sample.

Principle and Chemical reaction:-

Chlorine ions can be determined by Mohr’s method by titrating with standard

AgNO3 solution, using potassium chromate as indicator. Silver ion reacts with

chloride as well as chromate ions to form silver salts.

Ag+ + NO3

- + Cl

- → AgCl ↓ + NO3

- (Before end point)

2Ag+ +2 NO3

- + CrO4

2- → Ag2CrO4 + 2NO3

- (at the end point)

The solubility product of AgCl is at lower concentration of Ag+ than Ag2CrO4.

Hence AgCl is precipitated first. As soon as all Chloride ions are removed in the

form of AgCl the extra drop of AgNO3 reacts with the indicator forming brick red

Silver Chromate. (i.e S.P of Ag2CrO4 is greater than AgCl, Hence AgCl ppt first).

Procedure:-

Part - I: - Standardization of AgNO3 Solution.

1) Wash the burette, pipette and flask with water.

2) Take 10ml exact 0.1N KCl solution in the conical flask; add 2-3 drops of 5%

K2CrO4 indicator in the titration flask. Flask solution turns yellow.

3) Titrate this solution verses approx 0.1N AgNO3 solution from burette, with

constant shaking, till you get Brick- red precipitate.

4) Note down the reading. Take two more readings till the constant burette

reading is reached (X ml).

Observations:-

Burette : 0.1N AgNO3 solution

Pipette : 10ml 0.1N KCl solution

Indicator : K2CrO4

End point : Yellow to Brick-red.

Observation table:-

Sr.No. Volume of KCl

solution pipetted

out (ml).


X ml Final Initial Difference

1 10ml

2 10ml

3 10ml

Calculations:-

To determine exact normality of AgNO3

KCl Verses AgNO3

N1V1 = N2V2

0.1 x V1 = N2 x X

N2 = 0.1 x V1

X

N2 = ______ N.

Thus exact normality of given AgNO3 solution is _____ N.

Part-II:-Actual estimation of Chloride ions from given water sample.

1) Wash the burette, pipette and conical flask with water.

2) Fill burette with 0.1N AgNO3 solution.

3) Rinse the pipette with given water sample and then pipette out 10ml given

water sample (Chloride content) in titration flask. Add 2 drops of 5%

K2CrO4 indicator in titration flask. Flask solution turns yellow.

4) Titrate the above solution against 0.1N AgNO3 solution from burette, till you

get the end point (Yellow to brick-red).

5) Note down the reading. Take two more readings till the constant burette

reading is reached (Y ml).

Observations:-

Burette : 0.1N AgNO3 solution.

Pipette : 10ml Water sample (Chloride content)

Indicator : K2CrO4

End point : Yellow to Brick-red

Observation table:-

Sr.No. Volume of Water

sample pipetted

out ( ml).


Y ml Final Initial Difference

1 10ml

2 10ml

3 10ml

Calculations:-

1) 1000ml of 1N AgNO3 = 1000ml of 1N metal chloride

= 35.5gm of chloride ion

2) 1ml of 0.1N AgNO3= 35.5 x 0.1 gm of Chloride ion

1000

= 35.5 x 0.1x 1000 mg of Chloride ion

1000

= 3.55mg of Chloride ion

3) 1ml of 0.1N AgNO3= 3.55 gm of Chloride ion

Yml of N2 N AgNO3 = Y x N2 x 3.55 mg of Chloride ion

1 x 0.1

= (A) mg of Chloride ion

4) But, 10ml of water sample contains = (A) mg of Chloride ion

1000ml of water sample contains = 1000 x (A) mg of Chloride ion

10

= _____ ppm or mg of Chloride ions per lit.

Just like hardness, Chloride content is also measured in ppm.

1ppm = 1mg/Lit.

Result Table:-


1. Exact normality of AgNO3 solution(N2) ______N.

2.

Total Chloride content of water sample

______ ppm.

Conclusion:-The given sample of water can/can’t be used for ______________.

Tolerance limit of Cl- in ppm for different industries and domestic purposes are as

follows.

Washing Drinking Irrigation Textile Steam

roller

Sugar

500 250 100 100 100 20

Documents

Applied Chemistry I Lab Manual