Lab Manual of Engineering Chemistry - KTU B.Tech...

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Lab Manual of

Engineering ChemistryEngineering ChemistryEngineering ChemistryEngineering Chemistry

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I Water

Analysis

1 Determination of Total Hardness of Water by

Complexometric Titration with EDTA

2 Determination of Chloride ion in a given Water sample

by Argentometric Method (Mohr’s Method)

3 Determination of Dissolved Oxygen present in a given

Water sample by Iodometric Method (Winkler’s Method)

II Volumetric

Analysis

4 Determination of Percentage of Available Chlorine

present in Bleaching Powder sample

5 Determination of amount of Iron and the number of

Water molecules of Crystallization in Mohr’s salt

using Standard Potassium Dichromate solution

III Analysis of

Ores/Alloys

6 Determination of amount of Iron present in the

given Iron Ore/Alloy by Permanganometry

7 Determination of amount of Copper present in the

given Copper Ore/Alloy by Iodometry

IV Preparation of

polymers

8 Preparation of Urea-Formaldehyde resin

9 Preparation of Phenol-Formaldehyde resin

� Internal Continuous AssessmentInternal Continuous AssessmentInternal Continuous AssessmentInternal Continuous Assessment (Maximum Marks – 50 marks)

50% - Laboratory practical and record → Max = 25 marks

10% - Regularity in the class → Max = 5 marks

40% - Test → Max = 20 marks [Time: 3 hours]

Contribution of Marks from Daily Lab Work = 30 marks

Viva

[Marks = 10]

Performance in Lab

[Marks = 5]

Record

[Marks = 10]

Attendance

[Max = 5]

Total

Marks = 30

Contribution of Marks from Internal Lab Exam = 20 marks [Time: 3 hours]

Viva, Principle and Procedure

[Marks = 10]

Experiment [Marks = 10]

Total

Marks = 20

40 % for Tabulation

[Marks = 4]

60 % for Accuracy

[Marks = 6]

Viva = 2 Prin = 4 Proc = 4 Std = 2 Est = 2 Std = 3 Est = 3

3

Expt. No: 1111 Determination of Total Hardness of Water by

Complexometric Titration [EDTA Method]

Aim

To determine the total hardness of a given water sample by complexometric titration

(EDTA method)

Theory

Water which does not give ready and permanent lather with soap is called hard water.

Presence of calcium and magnesium salts in the form of bicarbonate, chloride and sulphate in water

makes water ‘hard’. Water free from soluble salts of calcium and magnesium is called soft water.

It gives lather with soap easily.

The property of water which restricts the lather formation with soap is called hardness.

It is of two types: (a) temporary hardness and (b) permanent hardness

Temporary Hardness: It is due to the presence of magnesium and calcium bicarbonates

[Ca(HCO3)2 and Mg(HCO3)2].

Permanent Hardness: It is due to the presence of soluble salts of magnesium and calcium in the form

of chlorides and sulphates in water (CaCl2, CaSO4, MgCl2 and MgSO4).

The unit used for expressing the hardness of water is parts per million (ppm). It is the number

of parts of calcium carbonate (CaCO3) equivalent hardness present in one million parts of water.

Eriochrome Black-T [EBT] is the indicator used in the determination of hardness by

complexometric titration with EDTA. Here, Eriochrome Black-T is a complex organic compound

[sodium-1-(1-hydroxy 2-naphthylato)-6-nitro-2-naphthol-4-sulphonate] and EDTA is a hexadentate

ligand [disodium salt of ethylenediamine tetraacetic acid].

OH

N N

NO2

HO

SO3Na

Eriochrome Black-T [EBT]

NH2C

H2CN

HOOCH2C

NaOOCH2C CH2COOH

CH2COONa

EDTA

[Disodium salt of ethylenediamine

tetraacetic acid]

4

Observations and Calculations

(a) Standardisation of EDTA solution with standard hard water

S.No Volume of standard

hard water in ml Burette Readings in ml Volume of

EDTA solution in ml Initial Final 1 2 3 4

Concordant Volume of EDTA solution, V1 =………ml

1 ml of Standard hard water = 1 mg CaCO3

V1 ml of EDTA solution = 20 ml of Standard hard water

= 20 mg CaCO3

1 ml of EDTA solution, = ��

��

= …………..mg CaCO3

(b) Estimation of Total Hardness of given water sample

S.No Volume of given

water sample in ml Burette Readings in ml Volume of

EDTA solution in ml Initial Final 1 2 3 4

Concordant Volume of EDTA solution, V2 =………ml

20 ml of given water sample = V2 ml of EDTA solution

= � × ��

��

1 ml of given water sample = � × ��

�� × ��

1000 ml of given water sample = � × ��

�� × �� × ��

= � �

× �� × ��

= � �

× ��

Total Hardness of given water sample = � �

× �� = ⋯ … … … . ��

5

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

coloured unstable complex with Ca2+ and Mg2+ ions of the sample water.

Ca2+/ Mg2+ + EBT → [Ca2+/ Mg2+ EBT]

from hard water Wine red coloured unstable complex

Now when this wine red-coloured solution is titrated against EDTA solution,

EBT in the unstable complex is replaced by EDTA to form a stable metal-EDTA complex and

liberates the free Eriochrome Black-T. At this point, the colour of the solution changes from wine

red to original blue colour which showing the end point of the titration.

[Ca2+/ Mg2+ EBT] + EDTA → [Ca2+/ Mg2+ EDTA] + free EBT

Wine red coloured Stable metal-EDTA complex Blue colour unstable complex (Colourless)

Apparatus:

Conical flask, Burette, Pipette, Beaker, Measuring flask

Reagents:

0.1M EDTA solution, Eriochrome Black-T indicator, Basic buffer solution (NH4OH and

NH4Cl), Standard hard water, Given water sample

Procedure:

(a) Standardisation of EDTA solution with standard hard water

Pipette out 20 ml of standard hard water in a washed conical flask. Add 5ml basic buffer

solution and 2-3 drops of Eriochrome Black-T indicator, the colour of the solution turns wine red.

Titrate this solution against EDTA solution taken in the burette until the colour changes from wine

red to clear blue at the end. The final reading of the burette is noted and the titration is repeated to get

concordant value.

(b) Estimation of Total Hardness of given water sample

Pipette out 20 ml of given hard water in a washed conical flask. Add 5ml basic buffer solution

and 2-3 drops of Eriochrome Black-T indicator, the colour of the solution turns wine red.

Titrate this solution against EDTA solution taken in the burette until the colour changes from wine

red to clear blue at the end. The final reading of the burette is noted and the titration is repeated to get

concordant value.

Result:

The total hardness of given water sample is………..ppm

6

Expt. No: 2222 Determination of Chloride ion in a given Water

sample by Argentometric Method (Mohr’s Method)

Aim

To determine the chloride ion of a given water sample by Argentometric method

(Mohr’s method)

Theory

Chlorides are present in water usually as NaCl, MgCl2 and CaCl2. Although chlorides are not

harmful as such, their concentrations over 250 ppm impart a peculiar taste to the water thus

rendering the water unacceptable for drinking purposes.

By argentometric method, chloride ions in a water sample (neutral or slightly alkaline) can be

determined by titrating it against standard silver nitrate (AgNO3) solution using potassium chromate

(K2CrO4) as an indicator. The pH should be in between 7-8. At higher pH, silver ions are

precipitated as silver hydroxide. At lower pH, potassium chromate indicator is converted to

potassium dichromate (K2Cr2O7).

Argentometric method is based on the precipitation titration in which silver nitrate solution is

released from the burette to the water sample which contains chloride ions and indicator.

The silver ions (from silver nitrate solution) react with chloride ions (from water sample) and

chromate ions (from indicator) to form white precipitate of silver chloride and red precipitate

of silver chromate.

�� + � ! ⟶ �� #$%&'(�)(*&�&'�'(+ �� + �)�,�! ⟶ ��)�,#-(.�)(*&�&'�'(+

Red colour formed because of formation of silver chromate disappears initially as the solution

contains high concentration of chloride ions.

��)�, + �� ! ⟶ �� + �)�,�!

When the concentration of chloride ions has decreased, the red colour starts disappearing

slowly and slowly on shaking and a stage is reached when all the chloride ions have formed silver

chloride. One extra drop of silver nitrate at this point reacts with potassium chromate and reddish

coloured silver chromate is formed.

��/�� + 0��)�, ⟶ ��)�, + �0/��

7

Observations and Calculations:

Titration with the Blank solution

S.No Volume of distilled

water in ml Burette Readings in ml Volume of

AgNO3 solution in ml Initial Final 1 2 3 4

Concordant Volume of AgNO3 solution, V1 =………ml

Titration with the Sample Water


water sample in ml Burette Readings in ml Volume of

AgNO3 solution in ml Initial Final 1 2 3 4

Concordant Volume of AgNO3 solution, V2 =………ml

Normality of standard AgNO3 solution, NA = �

1� = �, ��/

Volume of standard AgNO3 solution, VA = 3 � − �5

= ……………ml

Volume of given water sample, VW = …………… ml

Normality of given water sample, NW can be calculated from the normality formula,

i.e., NA x VA = NW x VW

Normality of given water sample, NW = /� × �

6

= ……………..N

Amount of chloride ions = /6 × 78. 6'9:�% 9)&;( = /6 × �1. ,1 g/Lit

= ……………..g/Lit

Amount of chloride ions in ppm = … … … … .× ��mg/Lit

= ………………..ppm

8

Apparatus:

Conical flask, Burette, Measuring flask, Beakers

Reagents:

Standard silver nitrate solution< /1�=, Indicator potassium chromate solution

Procedure:

Titration with the Blank solution

Transfer 50 ml of the distilled water in a conical flask and add 3-4 drops of indicator

potassium chromate solution. Slowly add standard silver nitrate solution from the burette and shake

the solution well. At the end point, light yellow colour starts changing to red colour. The titration is

repeated until a concordant volume V1 is obtained. The blank correction for the indicator should be

subtracted from the volume of the titrant obtained after titrating the sample solution as given below

Titration with the Sample Water

Transfer 50 ml of the given water sample in a conical flask and add 3-4 drops of indicator

potassium chromate solution. Slowly add standard silver nitrate solution from the burette and shake

the solution well. At the end point, light yellow colour starts changing to red colour and red colour

persists. The titration is repeated until a concordant volume V2 is obtained.

Result:

The amount of chloride ion in the given water sample is…………..ppm

9

Expt. No: 3333 Determination of Dissolved Oxygen present

in a given Water Sample by Iodometric Method (Winkler’s Method)

Aim

To determine the amount of dissolved oxygen (D.O.) in a given water sample by

Iodometric Method (Winkler’s Method)

Theory

Oxygen is poorly soluble in water. The solubility of oxygen of air in fresh water varies from

7.5 - 14.5 mg/Lit. Dissolved oxygen is needed for living organism to maintain their biological process.

It is an important factor in corrosion. Iodometric method (Winkler’s method) is used for determining

dissolved oxygen in water.

The principle involved in the determination of dissolved oxygen is to bring about the oxidation of

potassium iodide (KI) to iodine (I2) with the dissolved oxygen present in the water sample after adding

MnSO4, KOH and KI, the basic manganic oxide formed act as an oxygen carrier to enable the dissolved

oxygen in the molecular form to take part in the reaction.

>;?�, + �0�@ ⟶ >;#�@+� + 0�?�,

�>;#�@+� + �� ⟶ �>;�#�@+� Basic manganic oxide which on acidification gives

>;�#�@+� + @�?�, ⟶ >;?�, + �@�� + 3�5 �0A + @�?�, + 3�5 ⟶ 0�?�, + @�� + A�

The liberated iodine (I2) is titrated against standard sodium thiosulphate (Na2S2O3) solution using starch as

indicator3?'�)*% +A� ⟶ B C(*9 9C)(.*9�� (D5. A� + �/��?�� ⟶ /��?,�E + �/�A

Apparatus: Conical flask, Burette, Measuring flask, Beakers

Reagents:

Standard sodium thiosulphate solution< /1�=, Potassium iodide solution, starch solution as indicator

Procedure:

Take 100 ml of given water sample into a conical flask, and titrate slowly against N/50 standard

sodium thiosulphate solution (taken in the burette). When the colour of the solution is very light yellowish

add about 2 ml of freshly prepared starch solution, so the colour of the solution turned into blue. Continue the

titration till the disappearance of blue colour of the solution and note down the volume of the titrant used.

The titration is repeated until a concordant volume is obtained.

Result:

The amount of dissolved oxygen (D.O.) in a given water sample is……………..ppm

10



water sample in ml

Burette Readings in ml Volume of

Na2S2O3 solution in ml Initial Final

1

2

3

4

Concordant Volume of Na2S2O3 solution, V1 =………ml

Normality of standard Na2S2O3 solution, N1 = �

1� = �. ��/

Volume of standard Na2S2O3 solution, V1 = ……………ml

Volume of given water sample, V2 = ………….....ml

Normality of given water sample, N2 can be calculated from the normality formula,

i.e., N1 x V1 = N2 x V2

Normality of given water sample, N2 = /� × �

�

= ………………..N

Amount of Dissolved Oxygen = /� × 78. 6'9:�DF�(; = /� × G g/Lit

= ………………g/Lit

Amount of Dissolved Oxygen in ppm = … … … … . .× ��mg/Lit

= ………………..ppm

11

Expt. No: 4444 Determination of the Percentage of Available

Chlorine present in Bleaching Powder sample

Aim

Determine the percentage of available chlorine present in the given sample of bleaching

powder.

Theory

Bleaching powder is used as a bleaching agent and also as a disinfectant.

The main constituent of bleaching powder is calcium hypochlorite [Ca(OCl)2] which supplies

chlorine [Cl2] with dilute acids.

��#��H+� + ,@�H ⟶ ��H� + �@�� + ��H�

So the available chlorine is defined as the percentage of chlorine made available by bleaching

powder when treated with dilute acids. The available chlorine present in bleaching powder sample

is determined iodometricaliy by treating its solution with an excess of potassium iodide solution

in the acidic medium.

�H�! + �@� + �A! ⟶ A� + @�� + �H!

The liberated iodine (I2) is treated with sodium thiosulphate (Na2S2O3) solution using freshly

prepared starch solution as indicator to be added near the end point.

3?'�)*% +A� ⟶ B C(*9 9C)(.*9�� (D5 A� + �/��?�� ⟶ /��?,�E + �/�A

Apparatus:

Digital Balance, Burette, Conical flask, Measuring flask, Funnel, Glass rod, Beakers

Reagents:

Bleaching powder, Standard sodium thiosulphate solution< /��=, 10% Potassium iodide (KI)

solution, dilute acetic acid, freshly prepared starch solution as indicator

12


S.No

Volume of given

bleaching powder

solution in ml



1

2

3

4



�� = �. �/


Volume of given bleaching powder solution , V2 = ………….....ml

Normality of given water sample, N2 can be calculated from the normality formula,

i.e., N1 x V1 = N2 x V2

Normality of given bleaching powder solution, N2 = /� × �

�

= ……………………N

Amount of available chlorine = /� × 78. 6'9:�% 9)&;( = /� × �1. ,1 g/Lit

= ………………..g/Lit

Amount of available chlorine present in 100 ml of

the solution =

… … .× ��

= ………………..g

The percentage of available chlorine present in the

given sample of bleaching powder (BP) =

��9C;'9:�H$(&�%'9:BI × �� = … … … �

… … … � × ��

= …………..%

13

Procedure:

………….g of bleaching powder is accurately weighed in a weighing bottle.

It is transferred to a clean beaker and is ground to a thin paste with water. The mixture is allowed to

settle and the milky supernatant liquid poured into a 100 ml standard flask. The residue in the beaker

is ground with a little more water and the operation repeated until the whole of the sample has been

quantitatively transferred into the standard flask. Then the solution is made up to the volume.

The flask is shaken well for uniform concentration.

10 ml of the solution (use measuring flask) in a state of very fine suspension is taken into

a conical flask and add 10 ml of distilled water into it. Then add 10 ml of 10% potassium iodide (KI)

solution and 10 ml of dilute acetic acid. This solution [which contains liberated iodine (I2)] titrates

with standard sodium thiosulphate solution (taken in the burette) until the dark brown colour changes

to pale yellow. To this add, 2 ml of freshly prepared starch solution as indicator, so the colour of the

solution turned into blue. Continue the titration till the disappearance of blue colour of the solution

and note down the volume of the titrant used. The titration is repeated until a concordant volume is

obtained.

Result:

The percentage of available chlorine present in the given sample of bleaching powder

is…………..%

14

Expt. No: 5555 Determination of Strength of Iron and the number of Water

molecules of Crystallization in Mohr’s salt using Standard Potassium Dichromate solution

Aim

To determine the strength of iron and the number of water molecules of crystallization in

Mohr’s salt provided standard potassium dichromate solution (N/20), using diphenyl amine as

internal indicator.

Theory

Mohr’s salt is ferrous ammonium sulphate [FeSO4.(NH4)2SO4.6H2O]. For determination of

the amount of iron in the given solution of Mohr’s salt, a known volume of this solution is titrated

with standard potassium dichromate solution (K2Cr2O7) in a medium acidified with

dilute sulphuric acid. Potassium dichromate oxidises ferrous sulphate (FeSO4) present in Mohr’s salt

into ferric sulphate [Fe2(SO4)3].

JK(�� D&.�'&9;LMMMMMN K(�� + (!O × E

�)��P�! + �,@� + E(! -(.C*'&9;LMMMMMN ��)�� + P@��

/(')(�*'&9;:EK(�� + �)��P�! + �,@� ⟶ EK(�� + ��)�� + P@��

For finding out the end point, internal indicator diphenyl amine is used. At the end point, all

the ferrous ions present in the solution get completely oxidised to ferric ions by chromate ions and as

soon as a slight excess of potassium dichromate solution is added. It leads to the oxidation of

diphenyl amine which results in the formation of a blue coloured complex. This indicates the end

point of the titration.

NH

Diphenyl amine

Oxidation

with K 2Cr2O7

Blue Coloured complex

The number of water of water molecules of crystallization in Mohr’s salt can be calculated

from the following equation,

?')(;�'%9:%F.)�'(.R� '?')(;�'%9:�;%F.)�'(.R� ' = �G, + �GD

�G,

15


S.No

Volume of given

solution of Mohr’s salt

in ml


K2Cr2O7 solution in ml Initial Final

1

2

3

4

Concordant Volume of K2Cr2O7 solution, V1 =………ml

Normality of standard K2Cr2O7 solution, N1 = �

�� = �. �1/

Volume of standard K2Cr2O7 solution, V1 = ……………ml

Volume of given solution of Mohr’s salt, V2 = ………….....ml

Normality of given solution of Mohr’s salt, N2 can be calculated from the normality formula,

i.e., N1 x V1 = N2 x V2

Normality of given solution of Mohr’s salt, N2 = /� × �

�

= ………………..N

The strength of iron in the given sample of

Mohr’s salt = /� × 78. 6'9:A)9; = /� × 11. G1 g/Lit

= ………………g/Lit

The strength of anhydrated Mohr’s salt = /� × 78. 6' = /� × �G, g/Lit

= ……………….. g/Lit

The strength of hydrated Mohr’s salt = 20 g/Lit

StrengthofhydratedsaltStrengthofanhydratedsalt = 20g/Lit

… … . g/Lit = �G, + �GD

�G,

The number of water molecules of crystallization in Mohr’s salt, x

= ………………..

16

Apparatus:

Burette, Conical flask, Pipette, Measuring flask

Reagents:

Standard potassium dichromate solution< /1�= , dilute sulphuric acid, 1:1 phosphoric acid,

Indicator diphenyl amine

Procedure:

Pipette out 20 ml of given solution of Mohr’s salt into a conical flask, add 5 ml of dilute

sulphuric acid, 2-3 ml of 1:1 phosphoric acid and then two drops of diphenyl amine to this solution.

Run the potassium dichromate solution in small lots from the burette, shaking the conical flask after

each addition and also stirred at regular intervals. At the end point, the colourless solution becomes

deep blue. Note down the volume of the titrant used. The titration is repeated until a concordant

volume is obtained.

Result:

(i) The strength of iron in the given sample of Mohr’s salt is………………….g/Lit

(ii) The number of water molecules of crystallization in Mohr’s salt………….

17

Expt. No: 6666 Determination of the amount of Iron in

the given Iron Ore/Alloy by Permanganometry

Aim

To determine the amount of iron present in the given iron ore/alloy using potassium

permanganate (Permanganometry) provided standard Mohr’s salt solution (N/20).

Theory

Minerals which are naturally occurring chemical substances in the earth’s crust obtainable by

mining. Out of many minerals in which a metal may be found, only a few are viable to be used as

sources of that metal. Such minerals are known as Ores. Iron is the second most abundant metal in

the earth’s crust. The major ores of iron are Haematite (Fe2O3), Magnetite (Fe3O4), Siderite (FeCO3)

and Iron pyrites (FeS2).

Transition metals mix freely with each other in the molten state and on cooling a solid

solution of different metals results in the form of alloys. The alloy formation is explained on the

basis of similar sizes of atoms of these metals which allow the atoms of one metal to take up the

position in the crystal lattice of the other. Steal is essentially an alloy of iron and carbon.

Plain steel contains certain amount of C, Si, S, P and Mn apart from iron. For special purposes

varying amounts of other metals such as Cr, V, Mo, W, Ti, Ni, Co, Zr and Cu are added.

For determining the amount of iron in ore/alloy, the given ore/alloy sample is dissolved in

dilute sulphuric acid, when the iron present as dissolved as ferrous sulphate (FeSO4) and

hydrogen gas is evolved.

K( + @�?�, ⟶ K(?�, + @� ↑

The amount of ferrous ion (Fe2+) in the solution can be determined by a redox titration with

standard potassium permanganate (KMnO4) or potassium dichromate (K2Cr2O7). The ionic equations

of ferrous sulphate with acidified potassium permanganate are given below

JK(�� D&.�'&9;LMMMMMN K(�� + (!O × 1

>;�,! + G@� + 1(! -(.C*'&9;LMMMMMN >;�� + ,@��

/(')(�*'&9;:1K(�� + >;�,! + G@� ⟶ 1K(�� + >;�� + ,@��

18

Observations and Calculations

(a) Standardisation of KMnO4 solution with standard Mohr’s salt solution

S.No Volume of standard Mohr’s salt in ml, V1

Burette Readings in ml Volume of KMnO4 solution in ml Initial Final

1 2 3 4

Concordant Volume of KMnO4solution, V2 =………ml

Normality of standard Mohr’s salt solution , N1 =

�� = �. �1/

Volume of standard Mohr’s salt solution, V1 = …………….ml

Volume of KMnO4 solution, V2 = …………….ml

Normality of KMnO4 solution, N2 can be calculated from the normality formula,

i.e., N1 x V1 = N2 x V2

Normality of KMnO4 solution, N2 = /� × �

�

= ………………N

(b) Estimation of iron in the given iron ore/alloy

S.No Volume of given solution

of iron ore/alloy in ml Burette Readings in ml Volume of

KMnO4 solution in ml Initial Final 1 2 3 4

Concordant Volume of KMnO4 solution, V2a =………ml

Normality of KMnO4 solution, N2 = ……………..N

Volume of KMnO4 solution, V2a = …………….ml

Volume of given iron ore/alloy solution, V3 = …………….ml

Normality of given iron ore/alloy solution, N3 can be calculated from the normality formula,

i.e., N2 x V2a = N3 x V3

Normality of given iron ore/alloy solution, N3 = /� × ��

�

= ………………N

Strength of iron in the given iron ore/alloy solution = /� × 78. 6'9:A)9; = /� × 11. G1 g/Lit

= … …… … . . �/f&' The amount of iron present in the given sample of iron ore/alloy is

= … … … … . . �/f&' × �� f&'

= ……………….g

19

Apparatus:

Burette, Conical flask, Pipette, Measuring flask, Beakers

Reagents:

Standard Mohr’s salt solution< /��=, potassium permanganate solution, dilute sulphuric acid

Procedure:

(a) Standardisation of potassium permanganate solution

Pipette out 20 ml of the N/20 standard Mohr’s salt solution into a conical flask and add 10 ml

of dilute sulphuric acid. Then, titrate this solution slowly against the potassium permanganate

solution from the burette until a faint but permanent pink colour persists in the solution. Note down

the volume of the titrant used. Repeat the titrations until a concordant volume is obtained.

(b) Estimation of Iron in the given ore/alloy

The whole of the given iron ore/alloy solution is transferred into the 100 ml standard flask.

Make up the volume of the solution to 100 ml with distilled water and shake the solution thoroughly.

Pipette out 20 ml of the solution into a conical flask, add 10 ml of dilute sulphuric acid and titrate

against the potassium permanganate solution taken in the burette. The appearance of a faint but

permanent pink colour marks the end point. Note down the volume of the titrant used. Repeat the

titrations until a concordant volume is obtained.

Result:

The amount of iron present in the given sample of iron ore/alloy is…………………..g

20

Expt. No: 7777 Determination of the amount of Copper in

the given Copper Ore/Alloy by Iodometry

Aim

To determine the amount of copper present in the given copper ore/alloy provided standard sodium

thiosulphate solution (N/20).

Theory

The major ores of copper are Copper pyrites (CuFeS2), Malachite [CuCO3.Cu(OH)2] and Cuprite

(Cu2O). Brass is an alloy of copper and zinc. It may also contain small amounts of iron, lead,

tin or aluminium.

For determining the amount of copper in ore/alloy, the given ore/alloy sample is dissolved in nitric

acid. The excess acid is neutralized by drop wise addition of Na2CO3 solution until turbidity appears.

�C�� + �@� + ��! →�� + @�� + �C��#hC)i&.&'F+ This turbidity is removed by drop wise addition of acetic acid solution.

�C�� + ��@��@ → �� + @�� + �C�� + �@��!

The principle involved in the determination of amount of copper present in the given copper ore/alloy

is to bring about the oxidation of potassium iodide (KI) to iodine (I2) with copper ions (Cu2+).

��C�� + ,A! ⟶ ��CA + A�

The liberated iodine (I2) is titrated against standard sodium thiosulphate (Na2S2O3) solution using

starch as indicator3Starch +Il ⟶ Bluecolouredcomplex5. A� + �/��?�� ⟶ /��?,�E + �/�A Apparatus: Conical flask, Burette, Measuring flask, Beakers

Reagents: Standard sodium thiosulphate solution< /��=, Potassium iodide solution, starch solution as indicator

Procedure:

The whole of the given copper ore/alloy solution is transferred into the 100 ml standard flask.

Make up the volume of the solution to 100 ml with distilled water and shake the solution thoroughly.

Pipette out 20 ml of the solution into a conical flask, neutralize any free acid present in this solution

by adding sodium carbonate solution by drop wise till a faint permanent precipitate remains on shaking and

add dilute acetic acid drop wise until the precipitate dissolves. Add 10 ml of potassium iodide (KI) solution,

so the solution becomes brown and then titrate slowly against N/20 standard sodium thiosulphate solution.

When the colour of the solution assumes faint yellow colour of iodine, add about 2 ml of freshly prepared

starch solution, so the colour of the solution turned into blue. Continue the titration till the disappearance

of blue colour of the solution and note down the volume of the titrant used. The titration is repeated until

a concordant volume is obtained.

Result: The amount of copper present in the given sample of copper ore/alloy is…………………..g

21


S.No Volume of given solution of

copper ore/alloy in ml



1

2

3

4



�� = �. �1/


Volume of given solution of copper ore/alloy, V2 = ………….....ml

Normality of given solution of copper ore/alloy, N2 can be calculated from the normality formula,

i.e., N1 x V1 = N2 x V2

Normality of given solution of copper ore/alloy, N2 = /� × �

�

= ………………..N

Strength of copper in the given Cu ore/alloy solution = /� × 78. 6'9:�C = /� × E�. 1, g/Lit

= ………………g/Lit

The amount of copper present in the given sample of copper ore/alloy is

= … … … … . . �/f&' × �� f&'

= ……………….g

22

Expt. No: 8888 Preparation of Urea-Formaldehyde Resin

Aim

To prepare urea-formaldehyde (UF) resin

Theory

Urea-formaldehyde resin is prepared by the condensation polymerisation reaction between

urea and formaldehyde in neutral or acidic condition. Such resins are water soluble and hence are

used as textile finishing. They are also used in the paper industry to improve the wet strength of

paper. In the plywood industry they are used as adhesives. Such resins find uses in packaging,

accessories, unbreakable dishes, clock cases, etc.

C +O

NHCH2OH

NH2

CO

NH2

NH2

CO

H

H

+ CO

NHCH2OH

NHCH2OH

Urea Formaldehyde Monomethylol urea Dimethylol urea

CO

NHCH2OH

NH2

nCO

NCH2

NH2 n Monomethylol urea Linear UF resins

CO

NHCH2OH

NHCH2OH

nCO

NCH2

NCH2n

Dimethylol urea Cross linked UF resins

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Apparatus:

Beaker, Glass rod, Measuring flask, Funnel, Filter paper

Reagents:

Urea, 40% formaldehyde solution, Concentrated Sulphuric acid

Procedure:

Place 10 ml of 40% formaldehyde solution in a beaker. Add about 5g of urea while stirring

until a saturated solution is obtained. Add a few drops of concentrated sulphuric acid stirring

cautiously during the addition. All of a sudden a voluminous while solid mass appears in the beaker.

When the reaction is complete, wash the residue with water and dry the product and determine the

yield of the product.

Result:

The yield of the urea-formaldehyde (UF) resin is…………………g

24

Expt. No: 9999 Preparation of Phenol-Formaldehyde Resin

Aim

To prepare phenol-formaldehyde (PF) resin

Theory

Phenol - formaldehyde polymers (also called Bakelite) are the oldest synthetic polymers.

These are obtained by the condensation reaction of phenol with formaldehyde in the presence of

either an acid or a base catalyst. In the presence of acid catalyst, the reaction starts with the initial

formation of o-and/or p-hydroxymethylphenol derivatives, which further react with phenol to form

compounds having rings joined to each other through –CH2 groups. The initial product could be a

linear product – Novolac used in paints.

Novolac on heating with formaldehyde undergoes cross linking to form infusible solid mass

called Bakelite. It is used for making combs, phonograph records, electrical switches and handles of

various utensils.

25

Apparatus:

Beaker, Glass rod, Measuring flask, Funnel, Filter paper

Reagents:

Phenol, 40% Formaldehyde solution, Glacial acetic acid, Concentrated hydrochloric acid

Procedure:

Place 5 ml of glacial acetic acid and 2 ml of 40% formaldehyde solution in a beaker and add

about 2g of phenol. Wrap a cloth or towel loosely round the beaker. Add a few ml of concentrated

hydrochloric acid into the mixture carefully and heat it slightly. Within five minutes, a large mass of

pink plastic is formed. The residue obtained is washed with water, filtered, then dry the product and

determine the yield of the product.

Result:

The yield of the phenol-formaldehyde (PF) resin is…………………g

Lab Manual of Engineering Chemistry - KTU B.Tech...

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