lab manual of engineering chemistry - ktu b.tech...
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1
Lab Manual of
Engineering ChemistryEngineering ChemistryEngineering ChemistryEngineering Chemistry
2
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
S.No Volume of given
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
Observations and Calculations:
S.No Volume of given
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
Observations and Calculations:
S.No
Volume of given
bleaching powder
solution 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 = �
�� = �. �/
Volume of standard Na2S2O3 solution, V1 = ……………ml
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
Observations and Calculations:
S.No
Volume of given
solution of Mohr’s salt
in ml
Burette Readings in ml Volume of
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
Observations and Calculations:
S.No Volume of given solution of
copper ore/alloy 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 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
23
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