Advanced Practical Inorganic Chemistry For MSc SEM-III IV Students

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1 Advanced Practical Inorganic Chemistry-MSc SEM-III & IV- Dr Sakina .Z.Bootwala
WilsonCollege,Mumbai:400007
Advanced Practical Inorganic
Chemistry
For MSc SEM-III & IV students
Dr Sakina .Z.Bootwala
HOD, Department of Chemistry
Wilson College
Chowpatty, Mumbai:400007

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PREFACE
It gives me great pleasure to bring out the first edition of the book on “Practical in Inorganic
chemistry” written as per the new syllabus of MSc (SEM-III and SEM-IV) of University of
Mumbai which has been effective since June 2019.
The crucial role of practical work and experimentation in science curriculum is universally
accepted. This is more so in chemistry which is an experimental science. Most of the concepts in
this area are better understood by doing the practical work. The aims and objectives of practical
work in the laboratory can be stated as follows: 1. better understanding of scientific concepts and
principles. 2. Promotion of basic skills and competencies (procedural and manipulative skills,
observational skills, drawing skills, reporting and interpretation skills). 3. Awakening and
maintaining curiosity in the learning environment.
The new Practical Handbook has been introduced with the aim of providing the teachers and
students with necessary guidance for planning practical activities, engaging students effectively
in the teaching learning process and to promote students' practical skills in the discipline of
chemistry. I wish to make use of this opportunity to thank and express my appreciation to the
teacher friends, colleaques and students for their patronage and the faith shown on me.
I am also thankful to our publisher M/S Sheth Publishers Pvt. Ltd .Mumbai for all their effort to
bring out this book.
Mumbai Dr.Sakina. Z. Bootwala
Dated: January 2020.

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SYLLABUS (effective from June 2019)
SEMESTER-III
PSCHI3P1: Analysis of ores/alloys
1. Analysis of Brass alloy:
(i) Cu content by iodometric method, (ii) Zn content by complexometric method.
2. Analysis of Mangelium alloy:
(i) Al content by gravimetric method as basic succinate, (ii) Mg content by complexometric
method.
3. Analysis of Bronze alloy:
(i) Cu content by complexometric method, (ii) Sn content by gravimetric method.
4. Analysis of steel nickel alloy:
(i) Ni content by homogeneous precipitation method.
PSCHI3P2: Solvent Extraction
1. Separation of Mn and Fe using isoamyl alcohol and estimation of Mn
2. Separation of Co and Ni using n-butyl alcohol and estimation of Co
3. Separation of U and Fe using 8-hydroxyquinoline in chloroform and estimation of U
4. Separation of Fe and Mo using isoamyl alcohol and estimation of Mo
5. Separation of Cu and Fe using n-butyl acetate and estimation of Cu
PSCHI3P3: Inorganic Preparations
1. Preparation of V(oxinate)3
2. Preparation of Co(α-nitroso-β-naphthol)3
3. Preparation of Ni(salicylaldoxime)2

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4. Hexaamine cobalt (III) chloride
5. Preparation of Trans-bis (glycinato) Cu(II)
PSCHI3P4: Analysis of the following samples
1. Calcium tablet for its calcium content by complexometric titration.
2. Bleaching powder for its available chlorine content by iodometric method.
3. Iron tablet for its iron content colorimetry by 1,10-phenonthroline method.
4. Nycil powder for its Zn content complexometrically.
SEMESTER-IV
PSCHI4P1: Analysis of ores
1. Analysis of Galena ore
(i) Pb content as PbCrO4by gravimetric method,
(ii) Fe content by colorimetric method using 1,10 - phenanthroline.
2. Analysis of Zinc Blende ore
(i) Zn content by complexometric method.
(ii) Fe content by colorimetric method (Azide Method).
3. Analysis of Pyrolusite ore
(i) Mn content by complexometric method.
(ii) Acid insoluble residue by gravimetric method.
PSCHI4P2: Coordination Chemistry
1.Determination of Stability Constant of [Zn(NH3)4]2+
by potentiometry.
2. Determination of Stability Constant of [Ag(en)]+
by potentiometry.
3. Determination of CFSE values of hexa aqua complexes of Ti2+
and Cr3+
.

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4. Determination of Racah parameters for complex [Ni(H2O)6]2+
and [Ni(en)3]2+
.
PSCHI4P3: Analysis of the following Commercial Samples
1.Electral powder for Na content by flame photometry.
2.Fasting salt for Chloride content by conductometer.
3.Sea water for percentage salinity by Volhard’s method.
4.Soil for mixed oxide content by gravimetric method.
5.Fertilizer for Potassium content by flame photometry.
PSCHI4P4: Project Evaluation.

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CONTENTS
Sr.no Title Page no.
1 Introduction
Semester III
(PSCHI3P1: Analysis of ores/alloys
2 Analysis of brass alloy
3 Analysis of magnelium alloy
4 Analysis of bronze alloy
5 Analysis of steel nickel
PSCHI3P2: Solvent Extraction
6 Separation of cobalt & nickel
7 Separation of manganese & iron
8 Separation of uranium & iron
9 Separation of iron & molybednum
10 Separation of copper &Iron
PSCHI3P3: Inorganic Preparations
11 Preparation of V(oxinate)3
12 Preparation of Hg[co(SCN)4]
13 Preparation of Ni (salicyl aldoxime)2
14 Preparation of Cobalt Hexamine Chloride
15 Trans- bis (glycinato) Cu (II)
16 Preparation of Bis( α- Nitroso-β-Napthol)2 Cobalt (II)
PSCHI3P4: Analysis of the following commercial samples
17 Analysis of calcium tablet
18 Analysis of bleaching powder
19 Analysis of iron tablet for its iron content colorimetrically by 1,10-
phenanthroline method

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20 Analysis of nycil powder
Semester IV
PSCHI4P1: Analysis of ores
21 Analysis of Galena ore
22 Analysis of Zinc Blende ore
23 Analysis of Pyrolusite ore
PSCHI4P2: Coordination Chemistry
24 Determination of Stability Constant of [Zn(NH3)4]2+
by potentiometry.
25 Determination of Stability Constant of [Ag(en)]+
by potentiometry.
26 Determination of CFSE values of hexa aqua complexes of Ti2+
and
Cr3+
.
27 Determination of Racah parameters for complex [Ni(H2O)6]2+
and
[Ni(en)3]2+
.
PSCHI4P3: Analysis of the following Commercial Samples
28 Fasting salt for Chloride content by conductometer
29 Fertilizer for Potassium content by flame photometry.
30 Sea water for percentage salinity by Volhard’s method
31 Soil for mixed oxide content by gravimetric method
32 Electral powder for Na content by flame photometry

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INTRODUCTION
Ore is natural rock that contains desirable minerals, typically metals that can be extracted from
it. Ore is extracted from the earth through mining and refined, often via smelting, to extract the
valuable element or elements Metals are obtained from the earth’s surface, or from below earth’s
surface. Not all metals are found in their pure form. Some are mixed with impurities and some are a
mixture of various other elements. Metal ores are generally oxides, sulfides, silicates, native
metals such as copper, or noble metals such as gold.
Oxides:
Chromite (chromium),Coltan (niobium and tantalum),Columbite (niobium)Hematite (iron)
Ilmenite (titanium) Magnetite (iron) Pyrolusite (manganese) Tantalite (tantalum)
Uraninite (uranium).
Sulfides:
Acanthite (silver)Argentite (silver)Bornite (copper)Chalcopyrite (copper)Chalcocite (copper)Cin
nabar (mercury)Cobaltite (cobalt)Galena (lead)Molybdenite (molybdenum)Pentlandite (nickel)S
phalerite (zinc)
Carbonates:
Dolomite (magnesium)Magnesite (magnesium)Malachite (copper)
Others : Baryte (barium)Bauxite (aluminium)Beryl (beryllium)
Alloys :
An alloy is a mixture of two metals or even one metal and another substance, but the major
component being metal. The resulting mixture forms a substance with properties that often differ

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from those of the pure metals, such as increased strength or hardness .An alloy will retain all the
properties of a metal in the resulting material, such as electrical ,conductivity, ductility,
opaqueness and lustre . a combination of metals may reduce the overall cost of the material while
preserving important properties. In other cases, the combination of metals imparts synergistic
properties to the constituent metal elements such as corrosion resistance or mechanical strength.
Examples of alloys are steel, solder, brass , pewter, duralium, bronze etc .
Opening Of Ore/Alloy Sample: The chemical transformation of any ore/alloy sample into a
desired form so as to facilitate analysis of its constituents is known as opening of the sample
i.e., conversion of an ore or alloy sample into desired soluble or insoluble species The exact
choice of method for opening of an ore depends on the nature of the ore, the element of interest
and its chemical properties. The selection of the appropriate method of sample decomposition
is very important for correct analysis. There are mainly two methods to open an ore. They are
(1) wet method & (2) dry method.
Wet method constitutes the digestion of an ore sample with an acid or a mixture of acids to
convert the complex ore mineral into simple chemical compounds. The action of an acid or a
mixture of acids is supplemented by heat and pressure. This is normally preferred over dry
method
Dry method constitutes the fusion of an ore with a solid or mixture of solids to convert
complex minerals compound into simple chemical compounds. The reaction is effected by
melting the mass at elevated temperature. The fusion is carried out in variety of crucibles made
of platinum, gold, nickel, iron, zirconium, palladium, silica, porcelain etc. There are numbers
of chemical compounds used for fusion. The choice of fusion mixture depends on the nature of
the ore to be fused and the nature of the final product needed for further analysis.

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On the basis of chemical properties of an ore, various acids, combination of acids and acids are
generally employed for dissolution of an ore sample. Reactivity of various reagents are given
here:
Sr,No Reagents Metal ion to be estimated
1 water Alkali salts
2 HCl Carbonates and oxides ores of Fe,Ca,Mn,Mg,Sn ,Ti, Zn etc ,
3 HCl with oxidizing
agent
Sulphides/pyrites ores of Cu,Pb,Mo,Zn,Ni,Fe,As, etc
4 HNO3 Ores of Cd,Cu,Mo,Co,Ni etc
5 Aqua regia Ores of Cd,Mg,Rh,W etc
6 H2SO4 Ores of Al,Be,Mn,Pb,Th,Ti,V etc
7 H3PO4+ H2SO4 Fused Al2O3 and chrome ores
Decomposition process is generally increased by using very finely ground sample Acids should
never be added to a dry powder because of danger of loss due to ferocity of the reaction. The
sample should be moistened with water, the acid is added cautiously and the covered container
allowed to stand without additional heating till the completion of initial reaction
Determination of the content of common elements present in the stock solution.;
The following methods are used to determine the content of the elements present in Ores/alloys
sample stock solution.
1. Titrimetric Analysis
2. Gravimetric Analysis
3. Spectrophotometric

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TITRIMETRIC ANALYSIS
Titrimetric analysis is a method of analysis wherein a test substance is allowed to react with a
known standard solution in the presence of an indicator until the end point. An endpoint is one
where the test substance has been completely reacted with the analyzing regent. The reagent
added from burrete to estimate is called as a titrant. The substance underestimation is called
titrand.
The different types of titration are
A) Based on the method of titration:
They are of three types of titration based on the method used in the process of titration.
1. Direct titration: As the name indicates, it is a basic titration. A known amount of titrant is
added from a burrete to a titrand sample taken in a flask. Here one substance is analyzed for its
quantity by another substance of known volume and concentration.
2. In-direct titration: Theoretically it is converting a substance into acid and analyzing with a
base. (also vice-versa). This is a method extrapolated to use titration for non readily reactive
substances. A substance can be weakly acidic and so it does not permit for precise analysis by
direct titration. So first that substance is chemically altered to be more reactive in acidic or basic
form and then analyzed by adding a titrant.
3. Back titration: This method is also suitable for weakly reactive or non-reactive substance
estimation. Here a substance is allowed to react with excess and known quantity of a base or an
acid. The remnant excess base or acid is estimated by a known quantity of acid or base
receptively. It is called back titration as we are estimating a substance which was added by us.
B) Based on the nature of solvents and chemical reaction:
The titrations can also be classified based on the nature of solvents used. Based on the solvent
used they can be classified as aqueous and non-aqueous types. Further based on the nature of
chemical reaction they are classified as below

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1. Aqueous titrations:
• Acid-base titrations. ( neutralization): An unknown sample of acid is estimated with a known
quantity base or vice-versa. The result reaches a neutral point at pH-7 and in most cases, salt is
formed.
• Redox titrations: The full name is oxidation-reduction titration. Here a reducing agent is
allowed to react with an oxidizing agent till endpoint. The common oxidizing agents used are
Potassium permanganate, bromine, cerium salts, etc.
• Complexometric titrations: As the name indicates, the endpoint is seen by the formation of a
complex molecule. Here titrant and titrand react to form a complex till endpoint is reached. Once
complex is formed, the complex is stable and no further reaction takes place. The reaction
depends up the chelating agent’s ability to form a complex with the sample under test. Example
of the chelate is ethylene tetra-acetic acid (EDTA)sodium salt.
• Precipitation titrations: The reaction occurs by formation of a solid precipitate at the bottom of
the flask. Here titrant reacts with titrand to form an insoluble precipitate. Example for such
reaction is Silver nitrate with Ammonium chloride. The reaction forms a white precipitate of
silver chloride.
2. Non-Aqueous titrations.
These are the titration done for organic and medicinal compounds. So widely used in medicinal
chemistry and pharmaceutical labs.
a) Acid-base titrations. (neutralization): Here the reaction occur in organic solvents like
glacial acetic acid and the acid used is perchloric acid (HClO4)
b) Redox titrations. Here the reaction is done as iodometry or iodimetry. Iodine is used as an
oxidizing agent and a reducing agent is a thiosulfate.
c) Iodometry: Here the sample made to release iodine from within and this released iodine is
measure with sodium thiosulfate as reducing agent.
d) Iodimetry: Here the sample under test is measured with a known concentration of Iodine.
These are similar to aqueous titrations, but here instead of water as a solvent, an organic solvent
is used.

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GRAVIMETRIC ANALYSIS:
Gravimetric analysis is a method in analytical chemistry to determine the quantity of analyte
based on the mass of a solid. Example: Measuring the solids suspended in the water sample –
Once a known volume of water is filtered, the collected solids are weighed.
The principle of Gravimetric Analysis:
The principle behind the gravimetric analysis is that the mass of an ion in a pure compound and
can be determined. Later, used to find the mass percent of the same ion in a known quantity of an
impure compound.
Steps followed in the Gravimetric Analysis
1. Preparation of a solution containing a known weight of the sample.
2. Separation of the desired constituent.
3. Weighing the isolated constituent.
4. Computation of the amount of the particular constituent in the sample from the observed
weight of the isolated substance.
Types of Gravimetric Analysis
There are 4 fundamental types of gravimetric analysis. Of which, there are 2 common types
involving changes in the phase of the analyte to separate it from the rest of a mixture, resulting in
a change in mass.
1. Volatilization gravimetry: Volatilization Gravimetry involves separating components of
our mixture by heating or chemically decomposing the sample.

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2. Precipitation gravimetry: Precipitation Gravimetry uses a precipitation reaction to
separate one or more parts of a solution by incorporating it into a solid.
3. Electrogravimetry : Electrogravimetry is a method used to separate and quantify ions of
a substance, usually a metal.
4. Thermogravimetric :Thermogravimetric is a method of thermal analysis in which
changes in physical and chemical properties of materials are measured as a function of
increasing temperature or as a function of time.
SPECTROPHOTOMETRY:
Spectrophotometry is a method to measure how much a chemical substance absorbs light by
measuring the intensity of light as a beam of light passes through sample solution. The basic
principle is that each compound absorbs or transmits light over a certain range of wavelength.
This measurement can also be used to measure the amount of a known chemical substance.
An instrument that measures the amount of light (photons) in which a substance or
sample absorbed after the light has passed through it is referred to as spectrophotometer.
Determination of the amount of known chemical substance in concentration through the
measurement of detected intensity of light is one of the applications of spectrophotometry.

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A spectrophotometer is made up of two instruments: a spectrometer and a photometer. The
spectrometer is to produce light of any wavelength, while the photometer is to measure the
intensity of light. The spectrophotometer is designed in a way that the liquid or a sample is
placed between spectrometer and photometer. The photometer measures the amount of light that
passes through the sample and delivers a voltage signal to the display. If the absorbing of light
change, the voltage signal also changes.
The basic spectrophotometer instrument consists of a light source, a digital display, a
monochromator, a wavelength sector to transmit selected wavelength , a collimator for straight
light beam transmission, photoelectric detector and a cuvette to place a sample.
The intensity of light hitting the detector after passing through a “blank” solution is measured –
this is a solution that is identical to the sample but doesn’t contain the solute. This measurement
is called “I0”. Then the intensity of light hitting the detector after passing through the sample is
measured. This measurement is “I”.
The transmittance (the amount of light reaching the detector) is calculated as T= I / Io
The absorbance is calculated as A= - log10T = - log10 I / Io
The Beer-Lambert law is a linear relationship between the absorbance and the concentration,
molar absorption coefficient and optical coefficient of a solution: A= εcl ,where A is
absorbance, ε is molar absorption coefficient per moles per cm ,c is concentration in moles,l is
optical path length in cm
The molar absorption coefficient is a sample dependent property and is a measure of how strong
an absorber the sample is at a particular wavelength of light. The concentration is simply the
moles L-1
(M) of the sample dissolved in the solution, and the length is the length of the cuvette
used for the absorbance measurement and is typically 1 cm.
The Beer-Lambert law states that there is a linear relationship between the concentration and the
absorbance of the solution, which enables the concentration of a solution to be calculated by
measuring its absorbance. Using this calibration curve the concentration of an unknown solution

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can be determined by measuring its absorbance which is the main utility of the Beer-Lambert
Law.
SOLVENT EXTRACTION
Solvent extraction, is a method to separate compounds based on their relative solubilities in two
different immiscible liquids. Immiscible liquids are ones that cannot get mixed up together and
separate into layers when shaken together. These liquids are usually water and an organic
solvent. Solvent extraction is an extraction of a substance from one liquid into another liquid
phase. The most common use of the distribution principle is in the extraction of substances by
solvents, which are often employed in a laboratory or in large scale manufacturing.
Organic compounds and metal complexes are generally much more soluble in organic solvents,
like benzene, chloroform, and ether, than in water and these solvents are immiscible with water.
Organic compounds are then quite easily separated from the mixture with inorganic compounds
in aqueous medium by adding benzene, chloroform, etc. Upon shaking, these separate into two
layers. Since metal complexes have their distribution ratio largely in favor of the organic phase,
more of them would pass into a non-aqueous layer. Finally this non-aqueous layer is removed
and subjected to quantitative analysis by using spectroscopic technique .

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Partition Law or Distribution Law
This Law was given by Nernst. This law gives the relationship between the concentrations of a
given substance in two different phases in equilibrium with each other.
Suppose, if a small quantity of solute soluble in both the liquids is added then solute distributes it
self in two liquid & equilibrium is set up. Both the liquids must be immiscible with each other If
the concentration of the solute in two liquids are C1 & C2. According to this law,
C1/C2=K (const.)
k= distribution or partition coefficient
Factors affecting Distribution Coefficient-
1. The value of K depends upon the temperature , nature of solute , nature of two solvents .
2. It does not depend upon the amount of solute or solvents taken.

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If the solubilities of given solute in the two solvents at the given temperature are S1 &
S2 respectively, then
C1/C2 = S1/S2 =K
e.g Water & ether are immiscible with each other. Solute succinic acid is soluble in both of these
solvents. If small amount of succinic acid is added in a mixture . Then,
Concentration of succinic acid in water = Cwater
Concentration of succinic acid in ether = Cether
K = Cwater / Cether
C water is more than Cether so C water is taken in the numerator & C ether in the denominator.
For experimental verification of distribution law, different quantity of the solute are added in
different quantities of two solvents in many experiments . If the value of distribution coefficient
for each experiment remains the same. This shows that the data follows distribution law.
Conditions for Distribution Law
1. Both the solvents must be immiscible with each other.
2. It is applicable only for dilute solution.
3. Temperature remains constant throughout the experiment.
4. The molecular state of the solute in the two solvents should be the same.

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ANALYSIS OF BRASS ALLOY
AIM: To analyze the given sample of Brass alloy for its-
1} Copper content by Iodometric method.
2} Zinc Content by complexometric method.
REQUIREMENTS: Brass,10% KI,0.05N,K2Cr2O7,0.05N Na2SO3,2N CH3COOH,
Na2CO3,Starch indicator [freshly prepared],0.01M EDTA, Buffer pH=10,Eriochrome Black T
etc.
PREPARATION:
0.01M EDTA solution:
1000ml of 1M EDTA ≡ 372.24g of Na2EDTA
100ml of 0.01M EDTA≡ 0.372g of Na2EDTA
THEORY:
Brass is an alloy containing 63% of Copper and 37% of Zn.Alloy can be dissolved in
HNO3.The brass contains copper which can be estimated by Iodometric method. Cu+2
can
oxidize I-1
quantitatively from solution. The free H+
ion from the solution is neutralized by
sodium carbonate precipitated Cu(OH)2 must neutralized by acetic acid to release Cu+2
. The
liberated iodine can be titrated against standard sodium thiosulphate.
Zn can be estimated by complexometric titration after removal of Cu+2
using H2S gas.
Excess of H2S can be boiled off from the filtrate Zn form coloured complex with Eriochrome
Black T indicator which can be titrated against the standard EDTA solution at buffer pH=10.
PROCEDURE:
A] OPENING OF AN ALLOY:
1. Dissolve accurately weighed (0.3-0.5) or the supplied sample of the alloy in 10cm3
of
Conc. HNO3 in 250cm3
beaker and 10cm3
of distilled water and then heat it on sand bath.
2. Evaporate the clear solution nearly dryness.

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3. Cool the solution. Add 10cm3
of dilute HCl and again boil for few minutes.
4. Transfer the solution in 250cm3
standard measuring flask. Wash the beaker and collect all
washings in the same standard measuring flask.
5. Dilute it with distilled water to 250cm3
and shake well. This is the stock solution.
B] STANDARDISATION OF Na2S2O3:
1. Prepare 100cm3
of 0.05N K2Cr2O7 solution.
2. Pipette out 10cm3
of K2Cr2O7 solution in a conical flask. Add 5cm3
of conc. HCl to it.
3. Add one test tube of 10% KI solution to it. Shake well.
4. Titrate it against 0.05N Na2S2O3 from the burette using freshly prepared starch indicator.
END POINT: Blue to Green colour.
C] ESTIMATION OF COPPER:
of the stock solution in a conical flask.
2. Add anhydrous Na2CO3 pinch by pinch till a faint precipitate of Cu(OH)2 forms.
3. Dissolve the precipitate by adding 2N acetic acid drop wise with constant stirring of
flask, till a clear solution is obtained or effervescence of CO2 ceases.
4. Add one and half test tube of 10% KI solution to it and shake well.
5. Titrate the liberated I2 against 0.05N Na2S2O3 solution from the burette using freshly
prepared starch indicator.
END POINT: Blue to Colourless.
D] ESTIMATION OF ZINC:
of the stock solution in 250cm3
beaker. Add 10cm3
of conc HCl heat
and pass H2S gas to precipitate copper as CuS.
2. Filter the precipitate through an ordinary filter paper and collect the filtrate in 500cm3
beaker.

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3. Wash the precipitate with hot distilled water 2-3 times and collect all the washings along
with the filtrate.
4. Boil off the H2S completely. Cool and dilute it to 100cm3
in standard measuring flask.
of the diluted solution in a conical flask. Add 10cm3
of distilled water
to it.
6. Add 5-10 drops of 0.5% Xylenol orange indicator to it.
7. Add hexamine powder till colour of the solution becomes reddish violet or wine red
indicating that pH is raised to about 6.0.
8. Titrate it against 0.01M EDTA solution from the burette.
END POINT: Reddish Violet to Yellow colour.
REACTIONS:
1) CuCl2 + 4KI→ Cu2I2 + 4KCI + I2↑
I2 + 2Na2S2O3→Na2S4O6 + 2NaI
2)CuCI2 + H2S → CuS↓ + 2HCl
Zn+2
+ Na2C10H14O8N2 → 2Na+
+ Zn-C10H14O8N2
Zn-EDTA Complex
OBSERVATION AND CALCULATIONS:
STANDARDISATION OF Na2S2O3:
10cm3
of 0.05N K2Cr2O7 solution required x cm3
of Na2S2O3 solution.
Normality of Na2S2O3 solution= 10 X 0.0
x
=_____Y___N.
ESTIMATION OF COPPER:
25cm3
of the stock solution required__A___cm3
of ‘Y’N Na2S2O3 solution.

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250cm3
of the stock solution will require 10x_A___cm3
of Y’N Na2S2O3solution.
1000cm3
of 1N Na2S2O3 = 63.54g of Cu.
10x_A___cm3
of Y’N Na2S2O3 = 63.54 x10x A x Y = _______ g of Cu.
1000
10x__A___cm3
of Y’N Na2S2O3= ______g of Cu.
Percentage of Cu = Weight of Cu X 100
Weight of ore
=________%

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ESTIMATION OF ZINC:
10cm3
of diluted solution required ___A____cm3
of 0.05M EDTA solution.
⸫100cm3
of diluted solution will require 10x ___A____cm3
i.e.50cm3
of stock solution required __10A___cm3
⸫250cm3
of stock solution will require 5x __10A____cm3
1000cm3
of 1M EDTA = 65.58g of Zn.
5x ‘10A’cm3
of 0.05M EDTA solution = 65.58 x 5 x 10A x0.01g of Zn.1000
=_________g of Zn.
Percentage of Zn = Weight of Zn x 100
Weight of the ore
=_____________% of Zn.
RESULT:
1) Amount of Copper present in the given sample of Brass =_____g.
2) Percentage of Copper present in the given sample of Brass=___%.
3) Amount of Zinc present in the given sample of Brass =_____g.
4) Percentage of Zinc present in the given sample Brass=____%.

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ANALYSIS OF MAGNELIUM ALLOY
AIM: Determine the percentage of Aluminium and Magnesium in the given sample of
Magnesium alloy. (i) Al content by gravimetric method as basic succinate, (ii) Mg content by
complexometric method.
REQUIREMENTS: 4N H2SO4, Acetic acid, 10%Oxine solution, conc.HCl, Ammonium
acetate, Standard 0.01M EDTA solution, Buffer pH=10, Eriochrome Black T.
THEORY:
Magnelium is an alloy of 95% Al and 5% Mg metal. Al is amphoteric in nature and is difficult to
dissolve in a mineral acid/alkali. However, it can be dissolved in 4N sulfuric acid on heating.
Chloride and nitrate salt of Al are comparatively less soluble. Precipitation from a homogeneous
solution of aluminium sulphate by neutralisation using urea in presence of succinic acid leads to
the formation of a well-defined alumina precursor, basic aluminium succinate with excellent
free-flowing characteristics. The fitrate after removing aluminum as hydroxide can be
quantitatively analyzed for Mg content by complexometric titration. The Mg forms complex with
EDTA at buffer pH 10.
PROCEDURE:
1. Weigh accurately 500mg of alloy.
2. Add 30ml of 4N H2SO4 and keep it on sand bath.
3. Heat it till alloy dissolves completely.
4. Make up the solution to 250ml in a volumetric flask with distilled water.
A] ESTIMATION OF ALUMINIUM.
1. Pipette out 50 ml of diluted solution in a beaker. Add 25ml of water in the solution.
2. Add 0.200 gram of succinic acid and 20gram of urea
3. The solution was slowly heated to 363 K with constant stirring on hot water bath till the
pH reached 7.0
4. the formation of a basic aluminium succinate as a dense precipitate is obtained

25
5. The precipitate formed is kept for digestion for 30 mins.
6. Cool and filter the solution through previously weighed sintered Glass G4 crucible.
7. Dry the ppt at 1100
C till constant weight is obtained.
8. Calculate the amount of Al(OH)(C4H404). And hence the percentage of Al present in the
conversion factor.
B] ESTIMATION OF MAGNESIUM:
1. Pipette out 25ml of diluted stock solution in 250cm3
beaker.
2. Add 25ml of water and heat the solution. Add 3gm of NH4Cl in hot solution, stir till it
dissolve.
3. Add liquor NH3 to the hot solution till white gelatinous ppt obtained. Check the solution
should be alkaline.
4. Digest the above white ppt on a boiling water bath for 30mins.Filter the solution through
Whatmann’s filter paper 41. Collect the filtrate and washing in 100 cm3
standard flask.
5. Dilute it to 100ml in a standard flask.
6. Pipette out 10ml of diluted solution in a conical flask and add 10ml of distilled water.
7. Add 5 ml of buffer solution of pH=10 and a pinch of Erio-chrome Black T indicator.
8. Titrate it against standard 0.01 M EDTA solution till the colour changes wine red to pure
blue.
9. Record the constant burette reading and calculate the amount of Mg present in the
solution. Hence calculate the percentage of Mg.
REACTIONS:
2Al+3
+ 3H2SO4→ Al2(SO4)3 + 3H2↑
Al2(SO4)3 + 2C4H6O4 + 6NH4OH → 2Al(OH) (C4H404)+ 3(NH4 )2SO4+ 4H2O
Mg+2
+ Na2C10H14O8N2→ MgC10H14O8N2 + 2Na+2
OBSERVATION:
Weight of the Magnelium alloy =________mg

26
OBSERVATION AND CALCULATIONS:
A] ESTIMATION OF ALUMINUM:
1) Weight of empty sintered glass G4 crucible =______g.
2) Weight of the sintered glass crucible with Al(OH) (C4H404) = ______ g.
3) Weight of Al(OH) (C4H404) = ______g.
Let the weight of the residue Al(OH) (C4H404) be Y g.
50 cm3
of the diluted solution gave Y g of Al(OH) (C4H404).
250 cm3
of the diluted solution gave 5Y g of Al(OH) (C4H404).
128 g of basic aluminium succinate = 27g of Al.
5Y g of basic aluminium succinate = 27 x 5Y g of Al
128
= _______g of Al.
0.5g of alloy contains ______g of Al.
100g of the alloy contains = x 100
Weight of the alloy =___________% of Al.
B] ESTIMATION OF MAGNESIUM:
10 cm3
of diluted solution required ‘z’ cm3
100 cm3
of dilution solution required 10‘z’ cm3
25 cm3
of stock solution required 10‘z’ cm3
of 0.01M EDTA
250cm3
of stock solution required 25‘z’ cm3
of 0.01M EDTA
1000 cm3
of 1M EDTA = 24.32g of Mg.
25z cm3
of 0.01M EDTA = 24.32 x 25z x 0.01 g of M
1000

27
=________g of Mg.
Percentage of Mg = Weight of Mg x 100
Weight of the ore.
=___________%
RESULT:
1] Amount of Aluminum present in the given sample of alloy= _____g.
2] Percentage of Aluminum present in the given sample of alloy=___%.
3]Amount of Magnesium present in the given sample of alloy=____g.
4] Percentage of Magnesium present in the given sample of alloy=__%

28
ANALYSIS OF BRONZE ALLOY
AIM: To analyze the given sample of Bronze for it’s
1] Cu content by complexometric method.
2] Sn content by gravimetric method.
REQUIREMENTS: Bronze alloy, Acetic acid, distilled water, conc. HNO3, conc. HCl, buffer
pH=10, 0.05M EDTA, Fast sulphon black F indicator etc.
THEORY:
Bronze is an alloy containing 85-95% of Cu and 5-15% of Sn. Alloy can be dissolved in conc.
HNO3. Sn is precipitated as a crystalline hydrated meta-stannic acid SnO2.2H2O. The ppt can be
ignited at 7000
C to get definite composition of SnO2.The filtrate contains Cu, which can be
estimated by Iodometric method. Cu+2
can oxidize I-1
quantitatively from solution. The free H+
ion from the solution is neutralized by Na2CO3 and precipitated Cu (OH)2 must be neutralized by
acetic acid to release Cu+2
.The liberated iodine can be titrated against standard Na2S2O3 solution.
PROCEDURE:
A] OPENING OF AN ALLOY AND ESTIMATION OF Sn.
1) Weight accurately about 500mg of bronze alloy.
2) Add in 10 cm3
of conc. HNO3 and boil till it dissolves completely. Do not dry it completely.
3) Extract it with water and boil it for 10 minutes.
4) Filter it through quantitatively filter no.41.
5) Wash it with distilled water + 1:100 HNO3 and collect filtrate quantitatively in a 250 cm3
volumetric flask.
6) Collect the ppt. dry it in oven at 800
-1000
. Ignite it along the filter paper in previously weighed
silica crucible.
7) Heat for 30 mins, cool it and weigh the residue. Calculate the amount of Tin present in
sample.

29
B] ESTIMATION OF COPPER:
1) Pipette out 25 cm3
from the above filtrate.
2) Add 10cm3
of buffer solution of pH-10 and 3-4 drops of Fast sulphon black F indicator.
3) titrate it against 0.05M EDTA and the end point is blue to green.
4) Record the constant burette reading and calculate the amount of Cu present in the solution.
Hence calculate the percentage of Cu.
REACTIONS:
1) Cu(NO3)2 + Na2C10H14O8N2→ Cu-C10H14O8N2 + 2NaNO3
2) Sn + 4HNO3→ H2SnO3 + 4NO2 + H2O
3) H2SnO3→ SnO2 + H2O↑
OBSERVATIONS AND CALCULATIONS:
Weight of the Bronze alloy = mg.
A] ESTIMATION OF Sn:
1) Weight of empty silica crucible =________g.
2) Weight of crucible + residue of SnO2 =________g.
3) Weight of residue of SnO2 = _______ g.
Let the weight of residue SnO2 be X g
Now,
150.7 g of SnO2 = 118.7g of Sn.
X g of SnO2 = 118.7 x X g of Sn
150.7
= _______g of Sn.
‘W’ g of alloy contains -----------g of Sn.

30
100g of the alloy contains =Weight of Sn x 100
Weight of the ore
= ____% of Sn.
B] ESTIMATION OF Cu:
25cm3
of the stock solution requires ‘Y’ cm3
250cm3
of the stock solution required ‘10Y’ cm3
of the 0.05M EDTA solution.
1000cm3
of 1M EDTA ≡ 63.54g of Cu.
10Y cm3
of 0.05M EDTA ≡ 63.54 x 10Y x 0.05 g of Cu.
1000
≡ __________g of Cu.
Percentage of Cu ≡Weight of Cu x 100
Weight of alloy
≡ ________%
RESULT:
1] Amount of Tin present in the given sample of Bronze=______g.
2] Percentage of Tin present in the given sample of Bronze=____%.
3] Amount of Cu present in the given sample of Bronze=_______g.
4] Percentage of Cu present in the given sample of Bronze=____%.

31
ANALYSIS OF STEEL NICKEL
AIM:- Analyze the given sample of steel for its nickel content by homogenous precipitation
method.
REQUIREMENTS:- steel-nickel alloy, aqua regia, HCl ,solid NH4Cl, tartaric acid, 1% DMG in
alcohol . 1:1 NH3, etc.
THEORY:- Conc. HNO3 can dissolve the Ni from the steel. Ni can be Precipitated from the
solution by using alcoholic DMG solution at pH 7.Fe also forms ppt with DMG. Fe can be kept
in the solution by adding tartaric acid/citric acid which forms a stable salt with Fe. Ni-DMG
complex is more stable as compared to its tartarate complex. The Ni-(DMG)2 complex can be
filtered from G3 crucible and dried upto 110o
C.
PROCEDURE:-
1. Weigh accurately the given sample. Dissolve it in 20ml aqua regia. Evaporate it to dryness.
2. Dilute the solution upto 250ml in a standard measuring flask using distilled water.
3. Pipette out 50ml of the diluted sample solution in a 250ml beaker an add ½ test tube of dilute
HCl . Add 2 g of solid NH4Cl.
4. Add liq.NH3 in it with constant stirring till faint smell of ammonia persists.
5. Digest the ppt of ferric hydroxide in a water bath for about 30 mins and filter it using normal
filter paper.
6. Collect the washings and filtrate in a 500ml beaker.
7. To the filtrate add 1% alc.DMG solution and add 1:1 NH3 till complete precipitation.
8. Digest the ppt in a boiling water bath . filter it with G3 crucible and wash the ppt with water
till it get free from chlorides.
9. Dry and weigh the complex and hence calculate the Ni content.
Ni+2
+ 2DMG → Ni(DMG)₂
OBSERVATION AND CALCULATION:-

32
1) Weight of empty crucible- -------------- g
2) Weight of crucible + residue------------ g
3) Weight of residue- ------------ g
Let the weight of Ni-DMG complex be ‘Z’ g
Ni(DMG)2= Ni+2
288.69g of Ni(DMG)2 = 56.69g of Ni+2
Zg of Ni(DMG)2 = Z x 56.69 = g Ni+2
= Y g of Ni+2
288.69
This is the amount of Ni present in 50 ml of diluted solution.
Total amount of Ni present in 250ml of diluted solution=5xYg = A. g
Percentage of Ni = total amount of nickel x100
weight of ore
Percentage of Ni= %
RESULT;
1. Amount of Ni present in the given sample of steel= g
2. Percentage of Ni present in the given sample of steel= %

33
PAPER-II
SEPARATION OF COBALT & NICKEL
AIM:- Separation of cobalt & nickel in the given mixture using solvent extraction technique with
n-butyl alcohol & estimation of the amount of cobalt by colorimetric method.
REQUIREMENTS:- Cobalt sulphate soln (100ppm), nickel sulphate soln (100ppm), sat.KSCN
(40%), n-butyl alcohol.
THEORY:-
Cobalt forms dark blue coloured complex with large excess of SCN-
ion, while Ni does not form
such complex. The Co complex is highly soluble in n-butanol & Ni ion remains in aqueous
phase. n-butanol is an immiscible & heavier than water & also thermally stable. The blue colour
of Co-complex absorbs maximum at 620nm. The blue complex obeys Beer-Lambert’s Law in
the range of 1-25ppm of cobalt.
PROCEDURE:-
1. Preparation of 100ppm cobalt sulphate soln (soln 1) :-dissolve 0.476g of CoSO4.7H2O or
0.262g of CoSO4 in 1 ltr of D/W in standard measuring flask.
2. Preparation of 100ppm nickel sulphate soln (soln 2 ):- dissolve 0.236g of NiSO4 or
0.478g of NiSO4.7H2O iin small amount of D/W & dilute it to 1000ml in standard
measuring flask.
3. Preparation of calibration curve:- take 2,4,6,8,10cm3
of CoSO4 soln in 5 different
separting funnels.
4. Adjust the pH of the solution to 7 by adding 1 drop of NH3 (approximately).
5. Add 10ml of saturated KSCN soln& dilute it to 25ml with D/W. Shake well then add
10cm3
of n-butyl alcohol to each funnel & extract the complex into organic phase using
in a small flask. Repeat the above extraction with another 5ml of

34
n-butanol. Collect the organic phase(blue) & dilute it to 25ml with n-butanol. Measure
the absorbance at 620nm using n-butanol as blank. Plot the graph of O.D. v/s conc. of
cobalt in µgms.
6. Separation of mixture :-[sample- take 2.3cm3
of soln 1 & 1cm3
of soln 2 in 100cm3
beaker] transfer the given mixture into sample flask & wash with water. Add 10ml of
saturated KSCN. Shake well & carry out the steps as above. Measures its absorbance at
620nm & from the calibration curve calculate the amount of cobalt in the given mixture.
OBSERVATION:-
Wavelength used-620nm Blank solution- n-butanol.
OBSERVATION TABLE:-
Vol of
100ppm Co
soln in cm3
Vol of
sat.
KSCN
soln in
cm3
Vol of
D/W in
cm3
Final
vol in
cm3
Vol of n-
butanol
in cm3
Total
vol in
cm3
Conc. of
Co+2
in
µg
O.D.
2 10 14 25 10+5 25 8
4 10 13 25 10+5 25 16
6 10 12 25 10+5 25 24
8 10 11 25 10+5 25 32
10 10 10 25 10+5 25 40
Unknown 10 - 25 10+5 25
NATURE OF GRAPH:-

35

36
CALCULATION:- Separated organic phase-25cm3
From the graph of O.D. v/s Conc. of Co+2
in µg, the separated organic phase gave ‘a’ (unknown)
as its O.D. and this corresponds to ‘b’ppm of Co+2
From graph
Unknown concentration of Co+2
in the given solution = ‘b’ ppm
Therefore 1ppm = 1mg/1000ml
= 0.1mg/100ml
= 0.025mg/25ml
‘b’ppm = 0.025 x b
= ‘c’ mg in 25ml of diluted solution.
Thus, the given mixture contains ‘c’ mg of Co+2
RESULTS:-
Amount of cobalt in the given mixture is ‘c’ mg.

37
SEPARATION OF MANGANESE & IRON
AIM:- separation of manganese and iron in the given mixture using solvent extraction technique
with iso amyl alcohol and estimation of the amount of manganese by colorimetric method.
REQUIREMENTS:- Manganese solution (100 ppm),Fe+2
solution (100 ppm)
KIO4, KSCN (1M), phosphoric acid, HNO3 , iso amyl alcohol.
PREPARATION:-
1) Preparation of 100ppm Mn+2
solution.
Preparation of 500cm3
of 100ppm Mn+2
solution
1ppm = 1mg/1000ml
100ppm = 100mg/1000ml
= 50mg/500ml
54.983mg of Mn =169.02 mg of MnSO4.H2O
50 mg of Mn = 169.02 x 50/ 54.983
= 153.7mg of MnSO4.H2O
= 0.1537 g of MnSO4.H2O
Dissolve 0.154 g of MnSO4.H2O in distilled water & dilute to 500cm3
in std. measuring flask.
2) Preparation of 100ppm Fe+2
solution
1ppm = 1mg/1000ml
100ppm = 100mg/1000ml
55.897 mg of Fe = 392.13 mg of FAS (Fe+
)
50 mg of Fe = 392.13 x 50/55.897 mg of FAS (Fe+
)
= 0.351 g of FAS dissolve in 500 ml standard flask.

38
PROCEDURE:-
1. Preparation of 100ppm Manganese solution (soln.1):- dissolve 0.154g of MnSO4.H20 in a
small amount of distilled water & dilute it in 500ml standard measuring flask.
2. Preparation of 100 ppm of iron of iron solution (soln.2) :- dissolve 0.380g of ferrous alum
in a small amount of D/W it to 100ml in standard measuring flask.
3. Preparation of 1M KSCN Solution :- Dissolve 9.7g of KSCN in 100ml of D/W in
standard measuring flask.
4. Preparation of calibration curve :- take 1,2,3,4,5 ml of solution (1) in 5 different beakers.
To each beaker then add 5ml of H3Po4 & 1.0g of KIO4. Boil the solution till it turns violet
cool & dilute to 100ml in standard measuring flask .measure the absorbance of
coloredpermagnetic acid at 546 nm using D/W as blank. Plot the graph of O.D. v/s
concentration of Mn+2
in ppm.
SEPARATION OF MIXTURE- [ sample -25ml of soln.1&5ml of soln.2]
Dilute the given mixture of manganese & iron to 100cm3
with D/W. Pipette out 10ml of
this soln in a small flask& 10cm3
of 1M KSCN. Shake well. Then add 10ml of iso amyl
alcohol & shake for 10-15min . Allow it to stand for sometime. Separate the aqueous
layer containing Mn in small flask & repeat the extraction with 10ml of iso amyl alcohol.
Fe is separated into org.phase&Mn remains in aqueous phase. Collect the aq.phasein a
beaker. Add 5ml of H3PO4& 1.0g of KIO4& 25ml of D/W. Boil the solution for 10-15
min till we get violet colour (red to decolorize to violet) . Cool & dilute it to 100ml.
Measure the absorbance,
OBSERVATION: Wavelength used – 530 nm Blank solution – distilled water.
OBSERVATION TABLE:-

39
Vol of 100 ppm
Mn+2
soln in cm3
Vol of
phosphoric
acid in cm3
Wt. of
KIO4 in g
Final vol
in cm3
Concentrat
ion of
Mn+2
ppm
µg O.D.
1 5 1.0 100 1 100
2 5 1.0 100 2 200
3 5 1.0 100 3 300
4 5 1.0 100 4 400
5 5 1.0 100 5 500
unknown 5 1.0 100
CALCULATION:
From the graph of O.D v/s conc. Of Mn+2
in ppm in 10cm3
of mixture gave ‘a’ as its O.D. & this
O.D. corresponds i.e. ‘b’ ppm of Mn+2
. But 1ppm=1mg/1000cm3
.
‘b’ ppm= ‘b’mg/1000cm3
=b/10mg/100cm3
= ‘c’ mg /100 cm3
10cm3
of solution contains ‘c’mg of Mn+2
.
100cm3
of stock solution contains = c x 100/10 mg of Mn+2
= ‘d’ mg of Mn+2
Thus, the amount of manganese in the given mixture= mg.
RESULT:-
Amount of manganese in the given mixture = mg.

40
SEPRATION OF URANIUM & IRON
AIM:-Separation of U & Fe using 8-hydroxyquinoline in chloroform & estimation of
Uranium by Colorimetric Method.
REAGENT:-
1: 50 ppm 𝑈+6
soln :- Dissolve 0.052 gm of uranyl nitrate in D/w & dilute upto the mark in
500 𝑐𝑚3
std. measuring flask.
2: 0.02 M EDTA :- Dissolve 1.861gm of disodium salt of EDTA in D/W & dilute upto 250
mark in std. measuring flask.
3: 1% oxinesoln :- Dissolve 1gm of 8-hydroxyquinoline in 100ml chloroform in std.
measuring flask.
4: 50 ppm of 𝐹𝑒+3
soln :- Dissolve 0.0431 g of ferric ammonium sulphate in 6M H2SO4 in 100
cm3
std measuring flask.
5: Dilute NH3soln :- 1 ml of conc. NH3solution add 50ml of D/W. Mix it properly use this to
adjust pH.
THEORY:-
Uranium (VI) can be extracted in chloroform as an oxinate complex, the conc of complex ion
can be determine at pH-8.8 in presence of EDTA. The yellow oxinate complex absorbs at 400
nm. Interfering Fe+3
masked by adding excess of EDTA. Ammonia is added to adjust the pH of
solution.
PROCEDURE:-
A) PREPARATION OF CALIBRATION CURVE
1. Take 2,4,6,8,10 cm3
of 50 ppm U+6
in 5different separating funnel.
2. . Then add carefully 5-6ml of dil ammonia soln to maintain pH 8.8. This volume is
determined in advance.

41
3. Dilute the solution to 25ml with D/W & add 10ml 1% oxine solution to each funnel &
extract the complex into the organic phase by shaking separating funnel for 5-10 min with
occasionally releasing the vapour pressure developed.
4. Collect the organic phase in 25cm3
std vol. flask. Repeat the extraction with another 10 cm3
of 1% oxine shake well for few min. Collect the organic phase in same 25ml std vol. flask
& dilute it upto the mark with chloroform.
5. Measure the absorbance at 400 nm using D/W as blank. Plot the graph of absorbance v/s
conc of U+6
in ppm.
B) SEPARATION OF MIXTURE:-
1. Take unknown sample [ 5 cm3
of 50 ppm U+6
soln + 0. 2 ml of 50 ppm Fe+3
soln] in separating
funnel.
2. Add 5ml of 0.2M EDTA and 5-6ml 1:1 NH3to adjust the pH upto 8.8.
3. Transfer the solution to separating funnel and add 10cm3
of 1%oxine solution. Shake well and
allow to stand for few minutes.
4. Separate the lower organic layer.
5. Again add 10cm3
of 1%oxine solution and shake well, allow it to stand amd then separate the
organic layer in the 25cm3
seperating funnel.
6. Dilute the chloroform layer to 25cm3
with CHCl3.
7. Measure the absorbance of organic layer at 400 nm by using CHCl3 as blank.
8. From calibration curve, find the unknown concentration of U+6
and hence find the amount of
Uranium in the given solution.
OBSERVATION:-
Wavelength = 400 nm Blank = CHCl3

42
OBSERVATION TABLE:-
50 ppm
U+6
soln
cm3
0.02 M
EDTA
Cm3
Dil.NH3
Till pH-
8.8 (betw
5-6cm3
)
Dilute
upto
25ml 0f
D/W
1%
Oxinesolnin
CHCl3
Final
dilution
with
CHCl3
cm3
Final
conc
(ppm)
O.D
2 00 5.6 25 10+10 25 4
4 00 5.6 25 10+10 25 8
6 00 5.6 25 10+10 25 12
8 00 5.6 25 10+10 25 16
10 00 5.6 25 10+10 25 20
unknown 2.5 5.6 25 10+10 25
Calculation:-
From graph,
Concentration of U+6
in unknown soln is ‘x’ ppm
= x mg/1000cm3
1ppm = 1mg/1000cm3
= 0.1mg/100cm3
= 0.025mg/25cm3
‘x’ppm = x Χ 0.025mg
1000 cm3
of U+6
contains = x mg of U+6
25 cm3
of dil soln = ‘ y ’ mg of U+6
Y = x* 25/1000

43
= mg of U+6
RESULT:-
The amount of U+6
in unknown soln is _____ mg in 25cm3.

44
SEPERATION OF IRON & MOLYBEDNUM
AIM: - Seperation of Fe &Mo usingisoamyl alcohol & estimation of Mo.
REAGENTS: - 1) 10ppm Mo +6
soln : - Dissolve 0.092 gm of ammonium molybdate (NH4)
[Mo7O24] 4H2O 500cm3
of d/W.
2) Ferrous ammonium sulphate :-Disslove 1 gm of ferrous ammonium sulphate in 100cm3
of
0.01 N H2SO4 fiving 1% FAS.
3) 10% of SnCl2: - Dissolve 100gm of SnCl2 2H2O in 100cm3
of 1N HCl
4) 10% KSCN :- Dissolve 10 gm of KSCN in 100ml of D/W.
THEORY :- In ammonium molybdate (NH4)6[Mo7O24] 4H2O oxidation state of Mo is +6 when
treated with SnCl2 , Mo+6
is converted to Mo+5
. In +5 oxidation state Mo, potassium thiocynate is
added largely to form complex of type [Mo(SCN)5] which is red in colour. Isoamyl alcohol is an
oxygenated solvent which can be used to extract the complex. The colour depends upon the acid
concentration of thiocynate ion. The colour intensity is constant in the range of 2-10%. The
Molybdenum complex has max absorption at 465nm.
PROCEDURE:-
A] PREPARATION OF CALIBRATION CURVE
1. Take 0.5, 1, 1.5, 2, 2.5 cm3
of 10 ppm soln of Mo +6
in 5 different separating funnel.
2. Add 0.5, 1, 1.5, 2, 2.5 cm3
of D/W in each flask. Then add 2 cm3
of conc HCl (A.R. grade), 3
cm3
of KSCN soln in each separating funnel. Shake gently.
3. Add 3 cm3
of 10% SnCl2 then make the final volume upto 25 cm3
with D/W in each funnel, add
10 cm3
of isoamyl alcohol
4. Shake each separating funnel for 5 mins and then Allow it to stand for 5 min.

45
5. Separate the aqueous phase (lower) & collect the organic phase in 25cm3
std. measuring flask.
6 Again repeat with 10cm3
of isoamyl alcohol in the separating funnel ,shake well and collect in
the same 25cm3
std. measuring flask and dilute to 25 cm3
with isoamyl alcohol
6. Measure the absorbance at 465nm using isoamyl alcohol as blank. Plot the graph of absorbance
v/s conc of Mo+6
in ppm.
B. SEPARATION OF MIXTURE:-
1. Take unknown [1.7 or x ml of Mo+6
+ 0.5ml of Fe+2
soln] in 6th
separating flask.
2. Treat this mixture in same manner as mentioned in steps 2-6.
3. Measure the absorbance at 465 nm & from the calibration curve, calculate the amount of Mo+6
in given mixture.
OBSERVATION:-
Wavelength :- 465nm Blank :-isoamyl alcohol
Vol of Mo
+6
sol (cm3
)
D/W
cm3
Conc
HCl
cm3
KSCN
10
%
cm3
SnCl2
10%
cm3
Final
vol with
D/W
cm3
Isoamy1
Alcohol
cm3
O.D
0.5 0.5 2 3 3 15 10
1 1 2 3 3 14 10
1.5 1.5 2 3 3 13 10
2 2 2 3 3 12 10
2.5 2.5 2 3 3 11 10
unknown - 2 3 3 - 10

46
CALCULATION:-
From Graph,
Concentration of Mo+6
m unknown soln:- ‘a’ ppm
= ‘a’ mg/1000 cm3
1000 cm3
of 10 ppm solution = a mg of Mo+6
Therefore, 25 cm3
of solution = x mg of Mo+6
X= a×25/1000
= mg of Mo+6
RESULT:-
Amount of Mo-6
in unknown given mixture = _____mg.

47
Separation of Copper(Cu) &Iron(Fe)
AIM:-Separation of Cu & Fe using n-butyl acetate and estimation of Cu using colorimetric
Method.
REQUIREMENTS:-
1. Cupric sulphate pentahydrate (stock soln) for 100ppm Cu solution.
2. 25% citric soln (25 g of citric acid in 100ml of D/W)
3. 1:1 NH4OH
4. 4% EDTA (4 g of EDTA in a 100ml of D/W)
5. 0.2% Na di methyl di thiocarbamate (0.2 g of (Na-DDTC) in 100 ml d.w warm soln then filter
through cotton)
6. n-butyl acetate
7. 5 % H2SO4 (5 ml of A.R H2SO4 in 100 ml of D/W)
THEORY:-
Sodium diethyl dithiocarbamate (NA-DDTC) reacts with slightly acidic or ammonical solution
of Cu+2
in low concentration to produce a brown colloidal suspension of the cupric diethyl
dithiocarbamate. The suspension may be extracted with an organic solvent n-butyl acetate and
the coloured extract is determined spectrophotometrically at 530 nm (n-butlyacetate ) or 435nm
(CHCl3 or CCl4) many of the heavy metals give slightly soluble product (some white , some
coloured) with the reagent. The selectivity of the reagent may be improved by using masking
agent, particularly EDTA.
PREPARATION:-
1)Preparation of 100ppm Cu+2 solution.

48
Preparation of 500cm3 of 100ppm Cu+2
solution
1ppm = 1mg/1000cm3
100ppm = 100mg/1000cm3
= 50mg/500cm3
63.546 mg of Cu = 249.69 mg of CuSO4.5H2O
50mg of Cu = 249.69 ×50/63.546 = 196.46 mg of CuSO4.5H2O. OR
0.196gm//500 cm3
2) PREPARATION OF 100 PPM Fe+2SOLUTION
Preparation of 500cm3
of 100ppm Fe+2
solution.
1ppm = 1mg/1000 cm3
100ppm =100mg/1000cm3
= 50mg/500cm3
55.897 mg Fe = 392.13mg (FAS) Fe+2
50 mg of Fe = 392.13*50/55.897 = 350.76 mg FAS
= 0.351gm FAS (Fe+2
)/500cm3
H2O
PROCEDURE:-
PREPARATION OF CALIBRATION CURVE:-
1) Pipette out 1,2,3,4,5 ml of 10ppm Cu+2
solution in 100ml braker.
2) Add 5ml of 25% citric acid in each beaker.
3) Then add 1:1 NH3 in each beaker till pH is 8.5.
4) Add 15ml of 4% EDTA in each beaker and cool it at room temperature.
5) Transfer it to the separating funnel.
6) Add 10ml of 0.2 % Na directly dithiocarbamate solution in each of the separating funnel.

49
7) Shake well for 45minutes. A yellow brown colourdevelopes in the solution. Add 20ml of
n-butyl acetate in each of the funnel and shake for 30sec.
8) Organic layer will be of yellow colour. Cool and shake for 15sec and allow the phase to
separate.
10) Remove lower aqueous layer. Then add 20ml 0f 5% H2SO4 to each separating funnel.
Shake for 15sec and cool.
11) Separate the organic layer , collect it in 25ml standard measuring flask&dilute it to the
mark with n-butyl acetate.
12) Measure the absorbance of each flask at 530nm.
ESTMATION OF Cu:-
Take 5ml of 100ppm Cu+2
solution and 1ml 100 ppm of Fe+2
solution.Treat this solution in
above manner, and measure the absorbance and hence calculate the concentration of Cu
from the graph.
GRAPH:- Plot the graph of absorbance v/s cone, of Cu.
OBSRVATION:-
Vol
of100ppm
Cu+2
soln
in cm3
Vol of
5%
citric
acid soln
Adjust
pH to
Vol of
4%
EDTA
Vol of
0.2%
Na-
DDTC
Dilute
to
75ml
With
D/W
vol of
n- butyl
acetate
Vol of
5%
H2SO4
Conc
of
Cu+2
in
ppm O.D
1 5 8 -8.5 15 10 20 20 4
2 5 15 10 20 20 8
3 5 15 10 20 20 12
4 5 15 10 20 20 16
5 5 15 10 20 20 20
Unknown 5 15 10 20 20

50
CALCULATIONS:-
*From graph concentration of Cu+2
in unknown solution = __a___ ppm.
*1ppm = 1mg/1000ml
=0.1mg/100cm3
= 0.025mg/25cm3
‘a’ppm= ‘a’×0.025mg in 25 cm3
RESULT:- Amount of Cu+2
present in given unknown sample solutiom = -----mg
(5ml
Cu+2
+1ml
Fe+2
)

51
PAPER-III
PREPARATION OF V(oxinate)3
AIM: To prepare V (OX)3
REQUIREMENTS: NH4VO3 , 8-hydroxy-quinoline,conc HCl, 2N acetic acid.
THEORY: Proton of hydroxyl group on8-hydroxy quinoline is acidic. Thus, 8-hydroxy-
quinoline can form his derivative with vanadium.In which 8-hydroxy-quinoline acts as a
bidented ligand. Geometry of V (OX)3is octahedral.
PROCEDURE:
1. Take 0.5gm of Ammonium metavanadate;dissolve it
in one test tube of 1:1 HCl. If not dissolved add 1ml
of AR conc.HCl (take minimum amount of water).
2. In another beaker take 1gm of 8-hydroxy-quinoline.
Dissolveit 10cm3
of 2N acetic acid.
3. Add a solution of 8-hydroxy-quinoline in Ammonium
meta vanadate solution in small portion with
continuous stirring.
4. Keep this in ice bath and slowly add sodium acetate, till brown colour ppt is obtained
.add 1-2 drops more.
5. Cool and filter through whatmann filter paper no. 41.
REACTION:
NH4VO3+ 6HCl → VCl3 + NH4Cl + 3H2O + Cl2
VCl3 +3 → V(OX)3 + 3HCl

52
OBSERVATION:
Weight of NH4VO3= 0.5g Weight of 8-hydroxy-quinoline = 1g
CALCULATION:
Theoretical Yield
116.98g ofNH4VO3 = 432.48g of V(OX)3
0.5 g of NH4VO3= x g of V(OX)3
X g =
432.48 x 0.5
116.98
= 1.848g
Practical yield =
obtained weight
1.848
x100 =________%
RESULT:
Theoretical yield = 1.848 g
Practical yield = __________g
Percentage yield = ________%

53
PREPAPRATION OF Hg[Co(SCN)4]
AIM: To prepare Hg[Co(SCN)4]
REQUIREMENTS: CoSO4.7H2O, NH4SCN, HgCl2, etc.
THEORY:Hg [Co (SCN)4] is used as standard in magnetic moment method. It has high
magnetic susceptibility and henceforth a good standard for measuring
magnetic susceptibility of a para magnetic compound .This complex has
3 unpaired electrons so the complex is highly para magnetic.
Magnetic moment = [n(n+2)]1/2
= [3 (3+2)]1/2
= 3.87 BM
Hg[Co(SCN)4] the complex is tetrahedral so the splitting is ‘e’ and ‘t2’.
PROCEDURE:
1. Solution ‘A’ is to be prepared by dissolving 2g of Co(II) salt. CoSO4.7H2O/CoCl2in
10cm3
of water and then add 2.5g of NH4SCN to it. Boil the solution.
2. .Solution ‘B’ is prepared by dissolve of 3g of HCl2 into 25cm3
of water and 1cm3
of conc.
HCl . Boil the solution for 2-3mins till it dissolves.
3. Add hot solution ‘A’ to the boiling solution ‘B’ with constant stirring.
4. Blue colour ppt will form. Cool the solution and filter the precipitate through whatmann
filter paper no.41.
5. Wash the ppt with distilled water and then 1cm3
of ethanol. Air dry the precipitate and
weight the product.
REACTION:
CoSO4.7H2O + HgCl2 + 4NH4SCN → Hg[Co(SCN)4] + (NH4)2 SO4 + 2NH4Cl + 7H2O

54
CALCULATION:
Theoretical yield
280.93 g of CoSO4 = 491.52g of Hg [Co(SCN)4]
2g of CoSO4=
491.52 x 2.0
280.93
= 3.5g of Hg[Co(SCN)4]
Practical yield = ‘x’ g
Percentage yield =
practical yield
theoretical yield
x 100 =
x
3.5
x 100
RESULT:
Theoretical yield = ___________g
Practical yield= ___________g
Percentage yield =___________%

55
Preparation of Ni - (salicyl aldoxime)2
AIM: To prepare Ni (salicyl aldoxime)2
REQUIREMENTS: NiSO4.7H20, ethyl alcohol, dil. NH4OH, salicylic
aldoxime.
THEORY: NiSO4.7H20 reacts with salicycl aldoxime in presence
of base (NH4OH) to give Ni (salicycl aldoxime)2 complex. Base is
used to enhance deprotonation of ligand.
PROCEDURE:
1. Dissolve 0.3g of NiSO4.7H20 in 100ml of D/W.
2. Neutralizes this solution with very dil. NH4OH (use pH
paper).
3. To this add 0.531g of salicylic aldoxime dissolved in
minimum quantity of ethyl alcohol. Add very dil. NH4OH
drop wise till complete ppt.
4. Digest the ppt for 20min in boiling water bath. Filter through ordinary filter paper. Wash
it with D/W and followed with ethyl alcohol. Dry and weigh it.
REACTION: NiSO4.7H2O+ 2SAO [Ni(SAO)2] + H2SO4+ 7H2O
OBSERVATIONS:-
Weight of NiSO4.7H2O = 0.3g Weight of salicylic aldoxime = 0.531g
CALCULATIONS:-

56
Theoretical yield
280.69 g of NiSO4.7H2O = 330.97 g of Ni (salicylic aldoxime)2
0.3g of NiSO4.7H20= x
x =
330.97 x 0.3
280.69
= 0.353g
Practical yield
Weight of the product = ________
Percentage yield = ________ x 100
0.353
RESULT:
Theoretical yield = _________g
Practical yield = _________g
Percentage yield = _________%

57
Preparation of Cobalt Hexamine Chloride
AIM: To prepare cobalt hexamine chloride [Co (NH3)6]Cl3
REQUIREMNTS: NH4Cl, conc.NH3, CoCl2.6H2O, 30% H2O2 , conc.HCl, 6M HCl.
THEORY: CoCl2.6H2O reacts with liq. NH3 in NH4Cl to give [Co(NH3)6]Cl3. In CoCl2.6H2O
cobalt is in +2 oxidation state. When H2O2 is added to it Co2+
gets oxidized to Co3+
. Since
solution is highly basic concentrated HCl is used to neutralize it and precipitated out the product.
PROCEDURE:
1. Dissolve 1g of NH4Cl in 9cm3
of conc. NH3 in a beaker.
2. Add 2g of CoCl2.6H2O pinch wise with continuous
stirring to give yellow pink precipitate.
3. Add 10-15cm3
of 30% H2O2 at the rate of about per two
drops slowly per second with continuous stirring. Heat the mixture carefully on the water
bath.
4. When vigorous effervesces has ceased, cautiously add 6 cm3
of conc.HCl to give purple
ppt with pale blue green solution above it.
5. Heat the mixture at 700
C should not be very hot for 15mins using water bath. Cool in ice
bath. Filter through G3crucible and wash with 3 portions of 1.5 cm3
of cold water and 6M
HCl. Dry the product at 1000
C for 1 hr. Cool and weigh it.
REACTION: CoCl2.6H2O + NH3 [Co (NH3)6]Cl3
(excess)
OBSERVATIONS:
Weight of CoCl2.6H2O = 2g Weight of ammonium chloride = 1 g
Weight of empty crucible = ___________
Weight of crucible + residue = ___________
Weight of the product = ____________

58
CALCULATIONS:
THEORTICAL YIELD
237.933 g of CoCl2.6H2O = 267.433g of [Co(NH3)6]Cl3
2g of CoCl2.6H2O = x g of [Co(NH3)6]Cl3
=
2 𝑥 267.433
237.933
= 2.248 g
PERCENTAGE YIELD
=
Y
2.248
X 100 = _______%
RESULT:

59
TRANS- BIS (GLYCINATO) Cu (II)
AIM: To prepare trans- bis(glycinato) Cu(II)
REQUIREMENTS: glycine, CuCl2.2H2O, 4M NaOH solution etc.
THEORY: Glycine is a α-amino acid which exists as zwitter ion in solution.
NH2-CH2-COOH ↔ NH3
+
-CH2-COO-
In presence of a base NaOH glycine gets deprotonated and
forms a square planar complex with Cu(II).
REACTION:
CuCl2.2H2O + NH2-CH2-COOH →Cu(Gly)2+ 2HCl+ 2H2O
PROCEDURE:
1. Dissolve 0.5g of glycine in minimum quantity of distilled water.
2. In another beaker dissolve 0.5 to 0.6g of CuCl2.2H2O in distilled water in minimum quantity.
3. Add the glycine solution to the metal solution with constant stirring.
4. Then add 4M NaOH solution drop wise with stirring to get greenish colour product.
5. Check for complete precipitation with few more drops of NaOH solution.
6. Filter with filter paper no.41, wash with distilled water, dry and cool in ice bath and weigh the
product.
OBSERVATIONS AND CALCULATIONS:
Weight of the product = ___________
THEORETICAL YIELD:
170.546g of CuCl2.2H2O = 211.546 of trans- bis (glycinato) Cu (II)
∴x g of CuCl2.2H2O =
211.546 X x
170.546
= g of trans- bis (glycinato) Cu(II)
∴ Theoretical yield =_________g

60
PERCENTAGE YIELD:
% yield =
practical yield x 100
theoretical yield
% yield = ___________
RESULT:

61
PREPARATION OF Bis( α- Nitroso-β-Napthol)2 Cobalt (II)
AIM: To prepare Bis( α- Nitroso-β-Napthol)2 Cobalt (II)
THEORY: CoSO4 reacts with β-Napthol in presence of NaNo2 to give Co –( α- Nitroso-β-
Napthol)2 derivative .using NaNO2 is diazotizing agent.
N
O
O
N O
O
Co
PROCEDURE:
1. Take 1g of CoCl2.6H2O in 100cm3
beaker. Add minimum of water to dissolve the metal
salt.
2. Take 0.5g of β-Napthol and add 10cm3
of ethanol shake to dissolve. Add this solution to
the above cobalt solution.
3. Then add 4N sodium nitrite solution drop by drop to get red colour precipitate. Add few
drop of NaOH . Digest the precipitate in boiling water bath for 10-15 mins.
4. Filter the product through Whatman filter paper No.1 .wash the product with water. Dry
at 1000
C and weigh the product.
REACTION
CoSO4 + 2C10H7NO2 → Co(C10H5NO2)2 + H2SO4
CALCULATIONS:

62
THEORETICAL YIELD
237.939g of CoCl2.6H2O = 403g of complex
1g of CoCl2.6H2O =
403 x 1
237.939
= 1.639g
PRACTICAL YIELD= ________________g
PERCENTAGE YIELD=
Practical yield
1.693
x 10 =__________%
RESULT:

63
PAPER-IV
ANALYSIS OF CALCIUM TABLET
AIM: Determination of the amount of calcium in the given calcium tablet by complexometric
method.
REQUIREMENTS: Calcium tablet, EDTA (0.02M), buffer of pH=10, conc. HCl, Eriochrome
black T (EBT).
THEORY:
Calcium is an essential element in human body. Its deficiency can cause various disorder related
to bones. At elderly age, daily intake must be more than 100mg. Calcium tablets is a nutritional
supplements for many such person. Calcium is present as carbonates or penthathionates. The
powdered tablet can be dissolved in conc. HNO3. The excess of acid must be removed by
evaporation. The excess of acid will interfere with the complexometric titration. After filtration,
the Ca ion can be titrated with standard EDTA solution. (Eriochrome black T forms wine red
complex with Ca+2
at buffer pH=10). The ammonia buffer is necessary to neutralize the proton
released during titration. The Ca-EDTA complex being more stable than Ca-indicator complex, it
breaks and little excess of EDTA gives blue colour with indicator.
PROCEDURE:
1. Preparation of 0.02M EDTA: Dissolve 0.745g of EDTA in small amount of water and then
dilute it to 100cm3
with d/w.
2. Preparation of sample solution: Dissolve the given sample (0.3g) of calcium tablet in half
test tube of conc. HCl in 250cm3
beaker. Heat it (near to dryness) on a low flame in sand bath,
stirring the solution . Again add half test tube of conc.HCl and evaporate slowly near to dryness
and then add about 25cm3
of d/w. Boil the solution, cool and filter. Collect the filtrate and

64
washing in 100cm3
std. flask. Dilute the solution upto the mark with d/w. This is the sample
solution.
3. Estimation of calcium: Pipette out 10cm3
of sample solution. Then add about 10cm3
of d/w
and 5cm3
of buffer solution (pH=10). Add a pinch of EBT indicator and titrate it against 0.02M
EDTA solution. End point is indicated when the colour of solution just changes from wine red to
blue. From the constant burette reading, determine the amount of calcium present in the given
sample.
4. OBSERVATONS:
Weight of calcium tablet : 0.3g
Solution in burette : 0.02M EDTA
Solution in flask : 10cm3
of stock solution+10cm3
of d/w+5cm3
of buffer solution (pH=10)
Indicator : EB.
End point : wine red to blue
OBSERVATION TABLE:
Obs. No.
Initial reading
cm3
Final reading
cm3
Difference cm3
CBR
cm3
x cm3
CALCULATIONS:
0.3g of sample→ 100cm3
→10cm3
→0.02M EDTA
10cm3
of diluted solution = ‘x’ cm3
of 0.02M EDTA
100cm3
of solution = 10 × ‘x’cm3
of 0.02M EDTA
Now, 1000cm3
of 1M EDTA = 40.08g of Ca = 100g of CaCO3

65
⸫10x cm3
of 0.02M EDTA = 40.08×0.02×10x g of Ca
1000
= ‘a’ g of Ca
= 100×0.02×10x g of CaCO3
100
= ‘b’ g of CaCO3
Percentage of calcium = Amount of Ca obtained × 100
Weight of sample
= a × 100
(0.3g)
= ‘c’ %
RESULT:
1. 10cm3
of diluted solution of calcium requires ‘x’ cm3
of 0.02M EDTA
2. Amount of calcium in the given sample = ‘a’ g
3. Amount of calcium carbonate in the given sample= ‘b’ g
4. Percentage of calcium in the given sample = ‘c’ %

66
ANALYSIS OF BLEACHING POWDER
AIM: Determination of amount of chlorine in bleaching powder by iodometric method.
REQUIREMENTS: Bleaching powder, K2Cr2O7 (0.1N), KI (10%), 0.1N Na2S2O3.5H2O,
starch, etc.
THEORY:
The bleaching action of bleaching powder, Ca(ClO)2 depends upon available Cl2. In fact,
bleaching powder constantly loose Cl2 gas and its bleaching powder value decreases. It is
necessary to find the amount of available chlorine from the sample. A good sample of bleaching
powder should contain at least 30% of available Cl2. When the suspension of bleaching powder
is reacted by acetic acid in presence of KI, the liberated Cl2 reacts with KI solution giving free
iodine. The amount of free iodine is equivalent to the amount of available Cl2 in the sample of
bleaching powder.
Reactions:
OCl-
+ Cl-
+ 2H+
→ Cl2 + H2O
OCl-
+ 2I-
+ 2H+
→ Cl-
+ I2↑ + H2O
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
PROCEDURE:
1. Preparation of 0.1N K2Cr2O7 solution: Dissolve 0.49g of K2Cr2O7 in small quantity of d/w
and then dilute it in 100cm3
with d/w.
2. Preparation of 0.1N Na2S2O3 5H2O solution: Dissolve 6.2g of Na2S2O3 in small quantity of
d/w and then dilute it to 250cm3
with d/w.
3. Standardisation of Na2S2O3: Pipette out 10cm3
of 0.1N K2Cr2O7 solution into a conical flask
and add 10cm3
dil. HCl. Add 15cm3
of 10% KI and titrate it against 0.1N Na2S2O3 using starch

67
indicator (Starch is to be added towards the end of titration when the solution turns pale yellow).
End point is indicated when the colour of solution turns blue to light green. Thus, from constant
burette reading. Calculate the exact normality of Na2S2O3 solution.
4. Estimation of available chlorine: Transfer the given amount of bleaching powder (0.1-0.2g)
into a conical flask and add little d/w, 10cm3
of 10% KI and 10cm3
of glacial acetic acid. (Starch
is to be added towards the end of titration when the solution turns pale yellow). End point is
indicated when the solution turns blue to colourless. From the constant burette reading,
determine the amount of chlorine in the given bleaching powder.
OBSERVATIONS:
STANDARDISATION OF Na2S2O3:
Solution in burette : 0.1N approx Na2S2O3
of 0.1N K2Cr2O7 + 10cm3
dil.HCl +15cm3
10% KI
Indicator : Starch
End point : blue to light green
OBSERVATION TABLE:
Obs. No.
Initial reading
cm3
Final reading
cm3
Difference cm3
CBR
cm3
x cm3
CALCULATIONS:
Na2S2O3 = K2Cr2O7
N1V1 = N2V2

68
N1 = 0.1 × 10 = ‘b’ N
x
ESTIMATION OF CHLORINE:
Solution in burette : ‘b’ N Na2S2O3
Weight of bleaching powder : ‘c’ g
Solution in flask : Given powder + 10cm3
d/w +10cm3
glacial acetic acid + 10cm3
10% KI
Indicator : Starch
End point : blue to colourless
OBSERVATION TABLE:
Obs. No.
Initial reading
cm3
Final reading
cm3
Difference cm3
CBR
cm3
d cm3
CALCULATIONS:
‘c’ g of bleaching powder → d cm3
of ‘b’ N Na2S2O3
Now,
1000cm3
of 1N Na2S2O3 = 35.46g of Cl-
‘d’ cm3
of ‘b’ N Na2S2O3 = 35.46 × d × b g of chlorine
1000
= ‘e’ g of chlorine
Percentage of chloride(Cl-
) = Amount of chloride obtained× 100

69
Amount of bleaching powder
= e/c × 100
= A
Percentage of chlorine(Cl2) = ½ ×A = B
RESULT:
1. Amount of chloride present in the given sample of bleaching powder = ‘e’ g
2. Percentage of chloride in the given sample = A%
3. Percentage of chlorine in the given sample = B%

70
ANALYSIS OF IRON TABLET FOR ITS IRON CONTENT COLORIMETRICALLY BY
1,10-PHENANTHROLINE METHOD
AIM: To determine the amount of iron present in a sample of iron tablet by 1,10-phenanthroline
method.
REQUIREMENTS: Iron tablet, 100ppm of Fe+3
solution,
10% hydroxylamine hydrochloride, 0.25% 1,10-phenanthroline, 2M sodium acetate, d/w, etc.
THEORY:
Fe capsules contain Fe+2
ion as ferrous fumarates or gluconates or foliates or citrates. Such
capsules are generally prescribed to iron anemic patients. Fe (II) salt are soluble and can be
extracted in a GI tract easily. The recommended dietary allowances for Fe is 30mg/day.
Fe+2
forms orange-red colour Fe(Phen)3 complex with 1,10-phenanthroline in the pH range 2.8-
3.5 and absorbs maximum at λ= 480nm. Sodium acetate can give buffer pH in this range.
Hydroxylamine hydrochloride is a reducing agent and can reduce any Fe+3
to Fe+2
.
PROCEDURE:
A. Preparation of 10ppm Fe+3 (NH4Fe(SO4)2.12H2O) solution:
1ppm ≡ 1mg/1000ml
100ppm ≡ 100mg/1000ml
100ppm of Fe+3
≡ 100mg/1000ml
= 10mg/100ml
482.19mg of FAS (Fe+3
) ≡ 55.85mg Fe+3
10 × 482.19 ≡ 10mg Fe+3
55.85
i.e. 86.3mg of Fe(III)AS = 0.0863g of Fe(III)AS
Dissolve 0.0863g of Fe(III)AS in about 10ml of dil. H2SO4. Then dilute it upto 100cm3
with d/w.

71
This is 100ppm solution. Now, pipette out 10cm3
from it in another standard flask and dilute it to
100cm3
. This is 10ppm solution.
B. Preparation of stock solution:
1. Dissolve the given sample (0.300g) of Fe tablet in minimum quantity of conc. HNO3.
2. Heat it on a sand bath covering partly with a watch glass, evaporating it to dryness.
3. Repeat with about 3-4ml of conc. HNO3.
4. Add conc. HCl, heat it till no brown fumes persist.
5. Then add d/w, boil, cool and filter.
6. Collect the filtrate and washings in a 250cm3
standard measuring flask.
7. Dilute it upto the mark with d/w.
of this stock solution and dilute it to 100cm3
in a standard measuring flask.
9. Use 10cm3
of this solution for the colorimetric estimation as an unknown.
C. Preparation of 10% NH2OH.HCl: Dissolve 10g of hydroxylamine hydrochloride in
minimum quantity of d/w. Dilute it to 100cm3
D. Preparation of 2M sodium acetate solution: Dissolve 16.4g of sodium acetate in minimum
quantity of d/w. Dilute it to 100cm3
E. Preparation of 0.25% of 1,10-phenanthroline solution: Dissolve 0.25g of 1,10-
phenanthroline in minimum quantity d/w. Dilute it to 100cm3
F. Construction of calibration curve: Measure absorbance of each solution at λ = 480-520nm
for colorimeter and λ = 520nm for spectrophotometer.
G. OBSERVATIONS: Weight of Fe tablet (W) = ____ g
Sr.
No.
Volume
of10ppm
Fe+3
soln
cm3
Volume
of 10%
NH2OH.HCl
cm3
Volume
of 2M
Sodium
acetate
cm3
Volume
of 0.25%
1,10-
Phenanthrol
in cm3
Final
Volum
e
cm3
Final
Fe+3
conc.
ppm
OD

72
Graph:
Plot a graph of Absorbance vs conc. of Fe+3
.
CALCULATIONS:
From graph,
Conc of Fe+3
in unknown solution = ‘x’ ppm
10cm3
of diluted solution has ‘x’ ppm of Fe+3
100cm3
of diluted solution will have 10x ppm of Fe+3
Also,
10cm3
of stock solution have 10x ppm of Fe+3
250cm3
of stock solution will have 10x × 10ppm of Fe+3
Thus the original stock solution contains 10x × 10 ppm of Fe+3
= Y ppm
= Y mg/1000ml
Amount of iron in the sample = Z mg/250ml
Percentage of Fe present:
1. 5.0 10 01 10 100 0.5
2. 10.0 10 01 10 100 1.0
3. 15.0 10 01 10 100 1.5
4. 20.0 10 01 10 100 2.0
5. 25.0 10 01 10 100 2.5
6. 10ml of
unknown
10 01 10 100 ---

73
% of Fe = Z × 100 = ____ %
W
RESULT:
1. Amount of iron present in the given sample = _____ mg
2. Percentage of Fe present = ____ %

74
ANALYSIS OF NYCIL POWDER
AIM: Determination of amount of zinc in the given nycil powder by complexometric method.
REQUIREMENTS: Nycil powder, xylenol orange, hexamine powder, buffer solution (pH=10),
EDTA (0.02M), Eriochrome black T.
THEORY: Prickly heat nycil powder contains Zn as a constituent in the form of ZnO. HNO3
decomposes the organic matter and ZnO to their respective metal nitrates. The excess of acid
must be evaporated since it will interfere in the estimation of Zn. Zinc forms a stable complex
with EDTA subsequently. Zn can be estimated by using Eriochrome black T at pH=10.
PROCEDURE:
1. Preparation of 0.02M EDTA: Dissolve 0.745g of EDTA in 100cm3
of d/w in standard
measuring flask.
2. Preparation of stock solution: Dissolve the given sample (0.6-0.7g) of nycil powder in about
20cm3
of conc.HCl. Heat it on a sand bath. Evaporate it to dryness with constant stirring. Repeat
with another 10cm3
of conc.HCl. Add about 25cm3
d/w. Boil the solution. Filter. Collect the
filtrate and washings in 100cm3
standard flask. Dilute it upto the mark with d/w. This is the stock
solution.
3. Estimation of Zinc:
A) With xylenol orange indicator: Pipette out 10cm3
of stock solution. Then add about 25cm3
of d/w, pinch of xylenol orange indicator and hexamine powder till the colour of solution turns
wine red (pH=6) and titrate it against 0.02M EDTA. End point is indicated when the colour of
the solution changes from wine red to blue.
B) With EBT indicator: Pipette out 10cm3
of stock solution. Then add about 25cm3
of d/w and
5cm3
of buffer solution and add a pinch of Eriochrome black T and titrate it against 0.02M

75
EDTA. End point is indicated when the colour of the solution changes from wine red to blue.
From the CBR, determine the amount of zinc in the given sample.
OBSERVATIONS:
Weight of nycil powder : 0.6-0.7g
Solution in burette : 0.02M EDTA
stock solution + 25cm3
d/w + 5cm3
buffer solution (pH=10)
Indicator : EBT
End point : wine red to blue.
OBSERVATION TABLE:
Obs. No.
Initial reading
cm3
Final reading
cm3
Difference cm3
CBR
cm3
x cm3
CALCULATIONS:
0.6-0.7g of sample → 100cm3
→ 10cm3
→ 0.02M EDTA
10cm3
of diluted sample = ‘x’ cm3
of 0.02M EDTA
100cm3
of sample solution = 10 × ‘x’ cm3
of 0.02M EDTA
1000cm3
of 1M EDTA = 65.38g of Zn = 81.38g of ZnO
10x cm3
of 0.02M EDTA = 65.38 × 0.02 × 10x g of Zn
1000
= ‘y’ g of Zn

76
Percentage of zinc = Amount of Zn obtained × 100
Weight of sample taken
= y × 100 = a %
(0.6-0.7g)
10x cm3
of 0.02M EDTA = 81.38 × 0.02 × 10x g of ZnO
1000
= ‘z’ g of ZnO
Percentage of zinc oxide = Amount of ZnO obtained × 100
Weight of sample taken
= z × 100
(0.6-0.7g) = b %
RESULT:
1. 10cm3
of diluted sample requires ‘x’ cm3
of 0.02M EDTA.
2. Amount of zinc in the given sample = ‘y’ g.
3. Percentage of zinc in the given sample = a %.
4. Amount of zinc oxide in the given sample = ‘z’ g.
5. Percentage of zinc oxide in the given sample = b %.

77
SEM-IV
PAPER-I
ANALYSIS OF GALENA ORE
AIM: To analyze the given sample of galena ore for its
1. Pb content as PbCrO4 by gravimetric method using 5% K2CrO4.
2. Fe content by colorimetric method using 0.5% 1,10 phenanthroline.
REQIREMENTS: Conc. HNO3, 4N HNO3, 0.002M HCl, 100ppm of Fe(0.0865g of ferric alum
in 1000 cm3
of water).
THEORY: Galena is an ore of lead and iron. It is present in the form of sulphide. Conc. HCl is
used to dissolve the ore.Pb is estimated as PbCrO4 by gravimetric method using 5% K2CrO4.
Iron is present in trace amount and is therefore estimated by colorimetric method using 0.5%
phenanthroline, orange colored complex is formed. Calibration curve is prepared with 10ppm of
Fe soln & unknown is determined from it.
REACTIONS:
Pb+2
+ K2CrO4 → 2K+
+ PbCrO4
PROCEDURE:
Opening Of The Ore:
1. 0.5g of the galena ore was dissolved in 7cm3
of Conc. HNO3.
2. Heat it to dryness on sand bath.
3. Again 5cm3
of Conc. HNO3 was added and heated to dryness.
4. This was cooled to room temp. & then extracted with 10cm3
of dilute HNO3.
5. To this 15cm3
of 4N HNO3 & 15cm3
of distilled water was added & boiled for few
minutes.
6. It was then cooled & filtered through filter paper no. 41 & washed with 1:1 HNO3.

78
7. Filtrate & washings are collected in 250cm3
standard flask & diluted upto the mark using
distilled water. THIS IS STOCK SOLUTION.
A)Estimation of Pb content as PbCrO4 by gravimetric method using 5% K2CrO4.
1. Take 50cm3
of stock solution in to beaker & add to it 50cm3
of distilled water & heat.
2. Add 20cm3
5% K2CrO4 to it & heat it on water bath 90 min.
3. Cool & filter through G4 crucible.
4. Precipitate is washed with hot water & filtrate is checked for complete precipitation.
5. Precipitation is then dried in oven at 110˚C & weighed.
B) Estimation of Fe content by colorimetric method .
1. Preparation of 10ppm of Fe(III) solution.
1 ppm = 1mg / 1000 cm3
100ppm = 100mg/ 1000 cm3
= 10mg/ 100 cm3
482.19 mg FAS (III) = 55.85 mg of Fe.
10mg Fe+3
= 10 x 482.19 = 86.3 mg of Fe (III)
55.85
= 0.086 mg/ 100 cm3
=100ppm.
Now pipette out 10cm3
of 100ppm Fe+3
solution &dilute to 100 cm3
of water to get 10ppm
solution.
2. Prepare calibration curve for Fe (III) solution in the 10ppm solution of Fe+3
.
3. Add to this 10 cm3
of 10% NH2OH.HCl & 1 cm3
of 2M CH3COONa .
4. Then add 10cm3
of 0.25% of 1,10 phenanthroline in each flask.
5. Dilute it upto 100cm3
with water in standard measuring flask.
6. Measure the absorbance at 500nm.

79
of diluted stock solution in 100cm3
of standard measuring flask.
8. Repeat the same treatment given above for unknown solution & measure the absorbance
at 500nm.
Flask
no.
Volume
of 10ppm
Fe3+
ion
sol in cm3
Volume
of 10%
NH2OH.HCl
Volume
with 2M
CH3COONa
Volume of
0.25% 1,10
phenanthroline
Dilute
to
100cm3
Final
conc
ppm
OD
1 2 10 1 10 100 0.2
2 4 10 1 10 100 0.4
3 6 10 1 10 100 0.6
4 8 10 1 10 100 0.8
5 10 10 1 10 100 1.0
6 Unknown 10 1 10 100
OBSERVATIONS:
Weight of galena ore taken = X gm
A. Estimation of Pb as PbCrO4
Weight of empty crucible = a g
Weight of crucible + residue = b g
Weight of residue = (b-a) = c g
CALCULATIONS:
50 cm3
of stock solution contains c g of PbCrO4.
250 cm3
of stock contains c x 5 g of PbCrO4.

80
Now,
323.2g of PbCrO4 = 207.2 g of Pb
C x 5g of PbCrO4 =207.2 x C x 5
323.2
= A g of Pb
% of Pb = A x 100
X
= % of Pb.
B. Estimation of Fe
Wavelength used = 500 nm
Blank solution = Distilled water
From Graph,
Unknown solution contains __________ ppm of Fe+3
.
10ml of dil stock solution = ________ ppm
250ml of dil solution contains = x 250 = _________ mg
10
% = ____________ x 100 = _______ % = _______ g of Fe.
Wt of ore
RESULT:
1. Lead present in the sample of galena alloy = ______ %
2. Iron present in the sample of galena alloy = ______ %

81
ANALYSIS OF ZINC BLENDE ORE.
AIM: - Determine the amount of
1. Zn content by complexometric method.
2. Fe content by colorimetric method. (Azide method.) present in the given
sample of zinc blende ore.
REQURIMENTS: - Conc. HNO3, Standard EDTA, Buffer pH = 10, 1% Eriochrome Black T
indicator Standard 100 ppm Fe+2
solution, 1% Phenanthroline solution, 10% Hydroxylamine
hydrochloride, 2MSodium acetate, etc.
THEORY: - Sphalerite, also known as blende or zinc blende, is the major ore of zinc. When
pure (with little or no iron) it forms clear crystals with colours ranging from pale yellow to
orange and red shades , but as iron content increases it forms dark, opaque metallic crystals
known as Marmatite. Sphalerite may also contain considerable Mn.
Zinc Blende is a chief ore of Zn metal. It is mixed sulphide of Zn and Fe. Conc. HNO3
decomposes the sulphide ore to their respective metal nitrates. The excess of acid must be
evaporated since it will interfere in the estimationof Zn and Fe. Both these elements forms
complex with EDTA.Therefore Fe must be removed as a hydroxide prior to estimation of Zn.
The filtrate from the removal of Fe(OH)3 can be concentrated to remove excess of ammonia and
subsequently Zn can be estimated using Eriochrome Black Tindicator at pH 10. Fe(II) forms
orange red colour Fe(Ph)3 complex with 1:10 phenanthroline in the pH range of 2.8-3.5 and
absorbs maximum at λ = 480nm. Sodium acetate can give buffer pH in the range.
Hydroxylamine hydrochloride is a reducing agent and can reduce any Fe+3
to Fe+2
.
PROCEDURE: -
A)OPENING OF THE ORE: -
1. Weigh accurately 400mg of powdered ore.
2. Add 15ml of conc HNO3 or dilute HCl. Boil the content for 5 minutes.
3. After cooling add 25ml of water and boil again.

82
4. Filter through ordinary filter paper.
5. Collect the filtrate quantitatively in 250ml volumetric flask.
6. Make up to the mark with distilled water and use it for analysis of Zn and Fe.
B)ESTIMATION OF ZINC: -
1. Pipette out 50ml of diluted solution in a beaker and add 25ml of distilled water.
2. Add 5ml of conc. HNO3 and boil it for oxidation of Fe+2
to Fe+3
.
3. Add 1:1 ammonia solution till complete precipitation of Fe(OH)3 is over.
4. Filter through ordinary filter paper and collect the filtrate and washings quantitatively.
5. Concentrate the solution to 50ml and dilute it to 100ml in a volumetric flask.
6. Pipette out 25ml of the diluted solution in a conical flask and add 25ml of distilled water.
7. Add 5ml of buffer pH 10 and a pinch of 1% Eriochrome Black T indicator.
8. Titrate it against standard EDTA solution till wine red color change to blue.
9. Record the constant burette reading and calculate the amount of Zn present and calculate
the % of Zn.
C)ESTIMATION OF IRON: -
Preparation of 100ppm of Fe(III) solution.
1 ppm = 1mg / 1000 cm3
100ppm = 100mg/ 1000 cm3
= 10mg/ 100 cm3
482.19 mg FAS (III) = 55.85 mg of Fe.
10mg Fe+3
= 10 x 482.19 = 86.3 mg of Fe (III)
55.8
= 0.086 mg/ 100 cm3
=100ppm.
Now pipette out 10cm3
of 100ppm Fe+3
solution & dilute to 100 cm3
of water to get 10ppm
solution.

83
1. Prepare calibration curve for Fe(III) in the range 0.5-2.5ppm of Fe using appropriate
volume of 100ppm of stock solution of Fe(III).
2. Add 10ml of 10% w/v of hydroxylamine hydrochloride solution and 1ml of 2M sodium
acetate buffer solution.
3. Develop the colour using 10ml of 1% 1,10-phenanthroline solution {Fe(Ph)3 complex is
orange red in colour.}
4. Dilute the solution to 100ml in a volumetric flask.
5. Measure the absorbance against distilled water as a blank at λ= 500nm.
6. Plot the graph of absorbance against the concentration of Fe(III). {This is the calibration
curve.}
7. Pipette out 5ml of the diluted stock solution and develop the colour of Fe(Ph)3 as above.
8. Measure the absorbance and find the concentration of Fe(III) from the calibration curve.
9. Calculate the amount of Fe(III) present and hence the percentage of Fe in the given
sample of ore.
OBSEVATIONS AND CALCULATIONS: -
A)ESTIMATION OF ZINC:-
25cm3
of the diluted solution required x cm3
100cm3
of diluted solution will require 4x cm3
1000cm3
of 1M EDTA = 65.38g of Zn.
4x cm3
of 0.02M EDTA = 65.38 x 4X x 0.02 g of Zn.
1000
= ________ g of Zn.
50cm3
of stock solution contains ____________ g of Zn.
250cm3
of stock solution contains ___________ g of Zn.

84
= x 250 = ________ gof Zn.
50
Percentage of Zn = Weight of Zn x 100.= %
Weight of the ore.
B) ESTIMATION OF Fe: -
Blank = Distilled water.
λ = 500nm
Flask
No.
Volume
of
10ppm
Fe [cm3]
Volume of
10%
Hydroxyl
amine
hydrochloride
[cm3]
Volume of
2M
Sodium
Acetate
[cm3]
Volume of
phenan
-throline
[cm3]
Final
volume
[cm3]
Final
concentr
ation of
Fe
[ppm]
OD
1. 05 10 1 10 100 0.5
2. 10 10 1 10 100 1.0
3. 15 10 1 10 100 1.5
4. 20 10 1 10 100 2.0
5. 25 10 1 10 100 2.5
6. unknown 10 1 10 100 -

85
From graph,
Unknown concentration = ________ ppm.
= ________ mg.
‘z’ ml of stock solution = ________ mg.
250cm3
of stock solution = unknown conc. x 250
z
= ___________ mg of Fe.
= ________ g of Fe.
Percentage of Fe = Weight of Fe x 100
Weight of the alloy.
= ___________ %.
RESULT: -
1. Amount of zinc present in the given sample of Zinc Blende =_________ g.
2. Percentage of zinc present in the given sample of Zinc Blende =_________ %.
3. Amount of Iron present in the given sample of Zinc Blende =_________ g.
4. Percentage of Iron present in the given sample of Zinc Blende =_________ %.

86
ANALYSIS OF PYROLUSITE.
AIM: To analyze the given sample of pyrolusite for its
1. Manganese content by complexometric method.
2. Acid insoluble residue by gravimetric method.
REQURIMENTS: - 0.02M EDTA solution Buffer pH 10, standard 100ppm Fe+2
solution, 1%
Phenanthroline, 10% Hydroxylamine hydrochloride, 2M sodium acetate, distilled water etc.
THEORY: -
Pyrolusite is a mineral consisting essentially of manganese dioxide and is important as an ore of
manganese. It is a black, amorphous appearing mineral, often with a granular, fibrous, or
columnar structure, sometimes forming reniform crusts. It has a metallic lustre, a black or bluish-
black streak, and readily soils the fingers. The specific gravity is about 4.8. Its name is from the
Greek for fire and to wash, in reference to its use as a way to remove tints from glass. Pyrolusite
is the manganese ore. It may contain Mn and Fe in the form of their oxides along with other
metals silicates. Ore can be opened in conc HNO3 or aqua regia. Mn will pass in the filtrate as
soluble salt. Mn can be titrated against standard EDTA solution in presence of Fe using
thymolphthalexone indicator at pH 10.
Fe+2
forms orange red color Fe(Ph)3 complex with 1,10-phenanthroline in the pH range of 2.8-
3.5 and absorb at λ= 480nm. Sodium acetate can give buffer pH in this range. Hydroxylamine
hydrochloride is a reducing agent and can reduce any Fe+3
to Fe+2
PROCEDURE: -
A) OPENING OF THE ORE: -
1. Weigh 0.5g accurately the given amount of pyrolusite ore and transfer it in a 250cm3
beaker.
2. Open the ore with 10cm3
of aqua regia and heat it on a sand bath nearly to dryness.
3. Repeat the acid treatment twice with 5cm3
of aqua regia and evaporate it nearly to
dryness.
4. Extract the solution with 50cm3
of 2N HCl and boil the solution for few minutes.

87
5. Filter the solution through whatmann filter paper no. 41 and collect all the washings in
250cm3
of standard measuring flask.
6. Dilute it upto the mark with distilled water. This is used as stock solution.
REMOVAL OF Fe :-
1. Take 50cm3
of the stock solution in beaker. Add one test tube with distilled water, 5gm
NH4Cl and 1:1 NH3 dropwise with constant stirring till the precipitated as Fe(OH)3 .
2. Filter through ordinary filter paper & washing in 100cm3
of std measuring flask and
dilute upto the mark.
B)ESTIMATION OF Mn BY COMPLEXOMETRIC METHOD: -
of the diluted solution and add 5cm3
of buffer of pH = 10 & 3-4 drops
of Eriochrome black T indicator.
2. Titrate it against 0.01M EDTA solution.
END POINT: - Dark blue to light pink.
C)ESTIMATION OF ACID INSOLUBLE RESIDUE: -
1. Collect the paper containing acid insoluble obtained from the above opening of the ore.
2. Dry the filter paper on a cone. Incinerate it in previously weighed crucible
3. Determine the amount of residue.
OBSERVATION AND CALCULATIONS: -
ESTIMATION OF ACID INSOLUBLE RESIDUE: -
1. Weight of empty crucible = ________ g.
2. Weight of crucible + residue = _________ g.
3. Weight of residue = ________ g.

88
% of Acid insoluble residue = Weight of residue x 100
Amount of ore taken.
% of Acid insoluble residue = ------------ %.
ESTIMATION OF Mn: -
10ml of stock solution required ‘a’ ml of 0.01M EDTA solution.
100ml of the stock solution requires ’10 x a’ ml 0.01M EDTA solution.
Now 50cm3
of the stock solution required 10 ‘a’ cm3
of 0.01M EDTA.
250 cm3
of the solution required 5 x 10 x ‘a’ cm3
of 0.01M.
Now,
1000ml of 1M EDTA = 54.94g of Mn.
5 x 10 x a of 0.01M EDTA = 54.94 x 5 x 10 x a x 0.01 g of Mn.
1000
= g of Mn.
_______ g of ore contains _______ g of Mn.
Percentage of Mn = Amount of Mn x 100
Weight of ore.
Percentage of Mn = ________ %
RESULT: -
1. Amount of acid insoluble residue = _________ g.
2. % of acid insoluble residue = _________ %.
3. Amount of Mn present = __________ g.
4. % of Mn present = __________ %.

89
DETERMINATION OF STABILITY CONSTANT OF ZINC AMMONIA COMPLEX.
AIM:-To determine the stability constant ofzinc-Ammonia
[Zn(NH3)4]+2
Complex potentiometrically.
REQUIREMENTS:- 0.1M ZnSO4 ,2M NH4 OH solution, KNO3 salt Bridge, potentiometer,
0.1NsuccinicAcid, 0.1N NaOH, 0.1NHCl , phenolphthalein Indicator,etc.
THEORY:-Inaqueoussolution, Zn formsaquo complex onadditionofammoniaZn+2
forms most
stable complex with ammonia as [Zn(NH3)4]+2
.The dissociation of complex ion is represented
as,
Zn+2
+4NH3 ⇋[Zn(NH3)4]+2
Thus the stability constant is given by,
K= [Zn(NH3)4]+2
[Zn+2
][ NH3]4
Experimentally determine the value K for stability constant. Show that totally very
small of Zn+2
ions are produced by the dissociation of the complex. The stability of
the complex is quantitatively expressed by means dissociation which can be measured
by potentiometer.
PROCEDURE:-
PREPARATION OF 0,1M ZnSO4.7H2O SOLUTION:-
Dissolve 4.247g of ZnSO47H2O in little quantity of distilled water and dilute it to
250cm3
with distilled water.
PREPARATION OF 2M NH3 SOLUTION:-

Advanced Practical Inorganic Chemistry For MSc SEM-III IV Students

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