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40
Viva Voce in
Pharmaceutical Chemistry
CHLORIDE ION
Q 212 Give the identification tests for chlorides. Discuss the chemistry.
Ans: IP 2007 has given two tests for identification of chlorides.
1. Dissolve 2 mg of sample in 2 ml of water. Acidify with dilute
nitric acid, add 0.5 ml of silver nitrate solution. Shake and
allow to stand. A curdy white precipitate is formed, which is
insoluble in nitric acid but soluble, after being well-washed
with water, in dilute ammonia solution, from which it is
reprecipitated by the addition of dilute nitric acid.
2. Introduce into a test tube, substance equivalent to about
10 mg of chloride ion, add 0.2 g of potassium dichromate
and 1 ml of sulphuric acid. Place a filter-paper strip moistened
with 0.1 ml of diphenylcarbazide solution over the mouth of
the test tube, the paper turns violet-red.
The chromyl chloride evolved causes solution of diphenyl
carbazide to assume violet-red colour.
CITRATE ION
Q 213 How would you perform identification tests for the detection
of citrates? Elaborate underlying chemistry of tests.
Ans: IP 2007 gives 2 tests for identification of citrates.
1. To a neutral solution of substance under examination, add a
solution of calcium chloride. No precipitate is produced. Boil
Identification Tests for Anions and Cations 41
the solution. A white precipitate soluble in 6 M acetic acid is
produced.
2. Dissolve a quantity of substance containing about 50 mg of
citric acid in 5 ml of water. Add 0.5 ml of sulphuric acid and
3 ml of potassium permanganate solution. Warm until the
colour of the permanganate is discharged and add 0.5 ml of a
10% w/v solution of sodium nitroprusside in 1 M sulphuric
acid and 4 g sulphamic acid (H
2
NSO
3
H). Make alkaline with
strong ammonia solution, added dropwise until all the sulphamic
acid has dissolved. On addition of an excess of strong ammonia
solution, a violet colour, which turns violet-blue is produced.
Citrate ions decolourise potassium permanganate due to their
oxidation into carbon monoxide and acetone dicarboxylic acid,
which undergoes further partial decomposition into acetone
and carbon dioxide.
42
Viva Voce in
Pharmaceutical Chemistry
The intermediate acetone dicarboxylic acid is made to interact
with sodium nitroprusside in presence of sulphamic acid.
Violet colour changes to violet-blue.
GLUCONATE ION
Q 214 Discuss the identification tests for detection of gluconate
ions/gluconic acid.
Ans:
1. This test is described in IP 2007.
To 1 ml of a 3% w/v solution add 0.05 ml of ferric chloride
test solution. A yellow colour is produced due to formation
of ferric gluconate.
2. This test is described in IP 1985. To 0.75 g of sample in
7.5 ml of warm water, add 1 ml of glacial acetic acid and
1.5 ml of freshly distilled phenylhydrazine. Heat the
mixture on waterbath for 30 minutes. Allow to cool.
Scratch the inner surface of the tube with a glass rod, until
crystals of gluconic acid phenylhydrazide begin to form.
Set aside the mixture for ten minutes. Filter, dissolve the
precipitate in 10 ml of hot water and add a small amount
of decolourising charcoal. Filter the suspension. Allow the
filterate to cool and scratch the inner surface of the tube. White
crystals are obtained. Dry them and take the melting point.
Observe that the product melts at 200°C, with decomposition.
Identification Tests for Anions and Cations 43
IODIDE ION
Q 215 Discuss the identification tests along with chemistry for
iodides.
Ans:
1. Dissolve the substance containing about 4 mg of iodide
ion in 2 ml of water. Acidify with dilute nitric acid and add
0.5 ml of silver nitrate solution. Shake and allow to stand, a
curdy pale yellow precipitate is formed. Filter and wash with
water. Suspend the precipitate in 2 ml of water and add
1.5 ml of 10 M ammonia, the precipitate does not dissolve.
2. To 0.2 ml of solution of the substance under examination
containing about 5 mg of iodide ion per ml, add 0.5 ml of 1 M
sulphuric acid, 0.15 ml of potassium dichromate solution, 2 ml
of water and 2 ml of chloroform. Shake for few seconds and
allow to stand. The chloroform layer is violet or violet-red.
The liberated iodine dissolves in chloroform layer imparting
it violet-red colour.
3. To 1 ml of solution of the substance under examination
containing about 5 mg of iodide ion, add 0.5 ml of mercuric
44
Viva Voce in
Pharmaceutical Chemistry
chloride solution. A dark red precipitate is formed which is
slightly soluble in an excess of this reagent and very soluble
in an excess of potassium iodide solution.
The precipitate dissolves in excess of potassium iodide solution.
SULPHATE ION
Q 216 How would you identify sulphate ions?
Ans: IP 2007 describes three tests.
1. Dissolve 50 mg of substance in 5 ml of water. Add 1 ml of
dilute hydrochloric acid and 1 ml of barium chloride solution.
A white precipitate is formed.
2. Add 0.1 ml of iodine solution to the suspension obtained in
test A. The suspension remains yellow (distinction from
sulphites and dithionites) but is decolourised by adding
dropwise, stannous chloride solution (distinction from
iodates). Boil the mixture. No coloured precipitate is formed
(distinction from selenates and tungstates).
3. Dissolve about 50 mg of substance in 5 ml of water. Add
2 ml of lead acetate solution, a white precipitate, soluble in
ammonium acetate solution and in sodium hydroxide solution
is produced.
When sodium hydroxide is added to this precipitate, soluble
sodium hydroxy plumbate(II) is formed.
Identification Tests for Anions and Cations 45
THIOSULPHATE ION
Q 217 What are the identification tests for detection of thiosulphate
ions?
Ans: IP 2007 describes four tests for detection of thiosulphate ions.
1. Dissolve 0.1 g of substance in 5 ml of water. Add 2 ml of
hydrochloric acid. A white precipitate is formed which soon
turns yellow and sulphur dioxide, recognisable by its odour,
is evolved.
2. Dissolve 0.1 g of substance in 5 ml of water and add 2 ml of
ferric chloride test solution. A dark violet colour which quickly
disappears is produced.
On standing, the colour disappears due to formation of
tetrathionate and ferrous ions.
The overall reaction is:
3. Solution of thiosulphates decolourise iodine solution, the
decolourised solutions do not give the reactions of sulphates.
This is an oxidation–reduction reaction. Sodium thiosulphate
reduces iodine to colourless iodide.
4. Solutions of thiosulphates decolourise bromine solution. The
decolourised solutions give the reactions of sulphates.
46
Viva Voce in
Pharmaceutical Chemistry
GENERAL QUESTIONS
Q 218 What is volumetric or titrimetric analysis?
Ans: It is a quantitative chemical analysis which involves
determination of volume of a solution of known concentration
required to react quantitatively with the solution of substance
being analysed.
Q 219 What is standard solution?
Ans: The solution of accurately known strength is known as
standard solution.
Q 220 What is a titrant?
Ans: The solution, that is to be analysed is known as titrant.
Q 221 Where is the standard solution taken—in a flask or a burette?
Ans: The standard solution is taken in a burette.
Q 222 Where is the titrant taken?
Ans: The titrant is taken in the conical flask.
Q 223 What is titration?
Ans: The process of controlled addition of a standard solution to a
titrant with the purpose of determining its strength is known
as titration.
Q 224 Name different types of titrations.
Ans: Titrations can be classified into different types:
1. Neutralisation titrations.
2. Titrations in non-aqueous solvents.
3. Oxidation–reduction titrations.
4. Precipitation titrations.
5. Complexometric titrations.
Fundamentals of
Volumetric Analysis
3
46
Fundamentals of Volumetric Analysis 47
Q 225 What do you understand by the term ‘end point’ or stoichio-
metric point or equivalent point?
Ans: The point at which the titrant reacts with standard solution
in chemically equivalent amount is known as end point. At
this stage, sudden change in some property of reaction mixture
takes place, e.g. colour, precipitation, etc.
MOLECULAR WEIGHT AND EQUIVALENT WEIGHT
Q 226 Define molecular weight.
Ans: Molecular weight is sum of the atomic weights of the atoms
making up the molecule.
Q 227 What do you understand by ‘gram mole of substance’?
Ans: Gram mole of substance indicates molecular weight expressed
in grams.
Q 228 Define equivalent weight.
Ans: Equivalent weight of a substance is determined by the actual
reaction under consideration.
Neutralisation reaction: Equivalent weight of an acid or base is
that weight which contains 1.0079 g of replaceable hydrogen.
Equivalent weight of a base =
Molecular weight
Acidity
Equivalent weight of an acid =
Molecular weight
Basicity
Oxidation–reduction reaction: Here, the equivalent weight is
that weight which combines with 1.008 g of available
hydrogen or 8.008 g of available oxygen for use in oxidation
or reduction reaction under consideration.
Precipitation reaction: It is that weight of a substance that
combines or contains 1 g atom of a univalent metal.
Q 229 What is gram equivalent weight?
Ans: Equivalent weight expressed in grams is known as gram
equivalent weight.
Q 230 What is the equivalent weight of hydrochloric acid, nitric
acid, and acetic acid?
Ans: The equivalent weight of hydrochloric acid, nitric acid and
acetic acid is same as that of their molecular weight because
they contain only one replaceable hydrogen.
48
Viva Voce in
Pharmaceutical Chemistry
Acid Mol wt. Basicity Equi. wt
HCl 36.46 1 36.46
HNO
3
63.01 1 63.01
CH
3
COOH 60.05 1 60.05
Q 231 What is the equivalent weight of sulphuric acid and oxalic
acid?
Ans: Equivalent weight of sulphuric acid and oxalic acid is half of
their molecular weight because they contain two replaceable
hydrogens and their basicity is 2.
Acid Mol. wt Basicity Equi. wt.
H
2
SO
4
98.07 2 49.03
(COOH)
2
.
2H
2
O 126.07 2 63.03
Q 232 What is the equivalent weight of boric acid (H
3
BO
3
)?
Ans: The equivalent weight of boric acid is one-third of its
molecular weight as it has 3 replaceable hydrogens. Its
equivalent weight is 61.03/3 = 20.61.
Q 233 What is the equivalent weight of sodium hydroxide, pota-
ssium hydroxide, sodium bicarbonate, and potassium bicar-
bonate?
Ans: The equivalent weight of sodium hydroxide, potassium
hydroxide and potassium bicarbonate is same as their
molecular weight because all of them have acidity of one.
Base Mol. wt. Acidity Equi. wt.
NaOH 40.0 1 40.0
KOH 56.1 1 56.1
NaHCO
3
84.01 1 84.01
KHCO
3
100.12 1 100.12
Q 234 What is the equivalent weight of sodium carbonate and pota-
ssium carbonate?
Ans: The equivalent weight of sodium carbonate and potassium
carbonate is half the molecular weight of them because both
the compounds have acidity of 2.
Base Mol. wt. Acidity Equi. wt.
Na
2
CO
3
105.99 2 53
K
2
CO
3
138.21 2 69.110
Fundamentals of Volumetric Analysis 49
Q 235 How equivalent weights of oxidising or reducing agents are
calculated?
Ans: Equivalent weight of a reducing agent or oxidising agent
depends upon the number of electrons lost or gained in the
redox reaction.
Eq. wt. = Mol. wt./No. of electrons
Chemical Mol. wt. No. of e
–
gained or Eq. wt.
formula lost in redox reaction
KMnO
4
158.03 5 31.60
I
2
253.8 2 126.9
Na
2
S
2
O
7
.
5H
2
O 248.17 1 248.17
NORMALITY/ MOLARITY/ PREPARATION OF SOLUTIONS
Q 236 Name the terms used to express the strength of a solution.
Ans: Strength of a solution is expressed in terms of normality,
molarity or molality.
Q 237 What do you understand by the term normality?
Ans: Normality of a solution is the number of gram equivalents of
the substance dissolved per litre of the solution.
Normality =
N
umber of gram equivalents
Volume in litres
Q 238 How do you prepare 1 normal solution of a substance?
Ans: We dissolve 1 gram equivalent of the substance per litre of
solution to prepare 1N solution.
Q 239 Define ‘molarity’.
Ans: Molarity is the number of gram moles of the substance
dissolved per litre of the solution.
Molarity =
N
umber of gram moles
Volume in litres
Q 240 How would you prepare 1 molar solution of a substance?
Ans: One molar solution is prepared by dissolving 1 gram mole of
the substance per litre of solution.
Q 241 Define ‘molality’.
Ans: Molality is the number of gram moles of substance dissolved
per 1000 gram of solution.