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60
Viva Voce in
Pharmaceutical Chemistry
V
1
= Volume of the titrant
N
2
= Normality of standard solution
V
2
= Volume of standard solution
N
1
=
22
1
N
V
V
Q 304 How would you calculate molarity of the unknown sample
solution?
Ans: Molarity of the unknown sample solution is calculated using
the following formula.
Molarity =
W×M
V×F
where, W = Weight of primary standard
M = Standard molarity
V = Burette reading
F = Factor
Q 305 How the formula for calculating molarity is derived?
Ans: This formula is derived taking example of standardisation of
1M sodium hydroxide solution using potassium hydrogen
phthalate as primary standard.
• We calculated the amount of sodium hydroxide that needs to
be dissolved in given volume and prepared ‘m’ Molar sodium
hydroxide solution
• We have titrated ‘W’ gram of potassium hydrogen phthalate
• The burette reading is ‘V’ ml
As per IP the factor is
1 ml of 1 M sodium hydroxide 0.02042 g of C
8
H
5
KO
4
But in titration, we have used ‘m’ molar NaOH instead of 1 M
sodium hydroxide.
1 ml of ‘m’ Molar sodium hydroxide
0.02042 × m
1M
g of
C
8
H
5
KO
4
‘V’ ml of ‘m’ Molar sodium hydroxide
0.02042 × m × V
1M
At equivalence point, V ml correspond to ‘W’ weight of
potassium hydrogen phthalate.
Fundamentals of Volumetric Analysis 61
W =
0.02042 × m × V
1M
m =
W × 1M
V × 0.02042
=
Weight × Standard molarity
Burette reading × Factor
=
W×M
V×F
Q 306 How would you calculate percentage purity of given sample?
Ans: Percentage purity of given sample is calculated using the formula
% Purity =
V × m × F
M × W
×100
where,
V = Volume of the standard solution, i.e. burette reading
m = Exact calculated molarity of the prepared standard
solution.
F = Factor.
M = Standard molarity of the standard solution.
W = Weight of the sample to be assayed used in the titration.
Q 307 How the formula for calculating percentage purity is
derived?
Ans: Let us take example of assay of ammonium chloride where
‘W’ g of ammonium chloride is titrated with ‘V’ ml of ‘m’
molar sodium hydroxide solution. Factor given in IP is
1 ml of 0.1 M sodium hydroxide
0.005349 g NH
4
Cl
V ml of ‘m’ molar sodium hydroxide
0.005349×V×m
0.1
g NH
4
Cl
at equivalence point,
V ml
W g of NH
4
Cl
W g of sample contains =
0.005349 × V × m
0.1
g of NH
4
Cl
100 g of sample contains =
0.005349 × V × m
0.1 W
× 100 g of NH
4
Cl
% purity =
V × m × F
M × W
× 100
62
Viva Voce in
Pharmaceutical Chemistry
Q 308 How the factor is calculated?
Ans: For factor calculation, we must understand the reaction taking
place during titration. We must know the number of molecules
of reactants participating to produce number of molecules of
products. Let us take the example of
1. Standardisation of sodium hydroxide using potassium
hydrogen phthalate.
The reaction is
In this reaction, 1 molecule of potassium hydrogen phthalate
reacts with 1 molecule of NaOH.
C
6
H
5
KO
4
=
~
NaOH = 1000 ml of 1M NaOH
Molecular weight of potassium hydrogen phthalate is 204.2
204.2 g C
6
H
5
KO
4
=
~
1000 ml of 1M NaOH
0.2042 g C
6
H
5
KO
4
=
~
1 ml of 1 M NaOH
2. In the assay of borax, the reaction involved is
Na
2
B
4
O
7
.
10H
2
O + 2 HCl 4 H
3
BO
3
+ 2NaCl + 5 H
2
O
Na
2
B
4
O
7
.
10H
2
O = 2 HCl = 2000 ml of 1 N HCl
381.4 g Na
2
B
4
O
7
.
10 H
2
O = 4000 ml of 0.5 N HCl
381.4
4
g Na
2
B
4
O
7
.
10 H
2
O = 1000 ml of 0.5 N HCl
0.0953 g Na
2
B
4
O
7
.
10H
2
O = 1 ml of 0.5 N HCl
Q 309 What is the normality of sodium hydroxide, if 20 g of NaOH
is dissolved in 1000 ml of water?
Ans: Normality =
Wt. in
g
/litre
Equivalent wei
g
ht
20
= = 0.5 N
40
Q 310 Calculate the exact molarity of the NaOH.
Ans: Given: 40.3 g of sodium hydroxide is dissolved in 1000 ml of
water. The weight of potassium hydrogen phthalate is 4.9 g.
The burette reading is 25.3 ml. The factor is 1 ml of 1 M NaOH
0.2042 g of C
8
H
5
KO
4
.
Ans: Exact molarity of NaOH =
M × W
V × F
Fundamentals of Volumetric Analysis 63
=
1 × 4.9
25.3 × 0.2042
= 0.9484 M
NEUTRALISATION TITRATIONS
Q 311 What is neutralisation titration?
Ans: In neutralisation titrations, free bases react with standard acid
or free acids react with standard base to produce a salt and
water molecule.
Q 312 What is acidimetry?
Ans: The neutralisation of a base with standard acid is known as
acidimetry.
Q 313 What is alkalimetry?
Ans: The neutralisation of an acid with standard base is called
alkalimetry.
Q 314 Name the different theories of acid and base.
Ans: Different scientists have explained acids and bases in different
manner.
1. Arrhenius concept.
2. Bronsted Lowry concept.
3. Lewis concept.
Q 315 What is Arrhenius concept of acid and base? Give examples?
Ans: As per Arrhenius concept, acid may be defined as any
hydrogen containing substance which can give H
+
ions in
aqueous solution. While base may be any substance containing
hydroxy groups which are capable of providing hydroxy ions
in aqueous solution. Thus, HCl is an acid and NaOH is a base.
Q 316 What is Bronsted Lowry concept?
Ans: As per Bronsted and Lowry, an acid is a proton donor and a
base is a proton acceptor. So, hydrogen sulphate ion is an acid.
while bicarbonate ion is a base.
Q 317 What is Lewis concept of acid and base?
Ans: According to Lewis, any species (molecules, radical or ion)
that can accept electrons is called an acid. While any species
64
Viva Voce in
Pharmaceutical Chemistry
that can donate electron pair for the formation of coordinate
bond is called base, e.g.
Here, the hydrogen ion is accepting an electron pair and thus
it is an acid while ammonia is donating electron pair to form
a coordinate bond and thus it is base.
Q 318 What do you understand by the term strong acids?
Ans: These are the acids which dissociate or ionize completely, e.g.
HCl, HNO
3
.
These acids are able to donate protons with ease.
Q 319 What are weak acids?
Ans: These are the acids which ionize to limited extent, e.g. acetic
acid (CH
3
COOH), boric acid (H
3
BO
3
), carbonic acid (H
2
CO
3
),
phosphoric acid (H
3
PO
4
). The protons are not donated
with ease.
Q 320 What are strong bases?
Ans: The substances which completely dissociate in water to give
hydroxide ions are called strong bases, e.g. NaOH, KOH.
These bases accept protons with ease.
Q 321 What are weak bases?
Ans: These substances give hydroxide ions but they dissociate to
limited extent, e.g. barium hydroxide (BaOH
2
), ammonium
hydroxide (NH
4
OH). These bases cannot accept protons
with ease.
Q 322 What is monobasic or monoprotic acid?
Ans: The acids which dissociate in water to give only one proton
are called monoprotic acids, e.g. HCl, HNO
3
, CH
3
COOH.
Q 323 What is monoacidic base?
Ans: A base which can accept only one proton is called monoacidic
base, e.g. NaOH, KOH.
Q 324 What are dibasic/diprotic acids?
Ans: Those acids which dissociate to give two protons are called
dibasic acids, e.g. H
2
SO
4
Fundamentals of Volumetric Analysis 65
Q 325 What are tribasic/triprotic acids?
Ans: These are the acids which can donate 3 hydrogens, e.g.
phosphoric acid.
Q 326 Enlist five acids used in the laboratory.
Ans: Hydrochloric acid, nitric acid, sulphuric acid, acetic acid, and
boric acid.
Q 327 Enlist five bases used in the laboratory.
Ans: Sodium hydroxide, potassium hydroxide, ammonia, sodium
bicarbonate, and sodium carbonate.
Q 328 Write five neutralisation reactions.
Ans:
1. Standardisation of hydrochloric acid using sodium carbonate.
2. Assay of sodium bicarbonate using hydrochloric acid.
3. Standardisation of sodium hydroxide using succinic acid.
4. Assay of boric acid using sodium hydroxide.
5. Assay of ammonium chloride using sodium hydroxide.
Q 329 Why nitric acid is not commonly employed in acid–base
titration?
Ans: Nitric acid is not commonly employed because nitric acid is
highly corrosive and is capable of causing damage to tissues.
The gases evolved from spillage on heavy metals or on organic
materials are also exceedingly toxic.
Another reason is, in addition to its strongly acidic property,
it is an oxidising agent and a nitrating agent.
Q 330 In what type of titration, methyl orange is used as an indicator?
Ans: Methyl orange is used as an indicator in titrations of strong
acid with strong alkali and vice versa.
Q 331 In what type of titrations phenolphthalein is used as an
indicator?
Ans: Phenolphthalein is used as an indicator in titration of weak
acid with strong alkali.
Q 332 In the titration of weak bases with strong acids which
indicator is used?
Ans: Methyl red.
66
Viva Voce in
Pharmaceutical Chemistry
Q 333 Name primary standard used for standardisation of acid
solutions.
Ans: Anhydrous sodium carbonate AR grade is used for standar-
disation of acid solutions.
Q 334 Name primary standards used for standardisation of alkali
solutions.
Ans: Succinic acid AR or oxalic acid AR or potassium hydrogen
phthalate is used for standardisation of alkali solutions.
Q 335 What are neutralisation indicators chemically?
Ans: Neutralisation indicators are weak acids or weak bases which
have different colours in their conjugate base and acid
forms (two colour indicators), e.g. methyl orange, one colour
indicators have one form coloured with a colourless conjugate
form, e.g. phenolphthalein.
PREPARATION AND STANDARDISATION OF HYDROCHLORIC
ACID SOLUTION
Q 336 How would you prepare 200 ml of 0.1 N hydrochloric acid?
Ans:
1. We know that normality of concentrated hydrochloric acid is
11.5 N.
2. We apply allegation method to know the volumes of the conc.
HCl and water to be mixed to get the desired concentration.
So, addition of 0.1 ml of conc. HCl to 11.4 ml of water gives
11.5 ml of 0.1 N HCl.
We want to prepare 200 ml of 0.1 N HCl
11.5 ml of 0.1 N HCl =
~
0.1 ml of conc. HCl
200 ml of 0.1 NHCl =
~
?
=
200 × 0.1
11.5
= 1.739 ml
Add 1.739 ml of conc. HCl to water and make up the volume
to 200 ml. In the similiar manner we can calculate for the
hydrochoric acid of the other desired strengths.
Fundamentals of Volumetric Analysis 67
Q 337 Which primary standard is used for standardisation of the
hydrochloric acid?
Ans: Sodium carbonate is used as primary standard for standar-
disation of hydrochloric acid.
Q 338 How the primary standard solution (0.1 N) of sodium
carbonate is prepared?
Ans: The molecular weight of sodium carbonate is 106 g and acidity
is 2, so to prepare 1000 ml of 0.1 N standard solution, we must
dissolve 5.3 g of Na
2
CO
3
in water and dilute up to 1000 ml
with water.
Q 339 How exact normality of sodium carbonate primary standard
solution is calculated?
Ans:
The exact normality is calculated as follows:
1. We know that for preparing 1000 ml of 0.1N Na
2
CO
3
, 5.3 g
need to be dissolved in water to make up the volume to
1000 ml.
2. The exact weight of sodium carbonate transferred from the
weighing bottle is say 5.43 g
3. If 5.3 g 0.1 N
then, 5.43 g ?
=
5.43 × 0.1 N
5.3
= 0.1024 N
The generalisation can be made as
Exact normality =
Weight in g/litre
Equivalent weight
Q 340 Tell the procedure of standardisation of hydrochloric
acid.
Ans:
1. Take 10 ml of primary standard solution of sodium carbonate
in conical flask.
2. Add two drops of methyl orange solution.
3. Titrate with prepared hydrochloric acid solution till the colour
changes from yellow to pink.
4. Note the burette reading.
68
Viva Voce in
Pharmaceutical Chemistry
Q 341 Which indicator is used in the standardisation of hydroch-
loric acid?
Ans: Methyl orange is used as an indicator in standardisation of
hydrochloric acid solution.
Q 342 What is the colour change expected at end point in standardi-
sation of hydrochloric acid?
Ans: The colour changes from yellow to pink, when methyl orange
is used as indicator.
Q 343 Write chemical reaction occurring in standardisation of
hydrochloric acid?
Ans:
Q 344 How to prepare 0.1 M solution of hydrochloric acid?
Ans: As equivalent weight of hydrochloric acid is same as
molecular weight, 0.1 M solution is same as 0.1 N solution
and thus prepared in the same manner.
Q 345 How to calculate the normality of the prepared hydrochloric
acid solution?
Ans: Standardise the prepared hydrochloric acid with primary
standard solution of sodium carbonate. At the equivalence
point,
N
1
V
1
= N
2
V
2
where,
N
1
= Normality of Na
2
CO
3
solution
V
1
= Volume of Na
2
CO
3
solution
N
2
= Normality of hydrochloric acid
V
2
= Burette reading, i.e. volume of hydrochloric acid
N
2
=
11
2
N
V
V
PREPARATION AND STANDARDISATION OF
SODIUM HYDROXIDE SOLUTION
Q 346 How would you prepare 1000 ml of 1 m of sodium hydroxide?
Ans:
• Weigh 40 g of sodium hydroxide.
Fundamentals of Volumetric Analysis 69
• Dissolve it in about 100 ml of water.
• Dilute it to 1000 ml with water.
Q 347 How would you prepare 400 ml of 1 molar solution of sodium
hydroxide?
Ans: The molecular weight of sodium hydroxide is 40. So, if we
dissolve 40 g in water and make up the volume to 1000 ml,
we get 1 molar solution. We want to prepare 400 ml
1000 ml =
~
40 g
400 ml =
~
?
=
400 × 4
1000
= 16 g
Dissolve 16 g of sodium hydroxide in water and make up the
volume to 400 ml.
Q 348 Which are the primary standards employed in standar-
disation of sodium hydroxide solution? or How do you
standardise sodium hydroxide solution?
Ans: We can standardise sodium hydroxide solution by two methods:
1. By using potassium hydrogen phthalate as primary standard.
2. By using succinic acid as primary standard.
Q 349 Which indicator is used in standardisation of sodium
hydroxide solution?
Ans: Phenolphthalein is used as indicator in standardisation of
sodium hydroxide solution.
Q 350 What colour change is expected at the equivalence point
using phenophthalein as indicator?
Ans: Colourless to pink.
Q 351 Describe the procedure for standardisation of sodium
hydroxide solution by using potassium hydrogen phthalate.
Ans: Potassium hydrogen phthalate method I.P. 2010.
• Weigh accurately about 5 g of potassium hydrogen phthalate,
previously powdered and dried at 120° for 2 hours.
• Dissolve in 75 ml of carbon-dioxide free water.
• Add 0.1 ml of phenolphthalein solution.
• Titrate with sodium hydroxide solution until permanent
pink colour is produced.
• Factor: 1 ml of 1 M sodium hydroxide is equivalent to
0.2042 g of C
8
H
5
KO
4
.