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50
Viva Voce in
Pharmaceutical Chemistry
Q 242 How to prepare 1 molal solution?
Ans: One molal solution is prepared by dissolving 1 g mole of
substance per kilogram of the solution.
Q 243 What is the approximate normality of concentrated sulphuric
acid?
Ans: The approximate normality of concentrated sulphuric acid is
36 N or 18 M.
Q 244 What is the required percentage of H
2
SO
4
in concentrated
sulphuric acid?
Ans: The percentage of H
2
SO
4
in concentrated sulphuric acid
should not be less than 97% w/w of H
2
SO
4
.
Q 245 What is the approximate normality of concentrated
hydrochloric acid?
Ans: The approximate normality of concentrated hydrochloric acid
is 11.5 N.
Q 246 How much amount of HCl should be present in concentrated
hydrochloric acid as per IP?
Ans: Concentrated hydrochloric acid should have 35.0 to 38.0% w/w
HCl.
Q 247 What is the approximate normality of concentrated nitric
acid?
Ans: The approximate normality of concentrated nitric acid is 16 N
Q 248 What amount of HNO
3
should be essentially present in
concentrated nitric acid?
Ans: Concentrated nitric acid must contain 69–71% w/w of HNO
3
.
Q 249 What is the concentration of concentrated acetic acid?
Ans: Glacial acetic acid used in laboratory is approximately 17.5N.
It should contain not less than 99.0% w/w of C
2
H
4
O
2
.
Q 250 What is the strength of dilute hydrochloric acid?
Ans: Dilute hydrochloric acid contains about 10% w/w of HCl.
Q 251 How would you prepare dilute hydrochloric acid from the
concentrated hydrochloric acid?
Ans: Concentrated hydrochloric acid contains 35 g of HCl per
100 g of concentrated hydrochloric acid while dilute
hydrochloric acid contains 10 g of HCl per 100 g of dilute
hydrochloric acid. Using allegation method, we can
calculate the amounts of concentrated hydrochloric acid
Fundamentals of Volumetric Analysis 51
and water, which need to be mixed to produce the desired
volume of dilute hydrochloric acid.
Take ten parts of concentrated hydrochloric acid and mix it
with 25 ml of water to obtain 35 ml of dilute hydrochloric
acid.
Q 252 How would you prepare 100 ml of dilute hydrochloric acid
from concentrated hydrochloric acid?
Ans: From the above answer we know that 35 ml of dilute
hydrochloric acid contains 10 ml of concentrated hydrochloric
acid.
35 10
100 ?
=
00×10
35
= 28.57 ml of conc. HCl
Take 28.57 ml of concentrated hydrochloric acid and dilute to
100 ml with water.
Q 253 How would you prepare 100 ml of 1 N hydrochloric acid
from concentrated hydrochloric acid?
Ans: The normality of concentrated hydrochloric acid is 11.5 N.
Using allegation method,
Mix 1 part of concentrated acid with 10.5 parts of water to get
11.5 ml of 1 N hydrochloric acid.
For 11.5 ml 1N HCl 1 ml of conc. HCl
For 100 ml 1N HCl ?
=
100 1
8.69 ml
11.5
Dilute 8.7 ml of conc. HCl up to 100 ml with water to get 100
ml of dilute HCl.
52
Viva Voce in
Pharmaceutical Chemistry
Q 254 How would you prepare dilute nitric acid from concentrated
nitric acid?
Ans: Dilute nitric acid contains 10% w/w of HNO
3
. Concentrated
nitric acid contains 70% w/w of HNO
3
. Using allegation method:
Mix 10 parts of conc. nitric acid with 60 parts of water to get
70 ml of dilute nitric acid.
Q 255 How would you prepare 1000 ml of dilute nitric acid?
Ans: We know that
70 ml of dilute nitric acid 10 ml of conc. HNO
3
1000 ml of dilute nitric acid ?
=
1000 × 10
70
= 142.85 ml
Take 142.85 ml of conc. HNO
3
and dilute up to 1000 ml with water.
Q 256 What is fuming nitric acid?
Ans: Fuming nitric acid contains about 95% w/w of HNO
3
. It is a
fuming liquid.
Q 257 What is the concentration of glacial acetic acid?
Ans: Glacial acetic acid should contain not less than 99.0 % w/w
C
2
H
4
O
2
(CH
3
COOH).
Q 258 What is the normality of glacial acetic acid?
Ans: The approximate normality of glacial acetic acid is 17.5 N.
Q 259 What is acetic acid?
Ans: Acetic acid contains approximately 33% w/w of CH
3
COOH.
Q 260 What is the difference between glacial acetic acid and acetic acid?
Ans: Glacial acetic acid is three times more concentrated as
compared to acetic acid.
Q 261 How would you prepare acetic acid from glacial acetic acid?
Ans: Using allegation method.
Mix 33 parts of glacial acetic acid with 66 parts of water to get
99 parts of acetic acid.
Fundamentals of Volumetric Analysis 53
Q 262 How would you prepare 1000 ml of acetic acid?
Ans: 99 parts of acetic acid contain 33 parts of glacial acetic acid.
So, 1000 parts ?
=
1000 × 33
99
= 333
Dilute 333 parts of glacial acetic acid with water up to 1000 ml
to get 1000 ml of acetic acid.
Q 263 What is dilute acetic acid?
Ans: Dilute acetic acid contains approximately 6% w/w of
CH
3
COOH.
Q 264 How would you prepare 1000 ml of dilute acetic acid?
Ans: Glacial acetic acid contains 99% w/w of CH
3
COOH.
Mix six parts of glacial acetic acid with 93 parts of water to
get 99 ml of dilute acetic acid.
99 ml 6
1000 ml
?
=
1000 × 6
99
= 60.60 ml
Mix 60.60 ml of glacial acetic acid with water up to 1000 ml,
to get 1000 ml of dilute acetic acid.
Q 265 Give concentration of strong ammonia solution.
Ans: Strong ammonia solution contains 25% w/w of ammonia. The
strength is about 13.5 M.
Q 266 What is the concentration of ammonia in dilute ammonia
solution?
Ans: Dilute ammonia solution contains approximate 10% w/w of
NH
3
.
Q 267 How would you prepare dilute ammonia solution?
Ans: Using allegation method
54
Viva Voce in
Pharmaceutical Chemistry
Mix 10 parts of strong ammonia solution with 15 parts of water
to get 25 parts of dilute ammonia solution.
To prepare 1000 ml of dilute ammonia solution:
25 parts of dil. ammonia solution 10 parts of strong ammonia
solution.
1000 parts of dilute ammonia solution ?
=
1000 × 10
25
= 400 parts
Mix 400 parts of strong ammonia solution with 600 parts of
water to get 1000 ml of dilute ammonia solution.
Q 268 How would you prepare 1000 ml of 1N sodium hydroxide
solution?
Ans: The gram equivalent weight of sodium hydroxide is 40. So,
dissolve 40 g of sodium hydroxide in water and make up the
volume to 1000 ml with water.
Q 269 Is there any difference between 1 M sodium hydroxide and
1 N sodium hydroxide solution?
Ans: As the equivalent weight of sodium hydroxide is same as
molecular weight of sodium hydroxide, there is no difference
in two solutions.
Q 270 How would you preapre 1 M sodium carbonate solution?
Ans: The molecular weight of Na
2
CO
3
is 106. Dissolve 106 g in water
and make up the volume to 1000 ml.
Q 271 Is there any difference between 1 M and 1 N sodium
carbonate solution?
Ans: The equivalent weight of sodium carbonate is half of its
molecular weight because of the acidity number of 2. So, 1 M
sodium carbonate solution is equivalent to 2 N sodium
carbonate solution.
Q 272 How ‘x’ M or ‘x’ N reagent is prepared?
Ans: ‘x’ M or ‘x’ N reagent is prepared by multiplying with ‘x’ the
amount of reagent required for preparing 1 M or 1 N reagent.
STANDARD SOLUTIONS
Q 273 What is standard solution?
Ans: A solution of accurately known strength is called standard
solution.
Fundamentals of Volumetric Analysis 55
Q 274 How standard solution is prepared?
Ans: Standard solution is prepared
1. By direct weighing of the primary standard or
2. By standardizing the prepared solution with primary standard
solution.
Q 275 What is primary standard?
Ans: Primary standard is a chemical substance of high purity, used
for the standardization of other solution.
Q 276 What are the pre-requisites to become primary standard?
Ans: The primary standard should comply with following
requisites:
1. It sould be easily available in highly pure state.
2. It should be easily soluble in desired solvent.
3. Should be stable and should not decompose.
4. Should have high molecular weight.
5. Should react instantaneously with the solution to be standardized.
Q 277 Name a primary standard used for standardizing acid
solutions.
Ans: Sodium carbonate AR, is used for standardizing acid
solutions.
Q 278 Name a primary standard used for standardizing alkali
solutions.
Ans: Succinic acid AR, is used for standardizing alkali solutions.
Q 279 Name primary standards used for standardizing chemicals
used in redox reactions.
Ans: In redox reaction, oxalic acid AR is used for standardization
of KMnO
4
solution, and ferrous ammonium sulphate AR is
used for standardization of potassium dichromate solution.
Potassium dichromate is used for standardization of sodium
thiosulphate solution.
Q 280 What is used for standardization of silver nitrate solution?
Ans: Sodium chloride AR is the primary standard used for
standardization of silver nitrate solution.
Q 281 What is primary standard for standardization of sodium
chloride solution?
Ans: Sodium chloride solution is standardized using silver nitrate
AR as primary standard.
56
Viva Voce in
Pharmaceutical Chemistry
Q 282 In complexometric titrations, how disodium edetate is
standardized?
Ans: Grannulated zinc is used as primary standard for standardi-
zation of disodium edetate.
Q 283 What is primary standard solution?
Ans: The solution of accurately known strength of a primary
standard is known as primary standard solution. This solution
is used for standardization of other solutions.
Q 284 What are secondary standards?
Ans: Secondary standards are the substances which do not fulfill
the criteria set for primary standards.
Standard solutions of secondary standards cannot be prepared
by direct weighing.
Q 285 Name common secondary standards.
Ans: Common secondary standards are
1. Alkali hydroxides like sodium hydroxide, potassium hydroxide
2. Inorganic acids such as hydrochloric acid, nitric acid and
sulphuric acid.
Q 286 How secondary standard solutions are prepared?
Ans: Secondary standard solutions are prepared by dissolving
approximate weight/volume of the secondary standard in a
known volume of water and then the exact strength of the
solution is found by titrating it against standard solution of
suitable reagent/primary standard.
INDICATORS
Q 287 What is an indicator?
Ans: Indicators are complex organic substances which help in
detecting the equivalence point by bringing about change in
colour or appearance of precipitate in the reaction mixture.
Q 288 Why indicators bring about change in colour at end point?
Ans: Acid–base indicators change colour according to the hydrogen
ion concentration of the solution. Many times indicators are
weak acids or weak bases which have different colour in their
conjugate base and acid forms.
Q 289 What is indicator range?
Ans: The change in colour of an indicator from predominantly acid
colour (colour in acidic pH) to a predominantly alkaline colour
Fundamentals of Volumetric Analysis 57
(colour in basic pH) is not sudden and abrupt, but it takes
place within a small interval of pH (usually about two pH
units). This is called colour change interval or indicator range
of the indicator.
A sharp change in colour over a definite range of pH is known
as indicator range.
Q 290 Name commonly used indicators in neutralisation reaction.
Give their pH range, colour change and the type of titration
in which they are used.
Ans:
Indicator pH range Colour change Type of titrations
Phenolphthalein 8.3 to 10 Colourless to pink Weak acid with strong alkalies
Methyl red 4.2 to 6.3 Red to yellow Weak bases with strong acids
Methyl orange 2.9 to 4.6 Red to yellow Strong acid with strong alkali
or
Bromophenol blue 6.0 to 7.6 Yellow to blue Strong alkali with strong acid.
Methyl red and phenolphthalein can also be used with strong
alkali with strong acid and vice versa.
Q 291 Name commonly used indicators in redox titrations.
Ans: Redox titrations are further classified as permanganometry,
iodometry and iodimetry.
Type of titration Indicator Colour change
Permanganometry Acidic KMnO
4
solution Colourless to pink
Iodometry Starch mucilage Blue to colourless
Iodimetry Starch mucilage Blue to colourless
Q 292 Name the indicator used in precipitation titrations.
Ans: Precipitation titrations are further classified as Volhard’s
method and Mohr’s method.
Type of titration Indicator End point
Volhard’s method Ferric ammonium sulphate Appearance of reddish brown
colour
Mohr’s method Potassium chromate Appearance of red precipitate
Q 293 Name an indicator used in complexometric titrations.
Ans: In complexometric titrations Mordant black II is used and it
changes the colour from reddish pink to blue.
58
Viva Voce in
Pharmaceutical Chemistry
Q 294 How much indicator should be used in the titration?
Ans: Two to three drops of indicator should be used in the titration.
Q 295 When the indicator is avoided?
Ans: Indicator is avoided when titrant or standard solution posses
sharp colour in solution form.
Q 296 What is self-indicator?
Ans: Self-indicators are nothing but self-indicating reagent. Self-
indicators have sharp colours in solution form. Generally,
these are oxidising agents, e.g. potassium permanganate.
Q 297 Name three self-indicators.
Ans: Potassium permanganate, iodine solution, cerium(IV)
sulphate.
Q 298 What is the drawback of self-indicators?
Ans: The drawback of self-indicators is, the oxidising agent like
potassium permanganate is generally present in excess at
end point.
This can be taken care by determining indicator blank which
is substracted from the burette reading to get the accurate
amount of the oxidising agent used in titration.
Q 299 What are adsorption indicators?
Ans: Adsorption indicators are used in precipitation titration.
Adsorption indicators are adsorbed by the precipitate formed
and changes to produce a new colour.
Q 300 Name three adsorption indicators.
Ans: Fluorescein, dichlorofluorescein, tetrabromo fluorescein, and
tartrazine are adsorption indicators.
CALCULATIONS
Q 301 How would you prepare 1000 ml of 25 ppm chloride solution
using sodium chloride?
Ans: We need to dissolve that amount of sodium chloride which
corresponds to 25 mg of chloride, in 1000 ml of water.
We know that 58.5 g of NaCl 35.5 g of chloride
We need to calculate how many mg of sodium chloride
correspond to 25 mg (0.025 g) of chloride.
35.5 g of chloride 58.5 g of NaCl
0.025 g of chloride ?
Fundamentals of Volumetric Analysis 59
=
0.025 58.5
35.5
= 0.025 × 1.65
= 0.04125 g of sodium chloride
We need to dissolve 41.25 mg of sodium chloride in
1000 ml of water to get 25 ppm chloride solution.
Q 302 How would you prepare 1000 ml of 25 ppm chloride solution
from given 1% sodium chloride solution?
Ans: We know from above answer that
0.04125 g of NaCl needs to be dissolved in 1000 ml to get
25 ppm chloride solution. We have 1% solution, i.e. we have
dissolved 10 g of NaCl in 1000 ml.
0.04125 g 25 ppm
10 g ?
10
0.4125
× 25 6060 ppm
25 parts of 1% sodium chloride solution should be mixed with
6035 parts of water to get 6060 parts of 25 ppm chloride
solution.
6060 ml of 25 ppm sodium chloride solution corresponds to
25 parts of 1% sodium chloride solution, so 1000 ml of 25
ppm sodium chloride solution corresponds to
=
1000 25
4.1254 parts
6060
We should mix 4.12 ml of 1% sodium chloride solution with
995.88 ml of water to get 1000 ml of 25 ppm chloride solution.
Q 303 How would you calculate normality of the unknown sample
solution?
Ans: At the equivalence point, the number of gram equivalents of
the titrant and the standard solution are equal.
N
1
V
1
= N
2
V
2
where,
N
1
= Normality of titrant (unknown)