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17
Volumetric Analysis
3. Neutralization of acidic materials: Sodium hydroxide can also be used
as a neutralizing agent in petroleum refining. It is sometimes used as a
cleaner.
Caution
1. Dissolution of sodium hydroxide is highly exothermic and the resulting
heat may cause heat burns or ignite flammables.
2. Solid sodium hydroxide or solutions containing high concentrations of
sodium hydroxide may cause chemical burns, permanent injury or scar-
ring, and blindness.
3. Sodium hydroxide is extremely caustic and can react with fats and oils on
skin, in a reaction which creates salts. For this reason, sodium hydroxide is
very dangerous, and skin should be washed thoroughly with water follow-
ing any contact with this substance.
Result The prepared sodium hydroxide solution was found to
be_______________
Experiment 2 To prepare and standardize XN hydrochloric acid.
Primary standard Sodium carbonate (Na
2
CO
3
) and sodium bicarbonate
(NaHCO
3
)
Indicator: Methyl orange or methyl red
Method
s Preparation: Pipette out 85X mL of hydrochloric acid and make the
volume 1 L to give XN solution of hydrochloric acid.
Example: To prepare 0.1 N solution (X = 0.1), weigh 85 u 0.1 mL or
8.5 mL of hydrochloric acid for 1000 mL.
s Standardization: Weigh accurately 200300 mg of sodium carbonate
and dissolve it in minimum amount of water. Add 23 drops of
methyl orange indicator and titrate it with hydrochloric acid solution
until permanent light pink colour appears.
End point: Appearance of permanent light pink colour or yellow orange
colour.
Reaction involved
2HCl Na CO 2NaCl
Hydrochloric acid
23
Sodium carbonate
Sodium
+⎯→⎯⎯⎯
cchloride
2
Carbon dioxide
2
Water
CO H O++
18
Practical Pharmaceutical Analytical Chemistry
Calculation
N
W 1000
EV
=
×
×
where N is the normality of given hydrochloric acid solution, W is the weight
of primary standard, E is the equivalent weight of primary standard and V is the
volume of hydrochloric acid consumed, i.e. burette reading.
Safety
1. Hydrochloric acid in high concentrations forms acidic mist. Both the mist
and the solution have a corrosive effect on human tissue with the potential
to damage respiratory organs, eyes, skin, and intestines.
2. Upon mixing hydrochloric acid with common oxidizing chemicals, such as
bleach (NaClO) or permanganate (KMnO
4
), the toxic gas chlorine is produced.
Result The prepared hydrochloric acid solution was found to
be_______________
Experiment 3 To prepare and standardize XM of sulphuric acid.
Primary standard Sodium carbonate (Na
2
CO
3
) and sodium bicarbonate
(NaHCO
3
)
Indicator Methyl orange or methyl red
Method
s Preparation: Pipette out 27X mL of sulphuric acid and make the volume
1 L to give XN solution of sulphuric acid.
Example: To prepare 0.1 N solution (X 0.1), weigh 27 u 0.1 mL or
2.7 mL of sulphuric acid for 1000 mL.
s Standardization: Weigh accurately 200300 mg of sodium carbonate and
dissolve it in minimum amount of water. Add 23 drops of methyl orange
indicator and titrate it with hydrochloric acid solution until permanent
light pink colour appears.
End point: Appearance of permanent light pink colour or yellow orange colour
Reaction involved
HSO NaCO NaSO
Sulphric acid Sodium carbonate Sodium su
24 2 3 24
+⎯→⎯⎯⎯
llphate
Carbon dioxide Water
CO H O++
22
Calculation
N
W
EV
q
q
1000
19
Volumetric Analysis
where N is the normality of given sulphuric acid solution, W is the weight of
primary standard, E is the equivalent weight of primary standard and V is the
volume of sulphuric acid consumed, i.e. burette reading.
Result The prepared sulphuric acid solution was found to
be_______________
Experiment 4 To find out g/100 mL of sodium carbonate in the given sample.
Principle: Acidimetric determination
Molecular formula: Na
2
CO
3
Molecular weight: 106
Indicator: Methyl orange (3.14.4)
IP limit: The sample contains not less than 99% and not more than 105% w/w
of sodium carbonate calculated with reference to the dried substance.
Procedure
Weigh accurately about 0.5 g of sample, dissolve it in 20 mL of carbon dioxide
free water and titrate it with 0.5 M hydrochloric acid using 0.2 mL of methyl
orange solution as an indicator.
or
Pipette out 10 mL of Na
2
CO
3
solution in a conical flask. Add 23 drops of
methyl orange indicator and titrate it with previously standardized hydrochloric
acid solution until a permanent pink colour is obtained.
Reaction involved
22
23
HCl Na CO NaCl
Hydrochloric acid
Sodium carbonate
Sodium
+⎯→⎯⎯⎯
cchloride
Carbon dioxide Water
CO H O++
22
THEORY Sodium carbonate, also known as washing soda or soda ash is a sodium
salt of carbonic acid. The analysis of sodium carbonate is an example of a strong
acid–base reaction. The given sodium carbonate reacts with hydrochloric acid to
give neutral salt sodium chloride.
Equivalent factor
2HCl 1 Na CO
2 1000 mL of 1N HCl 2 23 1 12 3 16 g of Na C
23
2
≡
×≡×+×+×
()()()
{}
OO
2 1000 mL of 1N HCl 46 12 48 g of Na CO
2 1000 mL of 1N HCl 10
3
23
×≡++
×≡
()
66gof Na CO
1000 mL of 1N HCl 106/2 53 g of Na CO
1mL of 1N HCl 53
23
23
≡=
≡
()
//1000 g of Na CO 0.053 g of Na CO
23 23
()
=
20
Practical Pharmaceutical Analytical Chemistry
Calculation
1 mL of 1 N HCl { 0.053 g of Na
2
CO
3
a mL of 1 N HCl { (0.053 u a) g of Na
2
CO
3
a mL of yN HCl { (0.053 u a u y) g of Na
2
CO
3
Since 10 mL of Na
2
CO
3
solution contains (0.053 u a u y) g of Na
2
CO
3
1 mL will contain {(0.053 u a u y)/10} g of Na
2
CO
3
100 mL will contain {(0.053 u a u y u 100)/10} g of Na
2
CO
3
So
g mL of Sodium carbonate
Equivalent factor Burette reading
M
/100
u
u
oolarity HCl
()
u 100
10
Pharmaceutical uses
1. It is a food additive (E500) used as an acidity regulator.
2. It is used as anticaking agent.
3. It is used as raising agent.
4. It is used as a stabilizer.
Category: Electrolyte replenisher; Systemic acidifier
Result The given sodium carbonate solution was found to contain _______
g/100 mL of sodium carbonate.
Experiment 5 To find out g/100 mL of sodium bicarbonate in the given
sample.
IP followed: 1996
Principle: Acidimetric titration
Molecular formula: NaHCO
3
Molecular weight: 84
Indicator: Methyl orange (3.14.4)
IP limit: The sample contains not less than 99% and not more than 101.0%
w/w of sodium bicarbonate calculated with reference to the dried substance.
Procedure
Weigh accurately about 1.5 g of the sample, dissolve it in 50 mL of carbon
dioxide free water and titrate it with 1 M hydrochloric acid using 0.2 mL of
methyl orange solution as an indicator.
21
Volumetric Analysis
or
Pipette out 10 m of NaHCO
3
solution in a conical flask. Add 23 drops of
methyl orange indicator and titrate it with previously standardized hydrochloric
acid solution until a permanent pink colour is obtained.
Reaction involved
HCl NaHCO NaCl
Hydrochloric acid
3
Sodium bicarbonate
Sodium c
+⎯→⎯⎯
hhloride
2
Carbon dioxide
2
Water
CO H O++
THEORY Sodium bicarbonate is a base and for any reaction or titration it requires
an acid, i.e. hydrochloric acid. The equivalence point of sodium bicarbonate is
around pH 3.6 and hence methyl orange the pH value of which is 2.8-4 can be
used as an indicator.
Equivalent factor
HCl { NaHCO
3
{ 1H
1000 mL of 1 N HCl { {(1 u 23) (1 u 1) (1 u 12) (3 u 16)} g of NaHCO
3
1000 mL of 1 N HCl { (23 1 12 48) g of NaHCO
3
1000 mL of 1 N HCl { 84 g of NaHCO
3
1 mL of 1 N HCl { (84/1000) g of NaHCO
3
0.084 g of NaHCO
3
Calculation
1 mL of 1 N HCl { 0.084 g of NaHCO
3
a mL of 1 N HCl { (0.084 u a) g of NaHCO
3
a mL of yN HCl { (0.084 u a u y) g of NaHCO
3
Since 10 mL of NaHCO
3
solution contains (0.084 u a u y) g of NaHCO
3
1 mL will contain {(0.084 u a u y)/10} g of NaHCO
3
100 mL will contain {(0.084 u a u y u 100)/10} g of NaHCO
3
So
g mL of Sodium bicarbonate
Equivalent factor Burette reading
/100
u
uuuNormality HCl
()
100
10
Pharmaceutical uses
1. The aqueous solution of sodium bicarbonate is used as an antacid to
treat acid indigestion and heartburn.
2. It is used in metabolic acidosis such as chronic renal failure and renal
tubular acidosis.
3. It is used in urinary alkalinization for the treatment of aspirin overdose
and uric acid renal stones.
22
Practical Pharmaceutical Analytical Chemistry
4. It is used in treatment of hyperkalemia.
5. It is used as ingredient in some mouthwashes.
6. It is used as an antiseptic.
Category: Electrolyte replenisher; Systemic alkalizer
Result The given sodium bicarbonate solution was found to contain _______
g/100 mL of sodium bicarbonate.
Experiment 6 To find out g/100 mL of ammonia in the given sample.
Principle: Acidimetric and alkalimetric determination (back titration or
indirect titration)
Molecular formula: NH
3
Molecular weight: 17.03
Indicator: Methyl red
IP limit: The sample contains not less than 24.5% w/w and not more than
25.5% w/w of ammonia (strong ammonia).
Procedure
Weigh 1.5 g of ammonia in a flask containing 25 mL of 1 N sulphuric acid and
titrate the excess of acid with previously standardized sodium hydroxide solution
using methyl red as an indicator. The colour change at the end point is from
yellow to red.
or
Pipette out 10 mL of ammonia solution in a conical flask. Add 25 mL of 1 N
sulphuric acid. Add 23 drops of methyl red indicator in it and titrate it with
previously standardized sodium hydroxide solution until the colour changes
from yellow to red colour.
Reaction involved
2
324 4
2
4
NH H SO NH SO
Ammonia Sulphuric acid
Ammonium sulpha
+⎯→⎯⎯⎯
()
tte
Sodium hydroxide
Sulphuric acid Sodiu
NaOH H SO Na SO2
24 24
+⎯→⎯⎯⎯
mm Sulphate
Water
HO+ 2
2
THEORY Due to the volatile nature of ammonia, the ammonia solution is assayed
by back titration in order to avoid any loss in the solution during titration. In
this method excess of sulphuric acid is used, which reacts with ammonia to form
23
Volumetric Analysis
a salt, i.e. ammonium sulphate. The unreacted excess of the sulphuric acid will
react with sodium hydroxide.
Equivalent factor
1H
2
SO
4
{ 2NH
3
{ 2H
1000 mL of 1 N H
2
SO
4
{ {(1 u 14) (3 u 1)} g of NH
3
1000 mL of 1 N H
2
SO
4
{ (14 3) g of NH
3
1000 mL of 1 N H
2
SO
4
{ 17 g of NH
3
1 mL of 1 N H
2
SO
4
{ (17/1000) g of NH
3
0.017 g of NH
3
Calculation
N
1
V
1
N
2
V
2
where N
1
is the normality of sulphuric acid, V
1
is the volume of sulphuric acid
reacted with sodium hydroxide, N
2
is the normality of sodium hydroxide and
V
2
is the volume of sodium hydroxide reacted with sulphuric acid, i.e. burette
reading.
Amount of sulphuric acid reacted with ammonia (a) 25 V
1
1 mL of 1 N H
2
SO
4
{ 0.017 g of ammonia
a mL of 1 N H
2
SO
4
{ (0.017 u a) g of ammonia
a mL of yN H
2
SO
4
{ (0.017 u a u y) g of ammonia
Since 10 mL of ammonia solution contains (0.017 u a u y) g of ammonia
1 mL will contain {(0.017 u a u y)/10} g of ammonia
100 mL will contain {(0.017 u a u y u 100)/10} g of ammonia
So
g mL of Ammonia
Equivalent factor Amount of
sulphuric acid rea
/100 =
×
ccted with ammonia
Normality H SO×
()
×
24
100
10
Result The given ammonia solution was found to contain ________
g/100 mL of ammonia.
Experiment 7 To find out g/100 mL of zinc oxide in the given sample.
IP followed: 1985
Principle: Acidimetric and alkalimetric determination (back titration or indi-
rect titration)
Molecular formula: ZnO
Molecular weight: 81.38
24
Practical Pharmaceutical Analytical Chemistry
Indicator:
Methyl orange (3.14.4)
IP limit: The sample contains not less than 99.0% and not more than 100.5%
of zinc oxide calculated with reference to the ignited substance.
Procedure Weigh 1.5 g of zinc oxide and dissolve it in a mixture of 2.5 g of
ammonium chloride and 50 mL of 1 N sulphuric acid solution. Add 23 drops
of methyl orange indicator and titrate it with previously standardized sodium
hydroxide solution until the colour changes from pink to orange colour.
or
Pipette out 10 mL of zinc oxide solution (by using ammonium chloride) in a
conical flask. Add 50 mL 1 N sulphuric acid. Add 23 drops of methyl orange
indicator in it and titrate it with previously standardized sodium hydroxide
solution until the colour changes from pink to orange colour.
Reaction involved
ZnO H SO ZnSO H O
Zinc oxide
Sulphuric acid Zinc Sulphate
W
+⎯→⎯⎯⎯ +
24 4 2
aater
Sodium hydroxide
Sulphuric acid Sod
NaOH H SO Na SO2
24 24
+⎯→⎯⎯⎯
iium Sulphate
Water
HO+ 2
2
THEORY A back titration is done in which excess of sulphuric acid is used. Sulphu-
ric acid reacts with zinc oxide to form a salt, i.e. zinc sulphate. Excess of sulphuric
acid will react with sodium hydroxide. Ammonium chloride is used to solubilize
zinc oxide which is soluble in dilute mineral acid or solution of alkali hydroxides.
Ammonium chloride is also responsible for the prevention of precipitation in zinc
hydroxide both during the titration and also at the end point. For titration of weak
base against strong acid, the equivalence point comes near 5.28 and methyl orange,
methyl red, bromophenol blue or bromocresol green are the choice of indicators.
Equivalent factor
1H
2
SO
4
{ 1 ZnO { 2H
2 u 1000 mL of 1 N H
2
SO
4
{ {(1 u 65.38) (1 u 16)} g of ZnO
2 u 1000 mL of 1 N H
2
SO
4
{ (65.38 16) g of ZnO
1000 mL of 1 N H
2
SO
4
{ 81.38/2 g of ZnO 40.69 g of ZnO
1 mL of 1 N H
2
SO
4
{ (40.69/1000) g of ZnO 0.04069 g of ZnO
Calculation
N
1
V
1
N
2
V
2
where N
1
is the normality of sulphuric acid, V
1
is the volume of sulphuric acid
reacted with sodium hydroxide, N
2
is the normality of sodium hydroxide and
V
2
is the volume of sodium hydroxide reacted with sulphuric acid, i.e. burette
reading.
25
Volumetric Analysis
Amount of sulphuric acid reacted with zinc oxide (a) 50 V
1
1 mL of 1 N H
2
SO
4
{ 0.04069 g of ZnO
a mL of 1 N H
2
SO
4
{ (0.04069 u a) g of ZnO
a mL of yN H
2
SO
4
{ (0.04069 u a u y) g of ZnO
Since 10 mL of ZnO solution contains (0.04069 u a u y) g of ZnO
1 mL will contain {(0.04069 u a u y)/10} g of ZnO
100 mL will contain {(0.04069 u a u y u 100)/10} g of ZnO
So
g /100 mL of zinc oxide
Equivalent factor Amount of sulphuric
acid
=
×
rreacted with zinc oxide Normality
H SO 100
10
24
×
×
()
Pharmaceutical uses
1. It is used as mild astringent.
2. It is used as protective against eczema.
3. It is used to reflect UV rays, so it used as sunscreens.
4. It is used as component in tape (called zinc oxide tape) used by athletes as
a bandage to prevent soft tissue damage during workouts.
Category: Mild astringent; Topical protectant
Result The given zinc oxide solution was found to contain ________
g/100 mL of zinc oxide.
Experiment 8 To find out g/100 mL of lactic acid in the given sample.
IP followed: 1996
Principle: Acidimetric and alkalimetric determination (back titration or indi-
rect titration)
Molecular formula: CH
3
CH(OH)COOH
Molecular weight: 90
Indicator: Phenolphthalein (8.310)
IP limit: The sample contains the equivalent of not less than 88.0% and not
more than 92.0% w/w of lactic acid.
Procedure
Weigh accurately 1 g of lactic acid and add 10 mL of water to it. Add 20 mL of 1 M
sodium hydroxide solution with it and stopper the flask. Allow this to stand for
30 min and then titrate the excess of sodium hydroxide with 1 M of hydrochloric
acid using phenolphthalein as an indicator until pink colour disappears.
26
Practical Pharmaceutical Analytical Chemistry
or
Pipette out 10 mL of lactic acid solution in a conical flask. Add 50 mL of 1 M
sodium hydroxide. Titrate the excess of sodium hydroxide with 1 M hydrochloric
acid using phenolphthalein as an indicator until pink colour disappears.
Reaction involved
CH CH OH COOH NaOH CH CH OH
Latic acid
Sodium
hydroxide
So
33
()
+⎯→⎯⎯⎯
()
ddium lactate
Water
Hydrochloric acid Sodium
h
y
COONa H O
HCl NaOH
+
+
2
ddroxide
Sodium chloride
Water
NaCl H O⎯→⎯⎯⎯ +
2
THEORY A back titration done in which excess sodium hydroxide is used.
Sodium hydroxide will react with lactic acid to form a salt, i.e. sodium lactate.
Unreacted sodium hydroxide will react with hydrochloric acid.
Lactic acid is a mixture of lactic acid and its condensation products
(lactoyl-lactic acid and dilactide), containing the equivalent of not less than
87.5% w/w of lactic acid; USP limits to 8590% w/w.
Boiling 1 N sodium hydroxide solution neutralizes the lactic acid
and hydrolyses the condensation products which are always present in the
commercial acid. Being a hydroxyl acid, lactic acid is stronger than acetic acid
and the dissociation constant 1.55 u 10
4
is about 10 times great.
Since sodium hydroxide is being titrated with hydrochloric acid solution, it would
appear that methyl orange might be a suitable indicator. But the choice of indicator
is always governed by the weakest acid present in the system. In this case lactic acid is
liberated at the end point and hence phenolphthalein is used as an indicator.
Lactic acid cannot be titrated directly because quantitative reaction proceeds
rapidly only in the presence of excess reagent. Hence, back titration is preferred,
in which excess of alkali is back titrated with the help of an acid.
Equivalent factor
1 NaOH { 1 CH
3
CH(OH)COOH
1000 mL of 1 N NaOH { {(3 u 12) (1 u 6) (3 u 16)} g of
CH
3
CH(OH)COOH
1000 mL of 1 N NaOH { (36 6 48) g of CH
3
CH(OH)COOH
1000 mL of 1 N NaOH { 90 g of CH
3
CH(OH)COOH
1 mL of 1 N NaOH { (90/1000) g of CH
3
CH(OH)COOH
0.090 g of CH
3
CH(OH)COOH
Calculation
N
1
V
1
N
2
V
2