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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5986_Библиотеки_им_академика_М_И_Перельмана.pdf
57
Volumetric Analysis
IP limit:
Hydrogen peroxide solution (20 Vol) contains not less than 5% w/v
and not more than 7% w/v of hydrogen peroxide, corresponding to about 20
times its volume of available oxygen.
Procedure
Dilute 10 mL to 250 mL with water. To 25 mL of the dilution add 5 mL of 5 N
sulphuric acid and titrate with 0.1 N potassium permanganate to a permanent
pink colour.
Reaction involved
523
22 4 2 4
H O KMnO H SO
Hydrogen peroxide
Potassium permagnate Sulp
++
hhuric acid
Potassium sulphate Manganese sulph
K SO MnSO⎯→⎯⎯⎯ +
24 4
2
aate
Oxygen Water
OHO++58
22
THEORY The assay is based on oxidation-reduction titration. Sulphuric acid is used
as nascent oxygen generator. Hydrogen peroxide can act both as an oxidizing agent
as well as a reducing agent. If it is titrated with a stronger oxidizing agent like potas-
sium permanganate in solution acidified with sulphuric acid it gets oxidized to
water and oxygen. This forms the basis of the assay of hydrogen peroxide solution.
A fairly high concentration is used to reduce the danger of formation of manganese
dioxide which is an active catalyst for the decomposition of hydrogen peroxide.
Equivalent factor
1 mL of 0.1 N KMnO
4
{ 0.0001701 g of H
2
O
2
Calculation
1 mL of 0.1 N KMnO
4
{0.0001701 g of H
2
O
2
a mL of 0.1 N KMnO
4
{ (0.0001701 u a) g of H
2
O
2
a mL of yN KMnO
4
{ (0.0001701 u a u y) g of H
2
O
2
Since 10 mL of H
2
O
2
solution contains (0.0001701 u a u y) g of H
2
O
2
1 mL will contain {0.0001701 u a u y)/10} g of H
2
O
2
100 mL will contain {0.0001701 u a u y u 100)/10} g of H
2
O
2
So
g mL of Hydrogen peroxide
Equivalent factor Burette reading
/100 =
×
×
NNormality KMnO
4
100
10
()
×
Category: Antiseptic; Deodorant
Result The given hydrogen peroxide solution was found to contain _______
g/100 mL of hydrogen peroxide.
58
Practical Pharmaceutical Analytical Chemistry
Experiment 9 To find out g/100 mL of ferrous sulphate in the given sample.
Principle: Oxidationreduction titration
Molecular formula: FeSO
4
·7H
2
O
Molecular weight: 278
Indicator: Potassium permanganate (Self indicator)
IP limit: It contains not less than 98% and not more than 105% of ferrous
sulphate.
Procedure
Weigh accurately about 0.5 g of ferrous sulphate. Dissolve it in a mixture of
25 mL of dilute sulphuric acid and 25 mL of freshly boiled and cooled water.
Titrate the solution with previously standardized 0.02 M potassium permanga-
nate solution till a pink colour is obtained.
or
Pipette out 10 mL of given ferrous sulphate solution. To this add a mixture of
25 mL of dilute sulphuric acid and 25 mL of freshly boiled and cooled water.
Titrate the solution with previously standardized potassium permanganate
solution till a pink colour is obtained.
Reaction involved
10 2 8
4424
FeSO KMnO H SO
Ferrous Sulphate Potassium permanganate S
++
uulphuric acid
Potassium Sulphate Manganese su
K SO MnSO⎯→⎯⎯⎯ +
24 4
2
llphate
Ferric sulphate
Water
Fe SO H O+
()
+58
24
3
2
THEORY The assay is based on oxidation-reduction titration. Sulphuric acid is
used as an acidifying agent. Potassium permanganate oxidizes the divalent fer-
rous sulphate to trivalent ferric sulphate, i.e. Fe
2
(SO
4
)
3
. At the end point a slight
excess of potassium permanganate imparts pink colour.
Equivalent factor
Fe Fe e
Ferrous Ferric Electron
23++
⎯→⎯⎯⎯ +
KMnO
4
{ FeSO
4
·7H
2
O { 1e
1000 mL of 1 N KMnO
4
{ {(1 56) (1 u 32) (4 u 16) (7 u 18)} g of ferrous
sulphate
1000 mL of 1 N KMnO
4
{ (56 32 64 126) g of ferrous sulphate
59
Volumetric Analysis
1000 mL of 1 N KMnO
4
{ 278 g of ferrous sulphate
1 mL of 1 N KMnO
4
{(278/1000) g of ferrous sulphate 0.278 g of ferrous
sulphate
Calculation
1 mL of 1 N KMnO
4
{ 0.278 g of ferrous sulphate
a mL of 1 N KMnO
4
{ (0.278 u a) g of ferrous sulphate
a mL of yN KMnO
4
{ (0.278 u a u y) g of ferrous sulphate
Since 10 mL of ferrous sulphate solution contains (0.278 u a u y) g of ferrous
sulphate
1 mL will contain {0.278 u a u y)/10} g of ferrous sulphate
100 mL will contain {0.278 u a u y u 100)/10} g of ferrous sulphate
So
g mL of Ferrous Sulphate
Equivalent factor Burette reading
N
/100 =
×
×
oormality KMnO
4
100
10
()
×
Pharmaceutical use
1. Haematinic agent is used in the treatment of iron deficiency, e.g. Anaemia.
Category: Haematinic
Result The given ferrous sulphate solution was found to contain _______
g/100 mL of ferrous sulphate.
Experiment 10 To find out g/100 mL of ascorbic acid in the given sample.
IP followed: 1996
Principle: Oxidationreduction titration (iodimetric titration)
Molecular formula
O
O
HOH
2
C
OH
OH
HO
60
Practical Pharmaceutical Analytical Chemistry
Molecular weight:
176
Indicator: Starch solution
IP limit: It contains not less than 99.0% and not more than 100.5% of ascorbic
acid
Procedure
Weigh 0.1 g (100 mg) of ascorbic acid. To it add a mixture of freshly boiled
and cooled water and 25 mL, 1 M sulphuric acid. Titrate it with 0.05 M iodine
solution using starch as an indicator. At the end point violet colour will appear.
or
Pipette out 10 mL of ascorbic acid solution. To it add a mixture of freshly boiled
and cooled water and 25 mL of 1 M sulphuric acid. Titrate it with 0.05 M iodine
solution using starch as an indicator. At the end point violet colour will appear.
Reaction involved
O
O
HOH
2
C
OH
OH
HO
Ascorbic acid
I
2
Iodine
O
O
HOH
2
C
OH
O
O
Dehydroascorbic acid
2HI
Hydrogeniodide
THEORY The assay is based on oxidation-reduction titration. Sulphuric acid is
used as an acidifying agent. Iodine is a strong oxidizing agent. It oxidizes the
ascorbic acid to dehydro-ascorbic acid and itself reduces to hydrogen iodide.
During this reaction, water is boiled to remove carbon dioxide. This prevents
oxidation of ascorbic acid by carbonic acid.
L-form of ascorbic acid is pharmaceutically active.
Equivalent factor
1C
6
H
8
O
6
{ 2I
{ 1I
2
{2H
1C
6
H
8
O
6
{ 1I
2
1000 mL of 1 M I
2
{ {(6 u 12) (1 u 8) (6 u 16)} g of ascorbic acid
1000 mL of 1 M I
2
{ (72 8 96) g of ascorbic acid
1000 mL of 1 M I
2
{ 176 g of ascorbic acid
1 mL of 1 M I
2
{ (176/1000) g of ascorbic acid 0.176 g of ascorbic acid
Calculation
1 mL of 1 M I
2
0.176 g of ascorbic acid
a mL of 1 M I
2
{ (0.176 u a) g of ascorbic acid
a mL of yM I
2
{ (0.176 u a u y) g of ascorbic acid
61
Volumetric Analysis
Since 10 mL of ascorbic acid solution contains (0.176 u a u y) g of ascorbic
acid
1 mL will contain {0.176 u a u y)/10} g of ascorbic acid
100 mL will contain {0.176 u a u y u 100)/10} g of ascorbic acid
So,
g mL of Ascorboc acid
Equivalent factor Burette reading
Mola
/100 =
×
×
rrity I
2
100
10
()
×
Pharmaceutical use
1. Vitamin C is used in the treatment of scurvy.
2. It is used as an antioxidant.
Category: Antiscorbutic (Vitamin); Pharmaceutical aid (antioxidant)
Result The given ascorbic acid solution was found to contain _______
g/100 mL of ascorbic acid.
Experiment 11 To find out g/100 mL of potassium iodide in the given
sample.
IP followed: 1996
Principle: Oxidationreduction titration (Iodometric titration)
Molecular formula: KI
Molecular weight: 166
Indicator: Chloroform (Distribution indicator)
IP limit: It contains not less than 99.0% and not more than 100.5% of potas-
sium iodide, calculated with reference to the dried substance.
Procedure
Weigh accurately 0.35 g of potassium iodide and dissolve it in 10 mL of water. To
it add excess of hydrochloric acid (35 mL; !4 M) and 5 mL of chloroform. Titrate
it with 0.05 M potassium iodate till brown colour disappears in the chloroform
layer.
62
Practical Pharmaceutical Analytical Chemistry
or
Pipette out 10 mL of potassium iodide solution. To it add excess of hydrochloric
acid (35 mL; !4 M) and 5 mL chloroform. Titrate it with 0.05 M potassium iodate
till brown colour disappears in the chloroform layer.
Reaction involved
KIO KI HCl
Potassium iodide
Potassium iodide
Hydrochloric aci
3
26++
dd
Potassium chloride Iodine monochloride
Wa
KCl ICl H O⎯→⎯⎯⎯ + +333
2
tter
THEORY In assay of potassium iodide, during the titration, the yellow colour
of the solution changes to dark brown. This colour is due to the liberation of
iodine. As the titration proceeds, the solution becomes lighter in colour due
to the formation of iodine monochloride. In the presence of iodine, chloro-
form layer becomes brown and with the formation of brown colour disappears
slowly.
Another choice of indicator is amaranth. When the colour of the solution
becomes light yellow, 1 mL of amaranth is added. The titration is continued until
pink colour changes to pale yellow. Amaranth which is an azo dye is oxidized by
excess titrant at the end point to a colourless derivative.
In case of chloroform, vigorous shaking is necessary as the end point is
approached to ensure equilibrium between the indicator and titration solution.
The procedure can be modified in the presence of cyanide ions when
iodine monochloride is converted to iodine monocyanide which is more stable
and less susceptible than the former to hydrolyze. This was initially used but
although effective, is no longer acceptable because of the volatility and extreme
toxicity of hydrogen cyanide released.
Equivalent factor
1KIO
3
{ 2KI
1000 mL of 1 M KIO
3
{ 2 u {(1 u 39) (1 u 127)} g of potassium iodide
1000 mL of 1 M KIO
3
{ 2 u (39 127) g of potassium iodide
1000 mL of 1 M KIO
3
{ 2 u 166 g of potassium iodide
1 mL of 1 M KIO
3
{ (332/1000) g of potassium iodide 0.332 g of
potassium iodide
Calculation
1 mL of 1 M KIO
3
0.332 g of potassium iodide
a mL of 1 M KIO
3
{ (0.332 u a) g of potassium iodide
a mL of yM KIO
3
{ (0.332 u a u y) g of potassium iodide
63
Volumetric Analysis
Since 10 mL of potassium iodide solution contains (0.332 u a u y) g of
potassium iodide
1 mL will contain {(0.332 u a u y)/10} g of potassium iodide
100 mL will contain {(0.332 u a u y u 100)/10} g of potassium iodide
So
g mL of Potassium Iodide
Equivalent factor Burette reading
M
/100
u
u
oolarity KIO
3
100
10
()
u
Pharmaceutical use
1. It is used for prophylaxis and it is an antifungal
2. It is used as an antitussive and an expectorant
3. It is necessary for the formation of thyroid hormone.
4. It is used as a source of iodine in preoperative management of hyperthy-
roidism.
5. It is used in treatment of iodine deficiency disorders.
Category: Antithyroid; Antifungal; Expectorant
Result The given potassium iodide solution was found to contain _______
g/100 mL of potassium iodide.
Experiment 12 To find out g/100 mL of dimercaprol in the given sample.
IP followed: 1996
Principle: Oxidationreduction titration
Molecular formula
CH
2
SH
CHSH
CH
2
OH
Molecular weight: 124
Indicator: Starch mucilage
IP limit: It contains not less than 98.5% w/w and not more than 101.5% w/w
of dimercaprol.
Procedure
Weigh 0.1 g of dimercaprol and dissolve it in 40 mL of methanol and add 20 mL
of 0.1 N hydrochloric acid and 50 mL of 0.05 M iodine. Allow to stand for
10 min and titrate with 0.1 M sodium thiosulphate solution. Also perform a blank
titration. Difference in two titrations give the amount of iodine reacting with
dimercaprol.
or
64
Practical Pharmaceutical Analytical Chemistry
Pipette out 10 mL of dimercaprol solution. To it add 20 mL of 0.1 N
hydrochloric acid and 50 mL of 0.05 M iodine. Allow to stand for 10 min and
titrate it with 0.1 M sodium thiosulphate solution. Also perform a blank titration.
Difference in two titrations give the amount of iodine reacting with dimercaprol.
Reaction involved
CH
2
SH
CHSH
CH
2
OH CH
2
OH CH
2
OH
2I
2
Dimercaprol
Iodine Hydrogen iodide
4HI
CH
2
S
CHS
SCH
2
SCH
2
THEORY Determination depends upon the oxidation of the thiol groups with
iodine. The reaction is not reversible.
Equivalent factor
2I
2
{ 4I { 2 C
3
H
8
OS
2
1 C
3
H
8
OS
2
{ 2I
2 u 1000 mL of 1 N iodine
{ {(3 u 12) (8 u 1) (1 u 16) (2 u 32)}
g of dimercaprol
2 u 1000 mL of 1 N iodine
{ {36 8 16 64}g of dimercaprol
1000 mL of 1 N iodine
{ (124/2) g of dimercaprol 62 g of dimercaprol
1 mL of 1 N iodine { (62/1000) g of dimercaprol 0.062 g of
dimercaprol
Since for iodine 1 N 0.5 M
So
1 mL of 0.5 M iodine { 0.062 g of dimercaprol
1 mL of 1 M iodine { (0.062 u 2) 0.124 g of dimercaprol
Calculation
1 mL of 1 M iodine 0.124 g of dimercaprol
a mL of 1 M iodine { (0.124 u a) g of dimercaprol
a mL of yM iodine { (0.124 u a u y) g of dimercaprol
Since 10 mL of dimercaprol solution contains (0.124 u a u y) g of dimercaprol
1 mL will contain {(0.124 u a u y)/10}g of dimercaprol
100 mL will contain {(0.124 u a u y u 100)/10}g of dimercaprol
So
gm mL of Dimercaprol
Equivalent factor Burette reading
Mola
/100 =
u
u
rrity I
2
100
10
()
u
65
Volumetric Analysis
Category: Antidote in heavy metal poisoning; Metal complexing agent.
Result The given dimercaprol solution was found to contain _______
g/100 mL of dimercaprol.
Experiment 13 To find out g/100 mL of isoniazid in the given sample.
IP followed: 1996
Principle: Redox titration
Molecular formula: C
6
H
7
N
3
O isoniazid
N
CONHNH
2
Molecular weight: 137
Indicator: Methyl red
IP limit: It contains not less than 98.0% and not more than 101.0% of
C
6
H
7
N
3
O calculated with reference to the dried substance.
Procedure
Weigh accurately 0.25 g of isoniazid and dissolve it in water to produce 100 mL.
Pipette out 20 mL of the solution. To it add 100 mL water, 20 mL of hydrochloric
acid and 0.2 g of potassium bromide. Titrate the resulting solution with continuous
shaking with 0.0167 M potassium bromate solution using methyl red solution
until the red colour disappears.
or
Pipette out 10 mL of isoniazid solution. To it add 20 mL of water, 4 mL of
hydrochloric acid and 0.1 g of potassium iodide. Titrate the resulting solution
with continuous shaking with previously standardized potassium bromate
solution using methyl red solution until the red colour disappears.
Reaction involved
2KBrO
Isoniazid Potassium bromate
3H
2
O
N
CONHNH
2KBr
Isonicotinic acid
N
COOH
H
Potassium bromide
33
3N
2
WaterNitrogen
3
2
66
Practical Pharmaceutical Analytical Chemistry
THEORY Potassium bromate, an oxidizing agent causes the oxidation of isoniazid
into isonicotinic acid. At the end point excess of potassium bromate reacts with
potassium bromide to give bromine which decolourizes the methyl orange.
Equivalent factor
2KBrO
3
{ 3C
6
H
7
N
3
O
2 u 1000 mL of 1 M KBrO
3
{ 3 u {(6 u 12) (7 u 1) (3 u 14) (1 u 16)}
g of isoniazid
2 u 1000 mL of 1 M KBrO
3
{ 3 u {72 7 42 16}g of isoniazid 137 u 3 g
of isoniazid
2 u 1000 mL of 1 M KBrO
3
{ 411 g of isoniazid
1000 mL of 1 M KBrO
3
{ (411/2) g of isoniazid 205.5 g of isoniazid
1 mL of 1 N 1 M KBrO
3
{ (205.5/1000) g of isoniazid 0.2055 g of
isoniazid
Calculation
1 mL of 1 M KBrO
3
0.2055 g of isoniazid
a mL of 1 M KBrO
3
{ (0.2055 u a) g of isoniazid
a mL of yM KBrO
3
{ (0.2055 u a u y) g of isoniazid
Since 10 mL of isoniazid solution contains (0.2055 u a u y) g of isoniazid
1 mL will contain {(0.2055 u a u y)/10}g of isoniazid
100 mL will contain {(0.2055 u a u y u 100)/10}g of isoniazid
So
g mL of Isoniazid
Equivalent factor Burette reading
Molarit
/100 =
u
u
yyKBrO
3
100
10
()
u
Pharmaceutical use
1. It is used in treatment of tuberculosis
Category: Antitubercular
Result The given isoniazid solution was found to contain _______ g/100 mL
of isoniazid.
Experiment 14 To find out g/100 mL of metamizole (analgin) in the given
sample.
IP followed: 1996
Principle: Iodometric titration
Molecular formula: C
13
H
17
O
4
N
3
NaS