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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5466_Библиотеки_им_академика_М_И_Перельмана.pdf
107
Volumetric Analysis
and any Mg, which might be present as a contaminant, would not react. The
coloured Caindicator complex gives up Ca to EDTA, liberating the free
uncomplexed indicator, which is blue.
Calculation
M
1
u V
1
M
2
u V
2
where M
1
is the molarity of given disodium edetate solution, V
1
is the volume of
disodium edetate, i.e. burette reading, M
2
is the molarity of given zinc solution
and V
2
is the volume of zinc taken.
Result The prepared disodium edetate solution was found to be
_______________
Experiment 2 To find out g/100 mL of calcium carbonate (precipitated
chalk) in the given sample.
IP followed: 1996
Principle: Complexometric titration
Molecular formula: CaCO
3
Molecular weight: 100.09
Indicator: Calcon mixture
IP limit: It contains not less than 98.0% and not more than 100.5% of calcium
carbonate, calculated with reference to the dried substance.
Procedure
Weigh accurately 0.1 g of calcium carbonate and dissolve it in 3 mL of dilute
hydrochloric acid and 10 mL of water. Boil it for 10 min and cool, then dilute
it to 50 mL with water. Titrate with 0.05 M disodium edetate. After addition of
some amount of disodium edentate, add 8 mL of sodium hydroxide solution
and 0.1 g of calcon mixture and continue the titration until the colour of the
solution changes from pink to a full blue colour.
or
Pipette out 10 mL of calcium carbonate solution prepared in a mixture of
dilute hydrochloric acid and water. Boil it for 10 min. Cool, dilute it to 50 mL
with water. Titrate with 0.05 M disodium edetate. After addition of some volume
of disodium edetate, add 8 mL of sodium hydroxide solution and 0.1 g of calcon
mixture and continue the titration until the colour of the solution changes from
pink to a full blue colour.
108
Practical Pharmaceutical Analytical Chemistry
Reactions involved
2
32
HCl CaCO CaCl
Hydrochloric acid
Calcium carbanate
Calciu
+⎯→⎯⎯⎯
mm chloride Water Carbon dioxide
Calcium chloride
HO CO
CaCl I
++
+
22
2
nnCaIn
Ca In
Indicator
Calcium indicator complex
Calcium in
⎯→⎯⎯⎯ −
−
ddicator complex
Disodium e ate
CHNNa CHNCa+⎯→⎯⎯⎯ +
10 14 2 2 10 14 2
det
IIn
Indicator free
()
THEORY Calcium carbonate is practically insoluble in water and in alcohol. It dis-
solves with effervescence in dilute acetic acid, dilute hydrochloric acid and dilute
nitric acid. The indicator is added towards the end point and the solution is made
alkaline by the addition of sodium hydroxide solution. Calcon mixture gives pink
colour when it forms a complex with calcium carbonate when all the calcium car-
bonate reacted with disodium edetate the indicator (free form) gives blue colour.
Equivalent factor
1C
10
H
14
N
2
Na
2
{ 1CaCO
3
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ {(1 u 40) (1 u 12) (3 u 16)} g of
calcium carbonate
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ (40 12 48) g of calcium carbonate
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ 100 g of calcium carbonate
1 mL of 1 M C
10
H
14
N
2
Na
2
{ (100/1000) g of CaCO
3
0.1 g of
calcium carbonate
Calculation
1 mL of 1 M C
10
H
14
N
2
Na
2
0.1 g of calcium carbonate
a mL of 1 M C
10
H
14
N
2
Na
2
{ (0.1 u a) g of calcium carbonate
a mL of y M C
10
H
14
N
2
Na
2
{ (0.1 u a u y) g of calcium carbonate
Since 10 mL of CaCO
3
solution contains (0.1 u a u y) g of calcium carbonate
1 mL will contain (0.1 u a u y)/10) g of calcium carbonate
100 mL will contain (0.1 u a u y u 100)/(10) g of calcium carbonate
So
g mL of Calcium carbanate
Equivalent factor Burette reading
/100 =
×
×
MMolarity C H N Na
10 14 2 2
100
10
()
×
Pharmaceutical use
1. It is used as a dietary supplement of calcium.
2. It is used as an antacid.
3. It is used as phosphate binder for the treatment of hyperphosphatemia.
4. It is used as inert filler for tablets.
109
Volumetric Analysis
Category: Antacid; Pharmaceutical aid (excipient)
Result The given calcium carbonate solution was found to contain _______
g/100 mL of calcium carbonate.
Experiment 3 To find out g/100 mL of calcium gluconate in the given
sample.
IP followed: 1996
Principle: Complexometric titration
Molecular formula: C
12
H
22
CaO
14
·H
2
O
COO
COH
CH
2
OH
Ca
2
C
C
C
HO
OH
OH
H
H
H
H
Molecular weight: 448.40
Indicator: Mordant black II mixture
IP limit: It contains not less than 98.5% and not more than 102.0% of calcium
gluconate.
Procedure
Pipette out 10 mL of calcium gluconate solution and to it add 5 mL; 0.05 M
magnesium sulphate and 10 mL of strong ammoniaammonium chloride
solution and titrate with 0.05 M disodium edetate using mordant black II
mixture as an indicator until the colour of the solution changes from pink to
blue.
or
Weigh accurately 0.5 g of calcium gluconate and dissolve it in 50 mL of warm
water. Cool and add 5 mL; 0.05 M magnesium sulphate and 10 mL of strong
ammoniaammonium chloride solution and titrate with 0.05 M disodium
edetate using mordant black II mixture as an indicator until the colour of the
solution changes from pink to blue.
110
Practical Pharmaceutical Analytical Chemistry
Reactions involved:
Step I: When only magnesium is present and In is added
Mg In MgIn
Magnesium ion
Indicator
Complex Coloured
2+−
+⎯→⎯⎯⎯
()
Step II: Addition of Disodium Edetate (EDTA)
Ca H X CaX
Calcium ion
EDTA Calcium EDTA complex
2
2
22
+
−−
+
[]
⎯→⎯⎯⎯
[]
Step III: When all calcium is reacted then next drop of EDTA complexes
with magnesium
MgIn H X Mg X
Complex Coloured
EDTA Magnesium EDT
()
−−
+
[]
⎯→⎯⎯⎯ −
[]
2
22
AA Complex
Indicator
Second Colour
In+
−
()
THEORY This is a type of replacement titration. In this magnesium sulphate is
used to increase the sensitivity and thus gives a sharp end point.
In the beginning when no EDTA and calcium is present, magnesium forms
a complex with indicator. The indicator here shows one colour. Since, MgIn
complex is much more stable than CaIn complex, therefore, there is no change
occurring on addition of calcium to the reaction mixture. When EDTA is added
to it, it forms a complex with calcium. When all the calcium has been consumed,
the next drop of EDTA will break the complex of MgIn and complexes with
magnesium, i.e. thus freeing the indicator. Indicator at this point will show its
second colour which is the end point.
Equivalent factor
1C
10
H
14
N
2
Na
2
{ 1 C
12
H
22
CaO
14
·H
2
O
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ {(12 u 12) (1 u 22) (1 u 40) (16 u
14)18} g of C
12
H
22
CaO
14
·H
2
O
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ (144 22 40 224 18) g of calcium
gluconate
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ 448 g of calcium gluconate
1 mL of 1 M C
10
H
14
N
2
Na
2
{ (448/1000) g of C
12
H
22
CaO
14
·H
2
O 0.448 g
of calcium gluconate
Calculation
1 mL of 1 M C
10
H
14
N
2
Na
2
0.448 g of calcium gluconate
a mL of 1 M C
10
H
14
N
2
Na
2
{ (0.448 u a) g of calcium gluconate
a mL of y M C
10
H
14
N
2
Na
2
{ (0.448 u a u y) g of calcium gluconate
111
Volumetric Analysis
Since 10 mL of calcium gluconate solution contains (0.448 u a u y) g of
calcium gluconate
1 mL will contain (0.448 u a u y)/10) g of calcium gluconate
100 mL will contain (0.448 u a u y u 100)/(10) g of calcium gluconate
So
g/100mL of Calcium gluconate =
Equivalent factor Burette reading×
×
MMolarity C H N Na 100
10
10 14 2 2
()
×
Pharmaceutical use
1. It is used in the treatment of burns from HF as a gel or injection.
2. It is used in calcium deficiency.
3. It Is used as diuretic.
4. It is used as an anti-allergic agent.
5. It is used as a urinary acidifier.
6. It is used as dessicants for organic compounds.
Category: Calcium replenisher
Result The given calcium gluconate solution was found to contain _______
g/100 mL of calcium gluconate.
Experiment 4 To find out g/100 mL of calcium lactate in the given sample.
IP followed: 1996
Principle: Complexometric titration
Molecular formula: C
6
H
10
CaO
6
CH
3
CHCOO
OH
Ca
2
Molecular weight: 218.22 (anhydrous)
Indicator: Mordant black II mixture
IP limit: It contains not less than 98.0% and not more than 102.0% of calcium
lactate, calculated with reference to the dried substance.
112
Practical Pharmaceutical Analytical Chemistry
Procedure
Pipette out 10 mL of calcium lactate solution and to it add 5 mL; 0.05 M
magnesium sulphate and 10 mL of strong ammoniaammonium chloride
solution and titrate with 0.05 M disodium edetate using mordant black II
mixture as an indicator until the colour of the solution changes from pink to
blue.
or
Weigh accurately 0.5 g of calcium lactate and dissolve it in 50 mL of water.
Cool and add 5 mL; 0.05 M magnesium sulphate and 10 mL of strong ammonia
ammonium chloride solution and titrate with 0.05 M disodium edetate using
mordant black II mixture as an indicator until the colour of the solution changes
from pink to blue.
Reaction involved
Same as in experiment no. 4.
THEORY Same as in experiment no. 4.
Equivalent factor
1C
10
H
14
N
2
Na
2
{ 1 C
6
H
10
CaO
6
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ {(12 u 6) (1 u 10) (1 u 40) (16 u 6)}
g of C
6
H
10
CaO
6
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ (72 10 40 96) g of calcium lactate
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ 218 g of calcium lactate
1 mL of 1 M C
10
H
14
N
2
Na
2
{ (218/1000) g of C
6
H
10
CaO
6
0.218 g of
calcium lactate
Calculation
1 mL of 1 M C
10
H
14
N
2
Na
2
0.218 g of calcium lactate
a mL of 1 M C
10
H
14
N
2
Na
2
{ (0.218 u a) g of calcium lactate
a mL of y M C
10
H
14
N
2
Na
2
{ (0.218 u a u y) g of calcium lactate
Since 10 mL of calcium lactate solution contains (0.218 u a u y) g of calcium
lactate
1 mL will contain (0.218 u a u y)/10) g of calcium lactate
100 mL will contain (0.218 u a u y u 100)/(10) g of calcium lactate
So
g mL of Calcium lactate
Equivalent factor Burette reading
Mo
/100 =
u
u
llarity C H N Na
10 14 2 2
100
10
()
u
113
Volumetric Analysis
Pharmaceutical use
1. Used as antacid
2. Used in calcium deficiency
3. It is added to sugar free foods to prevent tooth decay
4. Used in remineralization of enamel
Category: Calcium replenisher
Result The given calcium lactate solution was found to contain _______
g/100 mL of calcium lactate.
Experiment 5 To find out g/100 mL of magnesium sulphate in the given
sample.
IP followed: 1996
Principle: Complexometric titration
Molecular formula: MgSO
4
·7H
2
O
Molecular weight: 246.47
Indicator: Mordant black II mixture
IP limit: It contains not less than 99.0% and not more than 100.5% of magne-
sium sulphate, calculated with reference to the dried substance.
Procedure
Pipette out 10 mL of magnesium sulphate solution and to it add 10 mL of strong
ammoniaammonium chloride solution and titrate with 0.05 M disodium
edetate using mordant black II mixture as an indicator until the colour of the
solution changes from pink to blue.
or
Weigh accurately 0.3 g of magnesium sulphate and dissolve it in 50 mL of
water. Add 10 mL of strong ammoniaammonium chloride solution and titrate
with 0.05 M disodium edetate using mordant black II mixture as an indicator
until the colour of the solution changes from pink to blue.
Reaction involved
Mg H X MgX
Magnesium ion
Disodium
++
−−−
+
[]
⎯→⎯⎯⎯
[]
2
edetate
Magnesiumm
edetate
complex
+
+
2H
114
Practical Pharmaceutical Analytical Chemistry
T
HEORY Magnesium sulphate is freely soluble in water and sparingly soluble in
alcohol. The indicator is added towards the end point and the pH of the solution
is kept at 10 by the addition of ammonia-ammonia chloride buffer. Mordant
black II mixture gives pink colour when they form complex with magnesium
sulphate when all the magnesium sulphate reacted with disodium edetate the
indicator (free form) gives blue colour.
Equivalent factor
1C
10
H
14
N
2
Na
2
{ 1 M gSO
4
·7H
2
O
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ {(1 u 24) (1 u 32) (4 u 16)
(7 u 18)} g of MgSO
4
·7H
2
O
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ (24 40 32 64 126) g of
magnesium sulphate
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ 246 g of magnesium sulphate
1 m L o f 1 M C
10
H
14
N
2
Na
2
{ (246/1000) g of MgSO
4
·7H
2
O 0.246 g
of MgSO
4
·7H
2
O
Calculation
1 mL of 1 M C
10
H
14
N
2
Na
2
0.246 g of magnesium sulphate
a mL of 1 M C
10
H
14
N
2
Na
2
{ (0.246 u a) g of magnesium sulphate
a mL of y M C
10
H
14
N
2
Na
2
{ (0.246 u a u y) g of magnesium sulphate
Since 10 mL of magnesium sulphate solution contains (0.246 u a u y) g of
magnesium sulphate
1 mL will contain (0.246 u a u y)/10) g of magnesium sulphate
100 mL will contain (0.246 u a u y u 100)/(10) g of magnesium sulphate
So
g mL of Magnesium Sulfate
Equivalent factor Burette reading
/100 =
×
×
MMolarity C H N Na
10 14 2 2
100
10
()
×
Pharmaceutical use
1. It is used in the treatment of hypomagnesemia.
2. It is used as a saline laxative.
3. Epsom salt is also available in a gel form for topical application in treating
aches and pains.
4. It is used to treat eclampsia in pregnant women.
5. It acts as a first aid for barium chloride poisoning.
6. Its paste has been used as agent for dehydrating boils, carbuncles and abscesses.
Category: Osmotic laxative; used in the treatment of electrolyte deficiency
115
Volumetric Analysis
Result The given magnesium sulphate solution was found to contain
_______ g/100 mL of magnesium sulphate.
Experiment 5 To find out g/100 mL of zinc sulphate in the given sample.
IP followed: 1996
Principle: Complexometric titration
Molecular formula: ZnSO
4
·7H
2
O
Molecular weight: 287.54
Indicator: Xylenol orange
IP limit: It contains not less than 99.0% and not more than 104.0% of zinc
sulphate, calculated with reference to the dried substance.
Procedure
Pipette out 10 mL of zinc sulphate solution in acetic acid. To it add 50 mg of
xylenol orange triturate and sufficient hexamine to produce violet-pink colour.
Add a further 2 g of hexamine and titrate with 0.05 M disodium edetate until the
colour of the solution changes to yellow.
or
Weigh accurately 0.5 g of zinc sulphate and dissolve it in 5 mL of 2 M acetic
acid and dilute it to 50 mL with water. To the solution add 50 mg of xylenol
orange triturate and sufficient hexamine to produce violet-pink colour. Add
a further 2 g of hexamine and titrate with 0.05 M disodium edetate until the
colour of the solution changes to yellow.
Reaction involved
Zn H X ZnX
Zincion
Disodiume ate Zince ate com
++
−−
+
[]
⎯→⎯⎯⎯
[]
2
det det pplex
H+
+
2
THEORY This is an example of direct titration. Initially zinc reacts with indicator
and at the end point free indicator gives its colour.
The hydrated form was historically known as ‘White vitriol’.
Equivalent factor
1C
10
H
14
N
2
Na
2
{ 1ZnSO
4
·7H
2
O
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ {(1 u 65) (1 u 32) (4 u 16)
(7 u 18)} g of ZnSO
4
·7H
2
O
116
Practical Pharmaceutical Analytical Chemistry
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ (65 32 64 126) g of zinc
sulphate
1000 mL of 1 M C
10
H
14
N
2
Na
2
{ 287 g of zinc sulphate
mL of 1 M C
10
H
14
N
2
Na
2
{ (287/1000) g of ZnSO
4
·7H
2
O
0.287 g of ZnSO
4
·7H
2
O
Calculation
1 mL of 1 M C
10
H
14
N
2
Na
2
0.287 g of zinc sulphate
a mL of 1 M C
10
H
14
N
2
Na
2
{ (0.287 u a) g of zinc sulphate
a mL of y M C
10
H
14
N
2
Na
2
{ (0.287 u a u y) g of zinc sulphate
Since 10 mL of zinc sulphate solution contains (0.287 u a u y) g of zinc
sulphate
1 mL will contain (0.287 u a u y)/10) g of zinc sulphate
100 mL will contain (0.287 u a u y u 100)/(10) g of zinc sulphate
So
g mL of Zinc sulfate
Equivalent factor Burette reading
Molar
/100 =
u
u
iity C H N Na
10 14 2 2
100
10
()
u
Pharmaceutical use
1. It is used as an astringent.
2. It s uaed as an emetic.
Category: Astringent
Result The given zinc sulphate solution was found to contain _______
g/100 mL of zinc sulphate.
2.6 Gravimetric Titration
Gravimetry is a type of quantitative analysis in which the constituents to be
determined is separated from all other components in the form of an insoluble
precipitate. This is the oldest and quite accurate among available quantitative
analysis measurement though tedious and time consuming.
Precipitation is a process which gives separable solid within liquid used to
separate solid in an aqueous solution.
Gravimetric involves the separation of the constituents (to be estimated) in
the form of insoluble compound of known composition.