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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5708_Библиотеки_им_академика_М_И_Перельмана.pdf
27
Volumetric Analysis
where N
1
is the normality of sodium hydroxide, V
1
is the volume of sodium
hydroxide reacted with hydrochloric acid, N
2
is the normality of hydrochloric
acid and V
2
is the volume of hydrochloric acid reacted with sodium hydroxide,
i.e. burette reading.
Amount of sodium hydroxide reacted with lactic acid (a) 50 V
1
1 mL of 1 N NaOH { 0.09 g of CH
3
CH(OH)COOH
a mL of 1 N NaOH { (0.09 u a) g of CH
3
CH(OH)COOH
a mL of yN NaOH { (0.09 u a u y) g of CH
3
CH(OH)COOH
Since 10 mL of lactic acid solution contains (0.09 u a u y) g of CH
3
CH(OH)
COOH
1 mL will contain {(0.09 u a u y)/10} g of CH
3
CH(OH)COOH
100 mL will contain {(0.09 u a u y u 100)/10} g of CH
3
CH(OH)COOH
So
g/ 100 mL of Lactic acid
Equivalent factor Amount of NaOH
reacted wi
=
×
tth Lactic acid Normality
H SO 100
10
24
×
×
()
Category: Pharmaceutical aid for sodium lactate injection
Result The given lactic acid solution was found to contain ________
g/100 mL of lactic acid.
Experiment 9 To find out g/100 mL of borax in the given sample.
Principle: Acidimetric determination
Molecular formula: Na
2
B
4
O
7
·10H
2
O
Molecular weight: 382
Indicator: Methyl red (4.26.3)
IP limit: The sample should not contain less than 99.0% and not more than
103.0% w/w of borax calculated with reference to the dried substance.
Procedure
Weigh 3 g of borax and dissolve it in 50 mL of water. Titrate it with previously
standardized 1 M hydrochloric acid using methyl red as an indicator. At the end
point a permanent red colour will appear.
or
Pipette out 10 mL of borax solution in a conical flask. Add 23 drops of methyl
red indicator and titrate it with previously standardized hydrochloric acid
solution until a permanent red colour is obtained.
28
Practical Pharmaceutical Analytical Chemistry
Reaction involved
2102
24 7 2
HCl Na B O H O NaCl
Hydrochloric acid
Borax
Sodium chl
+⋅⎯→⎯⎯⎯
ooride
Boric acid
Water
HBO HO++45
33 2
THEORY Borax, also known as sodium borate, sodium tetraborate or disodium
tetraborate, is a salt of boric acid. It is a direct acid-base reaction in which borax
react with hydrochloric acid to give a neutral salt, i.e. sodium chloride in addi-
tion to a weak acid—boric acid. Boric acid has a very low ionization constant
and has no effect on indicators.
Equivalent factor
2HCl { 1 Na
2
BO
4
O
7
·10H
2
O
2 u 1000 mL of 1 N HCl { {(2 u 23) (4 u 11) (7 u 16)
(10 u 18)} g of borax
2 u 1000 mL of 1 N HCl { (46 44 112 180) g of borax
2 u 1000 mL of 1 N HCl { 382 g of borax
1000 mL of 1 N HCl { (382/2 191) g of borax
1 mL of 1 N HCl { (191/1000) g of borax 0.191 g of borax
Calculation
1 mL of 1 N HCl 0.191 g of borax
a mL of 1 N HCl { (0.191 u a) g of borax
a mL of yN HCl { (0.191 u a u y) g of borax
Since 10 mL of borax solution contains (0.191 u a u y) g of borax
1 mL will contain {(0.191 u a u y)/10} g of borax
100 mL will contain {(0.191 u a u y u 100)/10} g of borax
So
g mL of boric acid
Equivalent factor Burette reading
Normali
/100
u
u
tty HCl
()
u 100
10
Pharmaceutical use
1. It is used externally in 12% solution as an eye wash.
2. It is used as an antiseptic and alkalizing agent.
3. It is used as an astringent for mucous membrane.
4. It is used as wet dressing for wounds.
Result The given borax solution was found to contain _________g/100 mL
of borax.
29
Volumetric Analysis
Experiment 10 To find out g/100 mL of boric acid in the given sample.
IP followed: 1996
Principle: Alkalimetric determination
Molecular formula: H
3
BO
3
Molecular weight: 61.81
Indicator: Phenolphthalein (8.310.0)
IP limit: The sample contains not less than 99.5% and not more than 100.5%
of boric acid calculated with reference to the dried substance.
Procedure
To a previously neutralized solution of glycerin (100 mL) and water (50 mL),
add 2 g of boric acid. Shake the mixture to dissolve (heat if necessary)
and titrate it with previously standardized 1 M of sodium hydroxide using
phenolphthalein as an indicator. At the end point, a permanent pink colour
will appear.
or
Pipette out 10 mL of boric acid solution in a conical flask. Add a mixture of
glycerin (2 mL) and water (15 mL). Add 23 drops of phenolphthalein indicator
and titrate it with previously standardized 1 M sodium hydroxide solution until
a permanent pink colour is obtained.
Reaction involved
Boric acidGlycerol
H
2
O
Water
H
3
BO
3
H
3
O
Hydronium ion
CH
2
OH
CHOH
CH
2
OH
HOC
B
COH
H
2
CO
OCH
2
CH
2
OHHOH
2
C
Glyceroboric acid
H BO NaOH NaBO
Boric acid
Sodium hydroxide
Sodium metabora
33 2
+⎯→⎯⎯⎯
tte Water
HO+ 2
2
THEORY Boric acid is a very weak acid and its dissociation constant is also very
less. In the presence of mannitol or glycerol the dissociation constant of boric
acid is increased owing to the formation of complex which is as strong as acetic
acid. Titration of the complex with phenolphthalein forms the basis of official
boric acid assay.
30
Practical Pharmaceutical Analytical Chemistry
Boric acid is esterified in the presence of polyhydric alcohols such as glycerol
and mannitol forming a monoprotic acid which is strong enough to give a
satisfactory end point. It is stated that glycerol is added to give glyceroboric acid.
At least 30% of glycerol or mannitol is needed to prevent hydrolysis of titratable
acid.
Mannitol and sorbitol are present in solid form. When they are added to
boric acid, they do not increase the volume of solution whereas glycerin does,
therefore, mannitol and sorbitols are preferred over glycerol.
Equivalent factor
1 NaOH { 1 H
3
BO
3
1000 mL of 1 N NaOH { {(3 u 1) (1 u 10.81) (3 u 16)}g of boric acid
1000 mL of 1 N NaOH { (3 10.81 48) g of boric acid
1000 mL of 1 N NaOH 61.81 g of boric acid
1 mL of 1 N NaOH { (61.81/1000) g of boric acid 0.06181 g of boric
acid
Calculation
1 mL of 1 N NaOH 0.06181 g of boric acid
a mL of 1 N NaOH { (0.06181 u a) g of boric acid
a mL of yN NaOH { (0.06181 u a u y) g of boric acid
Since 10 mL of boric acid solution contains (0.06181 u a u y) g of boric acid
1 mL will contain {(0.06181 u a u y)/10} g of boric acid
100 mL will contain {(0.06181 u a u y u 100)/10} g of boric acid
So
g mL of boric acid
Equivalent factor Burette reading
Normali
/100
u
u
tty HCl
()
u 100
10
Pharmaceutical use
1. It is used as a bacteriostatic and fungistatic.
2. It is used as an astringent.
3. It is used for it is antiseptic property.
4. It is externally used as buffer and antimicrobial in eye drops.
5. It is used as insecticide for cockroaches and black carpet bettle.
Category: Local antiinfective
Result The given boric acid solution was found to contain …………..
g/100 mL of boric acid.
31
Volumetric Analysis
Experiment 11 To find out g/100 mL of aspirin in the given sample
IP followed: 1996
Principle: Alkalimetric determination
Molecular formula:
OCOCH
3
COOH
Aspirin (Acetyl salicylic acid)
Molecular weight: 180
Indicator: Phenol red (6.88.4)
IP limit: It contains not less than 99.5% and not more than 100.5% of aspirin
calculated with reference to the dried substance.
Procedure
Accurately weigh 1.5 g and dissolve it in 15 mL of ethanol (95%) and further
add 50 mL of 0.5 N sodium hydroxide. Boil the solution on a water bath till it
starts bubbling. Cool and titrate the excess of alkali with 0.5 N hydrochloric
acid using phenol red as an indicator. Titrate it till yellow colour appears.
or
Pipette out 10 mL of aspirin solution in a conical flask. Add 50 mL of 0.5 N
sodium hydroxide solution and boil the solution on a water bath till it starts
bubbling. Cool and titrate the excess of alkali with 0.5 N hydrochloric acid
using phenol red as an indicator. Titrate it till yellow colour appears.
Reaction involved
OCOCH
3
COOH
Aspirin (acetyl
salicylic acid)
2NaOH
+
Sodium hydroxide
CH
3
COONa
Sodium acetate
OH
COONa
Sodium salicylate
+
Heat
NaOH excess HCl NaCl
Sodium Hydroxide
Hydrochloric acid So
()
+⎯→⎯⎯⎯
ddium Chloride
Water
HO+
2
THEORY Aspirin determination is an example of back titration. On boiling with
sodium hydroxide, aspirin leads to alkaline hydrolysis which causes formation
of sodium salts of salicylic acid and acetic acid. The excess sodium hydroxide is
reacted with hydrochloric acid.
32
Practical Pharmaceutical Analytical Chemistry
Equivalent factor
2 NaOH { 1 C
9
H
8
O
4
2 u 1000 mL of 1 N NaOH { {(9 u 12) (1 u 8) (4 u 16)} g of aspirin
2 u 1000 mL of 1 N NaOH { (108 8 64) g of aspirin
1000 mL of 1 N NaOH { (180/2) g of aspirin { 90 g of aspirin
1 mL of 1 N NaOH { (90/1000) g of aspirin { 0.09 g of aspirin
Calculation
1 mL of 1 N NaOH 0.09 g of aspirin
a mL of 1 N NaOH { (0.09 u a) g of aspirin
a mL of yN NaOH { (0.09 u a u y) g of aspirin
Since 10 mL of aspirin solution contains (0.09 u a u y) g of aspirin
1 mL will contain {(0.09 u a u y)/10} g of aspirin
100 mL will contain {(0.09 u a u y u 100)/10} g of aspirin
So
g mL of Aspirin
Equivalent factor Burette reading
Normality
/100
u
u
NNaOH
()
u 100
10
Pharmaceutical use
1. It is used as an analgesic.
2. It is used as an antipyretic.
Category: Analgesic; Antipyretic; Antirheumatic; Antithrombotic
Result The given aspirin solution was found to contain …………. g/100 mL
of aspirin.
Experiment 12 To find out g/100 mL of ammonium chloride in the given
sample.
IP followed: 1996
Principle: Alkalimetric determination
Molecular formula: NH
4
Cl
Molecular weight : 53.49
Indicator: Phenolphthalein (8.310.0)
IP limit: The sample contains not less than 99.0% and not more than 100.5%
of ammonium chloride calculated with reference to the dried substance.
33
Volumetric Analysis
Procedure
Weigh 0.1 g of ammonium chloride and dissolve it in 20 mL of water. To this
solution, add a mixture of formaldehyde (5 mL), previously neutralized with
dilute phenolphthalein solution and water (20 mL). Titrate it with previously
standardized 0.1 N sodium hydroxide using phenolphthalein as an indicator. At
the end point, a permanent pink colour will appear.
or
Pipette out 10 mL of ammonium chloride solution in a conical flask. Add
a mixture of formaldehyde (5 mL) and water (20 mL). Add 23 drops of
phenolphthalein indicator in it and titrate it with previously standardized sodium
hydroxide solution until a permanent pink colour is obtained.
Reaction involved
NH Cl H O NH OH H
4
Ammonium chloride
2
Water
4
Ammonium Hydroxide
++⎯→⎯⎯ CCl
NH OH HCHO
Hydrochloric acid
4
Ammonium hydroxide
Formaldehyde
+⎯⎯→⎯⎯ CH NH H O
CH NH
2
Methylenimine
2
Water
2
Methyleneimine
Polym
=+
=
eerizes
2
6
4
Hexamine
Hydrochloric acid Sodi
CH N
HCl NaOH
⎯→⎯⎯⎯⎯⎯
()
+
uum hydroxide Sodium chloride
2
Water
NaCl H O⎯→⎯⎯ +
THEORY Ammonium chloride is a salt of strong acid with a weak base, i.e. acidic
salt. It reacts with formaldehyde to liberate free acid which reacts further with
sodium hydroxide. Reaction of ammonium chloride with formaldehyde forms
methyleneimine which polymerizes to give hexamine.
Equivalent factor
1 NaOH { 1HCl { 1 N H
4
Cl
1000 mL of 1 N NaOH { {(1 u 14) (1 u 4) (1 u 35.49)} g of ammonium
chloride
2 u 1000 mL of 1 N NaOH { (14 4 35.49) g of ammonium chloride
1000 mL of 1 N NaOH { 53.49 g of ammonium chloride
1 mL of 1 N NaOH { (53.49/1000) g of ammonium chloride
{ 0.05349 g of ammonium chloride
Calculation
1 mL of 1 N NaOH 0.05349 g of ammonium chloride
a mL of 1 N NaOH { (0.05349 u a) g of ammonium chloride
a mL of yN NaOH { (0.05349 u a u y) g of ammonium chloride
Since 10 mL of ammonium chloride solution contains (0.05349 u a u y) g of
NH
4
Cl
34
Practical Pharmaceutical Analytical Chemistry
1 mL will contain {(0.05349 u a u y)/10} g of ammonium chloride
100 mL will contain {(0.05349 u a u y u 100)/10} g of ammonium chloride
So
g/100mL of ammonium chloride
Equivalent factor Burette reading
=
×
×××Normality NaOH 100
10
()
Pharmaceutical use
1. Ammonium ion plays an important role in the maintenance of acidbase
equilibrium of body, especially in acidosis. Thus, it acts as a system acidifying
agent. It is used in the treatment of severe metabolic alkalosis.
2. The diuretic effect of ammonium chloride is due to the formation of a
proton and a chloride ion, equivalent to hydrochloric acid. The hydrogen
ion reacts with the body buffers, mainly bicarbonate to form carbon diox-
ide and bicarbonates are displaced by chloride ion. It maintains urine at
an acid pH in the treatment of some urinary tract disorders. It potentiates
mercurial diuretics.
3. The expectorant action of ammonium chloride may be due to local irrita-
tion. The salt forms thin and increase the quantity of mucus.
4. It is used as mild astringent.
Category: Expectorant; Diuretic; Systemic acidifier
Result The given ammonium chloride solution was found to contain
…………. g/100 mL of ammonium chloride.
Experiment 13 To find out g/100 mL of ibuprofen in the given sample.
IP followed: 1996
Principle: Alkalimetric determination
Molecular formula: C
13
H
18
O
2
CH
3
H
COOH
Molecular weight: 206.28
35
Volumetric Analysis
Indicator:
Phenolphthalein (8.310.0)
IP limit: The sample contains not less than 98.5% and not more than 100.5%
of ibuprofen, calculated with reference to the dried substance.
Procedure
Weigh 0.4 g of ibuprofen and dissolve it in 100 mL of alcohol (95%). Titrate this
solution with previously standardized 0.1 N sodium hydroxide using phenol-
phthalein as an indicator. At the end point a permanent pink colour will appear.
Perform a blank titration and make any necessary correction.
or
Pipette out 10 mL of ibuprofen solution in a conical flask and add 25 mL of
ethanol. Add 23 drops of phenolphthalein indicator in it and titrate it with
previously standardized 0.1 N sodium hydroxide solution until a permanent
pink colour is obtained.
Reaction involved
Ibuprofen
{2-(4-Isobutylphenyl)propionic acid)
NaOH
Sodium hydroxide
H
2
O
Water
CH
3
H
COOH
CH
3
H
COONa
Sodium salt of Ibuprofen
{sodium 2-(4-Isobutylphenyl)propionate)}
THEORY Ibuprofen is a weak acid, derivative of acetic acid. Ibuprofen is practi-
cally insoluble in water, free soluble in ethanol (95%), chloroform, and acetone.
It ionizes completely in ethanolic solution and reacts with sodium hydroxide to
give sodium salt of ibuprofen.
Equivalent factor
1 NaOH { 1 C
13
H
18
O
2
1000 mL of 1 N NaOH { {(13 u 12) (18 u 1) (2 u 16)} g of ibuprofen
1000 mL of 1 N NaOH { (156 18 32) g of ibuprofen
1000 mL of 1 N NaOH { 206 g of ibuprofen
1 mL of 1 N NaOH { (206/1000) g of ibuprofen 0.206 g of ibuprofen
Calculation
1 mL of 1 N NaOH { 0.206 g of ibuprofen
a mL of 1 N NaOH { (0.206 u a) g of ibuprofen
a mL of yN NaOH { (0.206 u a u y) g of ibuprofen
Since 10 mL of ibuprofen solution contains (0.206 u a u y) g of ibuprofen
36
Practical Pharmaceutical Analytical Chemistry
1 mL will contain {(0.206 u a u y)/10} g of ibuprofen
100 mL will contain {(0.206 u a u y u 100)/10} g of ibuprofen
So
g mL of Ibuprofen
Equivalent factor Burette reading
Normali
/100
u
u
tty NaOH
()
u 100
10
Pharmaceutical use
1. It is used as analgesic
2. It is used as anti-inflammatory agent
Category: NSAID
Result The given ibuprofen solution was found to contain ______ g/100 mL
of ibuprofen.
Experiment 14 To find out g/100 mL of indomethacin in the given sample.
IP followed: 1996
Principle: Alkalimetric determination
Molecular formula: C
19
H
16
O
4
NCl
N
CH
3
O
CH
2
COOH
MeO
Cl
Molecular weight: 357.5
Indicator: Phenolphthalein (8.310.0)
IP limit: The sample contains not less than 98.0% and not more than 101.0%
of indomethacin, calculated with reference to the dried substance.
Procedure
Weigh 0.45 g of indomethacin and dissolve it in 75 mL of acetone. Titrate it
with 0.1 N sodium hydroxide solution using phenolphthalein as indicator.
At the end point a permanent pink colour will appear. Also, perform a blank