Добавил:
Sekretar
kiopkiopkiop18@yandex.ru
t.me/Prokururor I Вовсе не секретарь, но почту проверяю
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз:
Предмет:
Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5708_Библиотеки_им_академика_М_И_Перельмана.pdf
77
Volumetric Analysis
s Standardization: Pipette out 30 mL of 0.1 M silver nitrate solution in
a conical flask. Dilute to 50 mL with water, 2 mL; 2 M nitric acid and
2 mL of ferric alum (as an indicator) and titrate it with ammonium
thiocyanate solution until the colour becomes reddish yellow.
End point: Appearance of reddish yellow colour.
Reaction involved
AgNO + NH SCN AgSCN
3
Silver nitrate
4
Ammonium thiocyanate
Silve
⎯→⎯⎯
rr thiocyanate
White ppt.
43
Ammonium nitrate
4
Ammoni
+NHNO
NH SCN
()
uumthiocyanate
44
2
Ferric ammonium sulphate
+Fe NH SO Fe SC
()()
⎯→⎯⎯ NN+2NHSO
3
Ferric thiocyanate
Brick red colour
4
2
4
Ammonium s
() ()
()
uulphate
Calculation
M
1
u V
1
M
2
u V
2
where M
1
is the molarity of the given ammonium thiocyanate solution,V
1
is
the volume of ammonium thiocyanate, i.e. burette reading, M
2
is the molarity of
given silver nitrate solution and V
2
is the volume of silver nitrate taken.
Result The prepared silver nitrate solution was found to be
Experiment 3 To find out g/100 mL of sodium chloride in the given sample
by Mohr’s method.
Principle: Precipitation titration
Molecular formula: NaCl
Molecular weight: 58.5
Indicator: Potassium chromate (for Mohr’s method)
IP limit: It contains not less than 99.0% and not more than 100.5% of sodium
chloride, calculated with reference to the dried substance.
Procedure
Weigh accurately 1 g of sodium chloride and dissolve it in 100 mL of distilled
water. Pipette out 10 mL of solution and add two drops of potassium chromate as
an indicator. The solution was titrated with previously standardized silver nitrate
solution. At the end point brick red colour will be obtained.
or
Pipette out 10 mL of sodium chloride solution. The solution was titrated with
previously standardized silver nitrate solution using potassium chromate as an
indicator. At the end point brick red colour will be obtained.
78
Practical Pharmaceutical Analytical Chemistry
Reaction involved
AgNO NaCl AgCl
Silver nitrate
Sodium chloride
Silver chlori
3
+⎯→⎯⎯⎯
dde
White ppt
Sodium nitrate
Siver nitrate
P
NaNO
AgNO K CrO
.
()
+
+
3
324
ootassium chromate
Silver chromate
Brick red colour
Ag CrO⎯→⎯⎯⎯
(
24
))
+ 2
3
KNO
Potassium nitrate
THEORY Sodium chloride is a colourless crystal or white crystalline powder,
freely soluble in water. In this method, sodium chloride reacts with silver nitrate
directly to give white precipitate of silver chloride. At the end point excess of
silver nitrate reacts with potassium chromate to give brick red precipitate of
silver chromate.
Equivalent factor
1AgNO
3
{ 1 NaCl
1000 mL of 1 M AgNO
3
{ {(1 u 23) (1 u 35.5)} g of sodium chloride
1000 mL of 1 M AgNO
3
{ (23 35.5) g of sodium chloride
1000 mL of 1 M AgNO
3
{ 58.5 g of sodium chloride
1 mL of 1 M AgNO
3
{ (58.5/1000) g of NaCl 0.0585 g of sodium
chloride
Calculation
1 mL of 1 M AgNO
3
0.0585 g of sodium chloride
a mL of 1 M AgNO
3
{ (0.0585 u a) g of sodium chloride
a mL of y M AgNO
3
{ (0.0585 u a u y) g of sodium chloride
Since 10 mL of NaCl solution contains (0.0585 u a u y) g of sodium chloride
1 mL will contain (0.0585 u a u y)/10) g of sodium chloride
100 mL will contain (0.0585 u a u y u 100)/(10) g of sodium chloride
So
g mL of sodium chloride
Equivalent factor Burette reading
Mo
/100
u
u
llarity AgNO
3
100
10
()
u
Pharmaceutical use 0.9% w/v solution is isotonic with blood hence can be
used as a saline solution.
Category: Tonicity agent; Fluid and electrolyte replinisher
Result The given sodium chloride solution was found to contain
g/100 mL of sodium chloride.
79
Volumetric Analysis
Experiment 4 To find out g/100 mL of potassium iodide in the given sample
by Mohr’s method.
Principle: Precipitation titration
Molecular formula: KI
Molecular weight: 166
Indicator: Iodide free starch solution
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 0.3 g of potassium iodide and dissolve it in 10 mL of distilled
water. Add iodide free starch solution (10 mL, containing one drop of ethanolic
iodine solution in 50 mL) and titrate with previously standardized silver nitrate
solution until the blue colour is dispersed and a pale yellow solution remains.
or
Pipette out 10 mL of potassium iodide solution. The solution was titrated with
previously standardized silver nitrate solution using iodide free starch solution
(10 mL, containing one drop of ethanolic iodine solution in 50 mL) as an indicator.
At the end point blue colour is dispersed and a pale yellow solution remains.
Reaction involved
AgNO KI AgI K
Silver nitrate
Potassium iodide
Silver iodide
3
+⎯→⎯⎯⎯ + NNO
Potassium nitrate
3
THEORY Bromides and iodides cannot be titrated directly with silver nitrate by
Mohr’s method, since the colour of the precipitate interferes with the end point.
Potassium iodide can, however, be determined by direct titration with silver
nitrate using iodine and starch as an indicator. The titration is simply a double
decomposition. At the end point excess silver nitrate reduces the indicator,
iodine to, iodide ion, destroying the purple starch-iodide complex and leaving a
pale yellow solution.
Equivalent factor
1AgNO
3
{ 1KI
1000 mL of 1 M AgNO
3
{ {(1 u 39) (1 u 127)} g of potassium iodide
1000 mL of 1 M AgNO
3
{ (39 127) g of potassium iodide
1000 mL of 1 M AgNO
3
{ 166 g of potassium iodide
1 mL of 1 M AgNO
3
{ (166/1000) g of KI 0.166 g of potassium iodide
80
Practical Pharmaceutical Analytical Chemistry
Calculation
1 mL of 1 M AgNO
3
0.166 g of potassium iodide
a mL of 1 M AgNO
3
{ (0.166 u a) g of potassium iodide
a mL of y M AgNO
3
{ (0.166 u a u y) g of potassium iodide
Since 10 mL of KI solution contains (0.166 u a u y) g of potassium iodide
1 mL will contain {0.166 u a u y)/10} g of potassium iodide
100 mL will contain (0.166 u a u 100 u y)/10 g of potassium iodide
So
g mL of Potassium chloride
Equivalent factor Burette reading
/100 =
×
××
()
×Molarity AgNO
3
100
10
Pharmaceutical use
1. It is used as 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 source of iodine in preoperative management of hyperthy-
roidism.
5. It is used for prophylaxis and 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 5 To find out g/100 mL of sodium chloride in the given sample
by modified Volhard’s method.
IP followed: 1996
Principle: Precipitation titration
Molecular formula: NaCl
Molecular weight: 58.5
Indicator: Ferric ammonium sulphate (for modified Volhard’s method).
IP limit: It contains not less than 99.0% and not more than 100.5% of sodium
chloride, calculated with reference to the dried substance.
81
Volumetric Analysis
Procedure
Weigh accurately 0.1 g of sodium chloride and dissolve it in 50 mL of distilled water
followed by the addition of excess silver nitrate (50 mL; 0.1 M), nitric acid (5 mL;
2 M) and 2 mL of nitrobenzene or dibutylphthalate. Shake the contents and titrate
with previously standardized 0.1 M ammonium thiocyanate solution using ferric
ammonium sulphate as an indicator. At the end point brick red colour will be obtained.
or
Pipette out 10 mL of sodium chloride solution in a conical flask followed by
the addition of excess silver nitrate (50 mL; 0.1 M), nitric acid (5 mL; 2 M) and
2 mL of nitrobenzene or dibutylphthalate. Shake the contents and titrate with
previously standardized ammonium thiocyanate solution using ferric ammonium
sulphate as an indicator. At the end point brick red colour will be obtained.
THEORY This is the most common method used to determine halides/halogen
content in acid solution. In this method, halides are completely precipitated
as silver halide by excess of silver nitrate solution and determining amount of
unreacted silver nitrate solution by residual titration with standard ammonium
thiocyanate solution using ferric alum as an indicator.
The precipitate of silver chloride is separated by filtration or coated by some
organic solvents like nitrobenzene or butane.
Uses of nitrobenzene
Precipitate of silver chloride is slightly soluble and therefore it shall also react
with ammonium thiocyanate. Nitrobenzene forms a coating on the surface of
precipitated silver chloride and thus makes it completely insoluble and prevents
it from reacting with ammonium thiocyanate.
Dibutyl phthalate is preferred because nitrobenzene is carcinogenic in nature.
The precipitated silver chloride can also be removed by simple filtration
method to avoid reaction with ammonium thiocyanate. However, this method is
usually not followed because of the loss of mother liquor.
Uses of nitric acid
It is used to prevent hydrolysis of indicator, i.e. Fe(NH
4
)(SO
4
)
2
. It is also used to
provide common ion effect.
Reactions involved
AgNO NaCl AgCl
Silver nitrate
Sodium chloride
Silver chlori
3
+⎯→⎯⎯⎯
dde
White ppt
Sodium nitrate
Silver nitrate excess
NaNO
AgNO
.
()
(
+
3
3
))
+⎯→⎯⎯⎯NH SCN AgSCN
Fe NH
Ammonium thiocyanate
Silver thiocynate
4
444
2
4
3
()()
+SO NH SCN
Ferric ammonium sulphate
Ammonium thiocyanatee
Ferric thiocyanate
Ammonium sulphat
Fe SCN NH SO⎯→⎯⎯⎯
()
+
()
3
4
2
4
2
ee
82
Practical Pharmaceutical Analytical Chemistry
Equivalent factor
1AgNO
3
{ 1 NaCl
1000 mL of 1 M AgNO
3
{ (1 u 23) (1 u 35.5)} g of sodium chloride
1000 mL of 1 M AgNO
3
{ (23 35.5) g of sodium chloride
1000 mL of 1 M AgNO
3
{ 58.5 g of sodium chloride
1 mL of 1 M AgNO
3
{ (58.5/1000) g of NaCl 0.0585 g of sodium chloride
Calculation
1 mL of 1 M AgNO
3
0.0585 g of sodium chloride
1 mL of y M AgNO
3
{ (0.0585 u y) g of sodium chloride
Volume of silver nitrate taken 50 mL; y M
Since, M
1
V
1
M
2
V
2
where M
1
is the molarity of ammonium thiocyanate, M
2
is the molarity of silver
nitrate and V
1
is the amount of ammonium thiocyanate reacting with excess
silver nitrate, i.e. burette reading and V
2
is the amount of silver nitrate reacting
with ammonium thiocyanate.
Therefore,
Amount of silver nitrate reacting with sodium chloride (a) 50 V
2
Since
1 mL of y M AgNO
3
{ (0.0585 u y) g of sodium chloride
a mL of y M AgNO
3
{ (0.0585 u y u a) g of sodium chloride
Since 10 mL of NaCl solution contains (0.0585 u y u a) g of NaCl
1 mL will contain [(0.0585 u y u a)/10] g of sodium chloride
100 mL will contain [(0.0585 u y u a u 100}/10] g of NaCl
So
g mL of Sodium chloride
Equivalent factor Burette reading
/
-
100
50
u
(()
()
uuMolarity AgNO
3
100
10
Pharmaceutical use
0.9% w/v solution is isotonic with blood hence it a can be used as saline solution.
Category: Tonicity agent; Fluid and electrolyte replinisher
Result The given sodium chloride solution was found to contain
g/100 mL of sodium chloride.
83
Volumetric Analysis
Experiment 6 To find out g/100 mL of ammonium chloride in the given
sample by modified Volhard’s method.
Principle: Precipitation titration
Molecular formula: NH
4
Cl
Molecular weight: 53.5
Indicator: Ferric ammonium sulphate (for modified Volhard’s method)
IP limit: It contains not less than 99.5% and not more than 100.5% of ammo-
nium chloride, calculated with reference to the dried substance.
Procedure
Pipette out 10 mL of ammonium chloride solution in a conical flask followed by
the addition of excess silver nitrate (50 mL; 0.1 M), nitric acid (5 mL; 2 M) and
2 mL of nitrobenzene or dibutylphthalate. Shake the contents and titrate with
previously standardized ammonium thiocyanate solution using ferric ammonium
sulphate as an indicator. At the end point brick red colour will be obtained.
or
Weigh accurately 200 mg of ammonium chloride and dissolve it in 40 mL of
distilled water followed by the addition of excess silver nitrate (50 mL; 0.1 M),
nitric acid (3 mL 2 M) and 5 mL of nitrobenzene or dibutylphthalate. Shake
the contents and titrate with previously standardized ammonium thiocyanate
solution using 2 mL ferric ammonium sulphate as an indicator. At the end point
brick red colour will be obtained.
THEORY Same as in experiment number 5.
Reactions involved
AgNO NH Cl AgCl
3
Silver
nitrate
4
Ammonium
chloride
Silver
chlo
+⎯→⎯⎯
rride
White ppt.
43
Ammonium
nitrate
3
Silver
nitrate Ex
NH NO
AgNO
()
+
ccess
4
Ammonium
thiocyanate
Silver
thiocyanat
NH SCN AgSCN
()
⎯→⎯⎯+
ee
44
2
Ferric
ammonium
sulphate
4
Ammonium
thioc
Fe NH SO 3NH SCN
()()
+
yyanate
3
Ferric
thiocyanate
Bric redcolour
4
Fe SCN 2 NH⎯→⎯⎯
() ()
()
+
22
4
Ammonium
sulphate
SO
Equivalent factor
1AgNO
3
{ 1 NH
4
Cl
84
Practical Pharmaceutical Analytical Chemistry
1000 mL of 1 M AgNO
3
{ {(1 u 14) (1 u 4) (1 u 35.5)} g of ammonium
chloride
1000 mL of 1 M AgNO
3
{ (14 4 35.5) g of ammonium chloride
1000 mL of 1 M AgNO
3
{ 53.5 g of ammonium chloride
1 mL of 1 M AgNO
3
{ (53.5/1000) g of ammonium chloride
0.0535 g of ammonium chloride
Calculation
1 mL of 1 M AgNO
3
0.05349 g of ammonium chloride
1 mL of y M AgNO
3
{ (0.05349 u y) g of ammonium chloride
Volume of silver nitrate taken 50 mL; y M
Since M
1
V
1
M
2
V
2
where M
1
is the molarity of ammonium thiocyanate, M
2
is the molarity of silver
nitrate, V
1
is the amount of ammonium thiocyanate reacting with excess silver
nitrate, i.e. burette reading and V
2
is the amount of silver nitrate reacting with
ammonium thiocyanate.
Therefore,
Amount of silver nitrate reacting with ammonium chloride 50 V
2
Since
1 mL of y M AgNO
3
{ (0.05349 u y) g of ammonium chloride
(50 V
2
) mL of y M AgNO
3
{ (0.05349 u y) u (50 V
2
) g of ammonium
chloride
Since 10 mL of NH
4
Cl solution contains (0.05349 u y) u (50 V
2
) g of
ammonium chloride
1 mL will contain [{(0.05349 u y) u (50 V
2
)}/10] g of ammonium
chloride
100 mL will contain [{(0.05349 u y) u (50 V
2
) u 100}/10] g of NH
4
Cl
So
g mL of Ammonium chloride
Equivalent factor Burette readi
/100
50
=
u
nng
Molarity AgNO
()
()
uu
3
100
10
Pharmaceutical use
1. Ammonium ion plays an important role in the maintenance of acid
base equilibrium of the body, especially in acidosis. Thus, it acts as a
system acidifying agent. It is used in the treatment of severe metabolic
alkalosis.
85
Volumetric Analysis
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 dioxide
and bicarbonate which 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 layer and increase the quantity of mucus.
4. It is used as an mild astringent.
Category: Expectorant; Diuretic and systemic acidifier
Result The given ammonium chloride solution was found to contain
_______ g/100 mL of ammonium chloride.
Experiment 7 To find out g/100 mL of chlorbutol in the given sample by
modified Volhard’s method.
IP followed: 1996
Principle: Precipitation titration
Molecular formula: CCl
3
·C(CH
3
)
2
·OH, ½H
2
O
Molecular weight: 186.5
Indicator: Ferric ammonium sulphate (for modified Volhard’s method)
IP limit: It contains not less than 98% and not more than 101% of C
4
H
7
Cl
3
O,
calculated with reference to anhydrous substance.
Procedure
Weigh accurately 0.2 g and dissolve it in 5 mL of ethanol (95%). Add 5 mL of
sodium hydroxide solution and boil under a reflux condenser for 15 min. Cool,
dilute with 20 mL of water, add 5 mL of nitric acid, 1 mL of nitrobenzene or
dibutylphthalate and 50 mL of 0.1 M silver nitrate and shake vigorously for
1 min. Add 4 mL of ferric ammonium sulphate solution and titrate the excess
of silver nitrate with previously standardized 0.1 M ammonium thiocyanate
solution. At the end point brick red colour will be obtained.
or
refluxing with NaOH) in a conical flask followed by the addition of excess
86
Practical Pharmaceutical Analytical Chemistry
silver nitrate (50 mL; 0.1 M), nitric acid (5 mL; 2 M) and 2 mL of nitrobenzene
or dibutylphthalate. Shake the contents and titrate with previously standardized
ammonium thiocyanate solution using ferric ammonium sulphate as an indicator.
At the end point brick red colour will be obtained.
THEORY Organically, combined chlorine is converted by the hydrolysis with
sodium hydroxide to, ionic chloride which reacts with silver nitrate to give silver
chloride. The excess of silver nitrate reacts further with ammonium thiocyanate.
At the end point ferric alum reacts with ammonium thiocyanate to give brick
red colour.
Here ethanolic solution is used for the solubility of chlorbutol as it is freely
soluble in ethanol (95%) and slightly soluble in water.
Reaction involved
CCl C CH OH NaOH C OH C
Chlorobutol
Sodium
hydroxide
33
23
3
()
+⎯→⎯⎯⎯
()
.CCH OH NaCl
AgNO NaCl
Sodium
chloride
Silver nitrate
Sod
3
2
3
3
33
()
+
+
.
iium chloride
Silver chloride
White ppt
Sodi
AgCl NaNO⎯→⎯⎯⎯ +
()
33
3
.
uum
nitrate
Silver
nitrate Excess
Ammonium
thiocya
AgNO NH SCN
34
()
+
nnate
Slver thiocyanate
Ferric
ammonium
AgSCN
Fe NH SO
⎯→⎯⎯⎯
()()
44
2
ssulphate
Ammonium
thiocyanate
Ferric thi
NH SCN Fe SCN+⎯→⎯⎯⎯
()
3
4
3
oocyanate
Brick colour
Ammonium
sulphate
NH SO
()
+
()
22
44
Equivalent factor
3AgNO
3
{ 3NaCl { C
4
H
7
OCl
3
, ½H
2
O
3 u 1000 mL of 1 M AgNO
3
{ {(4 u 12) (1 u 7) (1 u 16) (3 u 35.5)
½ u 18} g of chlorbutol
3000 mL of 1 M AgNO
3
{ (48 7 16 106.5 9) g of chlorbutol
3000 mL of 1 M AgNO
3
{ 186.5 g of chlorbutol
1 mL of 1 M AgNO
3
{ (186.5/3000) g of chlorbutol
0.06216 g of chlorbutol
Calculation
1 mL of 1 M AgNO
3
0.06216 g of chlorbutol
1 mL of y M AgNO
3
{ (0.06216 u y) g of chlorbutol