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47
Volumetric Analysis
Experiment 1 To prepare and standardize 0.05 M iodine.
Primary standard Arsenic trioxide (As
2
O
3
)
Indicator: Starch solution
Method:
Preparation: Weigh accurately 14 g of iodine in an analytical balance. Add in
it 36 g of potassium iodide and 3 mL of hydrochloric acid. Dissolve it in mini-
mum amount of water and make up the volume to 1 L to give 0.05 M iodine
solution.
Note: Because iodine is practically insoluble in water, potassium iodide is
added to it which forms a complex with iodine, i.e. polyiodide which tends to
get solubilize in water. KI
3
behaves in solution as free iodine.
KI I KI
Potassium iodide
Iodine
Polyiodide
+⎯→⎯⎯⎯3
23
x Standardization: Accurately weigh 0.15 g of arsenic trioxide. To it add
20 mL sodium hydroxide solution and 40 mL of water. Add 0.1 mL of
methyl orange indicator and add hydrochloric acid drop-wise till the
colour changes to pink. To it add 2 g of sodium bicarbonate till the pink
colour disappears and add 50 mL of water to it. Titrate it with the iodine
solution. Towards the end point of titration add starch solution. Appearance
of violet colour indicates the end point.
End point: Appearance of violet colour.
Reaction involved
As O H O As O
Arsenic trioxide
Water
Arsenic pentaoxide
23 2 25
2+⎯→⎯⎯⎯ + 444He
Hydrogen ion electrons
+−
+
Ie I
iodine
electrons
Iodide
2
222+⎯→⎯⎯⎯
⎤
⎦
⎥
×
−−
As O H O I As O
Arsenic trioxide
Water Iodine
Arsenic pe
23 2 2 25
2++⎯→⎯⎯⎯
nntaoxide
Hydrogen ion Iodide
HI++
+−
44
NaHCO HI NaI
Sodium bicarbonate
Hydrogen iodide
Sodium iodi
3
+⎯→⎯⎯⎯
dde
Carbon dioxide Water
CO H O++
22
Hydrogen iodide is formed in the reaction. Because of strong reducing
properties of HI, the oxidation with iodine is reversible. The reaction is made
48
Practical Pharmaceutical Analytical Chemistry
to go completely to the right, i.e. to As
2
O
5
by removal of HI with NaHCO
3
.
NaOH or Na
2
CO
3
cannot be used to remove HI because they react with iodine
as follows:
63 5
23
NaOH I NaI NaIO
Sodium Hydroxide
Iodine
Sodium iodide
S
+⎯→⎯⎯⎯ +
oodium iodate
Water
HO+ 3
2
33 5
23 2
Na CO I NaI NaIO
Sodium carbanate
Iodine
Sodium iodide
+⎯→⎯⎯⎯ +
332
3
Sodium iodate
Carbon dioxide
CO+
In standardization, 20 mL of NaOH was added to give an alkaline condition
and thus dissolves As
2
O
3
. If NaOH would be present in the system then the
above reaction would take place and more of iodine is consumed resulting in
an error. Thus, NaOH is neutralized using hydrochloric acid and to check the
acidity methyl orange is added.
Calculation
N (W u 1000)/(E u V)
where N is the normality of given Iodine solution, W is the weight of primary
standard, i.e. As
2
O
3
, E is the equivalent weight of primary standard, i.e. As
2
O
3
and V is the volume of iodine consumed, i.e. burette reading.
Result The prepared iodine solution was found to be_______________
Experiment 2 To prepare and standardize xM of potassium iodate.
Primary standard Potassium iodide (KI)
Indicator: Starch solution or chloroform (example of distribution indicator)
Method
x Preparation: Weigh accurately 214x g of potassium iodate in an analytical
balance. Dissolve it in minimum amount of water and make up the volume
to 1 L to give xM of potassium iodate solution.
x Standardization: Because potassium iodate can be purchased in a high state
of purity (99.9%), a standard solution may be prepared by direct weighing.
However, solutions may be standardized by one of the following method:-
To a definite volume of potassium iodate solution, add excess of potassium
iodide and dilute hydrochloric acid. Titrate the liberated iodine with standard
sodium thiosulphate solution using starch solution as an indicator. Appearance of
violet colour indicates the end point.
49
Volumetric Analysis
or
Weigh accurately 350 mg of potassium iodide and dissolve it in 10 mL of
water. To it add excess of concentrated hydrochloric acid (35 mL) and 5 mL
chloroform. Titrate it with 0.05 M potassium iodate till brown colour disappears
in chloroform layer.
End point: Appearance of violet colour or disappearance of brown colour in
chloroform layer.
Reaction involved
KI HCl HI
Potassium iodide
Hydrochloric acid
Hydrogen iodide
+⎯→⎯⎯⎯ ++
+
KCl
KIO HI
Potassium chloride
Potassium Iodate
Hydrogen iodide
3
⎯⎯→⎯⎯⎯ + +33
2
22
22 3
IKIHO
Na S O
Iodine
Potassium iodide
Water
Sodium thhiosulphate
Iodine
Sodium iodide
INaSONaI+⎯→⎯⎯⎯ +
2246
2
or
KIO KI KCl
Potassium Iodide
Potassium Iodide
Potassium ch
3
23+⎯→⎯⎯⎯
lloride Iodine mono chloride
Water
ICl
HO
+
+
3
3
2
Calculation
N (W u 1000)/(E u V)
where N is the normality of given potassium iodate solution, W is the weight of
primary standard, i.e. KI, E is the equivalent weight of primary standard, i.e. KI
and V is the volume of potassium iodate consumed, i.e. burette reading.
Result The prepared potassium iodate solution was found to be
_______________
Experiment 3 To prepare and standardize 0.1 M of cerric ammonium nitrate.
Primary standard Arsenic trioxide (As
2
O
3
)
Indicator: Ferroin sulphate
Method
x Preparation: Take 56 mL of sulphuric acid and add 54.82 g of cerric
ammonium nitrate. Shake the contents for 2 min and add 100 mL of water
for five times. Shake the contents and finally make the volume to 1 L to get
0.1 M of cerric ammonium nitrate solution.
50
Practical Pharmaceutical Analytical Chemistry
x Standardization: Take 200 mg of arsenic trioxide and dry it at 105qC for
1 h. Add 25 mL of 8% w/v sodium hydroxide solution followed by the
addition of 100 mL water, 30 mL dilute sulphuric acid and 0.15 mL of
osmic acid. The mixture is titrated with cerric ammonium nitrate using
ferroin sulphate as an indicator. At the end point purple colour becomes
pale blue.
End point: Appearance of pale blue colour.
Reaction involved
As O NaOH NaAsO
Arsenic trioxide
Sodium hydroxide
Sodium a
23 2
+⎯→⎯⎯⎯
rrsenite
Sodium arsenite Water
Sodium a
NaAsO H O NaH AsO
22 24
2+⎯→⎯⎯⎯
rrsenate
He
Ce e Ce
++
×+
⎡
⎣
⎢
⎯→⎯⎯⎯
⎤
⎦
⎥
+−
+− +
24
4
43
Calculation
N (W u 1000)/(E u V)
where N is the normality of the given cerric ammonium nitrate solution, W is
the weight of primary standard, i.e. As
2
O
3
, E the equivalent weight of primary
standard, i.e. As
2
O
3
and V is the volume of cerric ammonium nitrate consumed,
i.e. burette reading.
Result The prepared cerric ammonium nitrate solution was found to
be__________
Experiment 4 To prepare and standardize 0.1 M of sodium thiosulphate.
Primary standard (i) Potassium dichromate (K
2
Cr
2
O
7
) and (ii) Potassium
iodate (KIO
3
)
Indicator: Starch solution
Method
s Preparation: Weigh 24.8 g of sodium thiosulphate (Na
2
S
2
O
3
·5H
2
O)
and dissolve it in 200 mL of freshly boiled and cooled water. Shake
the contents for 2 min and add remaining water to make the volume
to 1 L to get 0.1 M of sodium thiosulphate (Na
2
S
2
O
3
·5H
2
O) solution.
s Standardization:
1. With potassium dichromate: Weigh accurately 1.2 g of potassium
dichromate and dissolve it in 250 mL of distilled water. From the
solution, pipette out 25 mL and transfer it to an iodine flask. Dilute
51
Volumetric Analysis
the solution with 50 mL of water; add 2 g of potassium iodide
and 5 mL of concentrated hydrochloric acid. Stopper the flask
and allow it to stand for 5 min, preferably in dark. Dilute it with
50 mL of water and titrate it with sodium thiosulphate solution
until the colour becomes light yellow. Further to it add 1 mL of
starch solution. The solution becomes blue in colour. Continue the
titration until the blue colour disappears and green colour appears.
2. With potassium iodate: Weigh accurately 0.130.15 g of pure
dry potassium iodate and dissolve it in 25 mL of freshly boiled
and cooled water, add 2 g of iodatefree potassium iodide and
5 mL of 1 M sulphuric acid. Titrate the liberated iodine with
the thiosulphate solution with continuous shaking. Dilute it to
100 mL of water and add 1 mL of starch solution. The solution
becomes blue in colour. Continue the titration until the blue
colour disappears.
End point: Green colour (with K
2
Cr
2
O
7
) or colourless (with KIO
3
).
Reaction involved
SO H HSO S
23 3
−+ −
+⎯ →⎯⎯⎯ +
Sodium thiosulphate, a reducing agent is efflorescent in nature. Solutions of
thiosulphate prepared by conductivity water are perfectly stable. Ordinary
distilled water usually contains an excess of carbon dioxide which may cause
(as carbonic acid; H
2
CO
3
by reacting with water) a slow decomposition of
thiosulphate with the formation of sulphur. This decomposition may also be
caused by bacterial action.
Remedy
1. Use freshly boiled and cooled water (to expel carbon dioxide).
2. Add three drops of chloroform or 10 mg of mercurial iodide per litre (to
prevent bacterial action).
3. Add a small amount of sodium carbonate (to maintain pH between 9 and
10 at which bacterial activity is least).
KCrO KI HCl
Potassium dichromate
Potassium iodide Hydroch
227
614++
lloric acid
Iodine
Potassium chloride
CrCl I
KCl H O
⎯→⎯⎯⎯ +
++
23
8 7
32
2
WWater
Iodine
Sodium thiosulphate
S
I NaSO NaSO NaI
2223 246
2+⎯→⎯⎯⎯ +
oodium Iodide
Potassium dichromate in the acid solution is reduced by potassium iodide
to liberate an equivalent amount of iodine which in turn reacts with sodium
52
Practical Pharmaceutical Analytical Chemistry
thiosulphate. Free iodine forms a blue colour with starch. At the end point all
iodine reacts with thiosulphate which causes disappearance of blue colour. In
this titration the dichromate ion gets converted to green chromic salt which
gives green colour at the end point.
or
KIO KI H SO
Potassium
iodide
Potassium
iodide
Sulphuric
acid
324
53++ ⎯⎯→⎯⎯⎯ + +
+
333
24 2 2
2
KSO I HO
I
Potassium
sulphate
Iodine Water
Iodine
22
22 3 24 6
Na S O Na S O NaI
Sodium thiosulphate
Sodium
iodide
⎯→⎯⎯⎯ +
Reaction between potassium iodate and excess potassium iodide in acid
solution results in liberation of iodine which can be titrated with sodium
thiosulphate solutions. At the end point blue colour disappears due to the
conversion of free iodine into sodium iodide and no green colour appears (due
to absence of chromium ion) instead solution becomes colourless.
Calculation
N (W u 1000)/(E u V)
where N is the normality of given sodium thiosulphate solution, W is the weight
of primary standard, i.e. K
2
Cr
2
O
7
or KIO
3
, E is the equivalent weight of primary
standard, i.e. K
2
Cr
2
O
7
or KIO
3
and V is the volume of sodium thiosulphate
consumed, i.e. burette reading.
Result The prepared sodium thiosulphate solution was found to
be…………………
Experiment 5 To prepare and standardize 0.1 M of potassium permanganate.
Primary standard Oxalic acid (H
2
C
2
O
4
·2H
2
O) or sodium oxalate
(Na
2
C
2
O
4
) or arsenic trioxide (As
2
O
3
)
Indicator: No indicator; acts itself as an indicator (example of self indicator)
Method
s Preparation: Weigh 31.61 g of potassium permanganate (KMnO
4
) and
dissolve it in 1000 mL of distilled water to get 0.1 M of solution. Cover
the mouth of the flask with a watch glass and boil the solution for
1530 min. Allow the solution to cool and filter that through a funnel
containing a plug of glass wool or asbestos. Collect the filtrate and
store it in dark coloured bottle.
53
Volumetric Analysis
s Standardization: Weigh accurately 6.3 g of oxalic acid and dissolve it in
1 L of distilled water. Pipette out 20 mL of this solution, add 5 mL of
concentrated sulphuric acid and warm the reaction mixture to about
6070qC. Titrate the solution with given potassium permanganate
solution until a permanent pink colour appears.
End point: Appearance of pink colour.
Reaction involved
The reducing agents (traces of organic matter, etc.) present in the distilled water
may cause the decomposition of KMnO
4
into MnO
2
(manganese dioxide) which
catalyzes the formation of more MnO
2
by auto-decomposition of KMnO
4
. To
prevent this, solution of potassium permanganate is boiled. The traces of MnO
2
formed are further removed by filtration.
42
42
2
KMnO H O
Potassium permagagnate
Water
MnO
as catalyst
+⎯ →⎯⎯⎯⎯⎯ 44
43
22
KOH
MnO O
Potassium hydroxide
Manganese dioxide Oxygen
++
Potassium permanganate in acidified solution acts as an oxidizing agent. It
oxidizes oxalic acid into carbon dioxide.
5H C O 2KMnO 3H SO
224
Oxalic acid
4
Potassium permagnate
24
Sulphuri
++
ccacid
24
Potassium sulphate
42
Water
2
Ca
K SO 2MnSO 8H O 10CO⎯→⎯⎯⎯ + + +
rrbon dioxide
Ionic form
5C
2
O
4
2
2MnO
4
16H 2Mn
2
8H
2
O 10CO
2
−
+
−
+
++
++⎯→⎯⎯⎯
Half reaction
5C O 10CO 10e (Oxidation)
2MnO 16H 10e
24
2
2
4
−−
−+−
+
++
2Mn 8H O (Reduction)
2
2
+
+
Calculation
N
1
u V
1
N
2
u V
2
where N
1
is the normality of given potassium permanganate solution, V
1
is
the volume of potassium permanganate consumed, i.e. burette reading, N
2
is
the normality of given oxalic acid solution and V
2
is the volume of oxalic acid
solution taken.
54
Practical Pharmaceutical Analytical Chemistry
Result The prepared potassium permanganate solution was found to
be_______
Experiment 6 To prepare and standardize 0.1 M of potassium dichromate.
Primary standard Ammonium iron (II) sulphate
Indicator: Nphenyl anthranilic acid
Method
s Preparation: Weigh 29.46 g of potassium dichromate (K
2
Cr
2
O
7
) and
dissolve it in 1000 mL of distilled water to get 0.1 M solution.
To prepare 1 N the solution dissolve 49.03 g of potassium dichromate
(K
2
Cr
2
O
7
) in 1 L of distilled water.
s Standardization: To 50 mL of 0.1 M ferrous ammonium sulphate solution
add 20 mL of sulphuric acid and 0.5 mL of N-phenyl anthranilic acid.
Titrate the excess of the iron (II) salt with standard potassium dichromate
solution until the colour changes from green to restore red colour.
End point: Appearance of pink colour.
Reaction involved
23 7 68
22
3
25
2
2 7 4
28
2
22
Cr S O H O CrO HSO H
SO HO O
+− − −+
−
++⎯→⎯⎯⎯ + +
+⎯→⎯⎯⎯ ggFe HSO
()
++
+−3
4
4
Calculation
M
1
u V
1
M
2
u V
2
where M
1
is the normality of given potassium dichromate solution, V
1
is the
volume of potassium dichromate consumed, i.e. burette reading, M
2
is
the
normality of given ferrous ammonium sulphate solution and V
2
is the volume
of ferrous ammonium sulphate solution taken.
Result The prepared potassium dichromate solution was found to be
________
Experiment 7 To find out g/100 mL of copper sulphate in the given sample
by iodometric method.
Principle: Oxidationreduction titration
Molecular formula CuSO
4
·5H
2
O
55
Volumetric Analysis
Molecular weight:
249.54
Indicator: Starch indicator
IP limit: It should not contain less than 98.5% and not more than 101% of
copper sulphate calculated with reference to the dried substance.
Procedure
Dissolve 1 g of copper sulphate in 50 mL of water followed by the addition
of 3 g of potassium iodide, 5 mL of acetic acid and titrate the liberated iodine
with 0.1 N sodium thiosulphate using starch solution as an indicator. Continue
titration till a faint blue colour remains, add 2 g of potassium thiocyanate, stir well
and continue the titration till a milky white colour appears.
or
Pipette out 10 mL of copper sulphate solution in a conical flask. Add 3 g of
potassium iodide, 5 mL of acetic acid and titrate the liberated iodine with 0.1 N
sodium thiosulphate using starch solution as an indicator. Continue titration
till a faint blue colour remains, add 2 g of potassium thiocyanate, stir well, and
continue the titration till a milky white colour appears.
Reaction involved
24 2
424
CuSO KI K SO
Copper sulphate
Potasssium iodide
Potassi
+⎯→⎯⎯⎯
uum sulphate
Cupric iodide
Cupric iodide
CuI
CuI
+ 2
2
2
2
CCu I I
Na S O I
Cuprous iodide Iodine
Sodium thiosulphate
I
22 2
22 3 2
2
+
+
oodine
Sodium iodide
Cuprous iodide
Na S O NaI
Cu I KCNS
⎯→⎯⎯⎯ +
+
24 6
22
2
PPotassiumthiocyanate Cuprous thiocyanate
Sodi
CuCNS KI⎯→⎯⎯⎯ +22
uum iodide
THEORY The assay is based on iodometric titration. Copper sulphate in the pres-
ence of acetic acid reacts with potassium iodide to give cupric iodide. Cupric
iodide formed is unstable and it decomposes into cuprous iodide and iodine.
Formed iodine is titrated with standard sodium thiosulphate solution using
starch as an indicator. Starch is added towards the end point of titration.
The decomposition of cupric iodide to cuprous iodide is reversible. In the
presence of potassium thiocyanate, cuprous iodide gets converted into cuprous
thiocyanate which is more sparingly soluble than cuprous iodide. Potassium
thiocyanate is added towards the end of reaction to prevent adsorption of iodine
by the cuprous thiocyanate.
56
Practical Pharmaceutical Analytical Chemistry
Equivalent factor
2 Na
2
S
2
O
3
{ 2CuSO
4
·5H
2
O 1I
2
2e
1 N a
2
S
2
O
3
{ 1CuSO
4
·5H
2
O
1000 mL of 1 M Na
2
S
2
O
3
{ {(1 u 63.54) (1 u 32) (4 u 16) (5 u 18)} g of
CuSO
4
·5H
2
O
1000 mL of 1 M Na
2
S
2
O
3
{ (63.54 32 64 90) g of CuSO
4
·5H
2
O
1000 mL of 1 M Na
2
S
2
O
3
{ 249.54 g of CuSO
4
·5H
2
O
1 mL of 1 M Na
2
S
2
O
3
{ (249.54/1000) g of CuSO
4
·5H
2
O 0.249 g of
CuSO
4
·5H
2
O
Calculation
1 mL of 1 M Na
2
S
2
O
3
0.249 g of CuSO
4
·5H
2
O
a mL of 1 M Na
2
S
2
O
3
{ (0.249 u a) g of CuSO
4
·5H
2
O
a mL of yM Na
2
S
2
O
3
{ (0.249 u a)/y g of CuSO
4
·5H
2
O
Since 10 mL of CuSO
4
·5H
2
O solution contains (0.249 u a u y) g of
CuSO
4
·5H
2
O
1 mL will contain {(0.249 u a u y)/10} g of CuSO
4
·5H
2
O
100 mL will contain {(0.249 u a u y u 100)/10} g of CuSO
4
·5H
2
O
So
g mL of Copper sulphate
Equivalent factor Burette reading
Mo
/100 =
×
×
llarity Na S O
22 3
100
10
()
×
Pharmaceutical use
1. It is used as an insecticide in agriculture and wood industry.
Result The given copper sulphate solution was found to contain ________
g/100 mL of copper sulphate.
Experiment 8 To find out g/100 mL of hydrogen peroxide (20 Vol) in the
given sample.
Principle: Oxidationreduction titration
Molecular formula: H
2
O
2