60
Project – K
7. Double Titration
7.1
Acid - base double
titration
7.2
Oxidation Reduction titration
7.3
Calculation of double titration
61
7.1 Acid-Base Double Titration
Double Titration :
Titration is performed between the two solutions
of opposite nature. If A and B are the two
Solutions of same nature and normality or
concentration of one solution (suppose A ) is
known the concentration of solution B cannot be
determined directly after titration. In such cases
a solution having opposite nature is used is called
intermediate solution of both solutions is
performed by this. For example to determine the
concentration of unknown oxalic acid or
hydrochloric acid by standard solution of oxalic
acid. In such a case a solution having opposite
nature is used in called intermediate solution &
the titration.
is N/10 & prepare its 25 qubic cm. Solution.
Then according to the equation of normality :
Standard solution = Intermediate solution
At first the titration is performed between A
(standard solution) and C (intermediate
solution). So that the concentration of solution
C is determined. Now the titration is
performed between C and B and the
concentration of B is determined by
calculation. This kind of titration is called
‘Double Titration’.
Now find the normality of unknown solution
with the help of intermediate solution.
According to the equation of normality
More pure result are obtaind by the double
titration then the single (simple) titration
choice of indicator is the double titration –
In double titration the choice of indicator
depends on the nature of the solution of
titration flask. Add that indicator into the
solution of titration flask which does not
change the colour with the solution of titration
flask & produces the pink colour when the
solution of burrete is added at the last drop.
Formula used in the calculation of double
titration :
First find the normality of the intermediate
solution with the help of standard solution.
Suppose that normality of standard solution
N1 × V1 = N ′ × V2
1
10
∵
N1 =
∴
1
× 25 = N ′ × V2
10
or
N′ =
∴
V = 25
&
25
10 × V2
Normality of intermediate solution
( N ′) =
1 25
×
10 V2
..................... (a)
N ′V3 = H 2V4
N′
= Normality of intermediate solution =
25
10 V1 according to the equation (a)
= The Volume of intermediate solution
which has titration with 25 cubic cm. Solution
of unknown solution = 25 ml.
Put the above values into the equation of
normality Intermediate solution = Unknown
solution.
V3
25
× V3 = N 2 × 25
10 × V2
25 V3
1
×
×
= N2
10 V2 25
∴
,
1 V3
×
= N2
10 V2
Normality of unknown solution =
N V3
×
10 V2
/
62
were V2 = The volume of intermediate
solution which is used with the titration of 25
cubic cm standard solution
with the help. of standard solution of sodium
carbonate. After that find the normality of
unknown solution by performing titration of
sulphuric acid with unknown solution of sodium
carbonate.
Method : (i) To prepare a standard solution of
Na2CO3 Required quantity of
V3 =
The volume of intermediate solution which
is used of the titration of 25 cubic cm of solution
of unknown concentration.
Now, ∵ the concentration of standard solution
=
∴
Na2CO3 =
4 ×W × N
E
for N/10 solution = 53 × 1/ 10 × 1000 gm
the concentration in the normality of
unknown solution =
Equivalent weight × Normality × volume
1000
= 54/4 = 1.325 gm
4 × W × N V3
×
E
V2
1.325 gm Na2CO3 is required to prepare a 250
qubic cm. Solution of N/10 concentration of
Na2CO3 .
Concentration of
4 × N × measure × V3 × Value of
unknown solution =
 intermedia te



solution
which


 is used with



 unknown solution 


 The volume of intermedia te 


E × V2  solution which is used with 
 the known solution of



 Weight of 


 of subtance 
Concentration =
 volume of intermedia t

 solution which is used
 with unknown solution

 The volume of intermedia te 


V2  solution which is used with 

 the known solution


 weight of 
 × V3
4 × N × 
 substance 
Note : The above equation follows only when
the standard solution and unknown solution are
having same nature & same Volume.
Experiment No 1
Object : To prepare a standard solution of
sodium carbonate of N/10 concentration and
to find the concentration of sodium carbonate
of given concentration solution.
Principle : In this experiment the normality of
surface acid (intermediate solution) is determined





Take 1.325 gm Na2CO3 into the clean and dry
weighing tube and pour it into a flask of 250
cubic cm. capacity in which distilled water is
already present. Some crystals of the
substances are left on the wall of the funnel.
To pour these crystals into the funnel pour
some distilled water in it so that the all
substance of funnel goes to the flask, Remove
the funnel and stir the flask, that the substance
will completely dissolve. Pour that much of
water into the flask, which reaches to the
marked point of the flask.
Titration :
Clean the burette with distilled water and wash
it with the given sulfuric acid, note the reading.
Wash the pipette 2 times with the standard
solution of Na2CO3 and fill 25 ml. Na2CO3
solution into the titration flask by the
measuring flask. Add 2 drops of Methyl
orange so that the solution becomes yellow
colour. Place the titration flask on the burette
stand after putting it an the white paper and
gradually open the cork to allow the H 2SO4
to fall in it, fix the titration flask and stirr it.
At first the colour of the solution is pink after
sometime it will be disappeared. Now allow
to fall the solution drop by drop.
63
The drop at which the light read colour is not
disappeared & it becomes permanent. stop to
add the solution at that point & note the
reading of burette. This is the end point.
Throw the solution of the titration flask, clean it
to repeat the experiment upto the two same
readings of sulfuric acid is obtained.
After that perform the titration of sulfuric acid
(intermediate solution) with the solution of
Na2CO3 of unknown concentration according
to the above method.
Observation :
solution –
(B) Weight of weighing tube + oxalic acid =
8.5776 gm.
(C) Weighing tube + left oxalic and = 7.2532
gm.
∴
The weight of Na2CO3 pour into the flask
= 8.5776 – 7.2532 = 1.3244 gm.
Normality of solution of Na2CO3
4 ×W × N
E
=
=
=
The titration between intermediate solution (H 2SO4 )
and solution of (Na2CO3 ) of unknown concentration
S.
No.
Volume of
1.
Burette Volume
Na2CO3
Start
End
25.0 ml
0.0 ml
26.8 ml
2.
25.0 ml
0.0 ml
26.4 ml
3.
25.0 ml
0.0 ml
26.4 ml
Formula : (1) N1 × V1
4 × 1.3244 × N
53
=
(ii) Titration :
Table – 1
Standard solution ( Na2CO3 ) and intermediate solution
Solution Solution
(Na2CO3 ) (H 2SO4 )
(2)
Intermediate
Solution
concentration
Solution of
unknown
(H 2SO4 )
(Na2CO3 )
N2 V2
=
Volume of Reading
1.
2.
3.
2.0 ml
25.0 ml
25.0 ml
Na2CO3
0.0 ml
0.0 ml
0.0 ml
Burette
of Start
Volume
End of acid
25.8 ml
25.6 ml
25.6 ml
25.6 ml
N2 V2
Calculation : To calculate normality of
standard
solution
∵
Normality of Na2CO3 solution
=
4 ×W × N
E
=
5.2976
53
=
4 × 1.3244 × N
53
Standard
Solution
Solution
Intermeditate
H 2SO4
S.
No.
26.4 ml
N2 V2
(2) To calculate the normality of
intermediate solution (H 2SO4 )
N1V1
N 2 V2
=
5.2976
53
of acid
Standard Intermediate
(i) To prepare a standard
(A) Weight of the weighing tube = 7.240 gm.
∵
Table – 2
(Na2CO3 )
(H 2SO4 )
5.2976
× 25 = N 2 × 25.6
53
or
N2 =
5.2976 × 25
53 × 25.6
64
(3)
To calculate the normality of solution
of unknown concentration (Na2CO3 ) –
N 2 × V2
or
(4)
N3 × V3
=
Intermediate
Solution
Solution of unknown
concentration
(H 2SO4 )
(Na2CO3 )
5.2975 × 25 26.4
N3 =
×
53 × 25.6
25
6.
It each titration use some (less) indicator
both equal sides. It is difficult to see the
change in colour if you use less or more
indicator.
7.
The intensity of colour which is
represented by the indicator at the end
point in pair titration is same in both
titration.
= 0.103 N
To determine the concentration of
solution of unknown concentration –
Concentration (strength) – Normality
×Equivalent weight
Concentration of Na2CO3 solution 0.103
× 53
= 5.46 gm / litre
Result : (1) Normality of solution of
Na2CO3
= 0.103 N
(2) Concentration of solution of
Na2CO3
= 5.46 gm / litre
Indication for titration :
1.
Wash the pipette and burette with the
given solution.
2.
titration. It does not matter if the distilled
water is attached with the wall of the
titration flask. It is not necessary to dry
it because the certain amount of volume
is only pour into the flask by pipette.
So it don’t effect the result.
Note the reading for waiting few
seconds after stirring the solution when
the end point is found.
3.
Note the reading only when the changes
in color is appeared at end point, Do not
think about the intensity of colour.
4.
Do not take the mean of the different
volumes obtained in titration. But take
that Volume which is at least two times
same.
5.
It is necessary to wash the titration flask
with the distilled water after each
Experiment No – 2
Objective : You have a standard solution of
NaOH. Which is prepared by dissolving 0.8
gm subtance in 200 ml water. With its help
find the concentration of B solution of
Na2CO3 solution is normality solution of 500
ml. C solution of HCl in a standard solution
and phenolphthalein indicator.
Apparatus : As per the experiment no. 1
Equations of reaction : In Ist titration (Standard
solution NaOH and intermediate. solution HCl)
NaOH = Na + + OH −
HCl = H + + Cl −
NaOH + HCl = Na + + Cl − + H 2O
In IInd titration intermediate solution is HCl &
Solution of unknown concentrations is
Na2CO3 .
Result : phenolphthalein (indicator), solution of
NaOH, solution of Na2CO3 and intermediate
solution of acid.
Method :
As per the titration in the
experiment no 1.
65
(I)
S.
No.
1.
2.
3.
N1 × V1
Table – 3
Standard solution (NaOH) ‘A’ and
intermediate solution (HCl) ‘C’
Volume of
base
25.0 ml
25.0 ml
25.0 ml
Reading of Burette
Start
End
0.0 ml
27.7 ml
0.0 ml
22.3 ml
0.0 ml
22.3 ml
Volume
of acid
(II)
S.
No.
Volume of
base
Reading of Burette
Start
End
1.
2.
3.
25.0 ml
25.0 ml
25.0 ml
0.0 ml 25.8 ml
0.0 ml 25.4 ml
0.0 ml 25.4 ml
N 2 V2
=
Intermediate Solution of
unknown
concentration
3.
0.08 × 1000 0.8 × 10
=
200
2
25 N 25.4 25.5
×
=
= 0.114 N
22.3
25
2230
or
Normality of ‘B’ solution of
Na2CO3 = 0.114 N
4.
∴
To calculate the concentration –
Concentration = Normality × gm
equivalent weight
= 0.114 × 53
= 6.042 gm / litre
when 6.042 gm Na2CO3 is dissolved in
the 1000 ml solution .
The quantity of Na2CO3 dissolved in the
500 ml solution.
=
6.042 × 500
1000
Therfore concentration of NaOH
solution = 4 gm / litre.
=
6.042
2
Concentrat ion
Equivalent weight
To calculate the normality of
intermediate solution :
N 2 × 0.25
N2 =
= 0.8 × 4 gm
Normality =
2.
=
or
To calculate the normality of standard
solution when 0.8 gm NaOH is
dissolved in 200 ml water
=
To calculate the normality of solution
of unknown concentration :
= N 3 × V3
N 2 × V2
Standard
Solution of
solution
unknown
concentration
25 N
× 25.4
22.3
Calculation :
1.
Acid (HCl)
Normality of intermediate solution =
0.112 N
25.4 ml
N3 V3
N2 × V2
N
× 25 = N 2 × 22.3
10
N
25
25N
N2 =
×
=
10 22.3 22.3
N 2 = 0.112
or
Volume
of acid
Formula : (1) N1 V1
= N 2 V2
Standard Intermediate
Solution
Solution
(2)
Base (NaOH)
22.3 ml
Table – 4
Intermediate solution ‘C’ (HCl) and
solution of Na2CO3 ’B’
=
gm.
Result : (1) Normality of Na2CO3 solution
– 0.114 N
(2) Concentration into 500ml –
3.021 gm
Precaution : As per the experiment no. 1.
66
Experiment No – 3
Object : Add 12.6 gm oxalic acid into 250 gm
of water and add more water in it to prepare a
solution of Volume 2 litre. Determine the normality
of oxalic acid and its concentration into the 500
ml solution. With the help of standard solution,
the solution of phenolphthalein indicator and
caustic potash is intermediate solution.
S.
No.
Volume of
base
1.
2.
3.
25.0 ml
25.0 ml
25.0 ml
Reagents : Phenolphthalein indicator,
Standard solution oxilic acid, solution of
oxilic acid of unknown concentration and
solution of caustic potash.
=
N 2 × V2
Standard
solution
Intermediate
solution
(H 2C 2O4 )
(KOH)
2. N 2 × V2
=
N 3 × V3
Intermediate Solution of
unknown
concentration
2KOH + H 2C 2O4 = 2K + + C 2O4 − − + 2H 2O
Intermediate solution, solution of
unknown concentration.
Volume
of acid
Start
End
0.0 ml 25.8 ml
0.0 ml 25.4 ml
25.4 ml
0.0 ml 25.4 ml
Formula : 1. N1 × V1
Equation of Reaction :
A. H 2C2O4 + 2KOH = 2K + + C2O4 −− + 2H 2O
Standard solution, intermediate
solution.
B.
Reading of Burette
(KOH)
(H 2C 2O4 )
Calculation :
1.
To calculate the normality of standard
solution
When 12.06 gm oxalic acid is dissolved
into the litre solution.
Method :
Phenolphthalein indicator is used in the
titration therefore the solution of acid and base
is taken into the burette by pipette 1. Then
follow the above procedure.
12.5
= 6.3
2
gm / litre
Equivalent weight of oxalic acid = 63
Concentrat ion
Normality = Equivalent Weight
Table – 5
Standard solution (oxalic acid) and standard
solution (KOH)
S.
No.
Volume of
base
Reading of Burette
Start
End
1.
2.
20.0 ml
20.0 ml
0.0 ml 23.9 ml
0.0 ml 23.7 ml
3.
20.0 ml
0.0 ml 23.7 ml
Volume
of acid
23.7 ml
Table – 6
Standard solution (KOH) and Titration
between standard solution (KOH) and
solution of unknown concentration (H 2O2O4 )
=
2.
6.3
1
N
=
=
6.3 10 10
To calculate the normality of
intermediate solution
N 2 × V2
N3 × V3
=
Standard
Solution
Intermediate
Solution
(H 2C 2O4 )
(KOH)
N
× 20 = N 2 × 23.7
10
67
or
N2 =
20.0 × N
2.37 × 10
N2 =
N
= 0.084 N
237
N3 =
Concentration = Normality × gram equivalent
weight
Normality of intermediate solution (KOH) =
0.084 N
3.
To calculate the normality of solution
of unknown concentration :
N 2 × V2
=
Acid
27.3
= 0.115 N
273
N 3 × V3
= 0.115N × 63 = 7.245
= 7.245 gm / litre
When 6.049 gm of Na2CO3 is dissolved in
1000 ml solution
∴
Base
Concentration is 500 ml =
7.245 × 500
1000
= 3.62 gm.
20
× 27.3 = N 3 × 20
237
N3 =
20 27.3
×
273
20
N3 =
27.3
273
Result :
(1)
Normality of solution of oxalic acid = 0.115
N
(2)
Concentration of solution of oxalic acid is
500 ml =
3.62 gm.
7.2 Redox – Titration
KMNO 4
1.
– Oxilic acid titration :
heat upto the 60° – 70°C, Now titrate this
solution with the KMnO4 solution of
In acid medium ( is presence of H 2SO4 )
KMNO 4
acts as a
burette. The end point is indicated by light
according to
pink colour.
following reactions –
Experiment No. – 1
2KMnO 4 +3H 2SO4 →
K 2SO4 + 2MnSO 4 + 2H 2O + 5O
2.
The above produced oxygen oxidised
the oxalic acid.
acid and to find the concentration of given solution
of KMnO4 by titration.
Chemical Reaction : According to the above
principle
Indicator : KMnO4 is a self indicator.
COOH
5|
Object : To prepare a standard solution of oxalic
+ 5O → 5H 2O + 10CO2
COOH
Add dilute H 2SO 4 in oxalic acid and
(i)
To prepare a standard solution : The
required weight of oxalic acid to prepare
a 250ml, N/10 solution of oxalic acid.
68
63
= 6.3
10
∴
gm
gm/litre
Base of acid in 250 ml.
=
6.3 × 250
1000
=
∴
weight of oxalic acid in 1000 ml = 1.564
× 4 gm
1.575 gm
Take 1.575 gm oxalic acid in the weighing tube
and measure it. Now pour it into measuring flask
carefully with the help of funnel. Now again
measure the weighing tube. Pour the distilled
water on the funnel and put the substance in flask,
and dissolve it into water. in the flask upto the
mark & stirr the solution to make it homogeneous.
Note these observations into your practical note
book.
Titration :
Rinre the burette with given KMnO4 and fill
it with KMnO4 . Now fix the burette to the
stand. Take 20 ml solution of oxilic acid is
the conical flask with help of pipette. Add 5
ml, H 2SO4 in it and heat it upto 70.80°C. The
solution should not boil. Pour the solution of
KMnO4 from the burette into the hot so heated
solution gradually and stirres the solution. At
the time of starting colour of KMnO4 is
displaced but at the last point by adding one
drop of KMnO4 the light pink colour is
obtained permanently.
Write the observations of titration in form of
table.
Observation :
1.
Oxilic acid + weight of weighing tube
= 7.5642 gm
weight of empty weighing tube = 6.0002
gm
weight of oxilic acid in 250 ml = 1.5640
gm.
Normality of oxalic acid :
weight of oxalic acid in 250 ml = 1.564
∴
Therefore normality =
2.
6.256
N
N=
63
10.07
Titration of oxalic acid and KMnO4
Table – 1
S.
No.
Volume of
base
Reading of Burrete
Start
End
1.
2.
3
20.0 ml
20.0 ml
20.0 ml
0.0 ml 18.2 ml
0.0 ml 18.0 ml
0.0 ml 18.0 ml
Volume
of acid
18.0 ml
Calculation :
N1 × V1
N 2 × V2
=
Oxalic acid
KMnO4
N
N1 =
,
10.07
N2 = ?
V1 = 20 ml
V2 = 18 ml
or
∴
N1
× 20 = N 2 × 18
10.07
20
N
N2 =
×
= N / 9.063
18 10.07
∴
The normality of KMnO4 solution
= N / 9.063
Equivalent weight of KMnO 4 = 31.6
Thus concentration in gram / litre =
31.6
= 3.48 = gm / litre
9.063
Result : (1) Normality of KMnO4 solution =
N
9.063
(2) Concentration = 3.48 gm / litre
Precautions :
1.
The titration should be performed in the
hot oxalic acid solution.
69
2.
3.
4.
Do not add H 2SO 4 in less quantity. It
should be added in more quantity
otherwise the brown colour of MnO2 is
produced.
Read the up side surface of KMnO4
while taking the reading of burette
Place a glased tile or white paper below
the conical flask to see clearly last point
gradually.
KMnO4 Ferrous Ammonium Sulphate
Titration
Experiment – 2
Object : A standard solution of N/2 feerrous
ammonium sulphate is given. To find the
normality of given KMnO4 solution by the
titration
Principle :
burette the solution taken in to the conical flask
by KMnO4 N/10 ferrous sulphate.
Rinre the burette with KMnO4 solution and fill it
that. Take 25 ml ferrous sulphate solution in
conical flask or beaker with the help of pipette
and pour 5ml. H 2SO4 in it. Now titrate with
KMnO4 . First the colour of KMnO4 . is
disappeared, but at the last point light pink
colour will appcar. Repeat the titration and
obtain the 2 same readings of value of KMnO4
Observation
Table – 2
Volume of
ferrous
Sulphate
Reading of Burette
1.
25.0 ml
0.0 ml
21.9 ml
2 KMnO 4 + 3H 2SO4 →
2.
25.0 ml
0.0 ml
21.8 ml
K 2SO4 + 2MnSO 4 + 2H 2O + 50
3.
25.0 ml
0.0 ml
21.8 ml
It oxidises ferrous sulphate of ferrous
ammonium in ferrous sulphate. (NH 4 )2 SO 4
do not take parts in this reaction Acid sulphate
reacts with acidic KMnO4 in following way :
S.
No.
=
N 2 V2
Ferrous Sulphate KMnO4
N1 =
or
O4 + 8H 2SO4 + 10FeSO4 (NH 4 )2 SO4 →
2KMnO 4 + 10F2SO4 + 8H 2SO4 →
K 2SO4 + 2MnSO 4 + 5Fe
Ionization Equation :
2KMnO 4−
+
+ 16H + 10F2
2Mn
++
Titration method : The solution taken into the
V2 = 21.8 ml
N
× 25 = N 2 × 21.8
20
N
25
N2 =
×
20 21 .8
Concentration = Normality × Equivalent
weight
= 0.0573 ×31.6 = 1.812 gm/litre
(Equivalent weight of KMnO4 = 316)
Result :
(ous ) →
+ 10F2+ + + (i c ) + 8H 2O
N
, N 2 = 1 V1 25 ml ,
20
= 0.0573 N (Normality of KMnO4 )
1.
++
21.8 ml
N1V2
10F2SO4 (NH 4 )2 SO4 + 5H 2SO4 →
Because (NH 4 )2 SO 4 does not take part into
this reaction Therefore it can be written in
following way.
KMnO4
Calculation :
K 2SO4 + 2MnSO 4 + 2H 2O + 50
K 2SO4 + 2MnSO 4
End
Therefore 25 ml, N/20 ferrous sulphate = 21.8
ml KMnO4
2KMnO 4 + 3H 2SO4 →
5F2 (SO4 )3 + 5H 2O + 10(NH 4 )2
Start
Volume
of
2.
Normality of KMnO4 solution = 0.0573
N
Concentration of KMnO4 solution =
1.812 gm/litre
70
Precaution :
1.
H 2SO4 should be added into more
quantity otherwise the brown ppt of
MnO2 is produced.
2.
Do not heat the ferrous sulphate solution
at time of titration. Perform the titration
at the room temperature.
KMnO4 is coloured solution. Therefore
3.
see the upper surface at the time of
taking the reading of burette.
4.
Obtain the two same readings in titration.
5.
Do not use HCl or HNO3 in place of
H 2SO4 because the HCl is oxidised
itself and HNO3 is a oxidization
Substance.
6.
Place a glave tile or white paper below
the conical flask for observing the end
point gradually.
Oxidation Reduction Double Titration
Experiment No - 3
Object : The N/30 focus ammonia sulphate is
given. The normality and concentration of ferrous
sulphate is determined with its help. Potassium
permanganate is a intermediate solution.
Principle:
As per experiment no -2
Indicator - Self indicator
Method (i) The titration of ferrous ammonium
sulphate is performed by the potassium
permagnate solution. as per the
experiment no-2.
(ii) Titration of ferrous ammonium sulphate
of unknown concentration from
intermediate solution of KMnO4 :
Fill the burette with KMnO4 and note
its reading. Clean the pipette by distilled
water, wash it with ferrous ammonium
sulphate table then 25ml solution and
pour it into the titration flask. Pour dilute
H 2SO4 to fill the titration to half, then
gradually drop the solution of KMnO4
from the burette. Note the reading when
the permanent light pink colour appears
in the solution at the end point Repeat
the titration for equal volume of KMnO4
Observation :
1. The titration in standard solution of
ferrous ammonium sulpate and
intermediate solution of KMnO4 .
Table – 3
S. Solution of Reading Burette Used
No. foous
of
End
KMnO4
ammonium start
(ml)
solution
sulphate
(ml)
(ml)
1.
2.
3.
(ii)
25.0 ml
25.0 ml
25.0 ml
0.0 ml
0.0 ml
0.0 ml
24.8 ml
23.5 ml
23.5 ml
24.8
23.5
23.5
Proper
Used
KMnO4
(ml)
23.5
Titration in the solution of unknown
ferrous ammonium sulphate and
Intermediate KMnO4
Table – 4
S. Solution of
No. foous
ammonium
sulphate
Reading Burette Used
KMnO4
of
End
start
(ml)
solution
(ml)
(ml)
1.
25.0 ml
0.0 ml
24.0 ml
24.8
2.
25.0 ml
0.0 ml
23.9 ml
23.9
3.
25.0 ml
0.0 ml
23.9 ml
23.9
Proper
Used
KMnO4
(ml)
23.9
Calculation
1.
Standared
solution
(Ferrous
Ammonium Sulpate of normality N =
30)
2.
To calculate the noramality of
intermediate solution [KMnO4 ]
71
N1 × V1
=
Standard
Solution
N 2 × V2
Intermediate
Solution
N
× 25 = N 2 × 23.5
30
25
N
×
N2 =
30 23.5
Normality of intermediate solution
(N 2 ) =
3.
25 N
23.5 × 30
To calculate the normality of ferrous
ammonium sulphate of unknown
concentration : –
= N 3 × V3
N 2 × V2
Standard
Solution of unknown
Solution concentration
(F2SO4 (NH 4 )2 SO4 .6H 2O )
250 N
× 23.9 = N 2 × 25
23.5 × 3
250 N
23.5
N3 =
×
23.5 × 30
25
N3 =
N 23.9
23.9 N
×
=
30 23.5 30 × 23.5
Normality
N/20 oxalic acid. The solution of potassium
permanganate is intermediate solution.
Principle : As per the experiment no. 1
Indication : Self indicator [ KMnO4 solution]
Method : Take KMnO4 solution in burette and
measure oxalic acid by pipette. Its titration
method is according to the experiment no. 1.
Observation :
(i)
Titration of standard ferrous ammonium
sulphate solution and intermediate solution
of KMnO4 .
Table – 5
S. Solution of Reading Burette Used
KMnO4
No. of oxalic
of
End
acid
start
(ml)
solution
(ml)
(ml)
(ml)
1.
2.
3.
(ii)
solution
F2SO4 (NH 4 )2 SO4 ⋅ 6H 2O having unknown
concentration =
4.
of
23.9 N
30 × 23.5
Concentration of solution having
unknown concentration (Ferrous
ammonium sulphate)
Concentration = Normality × Equivalent
weight
=
23.9 N
× 392
30 × 23.5
= 12.77 gm/litre
Result : The concentration of given ferrous
ammonium sulphate is 12.77 gm / litre
Precautions : As per the experiment no - 2
Experiment – 4
Objectivet : To find the concentration of oxalic
acid having unknown concentration with the help
25.0 ml
25.0 ml
25.0 ml
0.0 ml
0.0 ml
0.0 ml
25.1 ml
24.8 ml
24.8 ml
25.1
24.8
24.8
Properly
Used
KMnO4
solution
v2 (ml)
24.8
Titration between the solutions of
unknown ferrous ammonium sulphate
and Intermediate KMnO4
Table – 6
S. Solution of Reading Burette Used
KMnO4
No. of oxalic
of
End
acid
start
(ml)
solution
(ml)
(ml)
(ml)
1.
2.
3.
25.0 ml
25.0 ml
25.0 ml
0.0 ml
0.0 ml
0.0 ml
23.4 ml
23.2 ml
23.2 ml
25.1
23.2
23.2
Properly
Used
KMnO4
solution
v2 (ml)
23.2
Calculation
1.
Normality of standard solution (oxalic
acid). N / 20
2.
To calculate Normality of intermediate
solution :
= N 2 × V2
N1 × V1
Standard
Intermediate
Solution
Solution
[C2H 2O4 ]
[KMnO4 ]
72
N
× 25 = N 2 × 24.8
20
25 N
N2 =
20 × 24.8
3.
N3 =
To calculate the normality of solution
of unknown concentration :
N 2 × V2
N3 × V3
=
Intermediate
concentration
[KMnO4 ]
[C2 H 2 O4 ]
4.
To calculate the concentration of oxalic
acid of unknown concentration :
Concentration = Normality × Equivalent
weight
= 0.0467 × 63
= 2.946 gm/litre
25 N
× 23.2 = N3 × 25
20 × 24.8
N3 =
Normality of oxalic acid of unknown
concentration = 0.0467 N
Solution of unknown
solution
23.2
N
20 × 24.8
Result : The concentration of given oxalic acid
in 2.946 gm/litre
25 N × 23.2
20 × 24 .8 × 25
Precaution : As per the experiment no. 1.
7.3 Calculations of double Titration
Example : In the experiment of titration to
neutralise 23.7 ml. deci normal solution of
H 2SO 4 you require 26.6 ml HCl acid to
neutrilize the 25 ml. NaOH and 25 ml NaOH
of its base. Calculate normality of, by
calculation
1.
To calculate the normality of
intermediate solution :
Standard solution (H 2SO4 ) = Deci
Normal solution =
Normality of intermediate solution
2.
To calculate the normality of solution of
unknown concentration.
N2 × N 2
Intermediate
solution
Solution of
unknown
concentration
[NaOH ]
[HCl ]
23.7
× 25 = N 3 × 26.6
250
N
10
N3 =
23.7 × 25
250 × 26.6
Standard Solution
Intermediate
Solution
N3 =
23.7
266
H 2SO4
NaOH
=
N
× 23.7 = N 2 × 25
10
N2 =
N 23.7
×
10 25
N3 × N 3
=
N 2 × V2
N1 × V1
23.7
250
=
0.884 N
Concentration = Normality × Equivalent
weight
= 0.088 × 36.5
= 3.22 = 3.22 gm/litre
Result :
1. Normality of HCl is 0.884 N
73
2. Concentration is 3.22 gm/ litre
Example 2 : To perform the titration of 25 ml.
NaOH of 2.09 N concentration from the
intermediate solution of HCl you require 20.8
ml. To neutralize 25 ml of KOH solution of
unknown concentration 21.5 ml, HCl solution
is required
To calculate the normality of KOH solution
and find the quantity of KOH dissolved in 3
litre solution.
Calculation :
1.
Normality of standard solution =
(NaOH ) . 09 N
2.
Calculation of Normality of
intermediate solution
N1 × V1
N 2 × V2
Base
Acid
0.09 × 25
N 2 × 20.8
N2 =
II.
=
Normality of KOH = 0.93 N
Concentration = Normality ×Equivalent
weight
= 0.093N × 56
= Concentration of 5.209 KOH
is 5.20 gm /litre
The 5.20 gm of solution is present in KOH
solution Solution present in 3 litre solution is
5.20 × 3 = 15.60 gm
Result :
1.
Normality of KOH = 0.093 N
2.
The quantity of solution 3 litre solution
of KOH.
Example 3: The standard solution 23.9 ml of
HCl solution 25 ml of standard NaOH Solution
when 22.7 ml HCl is require to neutralize 25%
of 25 ml Na2CO3 . Calculate the equivalent weight
of Na2CO3
I.
0.09 × 25
20.8
=
9 × 25
2080
=
225
2080
=
45
416
N1 × V1
=
[HCl]
43 × 9
4160
=
387
4160
Intermediate
Solution
Solution
N2 =
[KOH]
45 × 21.5
416 × 25
=
N 2 × V2
Standard
= 0.093 N
II.
HCl
N1 × V1
V2
=
N
25
×
20 23.9
=
25N
478
N 3 × V3
45
= 21.5 = N3 × 25
416
N3 =
=
NaOH
= 0.108 N
To calculate the normality of solution
of unknown concentration =
N 2 × V2
Calculate the normality of intermediate
solution (HCl):
To calculate the normality of Na2CO3
solution :
N 2 × V2
=
N 3 × V3
Intermediate
Na2CO3
Solution
Solution
25N
× 22.7 = N 3 × 25
478
74
N3 =
25N 22.7
×
478
25
Solution
Potassium
Permanganate
Normality of solution of Na2CO3 = 0.047 N
1
0.25% Concentration of Na2CO3 = 25 gm/litre
14.2
x 28.4
concentration
Ferrous Ammonium
sulphate
=
N3 x 10
= 2.5 gm/litre
Concentration = Normality × Equivalent weight
or
N3
Equivalent weight = Concentrat ion
=
Example 4: 14.2 ml of solution KMnO4 is
Sulphate is required for the titeration of 28.4 ml
solution of that KMnO4 solution then find the
quantity of ferrous ammonium sulphate in one litre
solution of ferrous ammonium sulphate
To calculate the normality of intermediate
solution:N1 x V1 = N2 x V2
Oxalic acid Potassium permanganate
0.1 x 10 = N2 x 14.2
N2 =
=
1.5 X 10.0
14.2
permanganate
-
14.2
There fore normality of KMnO4 =
N
14.2
II.
Calculation of normality of unknown
concentrated solution
N2 x V2
Intermediate
=
10
= 0.2N
250 ml. of solution have 4.9 gm. of
ferrous ammonium sulphate.
weight of ferrous ammonium sulphate
in 1000 ml. of solution
=
4.9 x 1000
250
normality
N3 x V3
Solution of unknown
1
10
= 78.4 gm/ltr.
Result
Concentration of ferrous ammonium sulphate
= 78.4 gm/ltr.
Example 5
4.9 gm of ferrous ammonium sulphate is
added to 250 ml of water to make a initial
solution. To neutralize 25 ml. of this solution,
22.4 ml. of KMnO4 solution is required. We
required 27.2 ml. of KMnO4 solution to
neutralize, a solution of ferrous ammonium
sulphate of unknown concentration. Find the
concentration of ferrous ammonium sulphate
solution.
-
1
x
= 0.2N x 392
Concentration of ferrous ammonium
sulphate
Result : Equivalent of Na2CO3 is 53.19
acid solution. If 10ml of ferrous ammonium.
14.2
Concentration=normality x equivalent
weight.
2.5
= 53.19
0.047
required for the titration of 10 ml, 0.1 N oxalic
28.4
2
Normality
=
1.
=
=
19.6
392
=
=
19.6 gm.
Concentration
Equivalent Weight
75
(Equivalent weight of ferrous ammonium
Sulphate = 392) = N/20
I.
To calculate the normality of
mitermediate solution
= N2 N2
N1 V1
Ferrous
Potassium
ammonium
permagnate
Sulphate
N
× 25
20
N2 =
II.
N 2 × 22.4
25N
20 × 22.4
Normality of KMnO4
Potassium
Permagnet
ferrous ammonium
sulphate
25N
× 27.2
20 × 22.4
=
N3 =
25N × 27.2
20 × 22 .4 × 25
27.2N
20 × 22.4
=
=
N3 × 25
27.2
= 0.0607 N
4480
To calculate the normality of solution
unknown concentration
Normality ferrous ammonium sulphate = 0.06.7
N Concentration of ferrous ammonium sulphate
=
N 2 × V2
N 3 × V3
= Normality × Equivalent weight
Intermediate
Solution of
= 0.0607 × 392
Solution
unknown
concentration
= 29.79 gm/litre
=
Result : 23.79 gm/litre
Exercise Question
Q.1 Give one example of self indication. How does it work ?
Q.2 What is a double titration ?
Q.3 What is Oxidation reduction Titration ?
Q.4 Describe the reaction between oxalic acid and potasium permaganate during titration ?
Q.5 Why ferrous ammonium sulphate is used in place of ferrous sulphate during titration ?
Q.6 What is the function of KMnO4 in Oxidation-Reduction titration ?
Q.7 Why the heated solution of oxalic acid in used in Oxidation-Reduction Titration ?
Q.8 Why HNO3 acid is not used in place of HCl in Oxidation-Reduction Titration ?
Q.9 Why the solution of potassium permaganate is taken into the burette during the titration between
ferrous ammonium sulphate and potassium permanganate ?
Q.10 Describe the reactions of the potassium permanganate during titration ?
Q.11 Sodium Carbonate is a salt, then why it is used as a base during titration ?
Q.12 Which solution should be take titration in the flask while using phinalptheiline indicator and why ?
Q.13 The burrete and pipetle are rinsed during titration, flask and beaker are not rinsed, why it is so ?
Q.14 Why standard solution of sulphuric acid and sodium. Hydroxide. Can’t be prepared by measuring
it ?
75
76
Q.15 What quantity of substance is required to prepare 250 ml, N/10 oxalic acid ?
Q.16 Prepare 250 ml. solution by dissolving 2.45 gm ferrous ammonium sulphate in acidic water. Find
its normality ?
Q.17 What quantity of substance is required to dissolve in 250 ml distilled water to prepare 0.2 ml
ferrous ammonium solution ?
Q.18 What quantity of substance is required to dissolve in 250 ml distilled water to prepare N/30
ferrous ammonium sulphate solution ?
Q.19 14.0 gm solvent is dissolved in 2.5 litre of caustic potash solution, 24.6 ml acid is spent for
the titration of 25ml intermediate solution of acilic acid. 19.7 ml acid of unknown
concentration was cansumed for the titration of 25ml solution of Caustic Soda. Find the
normality of Causitc Soda. Find the normality of Caustic Soda and it concentration in 1.5
ml solution.
Q.20 1.4175 gm acid is dissolved in 250 ml of any acid. Its atomic weight is 126 and basisity is
2. To neutrilize the 22.5 ml of its solution 25.00 ml is consumed. 25 ml of this is completely
neutralised by 20ml. Find the concentration is gm/litre
Answers
N
40
(15) 1.575
(16)
(19) 1.080, 4.8 gm/1.5 litre
(20) 4.96 gm / litre
(17) 19.6 gm
(18) 3.2667 gm