ESTIMATION OF ANILINE
Aim :
Determine the Amount of Aniline solution in kg / dm3
Given :
Bottle B : Na2S2O3 ( 0.1N )
Bottle C : Brominating Solution ( Approx.
0.1 N )
Bottle D : Aniline
Procedure :
Back Titration :
Take exactly 10 ml of
Brominating soln in a stoppered bottle and add about 10 ml of Aniline (by
pipette) . Add about 3 ml (¼ TT) conc. HCl and about 25 ml (2 TT) dist. water
Shake and wait for about 10 min. Then add about 10 ml 10% KI (1 TT). Titrate
immediately the liberated iodine against Na2S2O3.
Using starch indicator take 3 Readings
and call the const. reading as Y ml. Record the observations
BlankTitration :
Take exactly 10 ml of
Brominating soln in a stoppered bottle and add about 3 ml (¼ TT)
conc. HCl and add about 25 ml (2 TT) dist. water. Then add about 10 ml 10% KI
(1 TT). Titrate the liberated iodine against Na2S2O3.
Using starch indicator take 3 Readings
and call the const. reading as X ml. Record the observations and do the calculations.
RESULT :
Strength of
Aniline in G/L = ............. gms.
\kg / dm3 of Aniline = ____________.
ESTIMATION OF ANILINE
Blank
Titration Back
Titration
Soln. in Burette :
Na2S2O3(0.1 N) Soln.
in Burette : Na2S2O3(0.1 N)
Soln. in stoppered
Bottle : Soln.
in Stoppered Bottle :
10 ml Brominating
soln. 10
ml Brominating soln. + 10ml Aniline
+ 1/4 T.T. conc.
HCl +
2 TT Dist. H2O + 1/4 TT conc. HCl
+ 2 T.T. Dist.
Water Shake
and wait for 10 min + 1 T.T. 10% KI
+ 1 T.T. 10% KI
Indicator : Freshly prepared starch soln. (2 ml) Indicator : Freshly prepared starch
soln. (2 ml)
End Point : Blue
to colourless End
Point : Blue to white ppt
Reaction :
1) KBrO3 + 5KBr + 6 HCl ® 1)
5 KBr + KBrO3 + 6 HCl =
3 Br2 + 6 KCl + 3H2O 6 KCl + 3H2O + 3 Br2
2) Br2
+ 2 KI = 2KBr + I2 NH2 NH2
3) I2 + 2Na2S2O3 = Na2S4O6
+ 2NaI Br Br
2)
+ 3 Br2 = ¯ + HBr
Br
3)
Br2 + 2KI = 2KBr + I2
4)
I2 + 2Na2S2O3 = Na2S4O6
+ 2NaI
Burette Level
|
Pilot ml.
|
Burette
|
C.B.R. ml.
|
Burette Level
|
Pilot ml.
|
Burette
|
C.B.R. ml.
|
|||||
I
|
II
|
III
|
I
|
II
|
III
|
|||||||
Final
|
X ml.
|
Final
|
Y ml
|
|||||||||
Initial
|
0
|
0
|
0
|
0
|
Initial
|
0
|
0
|
0
|
0
|
|||
Difference
|
Difference
|
Calculations :
X ml of 0.1 N Na2S2O3
º Total Brominating soln.
Y ml of 0.1 N Na2S2O3 º Unused Brominating soln.
(X – Y) = V ml of Na2S2O3 º Used Brominating soln. by
10 ml Aniline
From Equations :-
2 Na2S2O3 º I2 º Br2 º C6H5NH2
\6 Na2S2O3
º 3I2 º 3Br2 º C6H5NH2
\6 Na2S2O3
, 5H2O º C6H5NH2
\6 x 156 gm Na2S2O3 5H2O º 93 gms of Aniline.
6 x 10,000 ml 0.1
N Na2S2O3
º 93 gms of Aniline.
\V ml of 0.1 N Na2S2O3
º V x 93 / 60,000 gms of
Aniline = ‘g’ gms
G/L of Aniline = ‘g’ x 100 gms.
\kg / dm3 of Aniline = ____________.