Benzoic Acid
Determination of distribution co-efficient of benzoic acid between ether and water .
PRINCIPLE :
When two solute is shaken up with two liquids
which are immiscible with each other , but
in both of which the solute is soluble, then the solute is soluble ,
then the solute is found to distribute itself between the two solvents in such
a way the ratio of the concentration of the solute in the two liquids is constant
at constant temperature .
If C1 and C2 are the
concentration of the solute in solvent 1 and solvent 2 respectively, then
C1
------------- =KD = Constant
C2
The ratio KD is known as the distribution co-efficient or
partition co-efficient .If the benzoic acid is shaken with ether and water it
will dissolve in both liquids . So KD may be expressed as
[BA] in organic
layer C2
KD =
----------------------------- = -------------- ( BA= Benzoic acid )
[BA] in aqueous layer C1
SV2 x
V V2
=---------
------------ = -------
V SV1 V1
Where,
C2 = Cone of benzoic acid in ether layer.
C1 = Cone of benzoic
acid in water .
V2 = Volume of NaoH
required to titrate ‘V’ ml of organic layer.
V1 = Volume of NaoH required to titrate ‘V’ ml of aqueous
layer .
S = Strength of NaOH .
CALCULATION-
At first we
prepare 0.1(N) oxalic acid solution and 0.1(N) NaOH solution. Then titrate to
find exact 0.1(N) of NaOH solution in the following way.
V1= V2 =
S1= S2
=
We know that, V1 S1 = V2 S2
V1 S1
=> S2 = ---------- = ---------------
=
V2
Let the
volume ‘V’ of exact 0.1(N) of NaOH solution
S V = S1V1
S1V1
=> V = ----------- = ------------- (y) V = (y)
ml =
S
Now we add ( 100-y) ml of water with ( y)
ml of NaOH which strength is S (N) and then we get 100ml of 0.1N NaOH solution.
Again we titrate organic layer and aqueous
layer with exact 0.1(N) NaOH solution. Then we get the following data –
APPARATUS :
i) Three Stoppered bottles
.
ii) Burette & Pipette .
iii) Conical flask
iv) Shaking machine .
CHEMICALS :
a) Solid benzoic acid
b) Ether
c) Oxalic acid
d)
Sodium hydroxide
e) Phenolphthalein .
PROCEDURE :-
i) Prepare 200ml solution of 0.1(N) benzoic acid in eather
.
ii)
Take solution in 250ml boottles in the following ratio .
Eather water
40 60
50
50
60
40
iii)Shake for one hour .
iv) Titrate organic and aqueous layer with
standardized sodium hydroxide solution .
Experimental data for Organic Layer :
No.of
Observation between ether & water
|
Volume of organic layer
|
N/10 NaOH solution in burette
|
Difference in volume V2
|
||
40 : 60
|
10
|
||||
50 : 50
|
10
|
||||
60 : 40
|
10
|
||||
Experimental data for aqueous Layer :
No.of
Observation between ether & water
|
Volume of aqueous layer
|
N/10 NaOH solution in burette
|
Difference in volume V1
|
||
40 : 60
|
10
|
||||
50 : 50
|
10
|
||||
60 : 40
|
10
|
||||
For 40 : 60ml ether & water :
V2
KD = ------------- = X1 = ----------- =
V1
For 50 : 50ml water & ether :
V2
KD = ------------- = X2 = ----------- =
V1
For 60 : 40ml ether & water :
V2
KD = ------------- = X3 = ----------- =
V1
x1 + x2
+ x3
Mean = ------------------ =
------------------- =
3
V2
= Organic Layer
V1 = aqueous Layer
RESULT :-
The distribution co-efficient or
partition co-efficient of benzoic acid between ether and water is ------
PRECAUTION :-
i) The bottles ware washed cleand and dried
carefully .
ii) The shaking was
contained at least for an hour .
iii) The titration process was done carefully
.
iv) The reading of
the burette was observed carefully and noted the end points of the pipette
pushed up to the bottomof the bottle with putting the thumb at the upper end of
the pipette and carefully observed that the aqueous layer not enter into the
pipette .
Benzoic Acid
Reviewed by M H Islam
on
10:27 AM
Rating:
Reviewed by M H Islam
on
10:27 AM
Rating:
