Determination of CO2
in soft drink
Principle:
Co2
gives the beverage its sparkle and tangy taste and prevents spoilage while it
has not been conclusively proved that carbonation offers a direct medical
benefit , carbonated beverages are used to alleviate postoperative nausea when
no other food can be tolerated as well as to be ensure adequate liquid intake.
Co2
is supplied to soft drink manufacture in either solid form (dry ice) or liquid
form maintained under approximately 1200psi pressure in heavy steel container. Light
weight steel containers are used when the liquid co2 is held under refrigeration.
In that case, the internal pressure is about 325psi. Carbonation is affected by
chilling the liquid and cascading it in thin layers over the series of plates
in an enclosure containing co2 gas under pressure. The amount of gas
water will absorb increases as the pressure is increased and temperature is
decreased.
Apparatus and chemicals:
Burette,
stand, beaker 100ml, 50ml, 0.1M NaOH, 0.05M oxalic acid, phenolphthalein,
pippete, glass rod etc
CHEMICAL EQUCATION:
(COOH)2
+ 2NaOH Ã Na2C2O4
+ H2O
H2CO3
+ NaOH Ã NaHCO3
+ H2O
A) STANDARIZATION OF NaOH BY
USING OXALLIC ACID.
PROCEDURE:
(1)
10ml of 0.1M
NaOH has been taken in the conical flask.
(2) Followed by 2 to 3 drops of phenolphthalein as
an indicator.
(3) 0.05M oxalic acid has been taken in the
burrette.
(4) Required amount of the oxalic acid to neutralize
the NaOH has been determined.
(5) Same process has been repeated for three times
and mean volume was determined.
(6) The initial pink colour of NaOH has been
converted into colourless on titrating it againt standard oxalic acid solution
taken in the burette.
(7) Hence, mean volume was noted.
OBSERVATIONS AND
CALCULATION:
Amount of the sample = 10ml
Molarity of NaOH used = ?M
Mean volume of oxalic acid = ?
|
Sr no
|
Initial volume
|
Final volume
|
Volume of oxalic acid used
|
|
01
|
0.0ml
|
10.0ml
|
10.0ml
|
|
02
|
10.0ml
|
20.0ml
|
10.0ml
|
|
03
|
20.0ml
|
30.0ml
|
10.0ml
|
Mean volume = 10+10+10/3= 10
1000ml of oxalic acid has no of moles = 0.05mol
1ml of oxalic
acid has no of moles = 0.05/1000
10ml of the oxalic acid has no of moles =
0.05/1000x10=0.0005mol
From equation, we have
(COOH)2 + 2NaOH Ã Na2C2O4 + H2O
No of moles of oxalic acid = no of moles of NaOH
0.0005 mol = 2x0.0005mol
= 0.001mol
10ml of NaOH contain no of moles = 0.001mol
1ml of NaOH contain no of moles = 0.001/10
1000ml of NaOH has no of moles = 0.001/10x1000=0.1M
So molarity of NaOH = 0.1M
(b)Determination of co2 in
soft drinks
Procedure:
(1) 20ml of the soft drink has been taken in a conical
flask.
(2) Followed by 2 to 3 drops of phenolphthalein as an indicator.
(3) Reaction mixture has been titrated with 0.1M NaOH
solution.
(4) 0.1M NaOH has been taken in the bRurette.
(5) Volume of the NaOH required to neutralize H2CO3 has
been determined.
(6) The colorless solution of soft drink has been
converted into pink on the end of the titration.
(7) Same process has been repeated for three times and
means volume has been noted.
OBSERVATIONS AND CALCULATIONS
Amount of the sample = 20ml
Molarity of NaOH used = 0.1M
Volume of the NaOH used =?
|
sr No
|
Initial volume
|
Final volume
|
Volume of the NaOH used
|
|
01
|
0.0ml
|
8.7ml
|
8.7ml
|
|
02
|
8.7ml
|
17.2ml
|
8.5ml
|
|
03
|
17.2ml
|
25.8ml
|
8.6ml
|
Mean=
8.7+8.5+8.6/3=8.6ml
1000ml of NaOH has no of moles = 0.1mol
1ml of NaOH has no of moles =
0.1/1000
8.6ml of NaOH has no of moles = 0.1/1000x8.6= 0.00086mol
From equcation, we have
H2CO3 + NaOH Ã NaHCO3 + H2O
No of moles of NaOH = no of moles of CO2/H2CO3
0.00086mol = 0.00086mol
So
20ml of the soft drink contains no of moles of co2 =
0.00086mol
1ml of the soft drink contain no of moles of co2 = 0.00086/20
1000ml of soft drink contains no of moles of co2 =
0.00086/20x1000=0.043mol
Now
Strength in g/L = Molarity x molecular weight
= 0.043 x 44 = 1.892g/L
= 1.892 x 1000mg/L
= 1892mg/L
RESULT:
Hence the given soft drink sample contains 1892mg/L of CO2.
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