Determination of total acidity in citrus fruits
Principle:
It is the sugar/acid
ratio which contributes towards giving many fruits their characteristic flavor
so is an indicator of commercial and organoleptic ripeness.
At the beginning of the
ripening process the sugar/acid ratio is low because of the low sugar content
and high fruit acid contents, this makes the fruit taste sour. During the repining
process, the fruit acids are degraded and sugar contents increases and
sugar/acid ratio achieves a higher value. Overripe fruits have a very low level
of sugar acids and therefore lack characteristic flavor.
This method of
titration involves the process to ascertain the amount of constituent substance
in the sample acid by using a standard counteractive reagent. For example an
alkali (NaOH)
As (NaOH) is a
secondary standard, therefore, we need to standardize it against some standard solution
(whose concentration is known) likewise oxalic acid.
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
A) STANDARDIZATION OF NaOH BY USING OXALIC 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 burette.
(4)
Required amount of oxalic acid to neutralize the NaOH has been determined.
(5) Same process has been repeated for three times and the mean volume was determined.
(6)
The initial pink color of NaOH has been converted into colourless on titrating
it against 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 total acidity in citrus fruits (orange)
Principle:
It is the sugar/acid ratio which contributes towards
giving many fruits their characteristic flavor so is an indicator of commercial
and organoleptic ripeness.
At the beginning of the
ripening process the sugar/acid ratio is low because of the low sugar content
and high fruit acid contents, this makes the fruit taste sour. During the repining
process, the fruit acids are degraded and sugar contents increases and
sugar/acid ratio achieves a higher value. Overripe fruits have a very low level
of sugar acids and therefore lack characteristic flavor.
This method of
titration involves the process to ascertain the amount of constituent substance
in the sample acid by using a standard counteractive reagent. For example an
alkali (NaOH)
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Apparatus and
chemicals
Burette, stand, beaker
100ml, 50ml, 0.1M NaOH, phenolphthalein,
pippete, glass rod etc
CHEMICAL EQUATION
PROCEDURE:
(1 Three fresh orange juices have been cut
into two halves and their extract was collected in a 500ml beaker by using a
squeezer.
(2 This orange has been filtered involving
no solid particles.
(3 10ml of this juice has been taken in a beaker
by means of pipette.
(4 10ml of the water has been added to dilute the
sample.
(5 Followed by 2 to 3 drops of phenolphthalein as
an indicator and its presence didn’t impart any color due to acidic media.
()
Sample was finally titrated against 0.1M NaOH
solution.
(7 Same procedure has been repeated for three
times and means volume was noted.
OBSERVATIONS AND CALCULATIONS
|
Sr
no
|
Initial
volume
|
Final
volume
|
Volume
of NaOH used
|
|
01
|
0.0ml
|
10.0ml
|
10.0ml
|
|
02
|
10.0ml
|
18.5ml
|
8.5ml
|
|
03
|
18.5ml
|
26.5ml
|
8.0ml
|
Mean = 10+8.5+8.0/3=8.8ml
1000ml
of NaOH has no of moles=0.1mol
1ml
of NaOH has no of moles = 0.1/1000
8.8ml
of NaOH has no of moles = 0.1/1000x8.8=0.00088mol
Now,
No of moles of NaOH =
no of moles of citric acid
3 = 1
1 = 1/3
0.00088 = 0.00088 x 1/3
= 0.000293mol
So,
20ml 0f the sample has
no moles = 0.000293
1ml of the sample has
no of moles = 0.000293/20
1000ml of the sample
has no moles = 0.000293/20 x 1000 = 0.01465M
Now,
Strength in g/L =
molarity x molecular weight
= 0.01465 x 192.124
= 2.815 g/L
Now,
Dilution factor =
total/original
= 20/10 =2
By multiplying with
dilution factor we get ,
Strength in g/L = 2.815
x 2 = 5.63 g/L
RESULT:
Hence the orange sample
juice contains 5.63g/L of citric acid.
(c) Determination of
total acidity in citrus fruits (lemon)
Apparatus, chemical
equation and chemicals are in accordance with part B of this experiment.
Procedure:
(1 Four fresh lemons have been cut into two
halves followed by the collection of extract in a 100ml beaker.
(2 This extract has been filtered and 3ml
of this extract has been taken in a beaker by means of pipette.
(3 Followed by 7ml of the water to dilute
the sample up to 10ml.
(4 2 to 3 drops of the phenolphthalein has
been added as indicator.
(5 Initially, colorless appearance of the
extract has been observed followed by pink color at the end of the titration
using 0.1M NaOH.
(6 The whole procedure has been repeated
for three times to obtain the mean volume.
OBSERVATIONS AND CALCULATIONS
|
Sr
no
|
Initial
volume
|
Final
volume
|
Volume
of NaOH used
|
|
01
|
0.0ml
|
10.0ml
|
10.0ml
|
|
02
|
10.0ml
|
20.3ml
|
10,3ml
|
|
03
|
21.0ml
|
31.1ml
|
10.1ml
|
Mean
= 10+10.3+10.1/3=10.1
1000ml
of NaOH has no of moles=0.1mol
1ml
of NaOH has no of moles = 0.1/1000
10.1ml
of NaOH has no of moles = 0.1/1000x10.1=0.00101mol
Now,
From
the equation,
No of moles of NaOH =
no of moles of citric acid
3 = 1
1 = 1/3
0.00101 = 1/3 x 0.00101
= 0.00034 mol
10ml
of sample has no of moles = 0.00034
1ml
of sample has no of moles = 0.00034/10
1000ml
of sample has no of moles= 0.00034/10 x 1000 = 0.034M
Strength
in g/L = molarity x molecular weight
= 0.034 x 192.124
= 6.53g/L
Now
Dilution
factor = total/ original
=10/3=3.33
Now
Strength
in g/L = 6.53 X 3.33 =21.77g/L
RESULT:
Hence
the given lemon sample contains 21.77g/L of citric acid.
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