Investigation to Determine the Percentage of Citric Acid in Lime Juice Essay Sample

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Introduction of TOPIC

Discussion and Method

A primary standard is a reagent that is extremely pure, stable, has no waters of hydration, and has a high molecular weight. This enables a concentration to be calculated from a direct weighing of its pure form.

NaOH is cannot be classed as a primary standard as a concentration cannot be calculated directly from a weighing. This is due to the fact that over time, in its pure form, NaOH will absorb atmosp

  1. Chemistry

heric water and react with atmospheric CO2. With NaOH solutions, again, over time, they will absorb atmospheric CO2 and will also slowly react with glass bottles in which it is stored. Therefore, before using NaOH as a standard in a titration, the solution must be standardised to determine the concentration which is being used.

Using good laboratory techniques, potassium hydrogen phthalate was weighed by difference into a conical flask. 25cm3 distilled water was added and was then mixed until the potassium hydrogen phthalate had dissolved. Phenolphtalein indicator was added and the solution was carefully titrated with an unknown concentration of NaOH solution until a pale-pink end point was reached. The titration was repeated for accuracy.

Results

Potassium hydrogen phthalate:

1. Weight by difference = 0.5525g

2. Weight by difference = 0.5089g

3. Weight by difference = 0.5193 g

Titration Data for NaOH

Initial Burette Reading (cm3)

Final Burette Reading (cm3)

Volume Used (cm3)

Rough Titration

(i)

0.1

27.0

26.9

1st Accurate Titration (ii)

0.4

25.1

24.7

2nd Accurate Titration (iii)

0.3

25.5

25.2

Standardisation Calculations

To calculate

1. Moles Standard = 10-3

Concentration NaOH (to 4 significant figures)

= 10-3

= 0.1006 moldm-3

1. 10-3

Concentration NaOH (to 4 significant figures)

= 10-3

= 0.1009 moldm-

3 1. 10-3 Concentration NaOH

(to 4 significant figures)

= 10-3

= 0.1009 moldm-3

Average of (ii) and (iii) for use in concentration of citric acid calculations

= 0.1009+0.1009 / 2

=0.1009 moldm-3

B – Titration of Lime Juice and % Citric Acid

Method

Using good laboratory techniques 5cm3 of lime juice was added to a conical flask and diluted with 25cm3 distilled water. Phenolphtalein indicator was added and the lime solution was titrated with the standardised NaOH solution until the permanent pale pink end point was reached. The titration was repeated for accuracy.

Results

Titration Data for Citric Acid (C6H8O7)

Initial Burette Reading (cm3)

Final Burette Reading (cm3)

Volume Used (cm3)

Rough Titration

(i2)

0.15

19.8

19.65

1st Accurate Titration (ii2)

0.50

20.20

19.70

2nd Accurate Titration (iii2)

0.20

19.9

19.70

NB: The reaction is 1:2 ratio, hence the mols of C6H8O7 are half that of NaOH used.

Calculations

(i2) Moles of NaOH = concentration of NaOH (0.1009 moldm-3) x titration volume (dm3)

=

= 1.983×10-3 moldm-3

Moles of C6H8O7 in 5cm3

= 1.983×10-3 x 0.5

=9.915×10-4 moldm-3

Mass of C6H8O7 in 5cm3 = 9.915×10-4 moldm-3 x RRM C6H8O7 (192)

= 9.915×10-4 x 192

=0.190

% amount is in 100 parts whereas the above amount is in 5 parts. The percentage C6H8O7 in the juice sample is therefore mass C6H8O7 x (100 ÷ 5)

= 0.190 x 20

= 3.8 % C6H8O7 in 5cm3 lime juice

(ii2) Moles of NaOH = concentration of NaOH (0.1009 moldm-3) x titration volume (dm3)

=

= 1.988×10-3 moldm-3

Moles of C6H8O7 in 5cm3

= 1.988×10-3 x 0.5

=9.94×10-4 moldm-3

Mass of C6H8O7 in 5cm3 = 9.94×10-4 moldm-3 x RRM C6H8O7 (192)

= 9.94×10-4 x 192

=0.191

% amount is in 100 parts whereas the above amount is in 5 parts. The percentage C6H8O7 in the juice sample is therefore mass C6H8O7 x (100 ÷ 5)

= 0.191 x 20

= 3.82 % C6H8O7 in 5cm3 lime juice

(ii3) Moles of NaOH = concentration of NaOH (0.1009 moldm-3) x titration volume (dm3)

=

= 1.988×10-3 moldm-3

Moles of C6H8O7 in 5cm3

= 1.988×10-3 x 0.5

=9.94×10-4 moldm-3

Mass of C6H8O7 in 5cm3 = 9.94×10-4 moldm-3 x RRM C6H8O7 (192)

= 9.94×10-4 x 192

=0.191

% amount is in 100 parts whereas the above amount is in 5 parts. The percentage C6H8O7 in the juice sample is therefore mass C6H8O7 x (100 ÷ 5)

= 0.191 x 20

= 3.82 % C6H8O7 in 5cm3 lime juice

To determine a final percentage, the average of the two accurate titrations is calculated as

(3.82+3.82) ÷ 2

=3.82%

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