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Investigatory Project on presence of Oxalate Ions in Guava

 

Investigatory Project on the Presence of Oxalate Ions in Guava Fruit at Different Stages of Ripening

Aim of the Project

 

The aim of the project is to estimate the amount of oxalate ions present in guava fruit at different stages of ripening.

 

Apparatus and Chemicals


1. 100 mL standard flask

2. Burette and pipette

3. A pestle and mortar

4. Chemical balance

5. N/20 KMnO4 solution

6. AR oxalic acid


Theory


The oxalate ions arc extracted from the fruit by boiling the pulp with dil. H2SO4. Oxalate is then converted to oxalic acid. The hot solution is then titrated against KMnO4 taken in the burette. KMnO4 solution is standardised using standard oxalic acid solution. The reactions involved in the estimation are

 

2KMnO4 + 3H2SO4  K2SO4 + 2MnSO4 + 3H2O + 5[O]


H2C2O4 + [O]  H2O + 2CO2

 

Procedure

 

Collect about 250 g of mature guava fruit. Weigh 50 g of fresh guava and crush it to a fine pulp using a pestle and mortar (keep the remaining fruit for ripening). Transfer the crushed pulp to a beaker and add about 50 mL dilute H2SO4 to it. Boil the contents for about 10 minutes. Cool and filter the contents to a 100 mL standard flask. Wash the beaker twice with small quantity of water and transfer to the standard flask through the same filter paper. Then make up the volume to 100 mL and shake well.


Pipette out 20 nil of the solution into into a conical flask and add about 20 mL of dilute sulphuric acid to it. Heat the mixture to around 60°C and after that titrate it against the N/20 KMnO4 solution. The end point is the appearance of permanent light pink colour. Repeat to get concordant titre value.


Repeat the above experiment with 50g guava fruit kept for 1, 2 and 3 days respectively. For standardisation of KMnO4, weigh accurately about 0.80 g of AR oxalic acid crystals using a chemical balance and transfer to a 250mL standard flask. Make up to 250 mL. Pipette out 20 mL of this solution into a conical flask and add about 20 ml of dilute sulphuric acid to it. Heat the mixture to around 60°C and after that titrate it against the N/20 KMnO4 solution. Repeat to get concordant values.


OBERVATIONS AND CALCULATIONS


Mass of H2C2O4 . 2H2O 250 ml = w1 g


Equivalent weight of oxalic acid = 63


Normality of oxalic acid solution = w1 x 4/63 = N1


Vol. of oxalic acid solution pipetted out = 20 mL


Vol. of KMnO4 required to react with 20 ml of oxalic acid solution = v1 mL


Vol of KMnO4 x Normality of KMnO4 = Vol. of oxalic acid x Normality of oxalic acid.


v1 x Normality of KMnO4 = 20 x N1


Normality of KMnO4 = 20 x N1/ v1 = N2

 

Fresh guava extract


Mass of guava fruit taken = 50.0 g


Vol. of guava extract pipetted out = 20 mL


Vol. of KMnO4 required = v2 mL


Therefore, Normality of oxalate ions in the guava extract =v2 x N2/20 = N3


Equivalent weight of oxalate ion (C2O42-) = 44


Therefore, Mass of oxalate ions in the whole of the prepared solution (i.e., 50 g guava fruit) = N3 x 44/10 g


Similarly, calculate the mass of oxalate ions in 50.0 g guava fruit kept for 1, 2 and 3 days.

 

Conclusion

 

Compare the mass of oxalate ions present and write a statement.

 

Viva Questions and Answers

 

1. How does KMnO4 react with oxalic acid?


Ans: Read the theory of the project.

 

2. What do you mean by volumetric analysis?


Ans: The quantitative analysis which involves exact measurement of volumes of reacting liquids is termed volumetric analysis.

 

3. What is a primary standard?


Ans: The substance whose standard solution can be prepared directly by weighing is known as a primary standard.

 

4. What are the primary standards used in permanganometry?


Ans: Crystalline oxalic acid H2C2O4 . 2H2O

Mohr’s salt FeSO4 . (NH4)2SO4 . 6H2O

 

5. Why do we heat the solution to 60°C when oxalic acid is titrated against KMnO4?


Ans: Heating is done to enhance the rate of reaction.

 

6. What is the equivalent weight of KMnO4?


Ans: 31.6 (Molecular mass/10)

 

7. The titration of oxalic acid against KMnO4 is slow in the beginning, but becomes fast as reaction proceeds. Why?


Ans: As the reaction proceeds, manganous sulphate is formed which acts as a catalyst and the reaction proceeds faster.

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