Study of the presence of oxalate ions in guava fruit at different stages of ripening.

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Solution

Equations involved
$2 K M n O_{4} + 3 H_{2} S O_{4} ⟶ K_{2} S O_{4} + 2 M n S O_{4} + 2 H_{2} O + 4 [O]$
$H O O C - C O O H .2 H_{2} O + [O] ⟶ 2 C O_{2} + 2 H_{2} O \times 5$
Add both reactions
$3 K M n O_{4} + 3 H_{2} S O_{4} + 5 H O O C - C O O H .2 H_{2} O ⟶ K_{2} S O_{4} + M n S O_{4} + 18 H_{2} O + 10 C O_{2}$
Procedure :
(1) Weigh $50 g$ of fresh guava and crush it to a fine pulp using pestle and mortar.
(2) Transfer the crushed pulp to a beaker and add about $50 m l$ dilute $H_{2} S O_{4}$ to it.
(3) Boil the content for about $10$ minutes. Cool and filter the contents in a $100 m l$ measuring flask.
(4) Make up the volume $100 m l$ by adding ample amount of distilled water.
(5) Take $20 m l$ of the solution from the flask and add $20 m l$ of dilute sulfuric acid to it.
(6) Heat the mixture to about $60^{o}$ C and titrate it against $\frac{N}{10} K M n O_{4}$ solution taking in a burette till the end point has an appearance of pink colour.
(7) Repeat the above experiment with 50 g of 1 day, 2 day and 3 day old guava fruit.

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