Active ingredient in phosphomidon

DETERMINATION OF ACTIVE INGREDIENT IN PHOSPHOMIDON

Principle

         Phosphomidon (Formula: C10H19O5 NPCI with molecular weight 299.698) along with its byproducts reacts with iodine in strongly alkaline medium (NaOH) consuming four equivalents of iodine. The byproducts alone, but not phosphomidan consume iodine in weak alkaline medium (NaCO3). The different titration thus gives the content of phosphamidon.

Reaction

2Na2S2O3+I2         →   Na2 S4O6 +2NaI

Materials required

a)      2N Sodium hydroxide, b) 2N Sodium carbonate, c) 5N HCL, IV) 0.1N Sodium thiosulphate, v) 0.1N Iodine, vi) 250 ml Volumetric flask, vii) 250 ml Conical flask, viii) Starch indicator and ix) Balance.

Procedure

§  Transfer exactly 1.5 g of phosphomidon into a 250ml volumetric flask, dissolve with distilled water and make up the volume.

§ 

Transfer 10ml of the above solution into 250ml conical flask and add 20 ml of 0.1N iodine solution and 20ml of 2N sodium hydroxide. Set aside in darkness for 30minutes at 25 +  2OC.

§  Add 20ml of 5N HCL and titrate the excess unreacted iodine against 0.1N sodium thiosulphate till the added starch is decolourised or until the disappearance of dark blue colour.

§ 

Transfer another 10ml of the original solution into a 250ml conical flask and add 5ml of 0.1N iodine and 5ml of 2N sodium carbonate solution. Set aside for 15min darkness at 25 +  2OC.

§  Then add 5ml of 5N HCL and titrate the excess unreacted iodine against 0.1N sodium thiosulphate till the added starch indicator is decolourised or until the disappearance of dark blue colour.

Note : It is essential that 0.1N iodine solution is added before the addition of alkali(sodium hydroxide or sodium carbonate) otherwise slight hydrolysis will occur vitiate the experiment.

§  Calculate the actual volume of 0.1N iodine consumed from the two titre values and from this the active ingredient content in phosphomidan.

Calculation

Volume of 0.1N Iodine taken for first titration

=

V1

Volume of 0.1N Na2S2O3 consumed in back titration

=

A ml

Actual volume of 0.1N iodine that react with phosphomidan and byproducts

=

V1- A=X ml

Volume of iodine taken for second titration

=

V2 ml

Volume of 0.1N Na2S2O3 consumed in back titration

=

B ml

Actual volume of 0.1N iodine that react with byproducts alone

=

V2     B=(Y ml)

Volume of iodine 0.1N iodine that react with phosphomidan alone

=

(V1A)- (V2- B) ml or (x-y) ml =q ml

Four equivalents of iodine react with 1 equivalent of phosphomidan i.e. 299.698 g of phosphomidan. Hence one equivalent of iodine reacts with 74.9245g of phosphamidon.

1000 ml of 1N iodine

=

74.925 g of phosphomidan

1ml of 0.1N iodine

=

0.074925 g of phosphamidon

q ml of 0.1N iodine

=

0.0074925 x q g of phosphomidan

This is present in 10 ml of solution

In 250 ml of solution

=

of phosphomidan

This is present in 1.5g of sample

In 100 g sample

=