How would I work out the weight of potassium iodate per 100g of sample by the following titre results?

1g of impure KIO3 was made to a standard solution by 250ml of water.

25ml was put into a conical flask with 10ml of H2SO4 and KI in excess (1g)

the end point was 23.7

I have more endpoints but I want to be able to do the rest myself... just need the breakdown of the calculations please :)

most appreciated :)

Update:

it was titred against 0.1M sodium thiosulfate solution!

1 Answer

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  • Anonymous
    8 years ago
    Favorite Answer

    IO3^- + 5I- + 6H+ ---> 3I2 + 3H2O

    then:

    3I2 + 6S2O3^2- ---> 6I- + 3S4O6^2-

    so: 6 moles S2O3^2- eqivalent to 1 mole IO3^-

    moles thiosulphate = 0.1 x 23.7/1000 = 0.00237 moles

    hence moles KIO3 is 0.00237/6 = 0.000395 moles

    this is in 0.1g impure KIO3 (because 250ml original solution and you took a 25 ml sample).

    Calculate mass of KIO3 present (mass = moles x RFM) (0.08453g)

    then % purity = (mass/0.1) x 100 (84.53%)

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