Chemistry

The mass of potassium iodate (KIO3) contained in an impure sample was determined by titration with sodium thiosulfate (0.1005 M). The impure KIO3 was dissolved in 50 cm 3 of water, an excess of potassium iodide (KI) and 5 cm^3 of dilute sulphuric acid was then added. At the starch endpoint, 30.65 cm^3 of thiosulfate solution had been added. Calculate the mass of KIO3 in the sample.

I - → ½I2 + e -

IO3- + 6H+ + 5e - → ½I2 + 3H2O

The liberated iodine reacts as follows:

I2 + 2e - → 2I-

2S2O3 2- → S4O62- + 2e-

Hint: 6 equivalents of S2O32- are equivalent to IO3-.

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  1. If you want this done with equivalents, then
    mL x M x (molar mass KIO3/6000) = grams KIO3

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  2. How do I find mL?

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  3. The problem tells you.
    30.65 cm^3 = 30.65 cc = 30.65 mL.
    1 cc = 1 mL = 1 cubic centimeter = 1 cm^3.

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  4. I got 0.1099g

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  5. Is that correry?

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