Asked by Gemma

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-.
Answered by DrBob222
If you want this done with equivalents, then
mL x M x (molar mass KIO3/6000) = grams KIO3
Answered by Gemma
How do I find mL?
Answered by DrBob222
The problem tells you.
30.65 cm^3 = 30.65 cc = 30.65 mL.
1 cc = 1 mL = 1 cubic centimeter = 1 cm^3.
Answered by Gemma
I got 0.1099g
Answered by Gemma
Is that correry?
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