mols S2O3^2- = M x L = ?
mols IO3^- = 1/6 * mols S2O3^-
grams KIO3= mols KIO3 x molar mass KIO3
Post your work if you get stuck.
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.
Type your answer to 4 significant figures (no units as it will not be able to be marked by Blackboard).
I - ¡æ ¨öI2 + e -
IO3- + 6H+ + 5e - ¡æ ¨öI2 + 3H2O
The liberated iodine reacts as follows:
I2 + 2e - ¡æ 2I-
2S2O3 2- ¡æ S4O62- + 2e-
Hint: 6 equivalents of S2O32- are equivalent to IO3-.
3 answers
How am I going to find the mols? Actualy I have corrected the units in my way of finding the answer. So far I have done the following:
30.65cm3/1000*0.1005= 3.080325x10-3(on the top, I don't know how to write it here)
What about next?
30.65cm3/1000*0.1005= 3.080325x10-3(on the top, I don't know how to write it here)
What about next?
My first line shows you how to find mols S2O3^2- and the second line shows how to convert to mols IO3^-.
5 answers