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Redox equilibria

Hey, can someone help me with the question please?
25.00cm3 of a solution of potassium iodate (V) was pipetted into a conical flask. 25.00cm3 of dilute sulphuric acid and 2g of potassium iodide (an excess) were added. The iodine liberated was titrated with 0.104 mol dm-3 sodium thiosulfate using starch as indicator. The mean titre was 23.20cm3 . Calculate the concentration of the potassium iodate (V) solution.
Thank you !
(edited 4 years ago)
Hi, do you know what the result is?
The first step you need to do is to write down two equations:

IO3- + 5I- + 6H+ ———> 3I2 + 3H2O
I2 + 2S2O3^2- ————> 2I- + S4O6^2-

Then you work out the moles for thiosulfate (S2O3^2-) which is concentration x volume so you do 0.104x(23.20/1000) = 0.0024128

Then you work out the moles for Iodine (I2) using the ratio 1:2 so you do 0.0024128/2 = 0.0012064

Then the final step is to work out the concentration of potassium iodate or iodate (IO3-) by doing moles/volume. We know the moles for iodate is 0.0012064 because the ratio is 1:1 and we know the volume which is 25cm3.

So you do 0.0012064/(25/1000) = 0.048 (2 s.f.) and that should be the answer I think
(edited 4 years ago)

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