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Topic: Determining purity of a potassium iodate sample  (Read 10049 times)

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Offline x58eno

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Determining purity of a potassium iodate sample
« on: February 14, 2010, 08:54:44 PM »
Background:
We were given a sample of KIO3 contaminated with NaCl. I measured out 0.9180g of the impure KIO3 and made up a 250mL solution with D.I. water. 25mL of this solution was then transferred to a separate flask where 1g of potassium iodide and 5ml of 2M sulfuric were added. This solution was titrated with a 250mL solution of 0.1M sodium thiosulfate.

The goal is to determine the purity of the potassium iodate. This is my first run in with a titration of any kind, so I'm honestly at a complete loss. Here's my attempt at a solution:

7mL Na2S2O3 x 0.1 mole Na2S2O3/1000mL Na2S2O3 x 1 mole I2/2 moles Na2S2O3 = 3.5x10-4 moles of I2

3.5x10-4 moles of I2 x 1 mole IO3/3 moles I2 x 174g IO3/1 moles IO3 = 0.0203g of IO3
_______________________________________________

This only gives a 2.2% purity, which I believe is incorrect since I earlier recrystallized 0.3170g from a 4g sample, which gives a 7.925% purity. As stated previously, I'm just at a complete loss for what to do. The attempted solution is from bits and pieces of information I was able to find on google. Any help would be appreciated.

Offline Borek

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Re: Determining purity of a potassium iodate sample
« Reply #1 on: February 15, 2010, 03:04:43 AM »
http://www.titrations.info/iodometric-titration

You are not completely off. Two things - first, you are interested in mass of POTASSIUM iodate, 174g is not molar mass of potassium iodate but just IO3- alone. Second, whatever mass you will calculate will be mass in 25 mL aliquot, but your sample was dissolved in 250 mL of water - take it into account.

I guess 7 mL was an average volume of the titrant used?
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Offline x58eno

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Re: Determining purity of a potassium iodate sample
« Reply #2 on: February 15, 2010, 11:15:19 AM »
Thanks for the reply. Ya, the 7ml was the average volume of the titrant.

Alright, so:
3.5x10-4 moles of I2 x 1 mole IO3/3 moles I2 = 1.17x10-4 moles of IO3

Dividing that by the volume of the aliquot, gives a concentration of: 4.67x10-3M

But since it was originally dissolved in a 250mL, we multiply by that to get the actual number of moles of IO3: 1.17x10-3 which is then multiplied by the MW (214) to give 0.2497g of KIO3 which in the end gives a 27.2% purity.

I know the calculated purity should be higher than that of the recrystallization (7.925%), but should there be such a discrepancy? 

Offline Borek

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Re: Determining purity of a potassium iodate sample
« Reply #3 on: February 15, 2010, 12:05:45 PM »
I know the calculated purity should be higher than that of the recrystallization (7.925%), but should there be such a discrepancy? 

All I can tell you is that 0.2497 g is a correct titration result. Plus minus significant figures, but that's another story.
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