Describe how would you prepare .005M solution of IO3^-1 starting with potassium iodate KIO3

Answer 1

I am thinking I need to work backwords on this problem so here is my math if anyone can confirm it for me that would be great.

.005M IO3-1 * .5 Lsolutions = .0025 moles of IO3-1

Then I took .0025 mol IO3= .0025 mol KIO3

Then I took .0025 mol KIO3 * 214.001 g/mol=.535g KIO3

pretty much I do not know if this is all the professor is looking for but I do not know where else to go from here.

Answer 2

Yes, the math is correct for 500 ml. of 0.005 M/l. solution. Since 500 ml is not specified in the question, you have to mention it somewhere in your answer.

Also, since this is a chemistry question, you need to describe fully how to do this, the aparatus required, the precautions necessary to fill up exactly to 500 ml, temperatures, etc.
You can read up articles on titration and how to prepare the solutions, for example:
http://en.wikipedia.org/wiki/Titration

Answer 3

Thank you