Question

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 <i>how</i> 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