Chemical Forums

Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Gman4554 on May 24, 2005, 02:08:36 PM

Title: Standardization of 0.05M KIO3
Post by: Gman4554 on May 24, 2005, 02:08:36 PM
The USP and other methods for the assay titration of Potassium Iodide involve Potassium Iodate as the tritrant. Normally, weigh 10.700gms in 1 liter of water for 0.05M solution. They offer no standardization of the titrant.

Is there a way to standardize this titrant to calculate the exact normality.

tks,
Gman4554
Siegfried-USA
Pennsville, NJ Pharmaceutical Division QC
Title: Re:Standardization of 0.05M KIO3
Post by: Borek on May 24, 2005, 03:31:14 PM
Could you describe the chemistry behind the titration? As far as I remember to determine iodides you use standard thiosulfate solution as titrant after iodate is added in excess to oxidise iodides to iodine in acidic solution.

No idea about the iodate standarization other than some reverse of that situation - add excess of iodide and some acid to known volume of iodate solution, then titrate iodine with thiosulfate,
Title: Re:Standardization of 0.05M KIO3
Post by: Gman4554 on May 25, 2005, 03:03:00 PM
There is a published method for the assay of KI in the ACS and it requires a direct titration of KI with 0.05M KIO3 using either CHCL3 or amaranth as the indicator.

It's not a thiosulfate titration of the excess Iodine.

The next drop of 0.05M KIO3 reacts with the external indicator to either destroy the viloet color in the CHCL3 or if amaranth is used, the color is red to yellow endpoint.

The half reaction for the CHCL3 method is 2I + IO3 + 6H + 6CL yields 3ICL2 +3H2O.

As you can see, the equivalent wt of KI in this reaction is 2 x it's mol wt (which is 166).
Therefore 332 or in millieqwt its 33.2.

The assay equation is %KI=ml's x 0.05 x 33.2 divided by spl wt in g.

Once again the 0.05M KIO3 is made up by weighing 10.7000gms of KIO3 in a liter of water. Mol wt of KIO3=214, 10.7/214=0.05M.

How would we know the exact strength of the KIO3.
Title: Re:Standardization of 0.05M KIO3
Post by: Borek on May 25, 2005, 06:04:15 PM
Once again the 0.05M KIO3 is made up by weighing 10.7000gms of KIO3 in a liter of water. Mol wt of KIO3=214, 10.7/214=0.05M.

How would we know the exact strength of the KIO3.

I just realized I am not sure what you are asking about. That's the price I am  paying for using language I don't know ;)

KIO3 is a standard substance - if it is dried before weighting and diluted to mark you know exact concentration, if that's what you ask about.

If the problem is with a ready solution of unknown concentration, I will try to use thiocyanate solution as I described before.