By labtech on Sunday, July 11, 2004 - 04:38 pm:

Hi:
Need a little help with a titration I did in the lab. I am testing for the assay of Polygylcol Acetate. The USP method call for 160mg of sample mixed with 6ml of .5N KOH, refluxed for 30 min. then tirated with .1N HCL.
I got 29.6 ml for a blank
and 9.9ml for 160.1mg of sample.
My question is how to I figure out the %PGA?
the only other information I have been given is the equil.factor for KOH which is 40.04mg.
%=((6 X .5)-(19.7 X .1)X40.04)/160.1=25.76%
I know my calculation is not correct. Any help wouuld be really appreciated.

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By Mark on Monday, July 12, 2004 - 11:12 am:

29.6-9.9
6 - -------- x 40.04 = mg PGA
5
5 in the denom comes from the KOH being 5x the HCl normality, so you have to correct the volume of titrant for it's concentration.

From this calc I got 157.8 mg, pretty close to the 160 mg originally taken for assay.

Regards,
Mark

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By Mark on Monday, July 12, 2004 - 11:14 am:

Sorry that formula looked OK until I said post the result, lets see if I can write it clearer:

6 - ((29.6 - 9.9)/5) x 40.04 = mg PGA

Regards,
Mark

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By Russ on Thursday, July 15, 2004 - 09:15 am:

I'm not sure I am following this. First, the volume and normality of the KOH is not really needed in the calculations. The meq KOH added is equal to the meq HCl found in the blank titration since 1 eq KOH is titrated by 1 eq HCl. Since one eq KOH reacts with one eq PGA
(AcOR + OH- ---> AcO- + ROH)
the meq KOH consumed = meq PGA (as I am not expecting the polyglycol alcohol to react with KOH)
Since meq KOH consumed =(N HCl)(mL blank - mL sample),
meq PGA = (N HCl)(mL blank - mL sample)
To convert to % PGA, you need a molecular weight.

(N HCl)(mL blank - mL sample)(PGA MW)(100) / (mg sample) = % PGA

The 40.04 is not the PGA molecular weight as the acetate MW is around 59, so I am not really sure what that value is. How was this described to you?
Did you use an automatic titrator or a visual indicator? Depending on your indicator, you might not be able to differentiate between excess OH- and the AcO- produced in the reaction with the caustic, causing significant "problems".

Sorry I can't help more but I guess I really should get back to work!

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By Consumer Products Guy on Saturday, July 17, 2004 - 11:37 am:

This is essentially the AOAC, ASTM, or AOCS Saponification Value test, using your equivalent weight of choice.