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Topic: Chemistry behind Epton Titration  (Read 2986 times)

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Offline ecnerwalc3321

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Chemistry behind Epton Titration
« on: January 07, 2013, 04:45:30 PM »
Sorry, I posted this also in the analytical chemistry forum but didn't get any response. 

Recently I performed a titration to determine the amount of SDS surfactant present in a solution.  To do this, I used Methylene blue as the titrant.  To me, the principle of this titration is intuitive but I don't understand why it occurs.   It basically goes like this:

SDS is soluble in water and not in chloroform.  Methylene blue is soluble in water and not in chloroform.  Chloroform and water form two distinct layers.  To determine the amount of SDS present, known amounts of Methylene blue dye is added (titrated) into a vial that contains known volume of chloroform and SDS in water.  Initially, the blue color is on the top aqueous layer.  However, after stirring for some (short) time, the blue color migrates to the chloroform layer which can then be analyzed using UV-vis.  Titration proceeds until no more blue migrates from the aqueous layer to the chloroform layer as evidenced by using UV-vis. 

My understanding is that the methylene blue is a cation that complexes with the anionic head of dodecyl sulfate.  The resulting complex is insoluble in water and thus migrates to the organic layer.  However, I was wondering, since all the ions are in aqueous solution, wouldn't the formation of methylene blue-dodecyl sulfate complex not be possible since all the ions are dissociated?  Or does the complex forms for a very short time near the interface and this is able to cause the migration of color?
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Offline sjb

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Re: Chemistry behind Epton Titration
« Reply #1 on: January 07, 2013, 05:04:33 PM »
Give people a chance to at least read the question. No need for repetition, locked here

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