KD calculations (Organic chem lab)

[ChemDoodle]

Chloroform is used to extract caffeine from water.If the distribution coefficient, KD=10.What relative volume of chloroform-water should be used to extract 90% of the caffeine present in one single extraction?

Now i know that KD=Co/Cw and ive solved a question on how much weight of an acid will be removed by extraction.But i'm lost here..don't i need the solubility of the products to solve this?

[Borek]

No, just assume it was not exceeded.

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[ChemDoodle]

Can u give me a hint on how to solve it? Like how to start?

[Borek]

Try to express the condition (90%) in terms of volumes of water and chloroform, and concentrations of caffeine in both. Name them Vc, Vw, Cc, Cw and use these symbols.

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[ChemDoodle]

Okkk..so the lab instructor didnt discuss ANY of this in the lab.She just mentioned the whole KD=co/cw thing,she didnt even stress on it..So i'm clueless..really.Sorry for being a pain..but a little more help,please?

[Borek]

I suppose these thing were not discussed just like multiplication table wasn't - you are expected to already know it :wink:

How many moles of caffeine in Vc of Cc chloroform solution?

How many moles of caffeine in Vw of Cw water solution?

How many moles total?

What does it mean '90% extracted'?

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[ChemDoodle]

But how am i supposed to know any of this?
n=c/v but i have none of these values.All i have is the 90% caffeine..so should i assume it to be 1L & take 90% of that?
Another thing, KD=Cc/Cw=10 in this case?

[Borek]

Don't worry about not knowing values - you are interested in ratio only, and it can be found using only symbols. However, if the symbols throw you off, try to assume some real values - say 10 mL and 0.01M.

And yes - KD=Cc/Cw=10.

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[ChemDoodle]

Thankyouu..i solved it :D Its 10:9?

[Borek]

10:9 or 9:10 :wink:

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