Equilibrium law
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When extracting caffeine from a liquid, why are two small additions of trichloroethane used as opposed to one larger one? Is there a worked example available?
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Jenkin writes .......
Using two small portions of the solvent is more efficient than using one large one; it will extract more of the caffeine.
When you carry out a solvent extraction, an equilibrium is established between the solute in the caffeine in the aqueous layer and that in the organic (trichloroethane) layer. There will be an equilibrium constant for this system:
Kc = [caffeine in C2H3Cl3]/[caffeine in water]
This constant is often called the distribution constant or partition constant. I don’t have a value for this particular example so I will use a different one as an illustration.
The distribution constant for iodine between CCl4 and water is 90. Compare the masses of iodine extracted from 1 dm3 of water containing 0.3 g of iodine with (a) 20 cm3 of CCl4 , and (b) two portions of 10 cm3.
(a) Suppose X is the mass of iodine extracted by the solvent. Then the mass of iodine left in the water is 0.3 – X g.
So the [iodine] in water is (0.3 – X) g dm-3, and the [iodine] in the CCl4 is X g in 20 cm3 or 50X g dm-3 (Note: 20 cm3 is 1/50th of 1 dm3)
So Kc = (50X)/(0.3 - X) = 90
solving the equation gives X = 0.193 g
(b) Using the same method for the first of the two 10 cm3 portions (the top line of the equation will be 100X this time) gives X = 0.142 g for the first extraction.
For the second portion, remember we are starting with 0.3 – 0.142 = 0.158 g of iodine in the water. Doing a similar calculation gives a value of 0.075 g for the mass of iodine extracted, so the total iodine extracted by the two portions is 0.142 + 0.075 = 0.217 g.
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updated: 13 April 2008
