Kd = (concn organic phase)/(concn aqueous phase) = 10
Let x = g in aqueous phase.
Then 5-x = g in benzene phase.
10 = [(5-x)/50 mL]/(x/150 mL)
Solve for x which will give you the amount (in grams) in the aqueous phase for the first extraction.I have approximately 1.15 g. For the second extraction, the organic phase starts with 1.15 so it becomes 1.15-x and the water is x, go through it again. Etc.
I found approximately 0.062 g in the water phase after three extractions. I didn't do the one extraction with a larger amount BUT it should not extract as well as three smaller ones.
You have 150 ml of a solution of benzoic acid in water estimated to contain about 5 g of acid. The distribution coefficient of benzoic acid in benzene and water is approx 10. Calculate the amount of acid that would be left in the water solution after three 50 ml extractions with benzene. Do the same calculation using one 150 ml extraction with benzene to determine which method is more efficient.
My ans:
k = (concentration of benzoic acid in benzene)/(benzoic acid conc in water) = 10
But I am not sure how to proceed from this. Any help appreciated
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