When compound A is partitioned between equal volumes of methylene chloride and water

Question

When compound A is partitioned between equal volumes of methylene chloride and water one finds that the partition coefficient is equal to 1. This means that the compound happens to be equally soluble in the two solvents and at equilibrium exactly 50% of the compound will be in each phase if their volumes are equal. If this were the case, how may separate extractions would be required with methylene chloride to successfully extract 98% of the compound from the aqueous layer. (Note assume you will extract x mL of aqueous solution with some number of x mL portions of methylene chloride). Show your work below.

Answer 1

If you recover 50% each time you do an extraction, then after the 1st extraction you have 50%

After the 2nd, you have 50% of the remaining 50%, or another 25% for a total of 75%.  Etc., etc.

The formula for such an event is fraction remaining = 0.5ⁿ where n is the number of extractions.

For 98% extraction, that would leave 2% as fraction remaining = 0.02

0.02 = 0.5ⁿ

log 0.02 = n log 0.5

-1.699 = -0.301 n

n = 5.6, so you would need to do 6 extractions.

 

Looking at it the long way

extraction #  amount recoverd

1..................50%

2.................25%

3.................12.5%

4.................6.25%

5................3.125%

6................1.563%

Add them all up to get 98.4%