Compound Y has a distribution coefficient of 4.0 when extracted from water with chloroform, with Y being more soluble in chloroform. How many 10.0 mL chloroform extractions would be required to extract at least 95% of Y from a 50.0 mL aqueous solution of water?

There are needed 3 extractions to extract at least 95%

Explanation:

The distribution coefficient of a compound is defined as the ratio in concentration of the compound in the organic solvent and the concentration in the aqueous solution:

K = Concentration organic solvent / Concentration in water

Assuming as initial amount of the organic solvent: 100% and X as the amount of Y that is extracted

First extraction:

4 = X / 10mL / (100-X) / 50mL

4 = 50X / 1000-X

4000 - 4X = 50X

4000 = 54X

X = 74.1%

In the first extraction, 74.1% of Y is extracted

And will remain: 100 - 74.1 = 25.9%

Second extraction:

4 = X / 10mL / (25.9-X) / 50mL

4 = 50X / 259-X

1036 - 4X = 50X

1036 = 54X

X = 19.2%

In the second extraction, 19.2% of Y is extracted

And will remain: 25.9 - 19.2 = 6.7%

Third extraction

4 = X / 10mL / (6.7-X) / 50mL

4 = 50X / 67-X

268 - 4X = 50X

268 = 54X

X = 5.0%

In the first extraction, 5.0% of Y is extracted

And are extracted:

74.1% + 19.2% + 5.0% = 98.3%

That means there are needed 3 extractions to extract at least 95%