What weight of the compound would be removed from a solution of X in 100 mL of water by single extraction with 100 mL of ether?

Question

The distribution coefficient, K, is 10 for compound X. What weight of the compound would be removed from a solution of X in 100 mL of water by single extraction with 100 mL of ether? And in contrast, what weight of compound X would be removed by two extractions using 50ml each time.

Answer 1

Kd = 10, which is the ratio of [X ether] / [X water]. Where the brackets indicate concentration.

Q aq, the fraction of X remaining in the aqueous solution, is equal to Vaq / (Kd * Vorg + Vaq)

For a single extraction with 100 ml of ether, Qaq would be = 100 / (10 * 100 + 100) = 0.091 or 9.1 % of our organic compound remains in the aqueous layer, which means we extracted 90.9%.

If the volumes of the aqueous and organic layers are the same for both extractions,

then the cumulative fraction of solute remaining in the aqueous layer after two extractions,

(Qaq)2, is:( Vaq / (Kd * Vorg + Vaq))^2

plugging in the 50 ml volume, (Qaq)2 equals (100 / (10 * 50 + 100))^2 = 0.027 or 2.7% of our organic compound remains in the aqueous layer, which means we extracted 97.3%.into the ether layer.

As you can see, multiple extractions with smaller volumes yield better results than a single extraction.

Hope this helps you understand better, let me now if you have any questions!