Question

Solute S has a partition coefficient of 6.2 between water (phase 1) and hexane (phase 2). 74.0 mL solution of S in water is extracted six times with 17.0 mL of hexane. Calculate the fraction of S remaining in the aqueous phase.

Answer 1

Step-by-step explanation:

The given data is as follows.

Partition coefficient, K = 6.2

Volume of phase 1,
`V_(1)` = 74.0 mL

Volume of phase 2,
`V_(2)` = 17.0 mL

So, after one extraction fraction of solute remaining is given as follows.

q =
`(V_(1))/(V_(1) + KV_(2))`

After 3 times extraction, fraction of S remaining is as follows.

q =
`[(V_(1))/(V_(1) + KV_(2))]^(3)`

Putting the given values into the above formula as follows.

q =
`[(V_(1))/(V_(1) + KV_(2))]^(3)`

=
`[(74.0 ml)/(74.0 ml + 6.2 * 17.0 ml)]^(3)`

=
`[(74.0 ml)/(179.4 ml)]^(3)`

= 0.0699

Thus, we can conclude that the fraction of S remaining in the aqueous phase is 0.0699.