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

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 = $\frac{V_{1}}{V_{1} + KV_{2}}$

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

              q = $[\frac{V_{1}}{V_{1} + KV_{2}}]^{3}$

Putting the given values into the above formula as follows.

               q = $[\frac{V_{1}}{V_{1} + KV_{2}}]^{3}$

                  = $[\frac{74.0 ml}{74.0 ml + 6.2 \times 17.0 ml}]^{3}$

                  = $[\frac{74.0 ml}{179.4 ml}]^{3}$

                  = 0.0699

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