Question

Please calculate the amount of a compound (A) that remains in 10 mL of an aqueous phase after two extractions with toluene (10 mL each) if KD = 2 (KD = partition ratio toluene / water). Show your calculation(s).

Answer 1

To determine the amount of compound A remaining in the aqueous phase after two extractions with toluene, we can use the partition coefficient ($K_D$) and the volumes of the aqueous and organic phases.

Given:

• Partition coefficient, $K_D = 2$ (toluene/water)

• Initial volume of aqueous phase, $V_{\text{aq}} = 10 \, \text{mL}$

• Volume of toluene used in each extraction, $V_{\text{tol}} = 10 \, \text{mL}$

• Number of extractions, $n = 2$

Assumption:

• Let the initial amount of compound A in the aqueous phase be $C_0$.

Calculation:

The fraction of compound A remaining in the aqueous phase after each extraction can be determined using the formula:

$$\text{Fraction remaining after one extraction} = \frac{1}{1 + K_D \times \frac{V_{\text{tol}}}{V_{\text{aq}}}}$$

Substituting the given values:

$$\text{Fraction remaining after one extraction} = \frac{1}{1 + 2 \times \frac{10}{10}} = \frac{1}{1 + 2} = \frac{1}{3}$$

After one extraction, $\frac{1}{3}$ of the compound remains in the aqueous phase.

After two extractions, the fraction remaining is:

$$\left( \frac{1}{3} \right)^2 = \frac{1}{9}$$

Therefore, after two extractions, $\frac{1}{9}$ of the initial amount of compound A remains in the aqueous phase.

Conclusion:

After two extractions with 10 mL of toluene each, approximately 11.11% of the initial amount of compound A remains in the 10 mL aqueous phase.

Answer 2