Question

100 mL of aqueous solution containing 2,4,6-trinitrotoluene (TNT) at an initial concentration of 0.05 ppm is shaken with 10 mL hexane. After the phase separation, it was determined that the organic phase has 0.4 ppm. What is the partition coefficient, and what is the percent extracted? If the equilibrium concentration (rather than initial concentration) is 0.05 ppm, this would simplify the calculation. What would then be the partition coefficient and percent extracted?

(Please answer or provide something for the second portion of the question as well, reasoning would also be appreciated)

Answer 1

The initial mg of solute is calculated:

mg = V * ppm = 0.1 L * 0.05 ppm = 0.005 mg

The mg in the organic phase is calculated:

mg org = 0.01 L * 0.4 ppm = 0.004 mg

The mg in the aqueous phase is calculated:

mg aq = 0.005 - 0.004 = 0.001 mg

The concentration in the aqueous phase is calculated:

ppm aq = mg / L = 0.001 mg / 0.1 L = 0.01 ppm

The distribution coefficient is calculated:

Kd = ppm org / ppm aq = 0.4 / 0.01 = 40

The extraction percentage is calculated:

% ext = 0.004 * 100 / 0.005 = 80%

In that case, 100% of the solute would have been extracted for the organic phase, and the distribution coefficient was 0.

If you liked the answer, please rate it in a positive way, you would help me a lot, thank you.