Question

223.6c -2 29.3 c Distribution coefficient KDi of caffeine in first extraction with 20 mL, use theoretical solubility of caffeine in water 2g/100mL. [I mark 7- Distribution coefficient Kp2 of after two extractions with 10mL each, use theoretical solubility of caffeine in water 2g/100mL [1 mark] 8-

Answer 1

Distribution coefficient (K_D1 and K_D2) will be the same, whatever may be the number of extractions. Only, the amount of compound that is extracted into the organic layer varies.

For example, let's say the solvent used for extraction is dichloromethane.

Let the mass of caffeine = 1 g, the volume of water used for extraction (V2) = 50 mL

The distribution coefficient of caffeine between water and dichloromethane = 4.6

(i) Volume of dichloromethane (V1) = 20 mL (single extraction)., i.e. n = 1

i.e. Final mass of caffeine in water = {V₂ / (V₂ + V₁*K_D)}ⁿ * Initial mass of caffeine in water.

Where n is the no. of extractions.

i.e. Final mass of caffeine in water = {50/(50 + 20*4.6)}¹ * 1 g

= 0.352 g

Therefore, the mass of caffeine extracted into dichloromethane in single extraction using 20 mL = 1-0.352 = 0.648 g

(ii) Volume of dichloromethane (V1) = 10 mL two extraction), i.e. n = 2

i.e. Final mass of caffeine in water = {V₂ / (V₂ + V₁*K_D)}ⁿ * Initial mass of caffeine in water.

i.e. Final mass of caffeine in water = {50/(50 + 10*4.6)}² * 1 g

= 0.271 g

Therefore, the mass of caffeine extracted into dichloromethane in single extraction using 20 mL = 1-0.271 = 0.729 g