Question

Q: what volume of octanol is required to extract 99.9% of pyrene present in 1l of water sample? assume thermodynamic equilibrium and a constant temperature of 25 degree celsius

Answer 1

Assume only one chemical form in each phase.

D(distribution ratio) = K_D = [S_org]/[S_aq]

Where S_org/aq is the solubility in mol/L in each phase.

In this case, for a single step extraction, the fraction remaining in the organic phase is 99.9% (or 0.999 as a simple fraction) :

K_D = (0.999/V_org) / (0.001/V_aq)

K_D = 999V_aq/V_org

V_org = 999V_aq/D

V_aq = 1 L

I don’t have a value for D (for the partition of pyrene in octanol-water ). Find in your book , replace and calculate.