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.

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