Name: Anonymous 2008-11-23 11:01
Hai guize, can anyone help me out here?
Supposedly extracting with 1 volume Vs is less efficient than extracting 2 times with volumes Vs/2. 3 volumes V/3 should produce even higher yield and so forth.
Can you show mathematically that this is true and by how much the yield is increased? The calculations for 2 volumes got a bit hairy for me, I'm not sure if my result makes sense and it looks like induction for n step extraction will get hairy.
Assuming partition coefficient K is equal to C2/C1 where C1 is concentration in the solvent you extract from, and C2 is ditto for the one you extract to, yield for 1 step is K*V2/(V1+K*V2).
For 2 steps I got yield=(K*Vs/2Vi)*((4*Vi-K*Vs)/(2*Vi+K*Vs)) where Vi is original solution you extract from, and Vs is your target solvent. (You take Vs liters of solvent, split it into two Vs/2 parts and serially extract with each).
Any tips would be apreciated.
Supposedly extracting with 1 volume Vs is less efficient than extracting 2 times with volumes Vs/2. 3 volumes V/3 should produce even higher yield and so forth.
Can you show mathematically that this is true and by how much the yield is increased? The calculations for 2 volumes got a bit hairy for me, I'm not sure if my result makes sense and it looks like induction for n step extraction will get hairy.
Assuming partition coefficient K is equal to C2/C1 where C1 is concentration in the solvent you extract from, and C2 is ditto for the one you extract to, yield for 1 step is K*V2/(V1+K*V2).
For 2 steps I got yield=(K*Vs/2Vi)*((4*Vi-K*Vs)/(2*Vi+K*Vs)) where Vi is original solution you extract from, and Vs is your target solvent. (You take Vs liters of solvent, split it into two Vs/2 parts and serially extract with each).
Any tips would be apreciated.