Hey folks!! This is my first post here...
The answer to the above question can be given by considering the 'law of distribution' by Nernst.
This law states that the rate of distribution of a dissolved substance between 2 immiscible phases is a constant, commonly labeled
K.
K is dependent of the employed volumes of both phases (and the temperature of the system), with the following correlation:
K = c1/c2 = (m1•v2)/(m2•v1)
c = concentration
m = amount of the dissolved substance (%)
v = volume
1 = organic phase
2 = aqueous phase
Consider the following example:
A substance has a K-value of 3, i.e. it is 3 times more soluble in the organic phase than in the aqueous one. If you like to extract now 95 % of the substance from the aqueous phase into the organic phase, one could calculate with the given formula that 6.3 times of the volume of the organic solvent is necessary to achieve this.
Hint: Using for the calculation: K=3; m1=95%, m2=5%, one gets in the end a equation that reads: 3 = (95•v2)/(5•v1). That gives resolved 95/15 = v1/v2 = 6.3If you take now just
one third of the volume of the organic solvent with respect to the aqueous one, one will extract 50% with the 1st extraction step (easy calculation, as
K was given with 3!). repeating this, one will extract further 25% (with respect to the starting amount); at the third time 12.5%, then finally ~6%.
Conclusion: After extraction 4 times with just a third of the volume (i.e. total volume of just 1.3 times the aqueous phase!), one gets 50% + 25% + 12.5% + 6% =
93.5% of extracted product.
So you can see that with several times little volume one can achieve the same result in comparison to extracting just a single time with a large volume.
NERNST FTW!
Peace!
Murphy
Muchas gracias