solvent extraction of a compound involves adding an immiscible solvent to the liquid which your product is in. if your product is in an aqueous solution then you would add a non-polar organic solvent and if your product is in an organic solvent then you would add water or an aqueous solution.

Distribution coefficient - $K_{\text{D}}$

the distribution of a solute between two immiscible solvents is in an equilibrium. we can define an equilibrium constant using the equation below

$$ K_{\text D}=\frac{[S_{\text{org}}]}{[S_\text{aq}]} $$

• solute is more soluble in the organic phase when $K_{\text D}>1$
• solute is just as soluble in aqueous and organic phase when $K_{\text D}>1$
• solute is more soluble in the aqueous phase when $K_{\text D}<1$

Extraction efficiency

following this equation, it is always better to extract a number of times with a smaller immiscible solvent than once with a large amount of solvent as each time you change the immiscible solvent you reduce the the concentration of solute in it and more solute changes phases

this can be shown mathematically with the equation below

$$ f_{\text{aq}}=\frac{[S_{\text{aq}}]}{[S_{\text{org}}]0}=\left( \frac{V{\text{aq}}}{K_{\text{D}}V_{\text{org}}+V_{\text{aq}}}\right)^i $$

• $f_{\text{aq}}$ - the fraction of the solute in the aqueous phase

• $[S_{\text{aq/org}}]$ - the concentration of solute in the phase

• $V_{\text{aq/org}}$ - the volume of solvent

• $i$ - the number of extractions performed

• worked example

we must make note of the law of diminishing return in solvent extraction of products. each time another extraction is performed, less product is removed. this must be though of in terms of economic, environmental and temporal impacts and limitations.

for a $K_{\text D}=9$ only three extractions, where $V_{\text{org}}=V_{\text{aq}}$, are needed to extract 99.9% of the solute

pH dependant extraction

if you have a compound that is ionisable (A) then you have 2 equilibriums at play: $\ce{HA_{\text{org}}<=>HA_{\text{aq}}<=> A- + H+}$. by making use of these equilibriums we can maximise the extraction.

if we alter the pH of our aqueous phase we can neutralise the dissociation and so only one species effectively exists in the aqueous phase. this pushes the equilibrium into the aqueous phase because as soon as $\ce{HA}$ goes from the organic to aqueous phase is is converted to $\ce{H+ + A-}$ and so the system can never reach equilibrium.

if the product is an acid then we can add base to the aqueous phase to neutralise it

if the product is basic then we can add acid to the aqueous phase to neutralise it