“IS MY PRODUCT IN THE AQUEOUS OR ORGANIC LAYER?”
To answer this question it is helpful to know the pKa of your product. You can then adjust the pH
of the aqueous layer in your separatory funnel in order to cause your product to be present in
whichever of these two layers you desire.
Electrically neutral organic substances with few polar functional groups tend to be soluble in the
organic layer.
Charged organic substances (anions or cations) tend to be soluble in the aqueous layer.
Thus a charge-neutral organic carboxylic acid will usually be found in the organic layer, whereas
the conjugate base, the carboxylate anion, will be found in the aqueous layer:
You can control which layer most of your product is found in by adjusting the pH of the aqueous
layer, because this will shift the position of the above equilibrium. A low pH (a high
concentration of H+) will shift the equilibrium to the left, and the carboxylic acid product to the
organic layer. A high pH (a low concentration of H+) will shift the equilibrium to the right, and
the carboxylic acid (as its conjugate, anionic base), to the aqueous layer.
We can use your knowledge of acid-base equilibria as a guide to tell us just how high or low we
need to adjust the pH to move most of the organic compound to either the organic or aqueous
layer.
A-  H +
[H + ][A - ]
Ka 
[HA]
Ka
[A - ]
=
[H + ] [HA]
 Ka 
[A - ]
log10  +   log10
[HA]
 [H ] 
HA
[A- ]
log10 K a  log10 [H ]  log10
[HA]
[A ]
pK a  pH  log10
[HA]
+
[A - ]
[HA]
[A - ]

[HA]
pH  pK a  log10
10(pH pKa )
The final equation, above, illustrates why it is sufficient to adjust the pH of the aqueous layer to
ca. 3 pH units above or below the pKa of the acid whose solubility is being manipulated. For
example, in an aqueous solution of pH 8, an acid with a pKa of 5 will be 99.9% A- and only 0.1%
HA.
10(85)  1000 
99.9
0.1
And in an aqueous solution at pH 2, an acid with a pKa of 5 would be 0.1% A- and 99.9% HA.
10(25)  0.001 
0.1
99.9
EXERCISE:
Develop a similar equation for the ionization of an organic base:
Answer:
10(pH-pKa ) =
[B]
[HB+ ]
Why in this case does a low pH (high H+ concentration) drive the organic base to the aqueous
layer (as opposed to the organic layer as was the case with an organic acid)?