Page 1 of 4
Chem201
Winter 2006
INTR O DU C TION TO ANA LY TI C A L SEPA R A TIO NS
Extraction is a separation technique, based on different solubility of solute(s) in two
immiscible solvents. This technique is used for the separation and isolation of different
solutes, concentrating of solute.
Since solvents used for the extraction are immiscible, they form a two-phase system (phase 1
and phase 2). Solute is partitioned between these phases:
S ( phase 1) ! S ( phase 2)
and this partition is governed by the corresponding equilibrium constant:
K=
[S]2
[S]1
If total amount of the solute in the system is m , then q is the fraction of it in phase 1 at
equilibrium and (1-q) is the fraction of the same solute at equilibrium in phase 2. Therefore,
concentration of the solute in the phase 1 is:
S1 =
qm
V1
and concentration of the solute in phase 2 is:
S2 =
(1! q )m
V2
Thus, the expression for the equilibrium is:
(1! q )m
[S]
(1! q )V1
V2
K= 2=
=
qm
[S]1
qV2
V1
If we solve this equation for q:
q=
V1
V1 + KV2
When n extractions are performed:
! V1 $ n
q =#
&
"V1 + KV2 %
Sample 1. Solute A is extracted by toluene (V2 ) from water (V1 = 100 mL). In this system K
value for the solute is equal 3. We can add 500 mL of toluene in one time, or 100 mL of
toluene 5 times to perform the extraction. What is the best scenario?
1.
q=
V1
100
=
= 0.062
V1 + KV2 100 + 3 ! 500
The fraction remaining in the aqueous phase is 6.2%.
Page 2 of 4
Chem201
Winter 2006
! V1 $ n !
$5
100
q =#
=
#
& = 0.00098
&
"V1 + KV2 % "100 + 3 '100 %
2.
The fraction remaining in the aqueous phase is 0.098%.
The conc lusion is, it is b ette r to do f e w sma ll extr a c tions than one big
extr a c tion.
E f f e c t of the pH
If solute is acid or base, its charge changes as pH is changed. Usually charged species are
more soluble in aqueous phase and non-charged – in organic phase.
Base in aqueous solution undergoes the following equilibria:
+
BH ! B + H
+
[B]aq [H + ]aq
Ka =
[BH + ]aq
The fraction of the neutral form present in aqueous phase is:
!B =
[B]aq
=
[B]aq + [BH + ]aq
[B]aq
Ka
=
+
[B]aq [H ]aq K a + [H + ]
[B]aq +
Ka
Neutral form is soluble in organic phase. And its distribution is governed by distribution
coefficient D :
D=
[B]org
concentration in organic phase
=
concentration in aqueous phase [B]aq + [BH + ]aq
It is similar to partition coefficient, but used instead when dealing with species present in
more than one chemic a l form .
When [BH + ] is expressed as function of the K a and placed into the expression for D, we can
end up with the following equation:
!
$
Ka
D = K#
= K' B
+ &
" K a + [H ] %
If acid has to be extracted then equation for the D is:
! [H + ] $
D = K#
= K' HA
+ &
" K a + [H ] %
Sample 2 . Partition coefficient (K) for the base, B, is 3. This base dissolved in water (pKa =
9), its concentration is 0.01M. 50 mL of this base is extracted with 100 mL of solvent. What
would be the forma l concentration of the base in aqueous phase at: (a) pH 10 and (b) pH 8?
For the p H 10:
Page 3 of 4
Chem201
Winter 2006
!
$ ! 10'9
$
Ka
D = K#
=
3
= 2.73
&
#
+
'9
'10 &
" K a + [H ] % "10 + 10 %
Since charged species are involved, we have to use D in place of K to find q:
q=
V1
50
=
= 0.15
V1 + KV2 50 + 2.73 !100
[B]aq = 0.15 ! 0.01 = 0.0015M
For the p H 8:
!
$ ! 10'9 $
Ka
D = K#
= 3# '9
= 0.273
+ &
'8 &
" K a + [H ] % "10 + 10 %
Since charged species are involved, we have to use D in place of K to find q:
q=
V1
50
=
= 0.65
V1 + KV2 50 + 0.273 !100
[B]aq = 0.65 ! 0.01 = 0.0065M
The conc lusion is, to extr a c t b ase w ith high e f f ic iency one has to inc r e ase the
pH o f the aqueous phase .
E f f e c t of the ch e lating agent
Selective chelating is a common practice to separate metal ions. However, charged chelated
ions are better dissolved in aqueous phase, which makes extraction impossible. Therefore,
special chelating agents are used for this purpose.
This time we should consider more equilibrium processes:
a. in aqueous phase
+
HL aq ! H aq + L aq
"
nL! + M n + " ML n
[H + ]aq [L" ]aq
Ka =
[HL ]aq
#=
[ML n ]aq
[nL! ]n aq [M n + ]
Each metal may complex with the ligand, but pH alters the selectivity.
b. between aqueous and organic phase
HL aq ! HL org
KL =
[HL ]org
[HL ]aq
Page 4 of 4
Chem201
ML n aq ! ML n org
KM =
Winter 2006
[ML n ]org
[ML n ]aq
Assuming that Mn+ is in aqueous phase and MLn is in organic phase, the distribution
coefficient is:
D=
total metal concentration in organic phase [ML n ]org
=
total metal concentration in aqueous phase [M n + ]aq
[ML n ]org = K M [ML n ]aq = K M ![L" ]n aq [M n + ]aq
Also,
[L! ]aq =
K a [HL ]aq
[H + ]aq
By substituting last equation into the previous:
" n
[ML n ]org = K M ! [L ]
aq
[M
n+
]aq = K M ![M
n+
# K [HL ] & n
]aq %% a + aq ((
$ [H ]aq '
If we put last equation into the expression for D:
" K [HL ] % n
K M ! [M ]aq $$ a + aq ''
" K [HL ] % n
# [H ]aq &
=
= K M !$$ a + aq ''
[M n + ]aq
# [H ]aq &
n+
D=
[ML n ]org
[M n + ]aq
Most of the HL is in organic phase, and therefore:
[HL ]aq =
[HL ]org
KL
By substitution we have final equation for D:
" K [HL ] % n K !K n [HL ]n org
D = K M !$$ a + aq '' = M n a (
[H + ]n aq
KL
# [H ]aq &
The conc lusion is, in o rde r to extr a c t m e tal io n w ith high e f f ic iency one has to
va ry the p H and concentr a tion of the ch e lating agent.