Extraction

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Ar e you r eal l y r eady?
Let ’s get set f or
a Separ at i on
Chemi st r y
l ec t ur e!
Sol vent Ex t r ac t i on
Extraction: transfer of a solute from
one phase to another.
Can use most any combination of
phases (solid, liquid, gas, supercritical
fluid)
Solvent extractions use two immiscible
liquids.
Typically aqueous/organic solvent combos
Sol vent Ex t r ac t i on
Organic solvents less dense than water
diethyl ether, toluene, hexane
Organic solvents more dense than water
chloroform, CCl
4
, dichloromethane
Like dissolves like so ideally, the
extracting solvent should be similar to
the solute (analyte)
Pr oper t i es of Ex t r ac t i on Sol vent s
Solvents used for extraction
1. immiscible with water (polarity)
2. high solubility for organic compound
3. relatively low boiling point (removal)
4. non-toxic, cheap, available
methylenechloride, diethyl ether, hexane, ethyl acetate
densities determine top or bottom
Ex t r ac t i on done by our mot her s
Solid-Liquid Extraction
brewing tea
percolating coffee
spices and herbs
shake
add second
immiscible
solvent
Separatory
funnel
Sol vent Ex t r ac t i on
Solute partitions between the two
phases
Sol vent Ex t r ac t i on
[S]
1
[S]
2
Phase 1
Phase 2
Theor y of Li qui d-Li qui d Ex t r ac t i on
Differential solubility in two immiscible solvents
Immisciblesolvents
organic product
impurity
Theor y of Ex t r ac t i on
If KD ~1, don’t get good separation.
Little separation: low yield
Ex t r ac t i on Theor y
Separation depends upon relative
solubility of the compound in each of the
two immiscible solvents.
(g/mL is solubility)
want K
D
>>>> 1 or <<<<<<1
( )
( ) water mL g
organic mL g
K
D
/
/
=
Sol vent Ex t r ac t i on
Equilibrium constant for this
partitioning is K (partition coefficient)
K=
[S]
2
[S]
1
Sol vent Ex t r ac t i on
Determination of solute concentration
in each phase
Define some variables:
V
1
& V
2
are volumes of solvents 1&2
m = total #of moles of solute (S) present
q = fraction of solute remaining in phase 1
at equilibrium
Sol vent Ex t r ac t i on
[S]
1
= qm/V
1
[S]
2
= (1-q)m/V
2
K=
[S]
2
[S]
1
qm/V
1
(1-q)m/V
2
= =
q/V
1
(1-q) /V
2
q =
KV
2
+ V
1
V
1
(1-q) =
KV
2
+ V
1
KV
2
fraction of S in: phase 1 phase 2
Rearrange:
Sol vent Ex t r ac t i on
Sol vent Ex t r ac t i on
If remove V
2
and extract V
1
with fresh
layer of V
2
, what fraction remains in V
1
?
Initial moles = m
after first extraction - qm
after second extraction - q(qm)=q
2
m
q(2) =
KV
2
+ V
1
V
1
2
Sol vent Ex t r ac t i on
q(n) =
KV
2
+ V
1
V
1
n
Fraction in V
1
after nextractions:
Sol vent Ex t r ac t i on
Example: Solute A has a partition
coefficient of 4.000 between hexane and
water. (K = [S]
hexane
/[S]
water
= 4) If
150.0 ml of 0.03000 M aqueous A is
extracted with hexane, what fraction of
A remains if:
Sol vent Ex t r ac t i on
a) one 600.0 ml aliquot of hexane is
used?
q =
4(600ml) + 150ml
150ml
= 0.05882 = 5.882%
#moles remaining
0.05882 (0.03M•0.150L) = 2.647x10
-4
moles
q =
4(100ml) + 150ml
150ml
= 0.0004115
6
#moles remaining
4.115 x 10
-4
(0.03M•0.150L) = 1.852x10
-6
moles
Sol vent Ex t r ac t i on
b) 6 successive 100.0 ml aliquots of
hexane are used?
Sol vent Ex t r ac t i on
Although same volume of hexane is
used, it is more efficient to do several
small extractions than one big one!
1 600 ml extraction extracts 94.12%
6 100 ml extractions extract 99.96%
B + H
2
O BH
+
+ OH
-
K
b
HA H
+
+ A
-
K
a
Generally, neutral species are more soluble
in an organic solvent and charged species
are more soluble in aqueous solution
Sol vent Ex t r ac t i on (pH ef f ec t s)
with organic acids/bases:
organic
HA H
+
+ A
-
K
a
aqueous
HA H
+
+ A
-
very little here, ions
have poor solubility
Sol vent Ex t r ac t i on (pH ef f ec t s)
Partitioning of organic acids between
two phases:
Sol vent Ex t r ac t i on (pH ef f ec t s)
When the solute (acid/base) can exist in
different forms, D (distribution
coefficient) is used instead of K
(partition coefficient)
Sol vent Ex t r ac t i on (pH ef f ec t s)
D =
total conc. in phase 2
total conc. in phase 1
D =
[HA]
org
[HA]
aq
+ [A
-
]
aq
HA
HA H
+
+ A
-
K
a
K
K
a
=
[H
+
][A
-
]
[HA]
[A
-
] =
K
a
[HA]
[H
+
]
Sol vent Ex t r ac t i on (pH ef f ec t s)
Substitute for [A
-
] in D eq. and
rearrange
D=
[HA]
[HA]
2
1
+
+
K HA
H
a
[ ]
[ ]
1
Sol vent Ex t r ac t i on (pH ef f ec t s)
D=
[HA]
[HA]
[HA]
[HA]
2
1
2
1
+
=
+





⎟ +
+
K HA
H
K
H
a
a
[ ]
[ ]
[ ]
1
1
D =
[HA]
[HA]
2
1
1+






+
K
H
a
[ ]
=K
Sol vent Ex t r ac t i on (pH ef f ec t s)
D =K 1+






+
K
H
a
[ ]
D=
K
1+






=
+
+
+
+
K
H
K H
H K
a a
[ ]
[ ]
[ ]
D
pH
[H
+
]=K
a
pH=pK
a
K
[H
+
]>>K
a
mainly
HA
[H
+
]<<K
a
mainly
A
-
Sol vent Ex t r ac t i on (pH ef f ec t s)
pH effect on D for organic acids
K
1
4 8
K
2
D
pH
Sol vent Ex t r ac t i on (pH ef f ec t s)
Example problem: Want to separate two
organic acids using a scheme based on pH.
Acid 1 (pKa= 4), Acid 2 (pKa= 8)
Acid 2 stays in
organic phase,
acid 1 is extracted
into aqueous phase
[H
+
] + K
a
D =
K K
a
D
pH
K
[H
+
]=K
a
pH=pK
a
[H
+
]>>K
a
mainly
BH
+
[H
+
]<<K
a
mainly
B
Sol vent Ex t r ac t i on (pH ef f ec t s)
Analogous treatment for organic bases
(proton acceptors, not KOH)
D
pH
K
acid
base
Sol vent Ex t r ac t i on (pH ef f ec t s)
In general:
Initial Aq. phase
Aq. Phase
Org. acid
Aq. Phase
Org. base
Ether Phase
Org. acid, Org. neutral
Ether Phase
Org. neutral
pH=1, extract with ether
extract with pH=12 Aq. Sol’n
Separ at e or gani c ac i d, base
and neut r al anal yt es
Cal c ul at i ons
Determine K
D
for extraction of solid
in the two solvents
) ( / ) (
) ( / ) (
mL water g unk
mL solvent g unk
K
D
=
Pr ac t i c e Cal c ul at i ons
Mass solid 120 mg
Total volume organic solvent 1.5 mL
Total volume water 1.5 mL
Mass solid extracted into
organic solvent
92 mg
Mass remaining in water 120-92 = 28 mg


K
D
= 92 mg/1.5 mL
28 mg/1.5 mL
= 3.3
Yes, I am f ul l , c onf used
Take a little break?
Yo, wis semene dhisik…..

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