Extraction
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Transcript of Extraction
Are you really ready?
Let’s get set for a Separation Chemistry lecture!
Solvent Extraction
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
Solvent Extraction
Organic solvents less dense than water diethyl ether, toluene, hexane
Organic solvents more dense than waterchloroform, CCl4, dichloromethane
Like dissolves like so ideally, the extracting solvent should be similar to the solute (analyte)
Properties of Extraction Solvents
Solvents used for extraction1. immiscible with water (polarity)2. high solubility for organic compound3. relatively low boiling point (removal)4. non-toxic, cheap, available
methylene chloride, diethyl ether, hexane, ethyl acetatedensities determine top or bottom
Extraction done by our mothers
Solid-Liquid Extractionbrewing teapercolating coffeespices and herbs
shake
add second immiscible
solvent
Separatoryfunnel
Solvent Extraction
Solute partitions between the two phases
Solvent Extraction
[S]1
[S]2
Phase 1
Phase 2
Theory of Liquid-Liquid Extraction
Differential solubility in two immiscible solvents
Immiscible solventsorganic productimpurity
Theory of Extraction
If KD ~ 1, don’t get good separation.
Little separation: low yield
Extraction Theory
Separation depends upon relative solubility of the compound in each of the two immiscible solvents.
(g/mL is solubility)want KD >>>> 1 or <<<<<<1
( )( )watermLgorganicmLgKD /
/=
Solvent Extraction
Equilibrium constant for this partitioning is K (partition coefficient)
K= [S]2[S]1
Solvent Extraction
Determination of solute concentration in each phaseDefine some variables:
V1 & V2 are volumes of solvents 1&2m = total # of moles of solute (S) presentq = fraction of solute remaining in phase 1
at equilibrium
Solvent Extraction
[S]1 = qm/V1
[S]2 = (1-q)m/V2
K= [S]2
[S]1 qm/V1
(1-q)m/V2= =q/V1
(1-q) /V2
q =KV2 + V1
V1 (1-q) =KV2 + V1
KV2
fraction of S in: phase 1 phase 2
Rearrange:
Solvent Extraction
Solvent Extraction
If remove V2 and extract V1 with fresh layer of V2, what fraction remains in V1?
Initial moles = mafter first extraction - qmafter second extraction - q(qm)=q2m
q(2) =KV2 + V1
V12
Solvent Extraction
q(n) =KV2 + V1
V1n
Fraction in V1 after n extractions:
Solvent Extraction
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:
Solvent Extraction
a) one 600.0 ml aliquot of hexane is used?
q =4(600ml) + 150ml
150ml= 0.05882 = 5.882%
# moles remaining0.05882 (0.03M•0.150L) = 2.647x10-4 moles
q =4(100ml) + 150ml
150ml= 0.00041156
# moles remaining4.115 x 10-4 (0.03M•0.150L) = 1.852x10-6moles
Solvent Extraction
b) 6 successive 100.0 ml aliquots of hexane are used?
Solvent Extraction
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 + H2O BH+ + OH-Kb
HA H+ + A-Ka
Generally, neutral species are more soluble in an organic solvent and charged species are more soluble in aqueous solution
Solvent Extraction (pH effects)
with organic acids/bases:
organic
HA H+ + A-Kaaqueous
HA H+ + A-
very little here, ions have poor solubility
Solvent Extraction (pH effects)
Partitioning of organic acids between two phases:
Solvent Extraction (pH effects)
When the solute (acid/base) can exist in different forms, D (distribution coefficient) is used instead of K (partition coefficient)
Solvent Extraction (pH effects)
D = total conc. in phase 2total conc. in phase 1
D =[HA]org
[HA]aq + [A-]aq
HA
HA H+ + A-Ka
K
Ka =[H+][A-]
[HA] [A-] =Ka [HA]
[H+]
Solvent Extraction (pH effects)
Substitute for [A-] in D eq. and rearrange
D =[HA]
[HA]
2
1 + +K HA
Ha[ ][ ]
1
Solvent Extraction (pH effects)
D =[HA]
[HA]
[HA]
[HA]
2
1
2
1+
=+
⎛⎝⎜
⎞⎠⎟+ +
K HAH
KH
a a[ ]
[ ] [ ]1
1
D =[HA][HA]
2
11+⎛⎝⎜
⎞⎠⎟+
KH
a
[ ]= K
Solvent Extraction (pH effects)
D = K1+⎛⎝⎜
⎞⎠⎟+
KH
a
[ ]
D =K
1+⎛⎝⎜
⎞⎠⎟=
++
+
+KH
K HH Ka a
[ ]
[ ][ ]
D
pH
[H+]=KapH=pKa
K[H+]>>Ka
mainlyHA
[H+]<<Ka
mainlyA-
Solvent Extraction (pH effects)
pH effect on D for organic acids
K1
4 8
K2
D
pH
Solvent Extraction (pH effects)
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+] + KaD =
K KaD
pH
K
[H+]=KapH=pKa
[H+]>>Ka
mainlyBH+
[H+]<<Ka
mainlyB
Solvent Extraction (pH effects)
Analogous treatment for organic bases (proton acceptors, not KOH)
D
pH
Kacid base
Solvent Extraction (pH effects)
In general:
Initial Aq. phase
Aq. PhaseOrg. acid
Aq. PhaseOrg. base
Ether PhaseOrg. acid, Org. neutral
Ether PhaseOrg. neutral
pH=1, extract with ether
extract with pH=12 Aq. Sol’n
Separate organic acid, base and neutral analytes
Calculations
Determine KD for extraction of solid in the two solvents
)(/)()(/)(
mLwatergunkmLsolventgunkKD =
Practice Calculations
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
KD = 92 mg/1.5 mL28 mg/1.5 mL
= 3.3
Yes, I am full, confused
Take a little break?Yo, wis semene dhisik…..