Solvent Extraction-LLE Popular technique For non - semi volatile organic compounds Partitioning the sample between two immiscible phases – Aqueous phase: Sample matrix – Organic phase: Organic solvent
Like dissolves like A(aq)
A(org) 11
LLE-Basic Theory Distribution Coefficient (Kd)
[A ]org KD = [A ]aq Kd is constant at a particular temperature 13
Separation by extraction Distribution equilibrium Partition of solute between two immiscible phases Aorg
Aaq
K=
( a A ) org ( a A ) aq
≈
[ A ] org [ A ] aq i
⎛ Vaq ⎞ ⎟ [A]o [ A ]i = ⎜ ⎜V K+V ⎟ aq ⎠ ⎝ org 14
Example K of I2 between organic solvent and H2O = 85. Find [I2] remaining in the aqueous layer after extraction of 50.0 mL of 1.00 x 10-3 M I2 with organic solvent of a) 50.0 mL, b) 2 x 25.0 mL, c) 5 x 10.0 mL 1
LLE-Basic Theory The fraction of analyte extracted (E) or Recovery (R)
Co Vo E =R = Co Vo + CaqVaq
(
E =R =
)
KDV
(1+ KDV)
V = Phase ratio; Vo/Vaq 16
LLE-Basic Theory Distribution Ratio (D) – If the analyte is partially dissociated in solution and exist as neutral species, free ions, ion-paired with counter ion Conc. of X in all chemical forms in the organic phase D= Conc. of X in all chemical forms in the aqueous phase
– Using D for KD
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LLE 10-1000 mL Phase ratio; 0.1 < V < 10 One-step extraction; Kd must be large, >10 – KOw (log KOw) Octanol-water partition coefficient
2-3 repeat extractions are required for quantitative recoveries
⎡ ⎤ 1 E = 1- ⎢ ⎥ ⎢⎣ (1+ K D V )⎥⎦
n
18
How many extraction will be necessary? KD = 10 V = 10 #1 E = 99.0% KD = 100 V=1 #1 E=99.0% KD = 1000 V = 0.1 #1 E=99.0%
KD = 10 V=1 #1 E = 90.9% #2 E = 99.2%
KD = 10 V = 0.1 #1 E = 50.0% #2 E = 75.0% #3 E = 87.5% #4 E = 93.8% #5 E = 96.9% #6 E = 98.4% #7 E = 99.2%
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How many extraction will be necessary? Typically, aqueous phase > organic phase – V = 0.05-0.1 (50:1000, 100:1000) KD = 1000 V = 0.1 #1 E = 99.0%
KD = 1000 V = 0.05 #1 E = 98.0% #2 E = 99.9%
KD = 100 V = 0.05 #1 E = 83.3% #2 E = 97.2% #3 E = 99.5%
Multiple extractions are more efficient 20
Single vs. Multiple Extraction KD
Single Single 1x150 mL 1x50 mL V=6.67 V=20
2nd extraction
3rd extraction
1x50 mL
1x50 mL
2x50 mL
1x50 mL
1x50 mL
3x50 mL
%E
%E
2nd %E
Add. %E
Cum. %E
3rd %E
Add %E
Cum. %E
1000
99.338
98.039
98.039
1.922
99.961
98.039
0.038
99.999
500
98.648
96.154
96.154
3.697
99.851
96.154
0.142
99.993
100
93.750
83.333
83.333
13.890
97.223
83.333
2.315
99.538
50
88.235
71.429
71.429
20.411
91.839
71.429
5.832
97.671
5
42.857
20.000
20.000
16.000
36.000
20.000
12.800
48.800 21
LLE The net amount of analyte extracted depends on the KD The net amount of analyte extracted depends on the Vorg/Vaq More analyte is extracted with multiple portions of extracting solvent than single portion of an equivalent volume of the extracting phase Recovery is independent of the concentration of the original aqueous sample
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Selection of Extraction Solvents Immiscible with water (Low solubility) Have polarity and H-bonding for good recovery of analytes (organic phase) Volatile for easy removal after extraction (pre-concentration if necessary) Compatible with method of analysis – GC – RP-HPLC
23
Solvent Modification Selectivity can be influenced by choices of additives affecting the equilibrium process – Adjusting pH – Ion pair Increase KD in organic phase – Chelating agent – Salting out
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Example Extraction of amine (Aniline) from aqueous sample – Buffer solution at least 1.5 pH units higher than its pKa, Amine will become un-ionized and will be extracted into the organic phase NH3+
+ H2O
NH2
+ H3O+
pKa = 4.6
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Typical Solvents used in LLE Aqueous solvent – – – – –
Pure water Acidic solution Basic solution High salt (Salting out) Complexing agents (ion-pairing, chelating and chiral agents) – Combination of two or more above
Water immiscible organic solvent – – – – – – – –
Diethyl ether Methylene chloride Chloroform Ethyl acetate Aliphatic ketones (C6+) Aliphatic alcohol (C6+) Toluene and xylenes Combination of two or more above 26
Emulsion Small droplets of organic phase floating in the immiscible aqueous phase Where the mass transfer occurs Vigorous shaking allows thorough interspersion between two immiscible phases; high efficiency Emulsion must be broken before collection Sample containing surfactants or fatty materials – slow breaking or incomplete breaking of emulsion
27
Breaking Emulsions Add salt to aqueous phase Use a heating-cooling extraction vessel Filter emulsion through a glass wool plug Filter the emulsion through phase separation filter paper Centrifuge Add a small amount of a different organic solvent 28
Continuous LLE KD is very small Kinetics of the extraction is slow Sample is large, requiring too many extractions Fresh organic solvent is recycled continuously in the form of droplets passing through the sample aqueous phase 29
Disadvantage – Time (18-24 hours) – Very volatile compounds can be lost – Thermally unstable compounds can be degraded 33
Disadvantage of LLE Time required; several successive extractions Seldom complete Use and disposal of large volumes of toxic organic solvents Formation of emulsion; hard to break emulsion Cumbersome glassware Labor-intensive Sample preconcentration is often required Not easily automated Cost 34
Extracting inorganic species Separating metals ions as chelates – Organic chelating agents react with metals to form uncharged complexes that are highly soluble in organic solvents 2HQ
MQ2
Organic phase
2HQ 2H+ + 2Q- + M2+ ↔ MQ2
Aqueous phase
Q = 8-hydroxyquinoline 35
Extracting inorganic species Overall equilibrium 2HQ (org) + M2+ (aq)
MQ2 (org) + 2H+ (aq)
[MQ 2 ] org [H + ] 2aq K′ = [HQ] 2org [M 2+ ] aq [HQ]org is present in large excess with respect to [M2+]aq + 2 [MQ ] [H ] aq 2 2 org K ′[HQ] org = K = [M 2+ ] aq [MQ 2 ] org K = +2 2+ [M ] aq [H ] aq 36
Extracting inorganic species [MQ 2 ] org K = +2 2+ [M ] aq [H ] aq Ratio of concentration of metal species in the two layers is inversely proportional to [H+]2aq K varies from metal ion to metal ion, makes it possible to selectively extract one cation from another by buffering aqueous solution
