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…..