Chem. 230

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Chem. 230 9/9 Lecture
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Announcements

• Specialized Topics Presentation
• I will have a sign up sheet available for next
  week
• I would advise you to find a partner by next week
• Next Tuesday Guest Lecturer, Dr. Justin
  Miller-Schulze, on Solid Phase Extraction
• List of Additional Resources Will be Posted on
  Website (some books, suggested journals and a
  website)

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Extractions Liquid to Solid

• Temperature in precipitation processes
• Example Acetic Acid and Water

Liquid path for cooling 50 acetic acid in water
Eutectic Point
Liquid solution
T
Ice CH3CO2H (l)
CH3CO2H(s) H2O(l)
Solid solution
0
100
X(CH3CO2H)
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Questions relatedto solid-liquid extractions

1. What are advantages and disadvantages of Soxhlet
   extractions?
2. Suggest a way to isolate a polar organic compound
   from ionic compounds in urine.
3. It is desired to isolate CN- from CO32- by adding
   Ag and monitoring Ag electrochemically.
   Assuming initial concentrations of CN- 1.0 x
   10-3 M and CO32- 5.0 x 10-3 M and given Ksp
   values for AgCN and Ag2CO3 are 2.2 x 10-16 and
   8.2 x 10-12, respectively, what would be the
   target Ag? Will this separation be very
   efficient?
4. Can acetic acid be isolated from all solutions in
   water by freezing it out?

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Additional Questions cont.

• A 1.00 L sample of sea water is analyzed for
  phenols. The 1.00 L sample is passed through a
  solid phase extraction cartridge to trap the
  phenols. Then 25.0 mL of methanol is used to
  remove the phenols and then reagents are added
  that convert the phenols to methoxyphenols. The
  methoxyphenols are extracted by adding 25 mL of
  water to the methanol and extracting with two
  successive 25 mL portions of hexane. The hexane
  portions are combined, evaporated, and
  redissolved in 2.0 mL of hexane. An aliquot is
  then determined by GC and found to contain 22.1
  mmol L-1 of a particular phenol. What was the
  original conc. of that phenol in sea water (in
  nmol L-1) if it is assumed that all transfers
  were 100 efficient? How could the sensitivity
  of the method be increased?
• The total NH3 (NH3 NH4) concentration of a
  water sample is determined by NH3 in the
  headspace above a sample. A water sample at a pH
  of 8.1 was found to have a headspace pressure of
  2.4 x 10-7 atm. If KH 62500 mol/atm m3 (at
  that T) and Ka(NH4) 5.6 x 10-10., calculate
  the total NH3 concentration in the sample

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Liquid-Liquid Extractions

• One of more common simple separations
• Often used to introduce partition theory
• Equipment is simple (separation funnel or vials
  syringes)
• Two liquids must be immiscible (form two distinct
  phases)
• Lower phase is more dense (usually water or
  chlorinated hydrocarbon)
• Most common with water (or aqueous buffer) and
  less polar organic liquids

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Liquid-Liquid Extractions

• Partition Coefficient
• Kp Xraffinate/Xextractant
• Kp depends on thermodynamics of dissolving X in
  two phases
• Most common rule for solubility is likes dissolve
  likes
• Water is typically one phase (but can use
  hexane/methanol)
• More polar compounds exist in greater
  concentration in water
• Koctanol-water values can be found in reference
  tables (octanol is assumed to be the raffinate)

X(org)
X(aq)
If sample starts in aq phase, aq phase is
raffinate, org is extractant
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Liquid-Liquid Extractions
Polar

• Effect of organic solvent on Kp
• Small or large Kp values mean effective phase
  transfer
• Less polar organic solvents (e.g. hexane) will
  give largest Kp values (assuming aqueous
  extractant phase) for non-polar analytes (e.g.
  octylstearate CH3(CH2)7OC(O)(CH2)17CH3), but
  smaller values for analytes of modest polarity
  (e.g. a phenol)
• For analytes of modest polarity, a more polar
  organic solvent will give a larger Kp value (e.g.
  ether or ethyl acetate)
• This can be seen by using a polarity scale
  (note a true polarity scale may be
  multi-dimensional) and looking at difference
  from solvent

H2O
phenol
ether
octyl-stearate
hexane
Non-Polar
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Liquid-Liquid Extractions

• Effect of organic solvent on Kp - Selectivity
• When separating two analytes, finding a solvent
  with better selectivity also is important
• Example n-butanol and benzyl acetate (hexane vs.
  benzene as organic phase)
• Ideally low Kp for analyte and high Kp for
  interferant
• a separation factor (Kp)A/(Kp)B (where (Kp)Agt
  (Kp)B)

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Liquid-Liquid Extractions

• The fraction of moles extracted depends on both
  Kp and volumes of phases
• k partition ratio or retention factor (in
  chromatography)
• For this example we will assume an organic
  raffinate
• k norg/naq
• k Kp(Vraf/Vext) Kp(Vorg/Vaq)
• Q fraction extracted (to extractant or aq.
  phase)
• Fraction left in raffinate 1 - Q
• Q 1/(1 k)

Vorg
X(org)
X(aq)
Vaq
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Liquid-Liquid ExtractionsDemonstrations

• Methylene Blue (charged organic dye)
• Which phase will it be in hexane or water?
• Iodine (I2 non-polar but polarizable
  compound)
• Which phase will it be in hexane or water?
• Will Kp (organic extractant) get larger or
  smaller by changing from water to methanol? Or by
  changing hexane to ethyl acetate?

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Liquid-Liquid Extractions

• To increase the fraction extracted, either Kp
  needs to be changed or the volume ratio
• Example If a compound is extracted from octanol
  to water and Kow 0.8, how much volume of water
  is needed if the compound is in 10 mL of octanol
  and the extraction should move 90 of the
  compound?

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Liquid-Liquid Extractions

• A different example, methylethylketone (MEK),
  CH3CH2COCH3, has Kow 24, if 25 mL of aqueous
  MEK is extracted with 10 mL of octanol, calculate
  Kp, Q, and 1 Q

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Liquid-Liquid Extractions

• Lets look at multiple extractions in more detail
• MEK example
• 91 MEK in 1st extraction to octanol
• 9 MEK left in water
• In 2nd extraction of octanol (2nd water portion),
  fraction of original in water Q(1 - Q)
  0.910.09 0.085, fraction of original in
  octanol Q2 0.82
• In 2nd extraction of water portion by octanol,
  fraction in octanol (1 Q)Q 0.085, fraction
  in water (1 Q)2 0.009
• If water is extracted twice with octanol and
  octanol fractions combined, fraction in octanol
  1 - (1 Q)2 0.99 (99)

Octanol (extractant)
Octanol transferred
Water (raffinate)
Water transferred
Fresh water added
Fresh octanol added
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Liquid-Liquid Extractions

• Efficiency of Multiple Extractions vs. Single
  Extractions
• Back to Previous example (compound with Kow 0.8
  before MEK) but using 3 successive 10 mL
  aliquots of water.
• Note successive extractions can be done in
  different ways (cross-current vs. counter
  current)
• In this example (assuming only 2 components), the
  10 mL water aliquots normally would be combined

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Liquid-Liquid Extractions
Crown ether (12-crown-4)

• Additional Equilibria
• Crown ether example
• Complexed metals become more organic soluble

Na
Crown ether added

• Equilibria
• Na(aq) L(aq) ? NaL(aq)
• Sodium is not ether soluble
• 3) NaL(aq) ? NaL(ether)
• 4) L(aq) ? L(ether)

Diethyl ether
Sodium conc. given by gray shading
water
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Liquid-Liquid ExtractionsAdditional Notes

• Crown ethers and extractions
• Extraction depends on
• Crown ether hole
• Size of metal cation
• Crown ether partition coefficients (rxn 3 and 4)

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Liquid-Liquid ExtractionsOther Methods for
Transferring Metals

• Metal Ligand Complexes
• Best transfer for neutral complexes with large
  organic ligands
• Most common ligands are L- form and bidentate (2
  bonds per ligand with metal)
• Reactions
• HL (aq) ? HL (org)
• HL (aq) ? H L-
• Mn nL- ? MLn (aq) (note this can be broken
  into n steps)
• MLn (aq) ? MLn (org)
• Best at intermediate pH and for metals with large
  complexation formation constants
• Ion Pairs (e.g. Na--O3S(CH2)5CH3)

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Liquid-Liquid ExtractionsAcidic/Basic Organics

• Ions have very small Kow values (assume 0)
• Form of acids and bases in neutral species
  depends on pH
• Distribution Coefficient KD
• KD Xtotal raffinate/Xtotal extractant
• Monoprotic acid example (HA extracted from
  organic to water)
• HA only organic phase species (a little
  different in text example)
• HA and A- possible in aqueous phase

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Liquid-Liquid ExtractionsAcidic/Basic Organics

• HA example continued
• Kp HAorg/HAaq constant
• KD HAorg/(HAaq A-) not constant
• Ka HA-/HA
• So KD HAorg/(HAaq KaHAaq/H)
• KD Kp/(1 Ka/H)
• At pH ltlt pKa (high H), KD Kp
• At pH gtgt pKa, KD ltlt Kp (better transfer to aq
  phase)

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Liquid-Liquid ExtractionsAcidic/Basic Organics

• HA example

KD Kp
logKD
Only HA present
HA and A- present
0
pH
14
pKa
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Liquid-Liquid ExtractionsAcidic/Basic Organics

• In general, only un-ionized formed will be
  organic soluble (applies to multi-functional
  compounds also)
• For weak bases, only base form B, not BH, will
  be organic soluble
• For weak bases, low pH increases partitioning
  into water

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Liquid-Liquid ExtractionsSome Questions

• How can the following compounds be separated?

Kow 40
Kow 130
Kow 15
pKa 3.0
pKa 9.5
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Liquid-Liquid ExtractionsSome Questions
1. Separate Comp. 1 in aq. phase from others at
pH 7
KD sketch
2. Separate comp. 2 in aq. phase from comp. 3 at
pH 13
3rd compound
2nd Compound
logKD
1st Compound
0
pH
14
3
9
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Liquid-Liquid ExtractionsSome Questions

• To extract Al3 to an organic phase with an HL
  type ligand, which complexation constant is most
  important?
• Amino acids can act as bidentate ligands for
  metal transfer. Why does the pH have to be
  greater than pKa2?

NH2
M2
CH2C6H5
-OCO
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Liquid-Liquid ExtractionsOne More Question

• Benzylamine (C6H5CH2NH2) is a weak base with a
  Kow of 12. The conjugate acid of benzylamine has
  a pKa of 9.35. Benzylamine is being extracted
  from 10 mL of water buffered to a pH of 8.00
  (raffinate phase) to octanol. (assume no other
  reactions occur in water or octanol)
• Determine the distribution coefficient.
• Calculate the fraction of benzylamine extracted
  into octanol if 20 mL of octanol is used?
• How could Q be increased?