Download Chem. 31 – 9/15 Lecture

Chem. 230 – 9/9 Lecture
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)
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
X(CH3CO2H)
100%
Questions related
to solid-liquid extractions
1.
2.
3.
4.
What are advantages and disadvantages of Soxhlet
extractions?
Suggest a way to isolate a polar organic compound
from ionic compounds in urine.
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?
Can acetic acid be isolated from all solutions in water
by freezing it out?
Additional Questions – cont.
5.
6.
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
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
Liquid-Liquid Extractions
• Partition Coefficient
Kp = [X]raffinate/[X]extractant
• 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
Liquid-Liquid Extractions
• 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 multidimensional) and looking at “difference” from
solvent
Polar
H2O
phenol
ether
octylstearate
hexane
Non-Polar
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
– α = separation factor = (Kp)A/(Kp)B
(where (Kp)A> (Kp)B)
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
Liquid-Liquid Extractions
Demonstrations
• 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?
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?
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
Liquid-Liquid Extractions
•
•
Let’s look at multiple extractions in
more detail
MEK example
Octanol
– 91% MEK in 1st extraction to
octanol
transferred
– 9% MEK left in water
– In 2nd extraction of octanol (2nd
water portion), fraction of original
in water = Q(1 - Q) = 0.91*0.09
= 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 Fresh water
added
octanol and octanol fractions
combined, fraction in octanol = 1
- (1 – Q)2 = 0.99 (99%)
Octanol
(extractant)
Water
(raffinate)
Water
transferred
Fresh
octanol
added
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
Liquid-Liquid Extractions
Crown ether
(12-crown-4)
• Additional Equilibria
– Crown ether example
– Complexed metals become
more organic soluble
O
O
Na+
O
Crown ether
added
Equilibria:
1) Na+(aq) + L(aq) ↔ NaL+(aq)
2) Sodium is not ether soluble
O
Diethyl ether
3) NaL+(aq) ↔ NaL+(ether)
4) L(aq) ↔ L(ether)
Sodium conc. given by
gray shading
water
Liquid-Liquid Extractions
Additional Notes
• Crown ethers and extractions
• Extraction depends on:
– Crown ether “hole”
– Size of metal cation
– Crown ether partition coefficients (rxn 3 and 4)
Liquid-Liquid Extractions
Other 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+ + LMn+ + 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)
Liquid-Liquid Extractions
Acidic/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 = [X]total raffinate/[X]total 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
Liquid-Liquid Extractions
Acidic/Basic Organics
HA example continued
• Kp = [HA]org/[HA]aq = constant
• KD = [HA]org/([HA]aq + [A-]) – not constant
• Ka = [H+][A-]/[HA]
• So KD = [HA]org/([HA]aq + Ka[HA]aq/[H+])
• KD = Kp/(1+ Ka/[H+])
– At pH << pKa (high [H+]), KD = Kp
– At pH >> pKa, KD << Kp (better transfer to aq phase)
Liquid-Liquid Extractions
Acidic/Basic Organics
• HA example
KD = K p
logKD
Only
HA
present
0
HA and
Apresent
pKa
pH
14
Liquid-Liquid Extractions
Acidic/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
Liquid-Liquid Extractions
Some Questions
• How can the following compounds be
separated?
O
O
O
CH3
OH
CH3
O
OH
O
Kow = 15
pKa = 3.0
Kow = 40
pKa ~ 9.5
Kow = 130
CH3
Liquid-Liquid Extractions
Some Questions
KD sketch
1. Separate Comp. 1
in aq. phase from
others at pH 7
2. Separate comp.
2 in aq. phase from
comp. 3 at pH 13
3rd compound
2nd Compound
logKD
1st Compound
0
3
pH
9
14
Liquid-Liquid Extractions
Some 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
-OC=O
Liquid-Liquid Extractions
One 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?