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Brown's Chemistry
Theoretical–Physical
Advanced Level
Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 4.1
4.1 Partition
equilibrium between two phases, partition coefficient, immiscible
liquids and solvent extraction
This page describes the
phenomena of a solute dissolving in two immiscible liquids in contact
with one another – known as a partition equilibrium. Simple calculations
are described using an equilibrium expression and the partition
coefficient. The theory behind multiple partition using a separating
funnel is also described and explained.
GCSE/IGCSE
Notes on reversible reactions and chemical equilibrium
Part 4 sub–index 4.1
Partition between two
phases * 4.2 Solubility product Ksp
& common ion effect *
4.3 Ion–exchange systems
Advanced Equilibrium Chemistry Notes Part 1. Equilibrium,
Le Chatelier's Principle–rules
* Part 2. Kc and Kp equilibrium expressions and
calculations * Part 3.
Equilibria and industrial processes * Part 4 sub–index
(this section): 4.1
Partition between two
phases * 4.2 Solubility product Ksp
and common ion effect *
4.3 Ion–exchange systems *
Part 5. pH, weak–strong acid–base theory and
calculations * Part 6. Salt hydrolysis,
Acid–base titrations–indicators, pH curves and buffers * Part 7.
Redox equilibria, half–cell electrode potentials,
electrolysis and electrochemical series
*
Part 8. Phase equilibria–vapour
pressure, boiling point and intermolecular forces
*
M = old fashioned shorthand for mol dm–3
4.1 Partition
equilibrium between two phases, partition coefficient, immiscible
liquids and solvent extraction
-
The word 'partition'
means a substance X is distributed between two phases in a
dynamic equilibrium.
-
It is a
heterogeneous equilibrium since the 'solute' is distributed
between two distinct phases.
-
The two phases may
be a gas and liquid–solution or, more likely, two immiscible liquids.
-
The basic
expression is:
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Kpartition
= [X(phase 1)] / [X(phase 2)]
-
Here the K
is called the partition
coefficient or distribution coefficient. If it
involves two immiscible liquids, K has no units.
-
If gas and a
solution are involved, then appropriate units must be used for the
partition .
-
e.g. CO2(g)
CO2(aq), where K = pCO2(g)
/ [CO2(aq)], so the units might be Pa mol–1
dm3
-
This balance is
observed in our environment with 0.03–0.04% of air being the weakly
acidic gas carbon dioxide.
-
There is a 2nd
'non–partition' homogeneous equilibrium derived from the 'dissolving' equilibrium,
and accounts for why even un–polluted rain is still very slightly acidic
at about pH 5.5 due to ....
-
CO2(aq)
+ H2O(l) HCO3–(aq)
+ H+(aq)
-
and note that if
one of the equilibrium positions is changed, then the other
equilibrium must change too, e.g. if the CO2 concentration in
air rises, the aqueous concentration of CO2 rises and so
does the hydrogen ion concentration etc.
-
However for all
the cases described below, the partition will involve the distribution
of a solute between two immiscible liquid phases, which is a more
likely and simpler situation to deal with.
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If the solute
is in the same molecular state in both liquid–phases, the
following simple partition equilibrium expression will apply:
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Kpartition
= |
[X(liquid 1)]
|
–––––––––– |
[X(liquid 2)] |
-
K is called
the partition/distribution coefficient and has no units
and is temperature dependent.
-
Both
concentrations must be in the same units e.g. molarity mol dm–3,
g dm–3, mg cm–3 or whatever.
-
If a substance is
added to a mixture which is soluble to a greater or lesser extent in
both immiscible liquids, on shaking and then allowing the mixture to
settle, the concentrations in each layer become constant. However, there
is continual interchange of solute between the liquid layers via the
interface i.e. a dynamic equilibrium is formed.
-
If more of the
substance X is added to the system, the solute will distribute
itself between the immiscible liquids so that the ratio of the solute concentrations remains the
same at constant temperature independently of the total quantity of X
in the same molecular state,
and that is essentially the partition equilibrium law.
-
Examples of
partition systems involving two immiscible liquids
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Iodine in
water/tetrachloromethane: I2(aq) I2(CCl4) : K = [I2(aq)]
/
[I2(CCl4)] = 0.0116
-
The non–polar
iodine is much more soluble in the non–polar organic solvent than
in the highly polar water solvent. The iodine cannot disrupt few
of the
strong intermolecular forces of hydrogen bonding between water
molecules.
-
This system
can be analysed by titrating extracted aliquots with standardised
sodium thiosulphate and starch indicator.
-
If the aqueous
solution is replaced by aqueous potassium iodide, the simple Kpartition
expression does not hold because of a 2nd homogeneous equilibrium
and both equilibria expressions must be satisfied:
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Ammonia in
water/trichloromethane: NH3(aq) NH3(CHCl3) : K = [NH3(aq)]
/ [NH3(CHCl3)] = 26
-
The highly
polar ammonia can hydrogen bond with the even more polar water
molecules, hence ammonia's much greater solubility in water than
the less polar trichloromethane.
-
This system
can be analysed by titrating extracted aliquots with standardised
hydrochloric acid and methyl orange indicator.
-
The small % of
the highly soluble ammonia that ionises in water can be reasonably
ignored in this case.
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Case studies
– uses of
partition
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Case study
4.1.1 Extraction and
purification of a reaction product
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Partition can
used to extract and help purify a desired product from a reaction
mixture and this technique is called solvent extraction and
employs the use of a separating funnel. The method
depends on the desired material being more soluble in one liquid
phase than another, though the removal of impurities may
involve chemical reactions.
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Suppose a
neutral solid organic compound (lets call it NOC) had been
prepared using aqueous reagents but was soluble in a non–polar
organic solvent which is less dense than water e.g. hexane or
ethoxyethane (ether) (lets call it NPS). After filtering
off any solid residues (if necessary), the NPS can be added to
dissolve the NOC to give an organic solution of the crude
product. This is shown as the upper yellow layer in the
diagram below with the lower grey aqueous layer.
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The mixture is shaken to extract the NOC into the
NPS and the two layers allowed to settle out.
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The aqueous
layer of impurities can be run off.
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This leaves
the NPS–NOC solution, but it will still contain impurities.
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You can then
repeat the sequence with pure water to extract any more water
soluble impurities.
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Organic
impurities can be dealt with in a variety of ways depending on
their nature e.g.
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If the NOC
solution contains an acidic impurity you can use an aqueous
solution of sodium hydrogencarbonate to extract it by a chemical
neutralisation reaction into the aqueous layer. A second
extraction with pure water is then needed to remove any residual
salts formed. In between steps 1 and 2 you need to invert the
separating funnel (with the stopper on!) and open the tap to
release any CO2 gas formed.
-
If the NOC
solution contained a base impurity, a dilute acid solution could
be used to extract it, followed by extraction with water etc.
-
Neutral
impurities can really only be separated from the NOC via
re–crystallisation (see below).
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After the
extractions, the NOC solution can be dried with a drying agent
(e.g. anhydrous sodium sulphate) and the solvent evaporated to
leave the solid, which can e then re–crystallised from a suitable
solvent.
-
There are a
whole series of permutations based on these ideas depending on
whether the organic compound is an acid or a base and whether it
is water soluble, where an organic solvent might be used to
extract the impurities.
-
In industry,
it possible to have multiple connected separating 'funnels'
running on a continuous basis to extract a desired material.
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See
calculation Q4.1.1
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Case study 4.1.2
Testing the absorption of harmful chemicals or pesticides by fatty
tissue.
-
Pesticide
activity is often linked to the transfer of pesticide molecules (PM)
from aqueous media to fatty tissue in cell membranes.
-
To be
effective in this way, the pesticide molecules must be much more
soluble in the very weakly or non–polar fat–'solvent' than in
water.
-
Octan–1–ol is
a weakly polar solvent and the long hydrocarbon chain that
simulates fat molecules quite well in terms of solvent properties.
-
Kow
= |
[PM(octan–1–ol)]
|
––––––––––––––– |
[PM(water)] |
-
In ecology
chemistry, the bioconcentration factor can be measured e.g.
-
bioconcentration factor |
PM concentration(animal) |
=
––––––––––––––––––––––––– |
PM concentration(water) |
-
and it is
found to correlate well with the Kow partition
coefficient values.
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Note that Kow
values can be so big, they are often quoted on a logarithmic
scale, lg Kow (lg = log10).
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For the now
banned pesticide DDT, lg Kow = 5.98, that
is Kow = 105.98 = 9.55 x 105,
and this factor of nearly a million clearly explains why DDT
builds up in an animal food chain.
-
On effect of this was that birds of prey like
kestrels, falcons, eagles etc. at the top of a food chain,
suffered the most grievously because DDT poisoning caused
diminished fertility in the adults and egg–shells were formed too
thinly and were easily broken. The egg production and chick
survival rate declined reducing the adult breeding population, and
some species declined to the point of extinction in many areas of
the UK. With the banning or severely restricted use of these
pesticides the populations of birds of prey have recovered.
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However,
laboratory analysis of animals (fish, birds etc.) is costly,
so the partition experiments offer a cheaper preliminary testing
situation.
-
Similar tests
with any potentially harmful chemical can contribute to the
potential toxicity profile of a substance.
-
Case study 4.1.3
Chromatography
-
Paper
chromatography, tlc, gas–liquid chromatography function because
the solute mixture being separated is distributed between a mobile
phase (solvent/carrier gas) and an immobile phase (paper/high
boiling liquid).
-
Example
calculations:
Part 4 sub–index
4.1 Partition between two
phases *
4.2 Solubility product Ksp
& common ion effect *
4.3
Ion–exchange systems
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