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GCE-AS-A2-IB Advanced Level Theoretical-Physical Chemistry revision notes
Equilibria Part 4
"Partition equilibria, solubility product and ion exchange"
GCSE
Notes on reversible reactions-equilibrium *
Advanced 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: 4.1
Partition between two
phases * 4.2 Solubility product Ksp *
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
* The K and ΔS-ΔG connection with EØcell
will be dealt with via new thermodynamics pages later, but an
example of a ΔS-ΔG calculation is given at the end of
the advanced kinetics pages and
ΔG for cells is
mentioned in Equilibria Part 7.
M = old fashioned shorthand for mol dm-3 *
EMAIL
query?comment
4.1 Partition
equilibrium between two phases
-
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:
-
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.
-
If the solute
is in the same molecular state in both liquid-phases, the
following simple partition equilibrium expression will apply:
-
|
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
-
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:
-
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.
-
Case studies - uses of
partition
-
Case study
4.1.1 Extraction and
purification of a reaction product
-
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.
-
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.
-
 The mixture is shaken to extract the NOC into the
NPS and the two layers allowed to settle out.
-
The aqueous
layer of impurities can be run off.
-
This leaves
the NPS-NOC solution, but it will still contain impurities.
-
You can then
repeat the sequence with pure water to extract any more water
soluble impurities.
-
Organic
impurities can be dealt with in a variety of ways depending on
their nature e.g.
-
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).
-
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.
-
See
calculation Q4.1.1
-
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.
-
Note that Kow
values can be so big, they are often quoted on a logarithmic
scale, lg Kow (lg = log10).
-
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.
-
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:
4.2 Solubility
Product
-
When a
sparingly soluble salt is mixed with water a dynamic equilibrium
is established in which salt is constantly dissolving and
crystallising at the same rate when the solution is saturated, and the
maximum constant concentration is achieved.
-
e.g. for calcium
sulphate: CaSO4(s)
+ aq Ca2+(aq) + SO42-(aq)
-
However, since the
concentration of water and the solid is effectively constant the
equilibrium expression is simplified to:
-
Ksp
= [Ca2+(aq)] [SO42-(aq)]
= 2.4 x 10-5 mol2 dm-3
-
Ksp
is called the solubility product of the ions concerned and
is constant at constant temperature for a saturated solution
i.e. when no more will dissolve.
-
The solubility
product for a sparingly soluble strong electrolyte is defined as the
product of the concentration of the ions raised to their appropriate
powers in a saturated solution at a specific temperature.
-
At the saturation
solution point, controlled by the Ksp expression, it doesn't matter how much solid you add, no more can
dissolve.
-
The solubility of
the sparingly soluble salt is governed by the Ksp
expression, i.e. whatever ion concentrations are present of the
compound, then the expression must be obeyed. (see
common ion effect below).
-
If on mixing
solutions containing the two constituent ions the Ksp
expression is exceeded, precipitation will take place until the
product of the ion concentrations equals the Ksp value. If
the Ksp expression is not exceeded, no precipitation will take place.
-
?
-
Other
solubility product expressions
-
Note: In the Ksp
expression, the
ion concentrations are raised to powers equal to their molar ratio in the compound:
-
silver chloride :
AgCl(s) + aq Ag+(aq) + Cl-(aq)
-
calcium carbonate:
CaCO3(s) + aq Ca2+(aq) + CO32-(aq)
(ΔH slightly -ve)
-
Ksp
= [Ca2+(aq)] [CO32-(aq)]
= 5.0 x 10-10 mol2 dm-6
-
Sea shells are
mainly made of calcium carbonate which is a tough mineral and will not
dissolve appreciably in water.
-
However, when the
sea creatures die and the remains sink to great depth in the oceans,
the shells will then dissolve because of changes in the position of
two equilibria. At great depths the pressure is much higher and the
temperature lower, both factors favouring more CO2
dissolving, therefore the following forward reaction is promoted,
-
CaCO3(aq)
+ CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3-(aq)
-
and secondly,
the lower temperature favours a higher Ksp value because
the dissolving process is slightly exothermic, so more of the
calcium carbonate will dissolve. There are no shells lying at the
bottom of the oceans! The limestone cliffs you see in the landscape
were formed in warm shallow tropical seas.
-
magnesium
hydroxide : Mg(OH)2(s) + aq Mg2+(aq) + 2OH-(aq)
-
silver
chromate(VI) : Ag2CrO4(aq) + aq
2Ag+(aq) + CrO42-(aq)
-
aluminium
hydroxide : Al(OH)3(s) + aq Al3+(aq) + 3OH-(aq)
-
antimony sulphide:
S2S3(s) + aq 2Sb3+(aq) + 3S2-(aq)
-
The
common ion effect
-
If a 2nd soluble
substance is added to the saturated solution of a salt, that has an
ion in common with the salt, this will cause precipitation of the
salt. This is because the Ksp expression is initially
exceeded, causing precipitation until the new concentrations satisfy
the Ksp expression Le Chatelier again!). This precipitation
is known as the 'common ion effect'.
-
Example
calculation 4.2.1
-
Example
calculation 4.2.2
-
If equal volumes
of aqueous 0.01 mol dm-3 sodium chloride and 0.005 mol dm-3
silver nitrate solution are mixed, show by calculation whether or not
a precipitate of silver chloride forms?
-
[NaCl(aq)]
= [Cl-(aq)] = 0.01 / 2 = 0.005 mol dm-3
-
[AgNO3(aq)]
= [Ag+(aq)] = 0.005 / 2 = 0.0025 mol dm-3
-
The ionic product
= [Ag+(aq)] x [Cl-(aq)] =
0.005 x 0.0025 = 1.25 x 10-5 mol2 dm-6
-
The ionic
product on mixing exceeds the Ksp value so precipitation
takes place.
-
Example calculation 4.2.3
-
For the
equilibrium: CaSO4(s) + aq Ca2+(aq) + SO42-(aq)
-
(a) What is
the solubility of calcium sulphate in a saturated solution of the salt
in g cm-3?
-
From the arguments
outlined in example 4.2.1(a)
-
[Ca2+(aq)]
= [CaSO4(aq)] =
√(2.0 x 10-5) =
4.47 x 10-3 mol dm-3
-
Ar's:
Ca = 40, S = 32, O = 16, Mr(CaSO4) = 136, 1 dm3
= 106 cm3
-
solubility = 4.47
x 10-3 x 136 = 0.608 g dm-3, so scaling down to
100 cm3
-
solubility
= 0.608 x 100/106 = 6.1 x 10-5 g/100 cm3
-
(b)
Assuming equal volumes of solutions are mixed, what is the minimum
concentration of sodium sulphate (Na2SO4)
solution that when added to a 1.0 x 10-4 mol dm-3
solution of calcium sulphate, will just begin to cause precipitation
of calcium sulphate?
-
Let [NaaSO4(aq)]
= [SO42-(aq)] = the theoretical
concentration of sodium sulphate at the point of precipitation.
-
On mixing the
equal volumes of solutions, the calcium sulphate concentration is
halved.
-
Therefore on
substituting into the Ksp expression on the point of
precipitation ...
-
2.0 x 10-5
= ((1.0 x 10-4)/2) x [NaaSO4(aq)]
-
therefore [NaaSO4(aq)]
= 2.0 x 10-5/0.5 x 10-4 = 0.4 mol dm-3
-
However, on
mixing, the sodium sulphate solution concentration must also be
halved,
-
therefore the
original sodium sulphate solution must be at least 0.8 mol dm-3
-
Example
calculation 4.2.4
-
Equal volumes of
0.025 mol dm-3 potassium bromide (KBr) and 0.005 mol
dm-3 lead(II) nitrate (Pb(NO3)2)solutions
were mixed.
-
(a) Write out (i)
the Ksp expression for lead(II) bromide and (ii) the ionic
equation for its precipitation.
-
(b) Show by
calculation if lead(II) bromide precipitates after mixing the
solutions.
-
[Br-(aq)]
= [KBr(aq)] = 0.025 / 2 = 1.25 x 10-2 mol dm-3
-
[Pb2+(aq)]
= [Pb(NO3)2(aq)] = 0.005 / 2 = 2.5 x
10-3 mol dm-3
-
The ionic
product = [Pb2+(aq)] [Br-(aq)]2
= 2.5 x 10-3 x (1.25 x 10-2)2 =
3.91 x 10-7 mol3 dm-9
-
The Ksp
value of 7.9 x 10-5 is NOT exceeded, so no precipitation
takes place.
4.3
Ion Exchange systems
-
Ion-exchange
materials have the capacity to hold ions in a dynamic equilibrium with
the same ions present in an aqueous solution.
-
They may be
synthetic polymer resins with immobile negative groups e.g. based on
the sulphonic acid group R-SO2O-H+(s)
acting as a cation exchanger. R represents the molecular
backbone of the polymer resin. (immobile,
exchangeable)
-
A resin with
immobile positive groups like R-N(CH3)3+Cl-(s)
can act as an anion exchanger.
-
Cation exchangers
occur naturally in the sheet structures of clay minerals in soil which
have excess immobile negative groups based on oxygen (e.g. clay-O-)
which hold cations like H+ or Ca2+.
-
How strongly
are ions held on the resin?
-
For singly
charged ions the binding order from strongest to weakest bound is:
-
For doubly
charged ions the binding order from strongest to weakest bound is:
-
For change in cation
charge: Not surprisingly, the general binding order from strongest
to weakest is M3+ > M2+ > M+ as the
increasing charge density of the hydrated ion increases, so will the
attraction of the ion to the immobile negative groups on the resin.
-
Effect of
cationic radius and extent of hydration for constant charge:
-
If you consider
the trends for Group 1 cations (M+) or Group 2 cations (M2+)
things don't seem to add up? The group trend is for increasing radius
down the group. This produces a decreasing charge density trend which
should result in weaker binding to the negatively charged resin.
However, the radius of the isolated ion does not count here, but what
does matter is the effective radius of the hydrated cation. The
smaller the ion, with its greater charge density, the greater its
attraction for water molecules and the larger the resulting hydrated
ion. Therefore the effective hydrated ionic radius actually decreases
down the group, and the effective surface charge density increases to
give the binding strength order.
-
Ion exchange case studies
-
Case study 4.3.1 Ion-exchange
processes are extremely important in soil chemistry.
-
Clay minerals are
based on sheets of linked silicate units.
-
Here the simple
tetrahedral silicate(IV) ion is SiO44- is linked
together via -O-Si-O- bonds in two dimensions, the resulting silicate
sheets have the general formula (Si2O52-)n
where n is very large number.
-
The excess
negative charge is balanced by various cations e.g. H+, K+,
Mg2+, Ca2+, Al3+ which are adsorbed
on or can fit in between the silicate sheets. The 'equation' below
shows how potassium ions might be exchanged with magnesium ions
-
One of the many
unfortunate consequences of acid rain from fossil fuel burning, is the
extra hydrogen ions will displace or wash out poisonous aluminium ions
from clay soils which are harmful to plants and animals.
-
Lime is added to
soil to reduce its acidity. The lime (calcium oxide) forms hydroxide
ions which will neutralise hydrogen ions held on the clay, so
increasing the pH. The hydrogen ions on the clay are replaced by
calcium ions, Ca2+.
-
Since clay
minerals act as cation exchangers, anions like chloride and nitrate
are readily washed out in rainwater, the latter ion from artificial
ammonium nitrate fertilisers can cause pollution problems like
eutrophication, though the ammonium cation is more likely to be
retained being a positive ion.
-
Radioactive
cations can be retained for quite some time in soil and only slowly
displaced and dispersed to non-harmful levels. Even now (2006) in
Northern England (Cumbria) sheep from a few farms cannot be sold for
meat because of radioactive caesium-137, strontium-90 and iodine-?
deposited on the soil they graze on. The radioactive contamination
came from rain containing radio-isotopes a few days after the Russian
Chernobyl nuclear reactor explosion in 1986. Caesium is more strongly
bound than most other singly charged cations and some M2+
cations too? All the caesium will eventually end up in the Irish Sea
and very diluted and harmless to aquatic life, but it takes time. The
equation below shows the adsorption of the caesium ions onto an
alumino-silicate sheets in clay by displacing less strongly held
potassium ions,
-
[clay-O]-K+(s)
+ Cs+(aq)
[clay-O]-Cs+(s)
+ K+(aq)
-
and the strontium
ion Sr2+ is also strongly binding and will also displace
other ions to remain in the soil for some time (see Mg2+...K+
exchange, 1st equation in section 4.3 above). However the radioactive
iodine is likely to end up as the iodide ion. I-, so, being
an anion, is more readily washed out of the soil by rainwater and not
retained by the negatively charged alumino-silicate sheets.
-
Case study 4.3.2
Removing hardness from water:
- Packs of ion
exchange resins can hold or release ions in an ion exchange process.
- Negative polymer resin columns hold hydrogen ions or sodium ions. These
can be replaced by calcium and magnesium ions when hard water passes down the column.
The more highly charged calcium or magnesium ions are more
strongly held on the negatively charged
resin. The freed or displaced hydrogen or sodium ions do not form a scum with soap
(see GCSE notes
on hard/soft water).
- e.g. 2[resin]-H+(s)
+ Ca2+(aq) [resin]-Ca2+-[resin](s)
+ 2H+(aq)
- or 2[resin]-Na+(s)
+ Mg2+(aq) [resin]-Mg2+-[resin](s)
+ 2Na+(aq) etc.
-
Case study 4.3.3 Water
purification:
- You can also use an ion-exchange resin to replace
negative ions by using a positively charged resin initially holding
hydroxide ions (OH-) e.g. to remove chloride (Cl-), nitrate (NO3-
and
potentially harmful) and sulphate ions (SO42-)
etc.
- [resin]+OH-(s)
+ Cl-(aq) [resin]+Cl-(s)
+ OH-(aq)
- [resin]+OH-(s)
+ NO3-(aq) [resin]+NO3-(s)
+ OH-(aq)
- 2[resin]+OH-(s)
+ SO42-(aq) [resin]+SO42-[resin]+(s)
+ 2OH-(aq) etc.
- Now, by using both a positive
ion-exchange resin (here) and a negatively charged ion-exchange resin
(see 4.3.2), you can completely de-ionise water
because the released hydrogen ions and hydroxide ions combine to form
very pure
water.
- H+(aq)
+ OH-(aq) ==> H2O(l)
- The ionic equation for
neutralisation.
- However, it will NOT remove
non-ionic substances like organic pesticides etc.
GENERAL
REVISION
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