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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.

(c) doc b 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:

    • 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) (c) doc b 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) (c) doc b 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.

    • Note that if chemical equilibrium is involved (like CO2 in water) the simple partition law may not apply even though K remains constant at constant temperature (see also example 1. below).

  • Examples of partition systems involving two immiscible liquids

    1. Iodine in water/tetrachloromethane: I2(aq) (c) doc b 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:

        • I-(aq) + I2(aq) (c) doc b I3-(aq) as well as I2(aq) (c) doc b I2(CCl4)

    2. Ammonia in water/trichloromethane: NH3(aq) (c) doc b 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.

  • TOP and LINKSCase 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.

        1. (c) doc b The mixture is shaken to extract the NOC into the NPS and the two layers allowed to settle out.

        2. The aqueous layer of impurities can be run off.

        3. 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.

        • If the NOC is a water insoluble liquid it may be purified using the same procedure without the addition of a NPS and after separation and drying in fractionally distilled to further purify it.

      • 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

    • TOP and LINKSCase 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.

          • (c) doc bOn 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.

    • TOP and LINKSCase 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:

    • Ex. Q4.1.1 20g of butanedioic acid (BDA) was shaken with a mixture of 100 cm3 ether and 100 cm3 water at 25oC. After titration with standard sodium hydroxide the concentration of the acid was found to be 0.024 mol dm-3 in ether and 0.16 mol dm-3 in water.

      • (a) Calculate the distribution coefficient KD for butanedioic acid between ether/water.

        • KD = [BDA(ether)] / [BDA(water)] = 0.024 / 0.16 = 0.15

      • (b) If 10g of BDA had been shaken with 50 cm3 of each solvent at 25oC, what value would expect for KD if the layers were again analysed?

        • 0.15, since the partition law states the ratio of concentrations remains constant at constant temperature.

      • (c) If 10g of butanedioic acid was dissolved in 50 cm3 of ether at 25oC, calculate how much of the acid can be extracted with 50 cm3 of water.

        • If we call x the mass in g extracted into water, V the liquid volumes, and expressing the concentrations in gcm-3, substitution into the partition expression gives ...

        • KD =

          [(10-x)/Vether]       (10-x)
          ---------------------- = ------- (since V's cancel out)
             [x/Vwater]            x
        • Rearranging: 0.15 x = 10 - x, so 1.15x = 10
        • therefore x = 10/1.15 = 8.7g of BDA extracted.

TOP and LINKS

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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