A LevelChemistryCIERevision Notes5. Physical Chemistry (A Level Only)5.5 Equilibria (A Level Only)5.5.9 Partition Coefficients

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Partition Coefficients (CIE A Level Chemistry)

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Partition Coefficient Formula

  • The partition coefficient (Kpc) is the ratio of the concentrations of a solute in two different immiscible solvents in contact with each other when equilibrium has been established (at a particular temperature)
  • For example, methylamine (CH3NH2) is dissolved in two immiscible solvents:
    • Water
    • An organic solvent

  • A separating funnel is shaken with the organic solvent and aqueous methylamine
  • The methylamine is soluble in both solvents, so when the mixture is left to settle an equilibrium is established
    • The rate of methylamine molecules moving from the organic layer into the aqueous layer is equal to the rate of molecules moving from the aqueous layer to the organic layer

CH3NH2(aq) ⇌ CH3NH2(organic solvent)

  • The value of its equilibrium constant is also called the partition coefficient

Kpc = fraction numerator stretchy left square bracket CH subscript 3 NH subscript 2 space open parentheses organic space solvent close parentheses stretchy right square bracket over denominator stretchy left square bracket CH subscript 3 NH subscript 2 space open parentheses aq close parentheses stretchy right square bracket end fraction

Equilibria - Partition Coefficient, downloadable AS & A Level Chemistry revision notes

The partition coefficient is the ratio of methylamine molecules in the organic and aqueous layer once equilibrium has been established

Calculating Partition Coefficients

  • The partition coefficient (Kpc) for a system in which the solute is in the same physical state in the two solvents can be calculated using the equilibrium expression

Worked example

Calculating the partition coefficient

100 cm3 of a 0.150 mol dm-3 solution of aqueous methylamine (CH3NH2) was shaken with 75.0 cm3 of an organic solvent and left in the separating funnel to allow an equilibrium to be established.

Only 50.0 cm3 of the aqueous layer was run off and titrated against 0.225 mol dm-3 hydrochloric acid (HCl) with an end-point of 14.1 cm3 of HCl.

Calculate the partition coefficient of methylamine the organic solvent and water.

Answer

  • Step 1: Write down the equilibrium equation:

CH3NH2 (aq) ⇌ CH3NH2 (organic solvent)

  • Step 2: Write down the equilibrium expression:
    • Kpc = fraction numerator open square brackets CH subscript 3 NH subscript 2 space open parentheses organic space solvent close parentheses close square brackets over denominator open square brackets CH subscript 3 NH subscript 2 space open parentheses aq close parentheses close square brackets end fraction
  • Step 3: Determine how many moles of CH3NH2 has reacted with HCl at the end-point:
    • At the end-point, all CH3NH2 (aq) has been neutralised by HCl (aq)

CH3NH2 (aq) + HCl (aq) → CH3NH3Cl (aq)

    • CH3NH2 and HCl react in a ratio of 1:1
    • Mol (HCl) = mol (CH3NH2) = 0.225 x 0.0141 = 3.18 x 10-3 mol
  • Step 4: Determine the number of moles of CH3NH2 present in the aqueous layer
    • Only 50.0 cm3 of the aqueous layer was used to titrate against HCl
    • Thus, 3.18 x 10-3 mol of CH3NH2 was present in only 50.0 cm3 of the aqueous layer
    • The number of moles of CH3NH2 in 100 cm3 aqueous layer is, therefore:
    • Mol (CH3NH2 aqueous layer) = 3.18 x 10-3 x 2 = 6.34 x 10-3 mol
  • Step 5: Determine the number of moles of CH3NH2 in the organic layer:
    • Mol CH3NH2 (organic layer) = mol CH3NH2 (total) - mol CH3NH2(aqueous layer)
    • Mol CH3NH2 (total) = 0.100 x 0.150 = 0.015 mol
    • Mol CH3NH2 (organic layer) = 0.015 - 6.34 x 10-3 = 8.67 x 10-3 mol
  • Step 6: Change the number of moles into concentrations:
    • Concentration (CH3NH2 in aqueous layer) = begin mathsize 14px style fraction numerator 6.34 cross times 10 to the power of negative 3 end exponent over denominator 0.100 end fraction end style = 0.063 mol dm-3 
    • Concentration (CH3NH2 in organic layer) = begin mathsize 14px style fraction numerator 8.67 cross times 10 to the power of negative 3 end exponent over denominator 0.075 end fraction end style = 0.116 mol dm-3
  • Step 7: Substitute the values into the Kpc expression:
    • Kpc = begin mathsize 14px style fraction numerator 0.116 over denominator 0.063 end fraction end style = 1.83
    • Since the value of Kpc is larger than 1, methylamine is more soluble in the organic solvent than in water

Factors Affecting the Partition Coefficient

  • The partition coefficient (Kpc) depends on the solubilities of the solute in the two solvents
  • The degree of solubility of a solute is determined by how strong the intermolecular bonds between solute and solvent  are
  • The strength of these intermolecular bonds, in turn, depends on the polarity of the solute and solvent molecules
  • For example, ammonia is more soluble in water than in an organic solvent such as carbon tetrachloride (CCl4)
    • Ammonia and water are both polar molecules that form hydrogen bonds with each other
    • Ammonia forms permanent dipole-induced dipole forces with the non-polar CCl4 molecules
    • Since these forces are much weaker than hydrogen bonding, ammonia is less soluble in CCl4

  • When Kpc is < 1 the solute is more soluble in water than the organic solvent
  • When Kpc is > 1 the solute is more soluble in the organic solvent than the water

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    Philippa

    Author: Philippa

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    Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener.