Explain how solvent polarity affects partitioning
Relate a partition coefficient to the polarities of the solute and the two solvents.
Section 25 · Equilibria
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 25. Use the equilibrium expressions and data stated in the question; written practice is self-checked and human subject-teacher review remains pending.
Before you begin
- Use AS equilibrium expressions, mole calculations, logarithms, and Brønsted–Lowry acid–base ideas.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is a useful first guide when predicting which solvent better dissolves a solute?
What you will learn
- Compare solute–solvent interactions qualitatively
- Relate polarity matching to relative equilibrium concentrations
- Explain why solvent choice changes Kpc
Solute and solvent polarity influence distribution
A solute tends to be more soluble in a solvent with which it can make favourable intermolecular interactions. Polar solutes are generally better stabilised by polar solvents; non-polar solutes are generally better stabilised by non-polar solvents.
If the solute is more strongly stabilised in the organic phase, its equilibrium concentration there may be relatively high and Kpc (organic/aqueous) larger. If it is better stabilised in water, the ratio may be smaller.
Use the actual species and stated solvent pair
Hydrogen bonding, dipole interactions and dispersion forces can all affect the relative solubility. Use polarity as a qualitative guide rather than assuming every organic solvent is non-polar or every solute follows one rule.
Kpc describes a particular solute–solvent pair at a particular temperature. Changing the solvent or temperature can change the coefficient; the ratio is not a universal constant for the solute alone.
Pause and explain
If a solute is more concentrated in water, what happens to Kpc=[org]/[aq]?
Worked example
A polar solute forms strong hydrogen bonds with water but weak interactions with a non-polar solvent. Predict which phase will have the greater equilibrium concentration and the trend in Kpc=[org]/[aq].
Show the worked solution
- Compare the solute's interactions with each solvent.
- Identify the phase that better stabilises the solute.
- Relate the concentration comparison to organic-over-aqueous Kpc.
Answer The aqueous phase is expected to contain the greater solute concentration, so [org]/[aq] and Kpc are expected to be relatively small for this solvent pair.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A non-polar solute interacts more favourably with a non-polar organic solvent than with water. Predict the relative phase concentrations and Kpc trend.
Give me a hint
Use the stated organic-over-aqueous convention.
Compare my reasoning
- The organic phase better stabilises the solute.
- Expect a higher organic equilibrium concentration.
- Relate the ratio to Kpc.
The organic phase is expected to have the greater solute concentration, so Kpc=[org]/[aq] is relatively large for that pair.
Look for these in your work
- Interaction comparison used
- Phase identified
- Ratio trend consistent
Two organic solvents have different polarities. Explain why the same solute may have different Kpc values when each is paired with water.
Give me a hint
Kpc depends on relative interactions with both solvents.
Compare my reasoning
- Changing the organic solvent changes solute–solvent interactions.
- The equilibrium concentration ratio can therefore change.
- Kpc belongs to a particular solvent pair and temperature.
The new solvent may stabilise the solute differently, changing its organic-to-aqueous concentration ratio. Kpc is specific to the solvent pair and temperature.
Look for these in your work
- Solvent interaction identified
- Ratio change explained
- Conditions specified
A learner says a large Kpc proves that the solute is non-polar. Explain why that conclusion is too strong.
Give me a hint
A coefficient compares two particular phases and can reflect several interactions.
Compare my reasoning
- Kpc is a relative concentration ratio.
- Both solvent identities and temperature affect it.
- A large value alone does not establish a complete molecular description.
A large organic/aqueous Kpc means the solute is relatively concentrated in that organic phase under those conditions. It does not by itself prove the molecule is non-polar; specific interactions and solvent choice also matter.
Look for these in your work
- Meaning of Kpc stated
- Solvent dependence included
- Inference qualified
Common mistakes
- Treating Kpc as a property of the solute alone
- Assuming every organic solvent is non-polar
- Using polarity without identifying the defined solvent ratio
- Presenting qualitative trends as exact numerical predictions
What can you explain now?
What is a useful first guide when predicting which solvent better dissolves a solute?
Compare with the explanation
Compare the polarity and intermolecular interactions
Relative solute–solvent interactions help explain the equilibrium distribution.
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Make it stick
How do polarity and intermolecular forces influence the concentration ratio without making Kpc universal?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.