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Colligative Properties — Notes

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1. Colligative properties; relative lowering of vapour pressure

Colligative properties of dilute solutions of non-volatile solutes depend on the number of solute particles, not their nature: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.

Relative lowering of vapour pressure (from Raoult’s law, p=p∘x1p = p^\circ x_1): p∘−pp∘=x2\dfrac{p^\circ - p}{p^\circ} = x_2. For dilute solutions x2≈n2n1=w2M1M2w1x_2 \approx \dfrac{n_2}{n_1} = \dfrac{w_2M_1}{M_2w_1}, giving M2=w2M1p∘w1(p∘−p)M_2 = \dfrac{w_2M_1p^\circ}{w_1(p^\circ - p)}.

Schematic: fewer solvent particles at the surface means fewer escape per second.

Formulas

  • Relative lowering of vapour pressure
    p∘−pp∘=x2≈w2M1M2w1\frac{p^\circ - p}{p^\circ} = x_2 \approx \frac{w_2 M_1}{M_2 w_1}

Key terms

Colligative property

A property of a dilute solution that depends only on the number of solute particles, not on their nature.

Relative lowering of vapour pressure

The lowering of vapour pressure divided by the vapour pressure of the pure solvent; equal to the mole fraction of the solute.

2. Elevation of boiling point

A liquid boils when its vapour pressure equals the external pressure; a solution’s vapour pressure is lower, so it boils at a higher temperature: ΔTb=Tb−Tb∘\Delta T_\text{b} = T_\text{b} - T^\circ_\text{b}.

Derivation: near Tb∘T^\circ_\text{b} the vapour pressure curves of solvent and solution are nearly parallel straight lines, so ΔTb∝Δp\Delta T_\text{b} \propto \Delta p. With p∘p^\circ constant, ΔTb∝Δp/p∘=x2≈n2/n1=mM1/1000\Delta T_\text{b} \propto \Delta p/p^\circ = x_2 \approx n_2/n_1 = mM_1/1000. Hence ΔTb=Kbm\Delta T_\text{b} = K_\text{b}m, with KbK_\text{b} the ebullioscopic (molal elevation) constant, in K kg mol⁻¹ (water 0.52, benzene 2.53). Thermodynamically Kb=R(Tb∘)2M11000 ΔvapHK_\text{b} = \dfrac{R(T^\circ_\text{b})^2M_1}{1000\,\Delta_\text{vap}H}.

Molar mass: M2=1000Kbw2ΔTbw1M_2 = \dfrac{1000K_\text{b}w_2}{\Delta T_\text{b}w_1}.

  • Pure solvent
  • Solution
Illustrative; over this small range the curves are nearly straight and parallel, which is the basis of the derivation.

Formulas

  • Elevation of boiling point
    ΔTb=Kb m=1000 Kb w2M2 w1\Delta T_\text{b} = K_\text{b}\,m = \frac{1000\,K_\text{b}\,w_2}{M_2\,w_1}

Key terms

Ebullioscopic constant

The elevation of boiling point produced by a 1 mol kg⁻¹ solution; depends only on the solvent.

3. Depression of freezing point

At the freezing point the solid and liquid solvent have equal vapour pressures. The solution’s vapour pressure is lower, so it equals that of the solid at a lower temperature: ΔTf=Tf∘−Tf\Delta T_\text{f} = T^\circ_\text{f} - T_\text{f}.

Derivation: as for boiling point, ΔTf∝Δp∝x2∝m\Delta T_\text{f} \propto \Delta p \propto x_2 \propto m, so ΔTf=Kfm\Delta T_\text{f} = K_\text{f}m, with KfK_\text{f} the cryoscopic (molal depression) constant (water 1.86, benzene 5.12, camphor 37.7 K kg mol⁻¹); Kf=R(Tf∘)2M11000 ΔfusHK_\text{f} = \dfrac{R(T^\circ_\text{f})^2M_1}{1000\,\Delta_\text{fus}H}.

Molar mass: M2=1000Kfw2ΔTfw1M_2 = \dfrac{1000K_\text{f}w_2}{\Delta T_\text{f}w_1}. Uses: antifreeze, salting roads, camphor (Rast) method.

  • Solid solvent (ice)
  • Liquid solvent
  • Solution
Illustrative curves. The solution freezes where its vapour pressure equals that of the solid solvent.

Formulas

  • Depression of freezing point
    ΔTf=Kf m=1000 Kf w2M2 w1\Delta T_\text{f} = K_\text{f}\,m = \frac{1000\,K_\text{f}\,w_2}{M_2\,w_1}

Key terms

Cryoscopic constant

The depression of freezing point produced by a 1 mol kg⁻¹ solution; depends only on the solvent.

4. Osmosis and osmotic pressure

Diffusion: spreading of particles from high to low concentration, no membrane. Osmosis: flow of solvent only through a semipermeable membrane (parchment, cellophane, copper ferrocyanide, animal bladder) from pure solvent or dilute solution into a more concentrated solution. Osmotic pressure π\pi: the excess pressure on the solution that just stops osmosis.

van’t Hoff laws: Boyle–van’t Hoff (π∝C\pi \propto C at constant TT); Charles–van’t Hoff (π∝T\pi \propto T at constant CC); Avogadro–van’t Hoff (equimolar solutions at the same TT have equal π\pi). Combined: π=CRT\pi = CRT, πV=nRT\pi V = nRT, M2=w2RTπVM_2 = \dfrac{w_2RT}{\pi V} — best for proteins and polymers (room temperature, large values).

Isotonic solutions have equal π\pi (0.9% NaCl and blood); a cell shrinks in a hypertonic solution and swells in a hypotonic one. Reverse osmosis: pressure greater than π\pi forces solvent out of the solution; used for desalination of sea water.

At equilibrium the pressure of the extra column of height h equals the osmotic pressure of the solution.

Formulas

  • Osmotic pressure
    π=CRT\pi = CRT
  • Molar mass from osmotic pressure
    M2=w2RTπVM_2 = \frac{w_2 R T}{\pi V}

Key terms

Semipermeable membrane

A membrane that allows solvent molecules but not solute particles to pass.

Osmotic pressure

The excess pressure that must be applied to a solution to prevent the flow of solvent into it through a semipermeable membrane.

Isotonic solutions

Solutions with the same osmotic pressure at the same temperature.

Reverse osmosis

Flow of solvent out of a solution through a semipermeable membrane when a pressure greater than its osmotic pressure is applied.

5. Abnormal molar masses and the van’t Hoff factor

Dissociation increases the number of particles (molar mass appears low); association decreases it (molar mass appears high, e.g. carboxylic acids dimerise in benzene).

i=observed colligative propertycalculated colligative property=normal molar massobserved molar massi = \dfrac{\text{observed colligative property}}{\text{calculated colligative property}} = \dfrac{\text{normal molar mass}}{\text{observed molar mass}}.

Modified equations: Δp/p∘=ix2\Delta p/p^\circ = ix_2; ΔTb=iKbm\Delta T_\text{b} = iK_\text{b}m; ΔTf=iKfm\Delta T_\text{f} = iK_\text{f}m; π=iCRT\pi = iCRT.

  • Dissociation into nn ions: i=1+(n−1)αi = 1 + (n-1)\alpha, so α=i−1n−1\alpha = \dfrac{i-1}{n-1}.
  • Association of nn molecules: i=1−α+αni = 1 - \alpha + \dfrac{\alpha}{n}, so α=1−i1−1/n\alpha = \dfrac{1-i}{1-1/n}.
Table of typical van’t Hoff factors
SoluteLikely i
Glucose, urea in water1
NaCl in waterup to 2
K₂SO₄ in waterup to 3
Ethanoic or benzoic acid in benzeneabout 0.5

Formulas

  • Dissociation
    i=1+(n−1)αi = 1 + (n-1)\alpha
  • Association
    i=1−α+αni = 1 - \alpha + \frac{\alpha}{n}

Key terms

van’t Hoff factor

Ratio of the observed colligative property to the value calculated assuming no dissociation or association.

Degree of dissociation

Fraction of the solute that dissociates into ions.

Exam tips

Common mistakes

  • Using the mass of solution instead of solvent for molality.
  • Adding 273 to a temperature difference.
  • Using mL instead of L, or a gas constant that does not match the pressure unit, in π = CRT.
  • Forgetting the van’t Hoff factor for electrolytes, or using n = 2 for K₂SO₄ (it is 3).
  • Mixing up hypotonic (cell swells) and hypertonic (cell shrinks).

How marks are usually awarded

  • Derivations (3 marks): the vapour pressure argument, the proportionality to x₂, and the final expression with the constant named.
  • Numericals: formula, substitution with units, answer with units; full working is expected in ISC.
  • van’t Hoff factor problems: calculated value, observed value, i, then α.