Solutions and Colligative Properties

1Raoult's law and ideal solutions

Raoult's law says each component of an ideal solution contributes vapour in proportion to how much of the liquid it is: p_A = x_A p_A°. Plot both partials against x_A and you get two straight lines crossing, and their sum is a third straight line running from p_B° at x_A = 0 up to p_A° at x_A = 1. A solution is ideal when the A–B attraction matches the A–A and B–B attractions it replaced, so ΔH(mix) = 0 and ΔV(mix) = 0 — benzene with toluene, hexane with heptane.

The vapour is not the liquid. Dalton's law turns the partial pressures into a vapour composition, y_A = p_A/p, and because the more volatile component contributes more than its share of the pressure, y_A comes out above x_A. That gap is the whole of fractional distillation.

Figure. Three straight lines, and they are not independent: at every x_A the total is exactly the sum of the two partials. The total therefore starts at p_B° = 50 on the left edge, where the liquid is pure toluene, and ends at p_A° = 150 on the right, where it is pure benzene. The marked point is the worked example at x_A = 0.40.

Reading the three lines

1. Each partialp_A climbs from 0 to p_A° as x_A runs 0 → 1, while p_B falls from p_B° to 0 over the same span.
2. The totalAdd them: p = p_B° + (p_A° − p_B°)x_A, a straight line joining the two pure vapour pressures.
3. The vapoury_A = p_A/p. Set it beside x_A and the more volatile component is always the enriched one.

2Relative lowering of vapour pressure

Make the solute non-volatile and only the solvent can contribute vapour, so Raoult's law collapses to p = x_solvent p°. Rearranged, (p° − p)/p° = x_solute: the relative lowering equals the mole fraction of the solute and depends on nothing else about it. Sugar, urea and glycerol at the same mole fraction all lower the vapour pressure by the same amount. This is the first, and the most direct, of the colligative properties.

In a dilute solution x_solute ≈ n_solute/n_solvent, and that turns a pressure reading into a molar mass: with masses w₁ of solvent (molar mass M₁) and w₂ of solute, M₂ = w₂ M₁ p° ÷ [w₁ (p° − p)]. Note which lowering appears where — the relative lowering is the dimensionless ratio, the absolute lowering p° − p is a pressure and grows with temperature as p° does.

Figure. Five arrows leave the pure surface and three leave the solution's, because 5 × 0.6 = 3: the escaping tendency is scaled by the solvent's mole fraction and by nothing else. The two arrows that are missing on the right are the surface sites the non-volatile solute is sitting on, and they contribute no vapour at all.

3Non-ideal solutions and deviations

Real mixtures rarely obey Raoult's law. If the A–B attraction is weaker than the A–A and B–B attractions it replaced, molecules escape more easily than the law allows and the measured total sits above the straight line at every intermediate composition. Mixing then costs energy, so ΔH(mix) > 0, and the loosely packed mixture expands, so ΔV(mix) > 0. That is positive deviation: ethanol with acetone, where acetone gets between the ethanol molecules and breaks up their hydrogen-bonded network.

Stronger A–B attraction gives the mirror image. The measured total falls below the line, mixing releases heat so ΔH(mix) < 0, and the tighter packing gives ΔV(mix) < 0. Chloroform with acetone is the standard case: a C–H···O hydrogen bond forms between the two that exists in neither pure liquid, so both escape less readily than they would alone.

Push either deviation far enough and the total-pressure curve develops an extremum — a maximum for positive deviation, a minimum for negative. At that composition the vapour and the liquid have the same composition, and the mixture is an azeotrope.

Figure. All three curves are pinned at the two ends, because a pure liquid has the vapour pressure it has whatever the other component would have done. Between the ends the positive curve lies above the dashed Raoult line and the negative curve below it. The maximum on the positive curve, at 71.25 mm Hg, beats both pure values of 60 and 40 — that is the azeotrope, and because it is a pressure maximum it is a boiling-point minimum.

Deciding which deviation you have

1. Compare the attractionsAsk whether the new A–B interaction is weaker or stronger than the A–A and B–B interactions it replaced.
2. Predict the heatWeaker A–B costs energy to make: ΔH(mix) > 0, endothermic, and the flask cools. Stronger A–B releases it and the flask warms.
3. Read the pressureEndothermic mixing lifts the observed total above the Raoult line; exothermic mixing drops it below.

4Azeotropes and why distillation stops

An azeotrope is a composition at which the vapour has the same composition as the liquid. Distil it and the distillate is identical to what stays behind, so no column, however tall, separates the pair. It is a property of the mixture at a given pressure, not a compound: change the pressure and the azeotropic composition moves.

Where the pressure curve has a maximum, the mixture is easier than either pure liquid to drive into the vapour, so it reaches 1 atm at a lower temperature. Maximum vapour pressure means minimum boiling point. Positive deviation therefore gives a minimum-boiling azeotrope: ethanol and water at 95.6% ethanol by mass, boiling at 78.2 °C, below both pure ethanol at 78.4 °C and water at 100 °C. Negative deviation, with its pressure minimum, gives a maximum-boiling azeotrope: 68% HNO₃ in water boils at 120.5 °C, above both.

The boiling-point curve is the pressure curve turned over. Not merely flipped in appearance — turned over for a reason: boiling happens where the vapour pressure reaches the applied pressure, and vapour pressure always rises with temperature, so a composition with more pressure needs less temperature. The extremum stays at the same composition and changes sense.

Figure. This is the previous figure's positive-deviation curve read as a boiling point instead of a pressure, and it is upside down. The pressure maximum at x_A = 0.625 has become the boiling-point minimum at x_A = 0.625 — same composition, opposite sense — and it lies below both pure boiling points. The dashed line is what an ideal pair does: boiling point sliding smoothly from one pure value to the other, with no extremum and therefore no azeotrope.

5Boiling-point elevation and freezing-point depression

A non-volatile solute lowers the solvent's vapour pressure at every temperature, and both of these properties follow from that one fact. The solution must be pushed hotter than the pure solvent before its vapour pressure reaches 1 atm, so the boiling point rises: ΔTb = i Kb m. It must be taken colder before its vapour pressure falls to that of the pure solid solvent, so the freezing point falls: ΔTf = i Kf m.

Both take molality — moles of solute per kilogram of solvent. Molality is settled by two weighings, so warming the solution cannot change it; molarity, per litre of solution, changes as soon as the solution expands. Kb and Kf belong to the solvent alone and say nothing about the solute: for water Kb = 0.52 and Kf = 1.86 K kg mol⁻¹, and for benzene Kf = 5.12 K kg mol⁻¹ — though exam papers conventionally hand you 4.9 for benzene, and you use the value the question gives.

Because Kf/Kb = 3.58 for water, the same solution shifts its freezing point 3.58 times as far as its boiling point. That, plus the fact that you can cool a sample without boiling anything away, is why cryoscopy is the standard laboratory method and ebullioscopy is not.

Figure. A solvent freezes where its liquid and solid curves cross, which for water is the right-hand mark. Raoult drags the whole solution curve below the pure-water curve, so it does not reach the ice curve until the left-hand mark: that gap is ΔTf. The solution drawn is deliberately concentrated — mole fraction of water 0.90 — because a 0.1 m solution's 0.372 °C would be invisible at this scale. The boiling end works the same way, with a horizontal 1 atm line off to the right, and the shift there is smaller by Kb/Kf = 0.28.

Both shifts from one cause

1. Vapour pressure dropsRaoult pulls the solution's vapour-pressure curve below the pure solvent's at every temperature.
2. Boiling is delayedThe lowered curve meets the 1 atm line further to the right, so the boiling point rises by i Kb m.
3. Freezing is advancedIt also meets the solid solvent's curve further to the left, so the freezing point falls by i Kf m.

6Osmosis and osmotic pressure

A semipermeable membrane lets solvent through and holds solute back. Put pure solvent on one side and a solution on the other and solvent crosses into the solution; osmotic pressure π is the pressure you would have to apply on the solution side to stop it. For a dilute solution π = i C R T, where C is the molarity — moles per litre of solution, not per kilogram of solvent. This is the one colligative property in the chapter written in molarity, which is also why it is the one that changes when the thermostat drifts.

Two solutions with equal π at the same temperature are isotonic. The more concentrated of an unequal pair is hypertonic and pulls water out of the other, which is what makes a red blood cell shrink in strong brine and burst in pure water. Apply more than π to the solution side and the solvent flows the other way: that is reverse osmosis, and it is how sea water is desalinated.

Osmotic pressure is the method of choice for macromolecules because it is huge next to the other three effects at the same concentration. A 1.0 × 10⁻⁴ M aqueous solution — at this dilution also 1.0 × 10⁻⁴ molal — stands up a couple of centimetres of water column, which you read with a ruler, while its freezing-point depression is about 2 × 10⁻⁴ °C, which no laboratory thermometer resolves.

Figure. With nothing pressing on either side, solvent moves towards the solution, so the solution's level climbs and the pure solvent's drops. Press on the solution harder than π and the flow reverses, which is the reverse osmosis in the prose above. It stops when the extra head h presses back exactly as hard as the osmosis pushes, and at that point π = ρgh — which is why an osmometer reads a length. The membrane is the dashed line: solvent passes it, solute does not, and that asymmetry is the whole mechanism.

7The van't Hoff factor

Colligative formulas count particles, so any solute that changes its particle count in solution needs a correction. The van't Hoff factor i is the observed colligative effect divided by the effect calculated for unchanged formula units — equivalently, the number of particles that one formula unit actually delivers.

i > 1 is dissociation. NaCl gives 2, K₂SO₄ gives 3, and Al₂(SO₄)₃ gives 2 Al³⁺ plus 3 SO₄²⁻, so 5. i < 1 is association: carboxylic acids hydrogen-bond into pairs in benzene, and complete dimerisation halves the particle count to i = 0.5. Partial change is measured by the degree α — α = (i − 1)/(n − 1) when one formula unit dissociates into n particles, and α = (1 − i)/(1 − 1/n) when n molecules associate into one.

Those integers are theoretical values, approached only at infinite dilution. At any concentration you can actually make up, oppositely charged ions spend part of their time paired and fewer particles are free than the formula promises, so the observed i falls short — slightly for NaCl, badly for Al₂(SO₄)₃, whose 3+ and 2− ions attract each other hardest of the four. Quote 5 for Al₂(SO₄)₃ as the limit, not as the measurement.

Figure. i = 1 + (n − 1)α for dissociation, so every dissociating solute leaves the dashed i = 1 line upwards with a slope of n − 1 and arrives at i = n when α = 1. Association runs the other way on the same axes: i = 1 − α/2 for dimerisation, ending at 0.5. Everything above the dashed line makes more particles than you weighed out and everything below it makes fewer. The marked point is the worked example's weak acid, barely off the line at α = 0.075.

8Abnormal molar mass

Every colligative measurement really returns a number of moles, and a molar mass is a mass divided by moles. So if the solute changed its particle count, the moles are wrong and the molar mass inherits the error exactly: M(observed) = M(formula)/i.

Dissociation makes i > 1, so it makes the measured molar mass come out too low. Association makes i < 1, so it makes it come out too high. The direction of the error is therefore a diagnosis, and the solvent decides which one you get: benzoic acid dimerises in benzene and reads near 244, while in water it behaves as a weak acid and reads a little below its formula mass of 122. Same solute, opposite abnormality.

Figure. Each bar is the measured molar mass as a fraction of the true formula mass, so the three are directly comparable even though the three substances are not. Every length is exactly 1/i times the same unit: 0.5 for NaCl, 1 for glucose, 2 for benzoic acid in benzene. Dissociation shortens the bar, association doubles it, and only a solute that neither dissociates nor associates measures what it weighs.