Solutions

Two beakers sit side by side under one sealed bell jar at constant temperature. The left holds 100 mL of pure water. The right holds 100 mL of sugar solution. Come back a week later.

Almost everyone predicts the same thing: both levels drop a little as vapour fills the jar, then nothing more happens.

What actually happens is that the left beaker is bone dry and the right one has overflowed. Every molecule of pure water has crossed over.

Nothing carried it across except the vapour. The jar settles at some pressure $p_{jar}$. Pure water is only satisfied at $p^{\circ }$, and $p_{jar}<p^{\circ }$, so it keeps evaporating. The solution is satisfied at $p<p^{\circ }$, and $p_{jar}>p$, so it keeps condensing. The traffic is one-way and it cannot stop until the left beaker has nothing left to give.

That is the whole chapter in one experiment. A solvent is more reluctant to leave a solution than to leave itself.

Boiling, freezing and osmosis are three further ways of asking the solvent to leave. All three record the same reluctance, which is why the four properties are called colligative, from the Latin for bound together.

And the reluctance is a counting effect. It depends on how many solute particles are present, never on what they are.

That single fact does three jobs:

1. Concentration, Briefly

The measures came with the mole concept. This chapter uses them constantly, so restate them precisely.

Every colligative formula in this chapter uses molality or mole fraction. The one exception is osmotic pressure, which uses molarity.

The reason is that these experiments change the temperature, and volume changes with temperature while mass does not. Osmotic pressure escapes because it is measured at one fixed temperature.

Trap. A boiling point elevation calculated from molarity is not slightly wrong. It is measuring a quantity that drifts while the experiment runs.

Two conversions are worth holding, since JEE asks for them directly. Take $M_A$ as the solvent molar mass, $M_B$ the solute molar mass and $d$ the solution density in g mL${}^{-1}$.

$x_B=\frac{m{M}_A}{1000+m{M}_A},\text{Molarity}=\frac{1000md}{1000+m{M}_B}$

Both follow from taking exactly 1 kg of solvent as the basis, which contains $1000/M_A$ mol of solvent and $m$ mol of solute.

Illustration 1

An aqueous glucose solution is 20.0 per cent by mass and has density 1.08 g mL${}^{-1}$. Find its molality, the mole fraction of glucose, and its molarity.

Take 100 g of solution as the basis: 20.0 g glucose and 80.0 g water.

$n_B=\frac{20.0}{180}=0.1111\text{mol},n_A=\frac{80.0}{18}=4.444\text{mol}$

$m=\frac{0.1111}{0.0800\mathrm{kg}}=1.39\mathrm{mol}\mathrm{k}{\mathrm{g}}^{-1},x_B=\frac{0.1111}{4.556}=0.0244$

The 100 g of solution occupies $100/1.08=92.6$ mL, so

$\text{Molarity}=\frac{0.1111}{0.0926\mathrm{L}}=1.20\mathrm{mol}{\mathrm{L}}^{-1}$

Molarity comes out below molality here because a litre of solution contains less than a kilogram of water once the glucose has taken up room.

Types of solution, and what dissolves in what

A solution is any homogeneous mixture. Either component may be solid, liquid or gas, giving nine combinations.

Like dissolves like. A polar solute dissolves in a polar solvent because the new solute-solvent attractions can pay for the ones broken. Non-polar in non-polar works for the same reason.

Solids. Solubility rises with temperature when dissolution is endothermic and falls when it is exothermic. That is Le Chatelier applied to a saturated solution. Cerium sulphate is the standard solid that becomes less soluble on heating.

Gases. Dissolution of a gas is always exothermic, so gas solubility always falls with rising temperature. Warm river water holds less dissolved oxygen than cold, which is the mechanism behind thermal pollution.

Henry's law

Henry's law. The partial pressure of a gas above a solution is proportional to its mole fraction in the solution.

$p=K_{H}x$

$K_H$ is the Henry constant for that gas in that solvent at that temperature.

Read the direction carefully. $K_H$ sits on the pressure side, so a large $K_H$ means a poorly soluble gas.

$K_H$ increases with temperature, which is the same statement as gases being less soluble when hot.

Illustration 2

A diver breathes air at the surface, where the partial pressure of nitrogen is 0.79 bar. Find the amount of nitrogen dissolved per litre of body water at 293 K, and how much extra dissolves at a depth where the total pressure is 4.0 bar. Take $K_H=76.48$ kbar.

$x=\frac{p}{K_H}=\frac{0.79}{76480}=1.03\times 10^{-5}$

One litre of water is 55.5 mol, and $x$ is tiny, so the amount of nitrogen is

$n=1.03\times 10^{-5}\times 55.5=5.7\times 10^{-4}\mathrm{mol}$

At depth the nitrogen partial pressure is four times larger, so $x$ and $n$ are four times larger: $2.3\times 10^{-3}$ mol per litre.

The extra $1.7\times 10^{-3}$ mol per litre is roughly 38 mL of gas at surface conditions, dissolved in every litre of the diver's body water. Ascend slowly and it leaves through the lungs. Ascend fast and it comes out as bubbles in the bloodstream, which is decompression sickness.

Divers avoid this by breathing helium-diluted air, and the table above says why: helium has the largest $K_H$ of the three, so least of it dissolves in the first place.

2. Vapour Pressure and Raoult's Law

Vapour pressure. The pressure of the vapour in equilibrium with its liquid in a closed container.

Add a non-volatile solute and it falls. Solute particles occupy part of the surface, so fewer solvent molecules are placed to escape, while the rate of return is unchanged. Equilibrium is restored at a lower pressure.

Raoult's law. For each volatile component, the partial vapour pressure equals its mole fraction in the liquid times its vapour pressure when pure.

$p_A=x_A{p}_A^{\circ }$

For two volatile liquids, add the partials by Dalton's law.

$p_{total}=x_A{p}_A^{\circ }+x_B{p}_B^{\circ }=p_B^{\circ }+x_A\left(p_A^{\circ }-p_B^{\circ }\right)$

That second form is worth memorising: total pressure is linear in $x_A$, running from $p_B^{\circ }$ at $x_A=0$ to $p_A^{\circ }$ at $x_A=1$.

For a non-volatile solute only the solvent contributes, and the first colligative property drops out in three lines.

$p=x_A{p}_A^{\circ }\text{and}{x}_A=1-x_B$

$\Rightarrow p=(1-x_B)p_A^{\circ }\Rightarrow p_A^{\circ }-p=x_B{p}_A^{\circ }$

$\Rightarrow \frac{p_A^{\circ }-p}{p_A^{\circ }}=x_B$

The relative lowering of vapour pressure equals the mole fraction of solute. Nothing about the solute's identity appears anywhere in that statement, which is the cleanest evidence that the property is colligative.

For a dilute solution $n_B\ll n_A$, so $x_B\approx n_B/n_A$, and a molar mass falls out.

$M_B=\frac{w_B{M}_A}{w_A}\cdot \frac{p_A^{\circ }}{p_A^{\circ }-p}$

Trap. The exact statement uses $x_B=n_B/(n_A+n_B)$. Only the molar mass formula drops the $n_B$ in the denominator. Use the exact form whenever the solution is not obviously dilute.

Illustration 3

What mass of a non-volatile solute of molar mass 60 g mol${}^{-1}$ must be dissolved in 200 g of water to lower the vapour pressure by 2.00 per cent?

A 2.00 per cent lowering means $x_B=0.0200$ exactly.

$n_A=\frac{200}{18}=11.11\text{mol},\frac{n_B}{n_B+11.11}=0.0200$

$\Rightarrow n_B=0.0200n_B+0.2222\Rightarrow 0.980n_B=0.2222\Rightarrow n_B=0.2268\text{mol}$

The mass required is $0.2268\times 60=13.6$ g.

Using the dilute approximation instead gives $n_B=0.2222$ mol and 13.3 g, about 2 per cent low. The size of the error is exactly the fraction dropped.

Vapour composition differs from liquid composition

The vapour is always richer in the more volatile component. That enrichment is the whole basis of fractional distillation.

$y_A=\frac{p_A}{p_{total}}=\frac{x_A{p}_A^{\circ }}{x_A{p}_A^{\circ }+x_B{p}_B^{\circ }}$

Write $y$ for vapour mole fractions and $x$ for liquid ones and keep them apart on the page. Conflating the two is the most common error in this section.

Illustration 4

Liquids A and B have pure vapour pressures 450 and 700 mmHg at the working temperature. A mixture of the two has a total vapour pressure of 600 mmHg. Find the liquid and vapour compositions.

Use the linear form.

$p_{total}=700+x_A(450-700)=700-250x_A=600\Rightarrow x_A=0.400$

$p_A=0.400\times 450=180\text{mmHg}\Rightarrow y_A=\frac{180}{600}=0.300$

The liquid is 40 per cent A and the vapour is 30 per cent A. B is the more volatile of the two, and it is B that the vapour is enriched in, from 60 per cent up to 70. The rule survives its test.

Illustration 5

Two ideal-solution measurements are made on the same pair of volatile liquids. At $x_A=0.25$ the total pressure is 0.30 bar; at $x_A=0.50$ it is 0.40 bar. Find both pure vapour pressures.

Total pressure is linear in $x_A$ with slope $D=p_A^{\circ }-p_B^{\circ }$ and intercept $p_B^{\circ }$.

$0.30=p_B^{\circ }+0.25D,0.40=p_B^{\circ }+0.50D$

Subtracting, $0.10=0.25D$, so $D=0.40$ bar. Then

$p_B^{\circ }=0.30-0.25(0.40)=0.20\text{bar},p_A^{\circ }=0.20+0.40=0.60\text{bar}$

Two data points fix a straight line, and the line's two endpoints are the two pure vapour pressures. No other information was needed.

3. Ideal and Non-Ideal Solutions

Ideal solution. One that obeys Raoult's law at every composition.

That requires solute-solvent interactions of the same strength as the solute-solute and solvent-solvent interactions they replace. Nothing is gained or lost by mixing, so

${\Delta }_{mix}H=0,{\Delta }_{mix}V=0$

Benzene with toluene, and hexane with heptane, come close. The molecules are similar in size and interact in the same way.

Two kinds of deviation

Positive deviation. New interactions weaker than the old ones. Molecules escape more easily than Raoult predicts, so the vapour pressure is higher.

Ethanol with water is the standard case. Ethanol hydrogen bonds strongly to itself, and inserting water breaks that network. Mixing is endothermic and the volume increases.

Negative deviation. New interactions stronger than the old ones. Molecules are held more tightly, so the vapour pressure is lower.

Chloroform with acetone is the standard case. A hydrogen bond forms between the chloroform hydrogen and the acetone carbonyl oxygen, an interaction present in neither pure liquid. Mixing is exothermic and the volume decreases.

Trap. The words positive and negative attach to the vapour pressure, and the enthalpy sign is the opposite of what the name suggests. Positive deviation means vapour pressure above the line and mixing that is endothermic. Say both halves out loud before answering.

Note what the dashed lines are doing in every panel. The two straight lines from each axis are the Raoult partials, and the third dashed line joining $p_B^{\circ }$ to $p_A^{\circ }$ is the ideal total. Deviation is measured against that third line, never against the axes.

Illustration 6

An equimolar mixture of two liquids whose pure vapour pressures are 120 and 180 mmHg is measured at 132 mmHg total. Classify the solution and predict the sign of ${\Delta }_{mix}H$, the sign of ${\Delta }_{mix}V$, and the type of azeotrope it could form.

Raoult predicts $0.5(120)+0.5(180)=150$ mmHg.

Observed 132 mmHg is below the prediction, so this is negative deviation.

Reading the chain backwards: lower vapour pressure means molecules held more tightly, which means the new solute-solvent interactions are stronger than the ones replaced. Energy is released, so ${\Delta }_{mix}H<0$. Molecules pulled closer together occupy less space, so ${\Delta }_{mix}V<0$. A deviation of this sign, if large enough, gives a maximum boiling azeotrope.

The deviation here is 12 per cent, which is large. This is behaviour of the chloroform-acetone kind.

Azeotropes

Azeotrope. A mixture that boils at constant composition, so distillation cannot separate it further.

At an azeotropic composition the vapour and the liquid have identical composition. There is then nothing left to enrich, and every further distillation stage returns exactly what it received.

This is why absolute alcohol cannot be made by simple distillation. Rectified spirit stops at 95.6 per cent and no number of plates in the column pushes it further.

Note that an azeotrope is not an exception to the enrichment rule. Enrichment follows from Raoult's law, and an azeotrope is precisely where the deviation has grown large enough to overturn Raoult's law.

4. The Four Colligative Properties

All four measure the same thing, and each is convenient over a different range.

Why they are linked. Every one traces back to the lowered vapour pressure, and that traces back to entropy. A solution is more disordered than the pure solvent, so solvent molecules are more reluctant to leave it, whether by evaporating, by freezing into an ordered crystal, or by diffusing across a membrane.

The phase diagram makes this visible in a single stroke. Drop the liquid's vapour pressure curve below the pure solvent's, and that one displaced curve now meets the 1 atm line further right and the solid curve further left.

The boiling point rises and the freezing point falls from one cause, and they move in opposite directions only because the solid curve rises more steeply than the liquid one.

Elevation of boiling point

Lowering the vapour pressure means a higher temperature is needed to push it back up to atmospheric.

$\Delta T_b=K_{b}m$

$K_b$ is the molal elevation constant, or ebullioscopic constant. It is the elevation produced by a one molal solution, and it depends only on the solvent.

Depression of freezing point

The freezing point is where solid and liquid have equal vapour pressure. Lower the liquid's and the equality moves to a lower temperature.

$\Delta T_f=K_{f}m$

Both constants come out of the same thermodynamics, with the solvent's molar mass $M_A$ in g mol${}^{-1}$ and the phase-change enthalpy in J mol${}^{-1}$.

$K_b=\frac{R{M}_A{T}_b^2}{1000{\Delta }_{vap}H},K_f=\frac{R{M}_A{T}_f^2}{1000{\Delta }_{fus}H}$

Put water's numbers in. For fusion, $T_f=273$ K and ${\Delta }_{fus}H=6008$ J mol${}^{-1}$, giving $K_f=1.86$. For vaporisation, $T_b=373$ K and ${\Delta }_{vap}H=40650$ J mol${}^{-1}$, giving $K_b=0.51$.

The formulas explain the pattern. $K_f$ beats $K_b$ for every solvent because ${\Delta }_{fus}H$ is far smaller than ${\Delta }_{vap}H$, and that division dominates the higher $T_b^2$ in the numerator.

Depression is therefore the preferred method in practice: the effect is larger for the same solution, and no heating is applied to decompose a fragile solute.

Illustration 7

Camphor has $K_f=39.7$ K kg mol${}^{-1}$. Dissolving 0.0250 g of an unknown compound in 0.500 g of molten camphor depresses the freezing point by 3.97 K. Find the molar mass.

$m=\frac{\Delta T_f}{K_f}=\frac{3.97}{39.7}=0.100\mathrm{mol}\mathrm{k}{\mathrm{g}}^{-1}$

$n_B=0.100\times 0.500\times 10^{-3}\mathrm{kg}=5.00\times 10^{-5}\text{mol}$

$M_B=\frac{0.0250}{5.00\times 10^{-5}}=500\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$

This is Rast's method, and the whole point is the size of $K_f$. Water would have given a depression of 0.186 K on the same molality, needing a far more sensitive thermometer and a hundred times more sample. A large $K_f$ buys precision on 25 milligrams.

Osmotic pressure

Osmosis. The movement of solvent through a semipermeable membrane from the dilute side to the concentrated side.

Osmotic pressure. The pressure that must be applied to the solution to stop it.

$\pi =CRT=\frac{n_B}{V}RT$

$C$ is the molar concentration. This is the one property that legitimately uses molarity, because the measurement happens at a single stated temperature.

Reverse osmosis. Apply a pressure greater than $\pi$ to the solution and the solvent is driven backwards through the membrane. This is how seawater is desalinated.

Why osmotic pressure wins for large molecules

Osmotic pressure is by far the largest effect for a given concentration, and it is read at room temperature.

A 1 per cent solution of a protein of molar mass 60000 depresses the freezing point by about 0.0003 K, which no ordinary thermometer resolves. The same solution gives an osmotic pressure of several millimetres of mercury, which is easy to measure.

That is why osmometry is standard for polymers and proteins, and why the thermal methods are reserved for small molecules.

Illustration 8

What concentration of sodium chloride is isotonic with 0.30 M glucose at 310 K? Take the van't Hoff factor of the salt as 1.86 at this dilution.

Isotonic means equal $\pi$, and at the same temperature that means equal $iC$.

$1\times 0.30=1.86\times C\Rightarrow C=\frac{0.30}{1.86}=0.161\mathrm{mol}{\mathrm{L}}^{-1}$

Converting to a mass concentration, $0.161\times 58.5=9.4$ g L${}^{-1}$, or about 0.94 per cent by mass per volume.

Clinical saline is made up at 0.9 per cent, which is 0.154 M. The agreement is the point: intravenous fluids are formulated to match blood's osmotic pressure, and getting it wrong bursts or shrivels red blood cells rather than merely diluting them.

Illustration 9

Dissolving 1.80 g of a non-volatile non-electrolyte in 90.0 g of water raises the boiling point by 0.0567 K. Taking $K_b=0.512$ K kg mol⁻¹, find the molar mass — then say why the freezing point would have been the better measurement.

Work back through the molality:

$m=\frac{\Delta T_b}{K_b}=\frac{0.0567}{0.512}=0.1107\mathrm{mol}\mathrm{k}{\mathrm{g}}^{-1}$

$n=(0.1107)(0.0900)=9.96\times 10^{-3}\mathrm{mol}\Rightarrow M=\frac{1.80}{9.96\times 10^{-3}}=181\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$

which is glucose.

Now the experimental point. For water $K_f=1.86$ against $K_b=0.512$, so the very same solution would have depressed the freezing point by

$\Delta T_f=(1.86)(0.1107)=0.206\mathrm{K}$

That is 3.6 times the signal, from identical starting material, and a temperature difference of 0.2 K is far easier to measure honestly than one of 0.06 K.

There is a second reason as well. Boiling the solution evaporates solvent, which concentrates it as the measurement proceeds and biases $\Delta T_b$ upward — so the error does not merely add noise, it pushes the molar mass systematically low. Freezing has no equivalent problem, which is why cryoscopy is the standard laboratory method and ebullioscopy is mostly an exam exercise.

5. Molar Mass and the van't Hoff Factor

Each property counts particles. Weigh the sample, count the particles, divide, and the answer is the mass per particle.

$M_B=\frac{K_b{w}_B\times 1000}{\Delta T_b{w}_A},M_B=\frac{K_f{w}_B\times 1000}{\Delta T_f{w}_A}$

with $w_B$ the solute mass and $w_A$ the solvent mass, both in grams. The osmotic version is $M_B=w_{B}RT/(\pi V)$.

When the answer comes out wrong

Measure the freezing point depression of sodium chloride solution and the molar mass comes out near 29, not 58.5.

The salt did not change. The count did. Each formula unit gives two ions, so the solution holds twice as many particles as assumed, giving twice the depression and therefore half the molar mass.

van't Hoff factor.

$i=\frac{\text{observed colligative effect}}{\text{effect calculated assuming no change}}=\frac{\text{normal molar mass}}{\text{observed molar mass}}$

Insert $i$ into all four formulas: $\Delta T_b=i{K}_{b}m$, $\Delta T_f=i{K}_{f}m$, $\pi =iCRT$, and relative lowering $=i{x}_B$.

Trap. The two ratios defining $i$ are inverted with respect to each other. Effects go on top, molar masses go on the bottom, because a bigger effect means a smaller apparent molar mass.

Dissociation and association

Carboxylic acids in benzene are the standard association case. Two molecules pair through a double hydrogen bond between the carboxyl groups, so $i$ approaches 0.5.

A solute can associate in one solvent and dissociate in another. Acetic acid dimerises in benzene and ionises in water, so the same substance gives $i$ near 0.5 in one and slightly above 1 in the other.

Behaviour belongs to the solution, not to the solute alone.

Illustration 10

A 0.0100 molal solution of acetic acid in water freezes 0.0194 K below pure water. Find the degree of dissociation and the acid dissociation constant. Take $K_f=1.86$ K kg mol${}^{-1}$.

Undissociated, the depression would be $1.86\times 0.0100=0.0186$ K.

$i=\frac{0.0194}{0.0186}=1.043$

Acetic acid gives two particles, so $n=2$ and $\alpha =(i-1)/(n-1)=0.043$. Only 4.3 per cent of the acid is ionised, which is why it is called weak.

Feeding that into the equilibrium expression, with $C=0.0100$,

$K_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{0.0100\times (0.043{)}^2}{0.957}=1.9\times 10^{-5}$

The accepted value is $1.8\times 10^{-5}$. A thermometer reading to the third decimal place has just measured an equilibrium constant, without a pH meter anywhere in the experiment. That is the reach of a counting method.

Summary

Colligative properties count solute particles and ignore everything else about them, which is why four different measurements answer one question and why any of them yields a molar mass.

Every formula except osmotic pressure uses molality or mole fraction, because volume changes with temperature and mass does not. Osmotic pressure is exempt because it is read at one fixed temperature.

Henry's law gives $p=K_{H}x$ for a dissolved gas, with a large $K_H$ meaning low solubility, and $K_H$ rising with temperature.

Raoult's law gives each volatile component's partial pressure as its mole fraction times its pure vapour pressure, and the total is linear in composition between the two pure values. For a non-volatile solute, the relative lowering of vapour pressure equals the solute's mole fraction.

Ideal solutions have zero enthalpy and volume change on mixing. Positive deviation means weaker new interactions, higher vapour pressure, endothermic mixing and a minimum boiling azeotrope. Negative deviation is the reverse throughout.

Boiling point rises by $K_{b}m$ and freezing point falls by $K_{f}m$, with $K_b=R{M}_A{T}_b^2/1000{\Delta }_{vap}H$ and $K_f=R{M}_A{T}_f^2/1000{\Delta }_{fus}H$. $K_f$ always exceeds $K_b$ because the enthalpy of fusion is much smaller than that of vaporisation, which is why depression is the preferred method.

Osmotic pressure is $CRT$, is the largest of the four effects, is measured at room temperature, and is therefore the only practical method for proteins and polymers.

An abnormal molar mass always means the particle count was wrong. The van't Hoff factor repairs all four formulas, exceeding 1 for dissociation and falling below 1 for association.

For dissociation into $n$ particles, $i=1+(n-1)\alpha$; for association of $n$ molecules, $i=1-(1-1/n)\alpha$. The same solute can do either, as acetic acid does in benzene and in water.