Ideal and non-ideal solutions

A test that every mixture can be given

Take two liquids. Mix them in many proportions and measure the total vapour pressure each time. Then plot it against the mole fraction of A. If Raoult's law holds, the points fall on a straight line.

Some pairs do give a straight line. Others bend above it or below it.

In the picture, the dashed line is what Raoult's law predicts. The solid curve is what is measured. Three cases appear, and each one tells us something about the molecules.

Ideal solutions

An ideal solution obeys Raoult's law at every composition and at every temperature. Two things are true for it:

• The pull between an A molecule and a B molecule equals the pull between two A molecules and between two B molecules.
• Because of this, mixing changes nothing else. The heat of mixing is zero, $\Delta_{mix}H = 0$, so the temperature does not change. The volume change is zero, $\Delta_{mix}V = 0$, so $50\ \text{mL}$ of A and $50\ \text{mL}$ of B make $100\ \text{mL}$.

Why should the heat be zero? To mix, an A molecule leaves other A molecules and moves in among B molecules. If the pulls are the same, the energy it gives up is the energy it takes.

Real mixtures are never perfectly ideal, but some come close. Their molecules are alike in size and in the kind of attraction:

• n-hexane and n-heptane,
• bromoethane and chloroethane,
• benzene and toluene.

Positive deviation

In some mixtures an A molecule is held less strongly by B molecules than by other A molecules. The same is true for B. So each molecule leaves the liquid more easily than Raoult's law expects.

• Each partial pressure is higher than $x\,p^\circ$, so the total pressure is above the Raoult line.
• To mix, strong attractions must be broken and weaker ones made. The mixture takes in heat: $\Delta_{mix}H > 0$, and the liquid gets colder.
• The molecules pack less well: $\Delta_{mix}V > 0$.

Ethanol and acetone are an example. Ethanol molecules hold each other with hydrogen bonds. Acetone molecules cannot join in, and they break some of those bonds. The pair of carbon disulphide and acetone also shows a positive deviation.

Negative deviation

In other mixtures an A molecule is held more strongly by B molecules than by other A molecules. Now the molecules leave less easily than Raoult's law expects.

• Each partial pressure is lower than $x\,p^\circ$, so the total pressure is below the Raoult line.
• Attractions get stronger on mixing, so heat is given out: $\Delta_{mix}H < 0$, and the liquid gets warmer.
• The molecules draw closer: $\Delta_{mix}V < 0$, so the volume of the solution is less than the sum.

Acetone and chloroform are an example. The hydrogen atom of chloroform forms a hydrogen bond with the oxygen atom of acetone, and neither liquid can do this alone. A mixture of nitric acid and water also shows a negative deviation.

The three cases together

• Ideal. A–B pull equal to A–A and B–B. $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$. Pressure on the Raoult line.
• Positive deviation. A–B pull weaker. $\Delta_{mix}H > 0$ and $\Delta_{mix}V > 0$. Pressure above the line.
• Negative deviation. A–B pull stronger. $\Delta_{mix}H < 0$ and $\Delta_{mix}V < 0$. Pressure below the line.

A way to remember it: weaker pull, more vapour, a colder mixture. Stronger pull, less vapour, a warmer mixture.

Dilute solutions are nearly ideal

Even a pair with a large deviation follows Raoult's law at one end of the graph. When the solvent A is almost pure, its A molecules meet mostly other A molecules. So the solvent obeys $p_A = x_A p_A^\circ$ as $x_A$ approaches $1$. The dissolved B follows Henry's law there instead. This is why Raoult's law works so well for the dilute solutions of the last lesson.

Measuring how far a solution departs

Compare the measured pressure with the Raoult value at the same composition. The percentage deviation is

An example. Pure A has $26.0\ \text{kPa}$ and pure B has $9.0\ \text{kPa}$. At $x_A = 0.50$ the measured pressure is $15.2\ \text{kPa}$.

• Raoult's law: $0.50 \times 26.0\ \text{kPa} + 0.50 \times 9.0\ \text{kPa} = 17.5\ \text{kPa}$.
• Deviation: $\dfrac{15.2\ \text{kPa} - 17.5\ \text{kPa}}{17.5\ \text{kPa}} \times 100 = -13\ \%$.

The pressure is below the Raoult value, so this is a negative deviation. The A–B attraction is stronger than the A–A and B–B attractions.

Set $d = 0$ and check that the curve lies on the dashed line. Then move $d$ to $1.0$ and to $-1.0$. At what composition is the gap the biggest?

Practise with new numbers

Try it in Python

The program takes four measurements of a pair of liquids and sorts each one as ideal, positive or negative.

Common mistakes

• Mixing up the signs. Positive deviation means a pressure above the Raoult line, and $\Delta_{mix}H$ is positive too.
• Thinking a higher vapour pressure means stronger A–B attraction. It means weaker attraction. The molecules leave more easily.
• Saying that a non-ideal pair never obeys Raoult's law. The solvent obeys it when it is almost pure.
• Calling a mixture ideal because both liquids look alike. Only the pressure and the heat and volume changes tell you.

Remember these

• Ideal: obeys Raoult's law everywhere, $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$.
• Positive deviation: weaker A–B pull, pressure above the line, $\Delta_{mix}H > 0$, $\Delta_{mix}V > 0$.
• Negative deviation: stronger A–B pull, pressure below the line, $\Delta_{mix}H < 0$, $\Delta_{mix}V < 0$.
• A nearly pure solvent obeys Raoult's law even in a non-ideal pair.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What are $\Delta_{mix}H$ and $\Delta_{mix}V$ for an ideal solution?

2. When acetone is poured into chloroform, the flask gets warm. What kind of deviation does the pair show?

3. A pair of liquids gives a total pressure above the Raoult line at every composition. Is the A–B attraction weaker or stronger than the others?

4. Why do n-hexane and n-heptane form an almost ideal solution?