Freezing Point Depression and Boiling Point Elevation

The freezing point and the boiling point are properties of solutions that depend on the number of solute particles added to the solution. These kind of properties are called colligative properties. The more solute particles there are in the solution the bigger the difference in these properties. So, which will have a bigger effect NaCl or sugar? From previous work we know that NaCl will provide two particles (Na⁺ and Cl^-) while sugar will only provide one (the sugar molecule). It is expected that NaCl will make a bigger difference.

When there are solute particles in the solution they will occupy space in the solution that would have been occupied by the solvent. This will influence the solvent's ability to solidify into its normal solid lattice at the freezing point and to change into the gas phase at the boiling point. The effect is to require more energy to be taken out in order to freeze (lowering the freezing point) and require more energy to be used in order to boil (raising the boiling point).

All solvents will not be affected the same way by the solute particles. Experiments have been done to determine numerical constants that account for the solvent's effect in changing the freezing and boiling points. For water the freezing point depression constant is K_f=1.86 °C/m and the boiling point elevation constant is K_b= 0.52 °C/m.

To determine the change in freezing point both the solvent and the number of particles in the solution must be taken into account. The number of particles can be expressed as a concentration unit. The concentration unit for these calculations is molal, which gives the number of particles per 1000 grams of water. The units for the constants given in the last paragraph are therefore degrees Celsius per molal (^oC/m, where m means molal). Those constants tell us the change in temperature due to the concentration when water is the solvent.

In order to calculate the change in freezing point we need to multiply the molality, which tells us about the number of particles, times the freezing point depression constant, which tells us the change in freezing point temperature per molality. The change in freezing point temperature, ΔT_f , is given by

ΔT_f = K_f m

A similar expression is found for the boiling point elevation, ΔT_b.

ΔT_b = K_bm

Here is an example: Suppose enough NaCl is added to water so that it has a particle concentration of 1.0 m. What will the freezing point of the new solution be?

ΔT_f = K_f m = (1.86 °C/m)(1.0 m) = 1.86 °C

Since the normal freezing point of water is 0 °C,
the new freezing point will be -1.86 °C.

Here is one for practice: What would the freezing point be if the solution was 2.0 m in solute particles? (Answer: -3.72 °C)

So, why do we add salt to the ice-water solution that is used to freeze ice cream?

Here are some practice problems for you to try.

1. Is bronze a solution? Explain.
2. If solute B changes the boiling point three times as much as solute A for the same amount of each, what can you say about the number of particles formed from A compared to the number of particles formed from B? Explain and give some sample calculations.
3. Why is CaCl₂ put on roads and sidewalks to melt the ice instead of NaCl?
4. Why don't you use straight antifreeze in your radiator?
5. Why do cooks put salt in the water before putting in the spagetti?
6. Why don't some popsicles get as hard as regular ice cubes?

Here are the answers.

Freezing Point Depression and Boiling Point Elevation Homework (Colligative6)