Depression of Freezing Point

What Freezing-Point Depression Means

The freezing point of a substance is the temperature at which its solid and liquid phases coexist in dynamic equilibrium — equivalently, the temperature at which the vapour pressure of the liquid equals the vapour pressure of the solid. Pure water freezes at 273.15 K; pure benzene at 278.6 K. Dissolve a non-volatile solute in either, and that temperature drops.

Let $T_f^{\circ}$ be the freezing point of the pure solvent and $T_f$ the freezing point of the solution. The quantity

$$\Delta T_f = T_f^{\circ} - T_f$$

is the depression in freezing point. Like the other colligative properties — relative lowering of vapour pressure, elevation of boiling point and osmotic pressure — its magnitude depends only on the number of solute particles dissolved, not on their chemical identity. A mole of glucose and a mole of urea, dissolved in the same mass of water, depress the freezing point equally.

The Vapour-Pressure Origin

The lowering of vapour pressure is the root cause. When a non-volatile solute is added, Raoult's law tells us the vapour pressure of the solvent in solution falls below that of the pure solvent. A solution freezes only when its vapour pressure becomes equal to the vapour pressure of the pure solid solvent. Because the solution's liquid-phase curve has been pushed downward, that equality is reached at a lower temperature.

The figure below shows this on a vapour-pressure-versus-temperature diagram. The solid-solvent curve is fixed. The pure-liquid curve meets it at $T_f^{\circ}$; the depressed solution curve meets it further to the left, at $T_f$. The horizontal gap between the two intersection points is $\Delta T_f$.

The Governing Equation

For a dilute (ideal) solution, the depression is directly proportional to the molality $m$ of the solution:

$$\Delta T_f \,\propto\, m \qquad\Longrightarrow\qquad \Delta T_f = K_f\,m$$

Here $K_f$ is the proportionality constant. Molality — moles of solute per kilogram of solvent — is used rather than molarity because it is independent of temperature, which matters when the temperature is itself the measured quantity. The relation parallels the elevation of boiling point, $\Delta T_b = K_b\,m$, with $K_f$ playing the role of $K_b$.

Because $\Delta T_f = K_f\,m$ is a direct proportionality with no intercept, a plot of the measured depression against molality is a straight line through the origin whose slope is exactly $K_f$. This is the graphical signature of an ideal dilute solution: double the molality and the depression doubles, while the slope — the cryoscopic constant — stays fixed.

The Cryoscopic Constant K_f

The constant $K_f$ is called the Freezing Point Depression Constant, the Molal Depression Constant, or the cryoscopic constant. It is the depression produced by a 1 molal solution — that is, by dissolving one mole of solute in 1 kg of solvent. It depends only on the nature of the solvent and carries the SI unit $\text{K kg mol}^{-1}$.

Theoretically, $K_f$ is fixed by the solvent's freezing point and enthalpy of fusion:

$$K_f = \frac{R\,M_1\,T_f^{2}}{1000 \times \Delta_{\text{fus}}H}$$

where $R$ is the gas constant, $M_1$ the molar mass of the solvent, $T_f$ its freezing point in kelvin and $\Delta_{\text{fus}}H$ its enthalpy of fusion. The table lists $K_f$ for common solvents alongside their freezing points; benzene and cyclohexane have notably large constants, which is why they are favoured for cryoscopic molar-mass work.

Finding Molar Mass

If $w_2$ grams of a solute of molar mass $M_2$ are dissolved in $w_1$ grams of solvent, the molality is

$$m = \frac{w_2 / M_2}{w_1 / 1000} = \frac{w_2 \times 1000}{M_2 \times w_1}$$

Substituting into $\Delta T_f = K_f\,m$ and rearranging for the molar mass gives the working formula:

$$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$$

To determine an unknown molar mass, one measures $\Delta T_f$ for a known mass of solute in a known mass of solvent and reads off $K_f$ from a table. Because $K_f$ values for solvents such as benzene and cyclohexane are large, even small amounts of solute give a measurable depression, making the method sensitive.

Worked Examples

Electrolytes and the van't Hoff Factor

The relation $\Delta T_f = K_f\,m$ holds for non-electrolytes that neither dissociate nor associate. When a solute dissociates (an electrolyte) or associates (such as a carboxylic acid dimerising), the effective number of particles changes, and the equation is corrected with the van't Hoff factor $i$:

$$\Delta T_f = i\,K_f\,m$$

For $\ce{NaCl}$, complete dissociation gives $i \approx 2$; for $\ce{Na2SO4}$, $i \approx 3$. Each extra particle adds to the depression, which is why dissociating salts are doubly effective at lowering freezing points. Conversely, benzoic acid in benzene dimerises ($i < 1$) and shows a smaller depression than its formula would predict — a standard route to calculating the degree of association.

Applications

Beyond the de-icing and antifreeze examples, depression of freezing point underpins cryoscopy — the experimental determination of molar masses from measured depressions — and it explains why sea water freezes below 0 °C. The table summarises where each governing quantity comes into play.