One of the most familiar consequences of dissolving a solute in a solvent is that the freezing point of the solvent decreases.
Examples include spreading salt on icy roads and adding antifreeze to an automobile radiator. In both cases, the dissolved solute lowers the temperature at which the solvent freezes.
To understand why this occurs, we must examine the effect of the solute on the chemical potential of the solvent.
A pure liquid freezes when the chemical potentials of the liquid and solid phases are equal:
\[ \boxed{ \mu_A^{(l)} = \mu_A^{(s)} } \]
At temperatures above the freezing point,
\[ \mu_A^{(l)} < \mu_A^{(s)} \]
and the liquid phase is favored.
At temperatures below the freezing point,
\[ \mu_A^{(l)} > \mu_A^{(s)} \]
and the solid phase is favored.
The freezing point therefore occurs at the temperature where the chemical potentials of the two phases become equal.
Consider a solution in which component A is the solvent and component B is a nonvolatile solute.
The chemical potential of the solvent in an ideal solution is
\[ \mu_A^{(soln)} = \mu_A^\circ + RT\ln x_A \]
Since the mole fraction of the solvent is always less than one,
\[ x_A < 1 \]
and therefore
\[ \ln x_A < 0. \]
Consequently,
\[ \boxed{ \mu_A^{(soln)} < \mu_A^\circ } \]
The dissolved solute lowers the chemical potential of the solvent.
The solid phase consists of pure solvent, so its chemical potential is essentially unaffected by the presence of the solute.
The liquid phase, however, experiences a reduction in chemical potential:
\[ \mu_A^{(soln)} < \mu_A^{(l)} \]
The result is shown schematically below:
Pure solvent: \[ \mu_A^{(l)} = \mu_A^{(s)} \]
Solution: \[ \mu_A^{(soln)} < \mu_A^{(s)} \]
The liquid phase is therefore stabilized relative to the solid phase.
To restore the condition
\[ \mu_A^{(soln)} = \mu_A^{(s)} \]
the temperature must be lowered until the chemical potential of the solid decreases enough to match that of the solution.
Thus, the freezing point of the solvent is depressed.
Following this argument and using the temperature dependence of chemical potential,
\[ \left( \frac{\partial \mu}{\partial T} \right)_p = -S \]
it can be shown that
\[ \boxed{ \Delta T_f = \frac{R(T_f^\circ)^2} {\Delta H_{\mathrm{fus}}} x_B } \]
where:
The quantity
\[ K_f = \frac{R(T_f^\circ)^2} {\Delta H_{\mathrm{fus}}} \]
is called the cryoscopic constant of the solvent.
In practice, freezing-point depression is usually written as
\[ \boxed{ \Delta T_f = K_f m } \]
where \(m\) is the molality of the solute.
The same reduction in solvent chemical potential is responsible for boiling-point elevation.
Boiling point elevation can be modeled using \( K_b \), which is called the ebullioscopic constant, as follows:
\[ \Delta T_b = K_b m \]
Where \(\Delta T_b \) is the increase the boiling point of the solvent, \( K_b \) is the ebullioscopic constant, and \( m \) is the concentration of the solute in mol solute per kg solvent (molality).
In both cases, the addition of a solute lowers the chemical potential of the solvent and stabilizes the liquid phase.
For freezing, the temperature must be lowered to restore equilibrium between the liquid and solid phases.
For boiling, the temperature must be increased to restore equilibrium between the liquid and vapor phases.
Big picture: Freezing-point depression occurs because dissolved solutes lower the chemical potential of the solvent, making the liquid phase more stable than the solid phase. Boiling-point elevation arises from exactly the same thermodynamic principle: lowering the chemical potential of the solvent favors the liquid phase and shifts the temperature at which phase equilibrium occurs.
A \(2.00\ \mathrm{g}\) sample of naphthalene is dissolved in \(250.0\ \mathrm{g}\) of toluene. The freezing point of the solution is depressed by
\[ \Delta T_f = 0.222^\circ\mathrm{C}. \]
The cryoscopic constant of toluene is
\[ K_f = 3.55\ ^\circ\mathrm{C\,kg\,mol^{-1}}. \]
Use these data to calculate the molar mass of naphthalene.
Freezing-point depression is given by
\[ \Delta T_f = K_f m \]
Solving for molality:
\[ m = \frac{\Delta T_f}{K_f} \]
Substitute the values:
\[ m = \frac {0.222^\circ\mathrm{C}} {3.55\ ^\circ\mathrm{C\,kg\,mol^{-1}}} = 0.0625\ \mathrm{mol\,kg^{-1}} \]
Molality is defined as
\[ m = \frac{\mathrm{mol\ solute}} {\mathrm{kg\ solvent}} \]
The mass of toluene is
\[ 250.0\ \mathrm{g} = 0.2500\ \mathrm{kg} \]
Therefore,
\[ \mathrm{mol\ solute} = (0.0625\ \mathrm{mol\,kg^{-1}}) (0.2500\ \mathrm{kg}) \]
\[ \mathrm{mol\ solute} = 0.0156\ \mathrm{mol} \]
The molar mass is the ratio of mass to moles:
\[ M = \frac{\mathrm{mass}} {\mathrm{mol}} \]
Substituting:
\[ M = \frac {2.00\ \mathrm{g}} {0.0156\ \mathrm{mol}} \]
\[ \boxed{ M = 128\ \mathrm{g\,mol^{-1}} } \]
Interpretation: The calculated molar mass is \(128\ \mathrm{g\,mol^{-1}}\), which agrees with the known molar mass of naphthalene, \(C_{10}H_8\). Freezing-point depression is therefore a useful experimental method for determining molar masses of nonelectrolyte solutes.
| Question 1 |
|---|
| A \(1.50\ \mathrm{g}\) sample of an unknown nonelectrolyte is dissolved in \(200.0\ \mathrm{g}\) of benzene. The freezing point of the solution is depressed by \[ \Delta T_f = 0.300^\circ\mathrm{C}. \] The cryoscopic constant of benzene is \[ K_f = 5.12\ ^\circ\mathrm{C\,kg\,mol^{-1}}. \] Calculate the molar mass of the solute. |
Show Answer\[ m = \frac{\Delta T_f}{K_f} = \frac{0.300}{5.12} = 0.0586\ \mathrm{mol\,kg^{-1}} \] \[ n = (0.0586) (0.2000) = 0.0117\ \mathrm{mol} \] \[ M = \frac{1.50\ \mathrm{g}} {0.0117\ \mathrm{mol}} = 128\ \mathrm{g\,mol^{-1}} \] \[ \boxed{ M = 128\ \mathrm{g\,mol^{-1}} } \] |
| Question 2 |
|---|
| A \(3.00\ \mathrm{g}\) sample of an unknown nonelectrolyte is dissolved in \(250.0\ \mathrm{g}\) of water. The freezing point of the solution is lowered by \[ \Delta T_f = 0.372^\circ\mathrm{C}. \] The cryoscopic constant of water is \[ K_f = 1.86\ ^\circ\mathrm{C\,kg\,mol^{-1}}. \] Calculate the molar mass of the solute. |
Show Answer\[ m = \frac{0.372}{1.86} = 0.200\ \mathrm{mol\,kg^{-1}} \] \[ n = (0.200) (0.2500) = 0.0500\ \mathrm{mol} \] \[ M = \frac{3.00\ \mathrm{g}} {0.0500\ \mathrm{mol}} = 60.0\ \mathrm{g\,mol^{-1}} \] \[ \boxed{ M = 60.0\ \mathrm{g\,mol^{-1}} } \] |
| Question 3 |
|---|
| A \(2.50\ \mathrm{g}\) sample of an unknown nonelectrolyte is dissolved in \(100.0\ \mathrm{g}\) of carbon tetrachloride. The boiling point of the solution is elevated by \[ \Delta T_b = 0.502^\circ\mathrm{C}. \] The ebullioscopic constant of carbon tetrachloride is \[ K_b = 5.02\ ^\circ\mathrm{C\,kg\,mol^{-1}}. \] Calculate the molar mass of the solute. |
Show AnswerFor boiling-point elevation, \[ \Delta T_b = K_b m \] Therefore, \[ m = \frac{0.502}{5.02} = 0.100\ \mathrm{mol\,kg^{-1}} \] \[ n = (0.100) (0.1000) = 0.0100\ \mathrm{mol} \] \[ M = \frac{2.50\ \mathrm{g}} {0.0100\ \mathrm{mol}} = 250\ \mathrm{g\,mol^{-1}} \] \[ \boxed{ M = 250\ \mathrm{g\,mol^{-1}} } \] Notice that the calculation is nearly identical to the freezing-point depression problems. The only difference is the use of the ebullioscopic constant \(K_b\) instead of the cryoscopic constant \(K_f\). |
Big picture: Solubility and colligative properties are direct consequences of the effect dissolved solutes have on the chemical potential of the solvent. By lowering the solvent's chemical potential, solutes stabilize the liquid phase, leading to increased solubility, freezing-point depression, and boiling-point elevation. These effects provide powerful experimental tools for studying solutions and determining molar masses.