Question

The freezing-point depression of a \(0.091-m\) solution of \(\mathrm{CsCl}\) is \(0.320^{\circ} \mathrm{C}\). The freezing-point depression of a \(0.091-\mathrm{m}\) solution of \(\mathrm{CaCl}_{2}\) is \(0.440^{\circ} \mathrm{C}\). In which solution does ion association appear to be greater? Explain.

Short answer

Ion association appears to be greater in the \(\mathrm{CaCl}_{2}\) solution. This is because the observed van't Hoff factor for \(\mathrm{CaCl}_{2}\) is farther from its theoretical value (3) than the observed van't Hoff factor for \(\mathrm{CsCl}\) is from its theoretical value (2).

Step-by-step solution

Calculate the theoretical freezing-point depression for both solutions

First, we need to calculate the theoretical freezing-point depression for both solutions. For an electrolyte, the freezing-point depression, \(\Delta T_f\), is given by: \[\Delta T_f = K_f \times molality \times i\] where \(K_f\) is the freezing-point depression constant, \(molality\) is the solution's molality, and \(i\) is the van't Hoff factor. From the problem, we know the molality is the same for both solutions (\(0.091-m\)). The theoretical van't Hoff factors for \(\mathrm{CsCl}\) and \(\mathrm{CaCl}_{2}\) are 2 and 3, respectively, because \(\mathrm{CsCl}\) dissociates into 2 ions and \(\mathrm{CaCl}_{2}\) dissociates into 3 ions. Thus, we have: Theoretical $\Delta T_f (\mathrm{CsCl}) = K_f \times 0.091mol \times 2\] Theoretical $\Delta T_f (\mathrm{CaCl}_{2}) = K_f \times 0.091mol \times 3\]

Calculate the experimentally observed van't Hoff factors

Now, we need to determine the experimentally observed van't Hoff factors for both solutions: For \(\mathrm{CsCl}\): \[i_{CsCl} = \frac{\Delta T_{f_{observed}} (\mathrm{CsCl})}{\Delta T_{f_{theoretical}} (\mathrm{CsCl})} = \frac{\Delta T_{f_{observed}} (\mathrm{CsCl})}{0.091mol \times K_f \times 2}\] For \(\mathrm{CaCl}_2\): \[i_{CaCl_2} = \frac{\Delta T_{f_{observed}} (\mathrm{CaCl}_{2})}{\Delta T_{f_{theoretical}} (\mathrm{CaCl}_{2})} = \frac{\Delta T_{f_{observed}} (\mathrm{CaCl}_{2})}{0.091mol \times K_f \times 3}\]

Compare the observed and theoretical van't Hoff factors

We compare the observed van't Hoff factors obtained in Step 2 with their theoretical values. For \(\mathrm{CsCl}\): \[i_{CsCl_{observed}} < 2\] For \(\mathrm{CaCl}_2\): \[i_{CaCl_{2_{observed}} < 3\] The solution where the observed van't Hoff factor is farthest from the theoretical value is the one with greater ion association.

Determine which solution has greater ion association

After comparing the observed van't Hoff factors with the theoretical values, we find that the difference is greater for the \(\mathrm{CaCl}_{2}\) solution. Thus, ion association appears to be greater in the \(\mathrm{CaCl}_{2}\) solution.

Key concepts

van't Hoff factor

The van't Hoff factor, represented by the symbol 'i', is a crucial concept to consider when discussing solutions and their boiling and freezing points. It refers to the number of particles into which a compound dissociates in solution. For instance, a compound like table salt (NaCl) typically has a van't Hoff factor of 2 because it dissociates into two ions, Na+ and Cl-.

However, it is important to note that the van't Hoff factor is not always a whole number or equal to the number of particles a formula suggests should be present. This discrepancy can be due to ion association or incomplete dissociation in solution. When we compare the theoretical and observed freezing-point depressions, the difference gives us insight into the extent of ion association - the more significant the difference, the greater the presence of ion associations. Thus, van't Hoff factors are central in assessing the behavior of electrolytes in solutions.

colligative properties

Colligative properties are specific physical properties of solutions that depend on the concentration of dissolved particles, not on their chemical identity. These properties include boiling-point elevation, freezing-point depression, vapor-pressure lowering, and osmotic pressure. Freezing-point depression, which is the focus of our exercise, occurs because the presence of solute particles interferes with the formation of the crystal lattice structure of ice, necessitating a lower temperature to achieve the solid state.

A key factor in calculating the extent of freezing-point depression is the molality of the solution, which measures the number of moles of solute per kilogram of solvent. Understanding colligative properties helps us to determine the molar mass of substances and to investigate the interactions between particles in a solution.

ion association

Ion association refers to the phenomenon where dissolved ions in solution tend to recombine to form a certain percentage of the associated, or undissociated, species. This association can occur due to interionic attractions, particularly in solutions of high ionic strength or concentration. A greater degree of ion association would result in a smaller observed van't Hoff factor compared to the expected theoretical value for a given compound, as fewer ions are present to contribute to the colligative property being observed.

In our textbook exercise, we compare the observed freezing-point depressions for CsCl and CaCl₂ solutions with the same molality. The differences between the observed and theoretical van't Hoff factors indicate the level of ion association, suggesting that CaCl₂ shows greater ion association because the difference in its factor is larger.

molality

Molality is a measure of concentration used frequently in the study of colligative properties. Defined as the number of moles of solute per kilogram of solvent, molality (m) is a unit that is particularly useful because it does not change with temperature, unlike volume-based measurements which can expand or contract with temperature variations.

In calculations involving freezing-point depression or boiling-point elevation, molality serves as a crucial factor. It helps us collaborate with the van't Hoff factor and the particular property constant (like the freezing-point depression constant, K_f) to determine the expected change in temperature. In our exercise, both CsCl and CaCl₂ have the same molality, allowing us to isolate the impact of ion association and the van't Hoff factor when comparing their freezing-point depressions.