Question

Calculate the freezing point of a 0.100 m aqueous solution of \(\mathrm{K}_{2} \mathrm{SO}_{4},\) (a) ignoring interionic attractions, and (b) taking interionic attractions into consideration by using the van't Hoff factor (Table 13.4 )

Short answer

\(a) \) Ignoring interionic attractions, the freezing point of the 0.100 m aqueous solution of K2SO4 is approximately -0.186 °C. \(b) \) Taking interionic attractions into consideration using the van't Hoff factor (i = 2.9), the freezing point is approximately -0.5394 °C.

Step-by-step solution

The formula for calculating the freezing point depression is as follows: ΔTf = Kf * m * i Where ΔTf is the freezing point depression, Kf is the freezing point depression constant, m is the molality of the solution, and i is the van't Hoff factor. The van't Hoff factor, i, takes into account the effect of interionic attractions, and it equals 1 for non-dissociating solutes (i.e., for part (a)) or a value from the table for dissociating solutes (i.e., for part (b)). ##Step 2: Calculate the freezing point depression (ignoring interionic attractions)##

Since we're ignoring interionic attractions, we'll use i = 1. Using the known values in the formula, we get: ΔTf = Kf * m * i ΔTf = (1.86 °C/molal) * (0.100 molal) * (1) ΔTf = 0.186 °C ##Step 3: Calculate the new freezing point (ignoring interionic attractions)##

Since the freezing point of pure water is 0°C, we can calculate the new freezing point, Tf, by subtracting the freezing point depression (ΔTf) from the pure water freezing point: Tf = 0 °C - ΔTf Tf = 0°C - 0.186 °C Tf ≈ -0.186 °C So the freezing point of the solution, ignoring interionic attractions, is approximately -0.186 °C. ##Step 4: Find the van't Hoff factor for K2SO4 from Table 13.4##

From Table 13.4, we find that the van't Hoff factor for K2SO4 is i = 2.9. ##Step 5: Calculate the freezing point depression (considering interionic attractions)##

Now we'll use the van't Hoff factor i = 2.9 for K2SO4 in the freezing point depression formula: ΔTf = Kf * m * i ΔTf = (1.86 °C/molal) * (0.100 molal) * (2.9) ΔTf ≈ 0.5394 °C ##Step 6: Calculate the new freezing point (considering interionic attractions)##

Using the freezing point depression from Step 5, we can calculate the new freezing point, Tf: Tf = 0 °C - ΔTf Tf = 0°C - 0.5394 °C Tf ≈ -0.5394 °C So the freezing point of the solution, taking interionic attractions into consideration, is approximately -0.5394 °C.

Key concepts

Colligative Properties

Colligative properties are unique in that they depend on the number of particles dissolved in a solvent and not on the nature of the chemical species present. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. In the context of freezing point depression, the more solute particles that are present in a solution, the lower its freezing point will be. This happens because the solute particles disrupt the formation of the solid crystal lattice, making it harder for the solvent to solidify.

In our exercise, the presence of \( \mathrm{K}_{2}\mathrm{SO}_{4} \) in water causes the solution to freeze at a lower temperature than pure water. Understanding colligative properties can help students to figure out how solutions will behave compared to their pure solvent counterparts, like water.

van't Hoff factor (i)

The van't Hoff factor, represented by the symbol \(i\), quantifies the number of particles into which a compound dissociates in solution. For substances that do not dissociate, such as sugar in water, the van't Hoff factor is 1. However, for ionic compounds, this factor can be greater because they split into ions when they dissolve. For instance, \(\mathrm{K}_{2}\mathrm{SO}_{4}\) ideally dissociates into three ions - two potassium ions and one sulfate ion - hence, the van't Hoff factor would ideally be 3.

Why is it important?

The van't Hoff factor allows us to calculate the actual freezing point depression of a solution, considering the effect of ionic dissociation which is vital for accurate predictions in real-world applications, such as anti-freeze in cars.

Molality (m)

Molality is a measure of the concentration of a solute in a solution and is defined as the number of moles of solute per kilogram of solvent. This unit of concentration is temperature-independent, which is helpful in scenarios where temperature changes, as seen with freezing or boiling points. Unlike molarity, which is volume-based, molality is mass-based and therefore unaffected by changes in volume due to temperature variance.

Application in Calculations:

In the given exercise, the molality of the \(\mathrm{K}_{2}\mathrm{SO}_{4}\) solution is 0.100 m, indicating there are 0.100 moles of \(\mathrm{K}_{2}\mathrm{SO}_{4}\) per kilogram of water. This precise value is crucial when using the colligative property formula to determine the extent of freezing point depression.

Interionic Attractions

Interionic attractions refer to the forces that ions exert on each other in a solution. Ions in solution are surrounded by a shell of solvent molecules but also interact with other ions through electrostatic forces. These attractions can reduce the number of free ions in solution, which may result in an actual van't Hoff factor (i) that is lower than the expected value for complete dissociation.

In real solutions, like our \(\mathrm{K}_{2}\mathrm{SO}_{4}\) example, these attractions cause a discrepancy between the ideal behavior and actual observations. Understanding interionic attractions helps explain why the actual freezing point depression might differ from theoretical predictions, which is essential in fields requiring precise calculations, such as pharmacology or materials science.