Question

The colligative properties of electrolytes require a slightly different approach than the one used for the colligative properties of non- electrolytes. The electrolytes dissociate into ions in solution. It is the number of solute particles that determine the colligative properties of a solution. The electrolyte solutions, therefore show abnormal colligative properties. To account for this effect we define a quantity; called the van't Hoff factor which is given by [solution] \(i=\) \(\frac{\text { Actual number of particles in solution after dissociation }}{\text { Number of formula units initally dissolved in solution }}\) \(\mathrm{i}=1\) (for non - electrolytes); \(\mathrm{i}>1\) (for electrolytes, undergoing dissociation) \(\mathrm{i}<1\) (for solute, undergoing association) \(0.1 \mathrm{M} \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) is \(60 \%\) ionized. What will be its van't Hoff factor? (a) \(3.4\) (b) \(1.7\) (c) \(2.4\) (d) \(2.2\)

Short answer

The van't Hoff factor is 3.4, option (a).

Step-by-step solution

Understand the dissociation

The compound \( \mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_{6}] \) dissociates into \( 4 \) \( \mathrm{K}^+ \) ions and \( [\mathrm{Fe}(\mathrm{CN})_{6}]^{4-} \), totaling \( 5 \) ions when fully ionized.

Identify the degree of ionization

The compound is \( 60\% \) ionized, meaning out of the total possible dissociation, only \( 60\% \) is achieved. Therefore, \( 60\% \) of the \( 5 \) ions are actually present in the solution.

Calculate the actual number of particles

Calculate the actual number of particles in solution: \( 1 - 0.6 = 0.4 \) still as undissociated formula units, and \( 0.6 \times 5 = 3 \) particles from dissociation. Total actual particles = \( 1 (undissociated) + 3 (dissociated) = 4 \).

Compute the van't Hoff factor

The van't Hoff factor \( i \) is calculated using the formula: \[ i = \frac{\text{Actual number of particles}}{\text{Number of formula units initially dissolved}} \]. Thus, \( i = \frac{4}{1} = 4 \times 0.6 + 1 \times 0.4 = 3.4 \).

Choose the closest answer

The calculated van't Hoff factor for \( \mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_{6}] \) is \( 3.4 \), so the answer is \( \text{(a) } 3.4 \).

Key concepts

Colligative Properties

When studying solutions, colligative properties are incredibly important. These properties depend on the number of solute particles within a solution, rather than the type of particles. This means the properties only change with the amount of dissolved solute. Adding more particles will affect boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure.
- Boiling Point Elevation: More solute particles cause the boiling point of a solution to increase, requiring more energy to achieve the gaseous state.
- Freezing Point Depression: Solute particles interfere with the formation of a solid, lowering the freezing point.
- Vapor Pressure Lowering: More solute particles result in fewer solvent molecules escaping into the vapor state, reducing vapor pressure.
- Osmotic Pressure: The pressure required to stop solvent molecules from passing through a semipermeable membrane into the solution increases with more solute particles.
Whether they are electrolytes or non-electrolytes, the number of particles present directly influences these properties.

Electrolyte Dissociation

Electrolytes are compounds that, when dissolved in water, dissociate into ions. This dissociation is key to understanding how they impact colligative properties. Unlike non-electrolytes, which dissolve as neutral molecules, electrolytes form charged particles that can conduct electricity.
Strong electrolytes fully dissociate into their ions, while weak electrolytes only partially dissociate. For example, the compound \( \mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_{6}] \) dissociates into 4 potassium (\( \mathrm{K}^+ \)) ions and one \([\mathrm{Fe}(\mathrm{CN})_{6}]^{4-}\) ion, making a total of 5 ions per formula unit when fully dissociated.
This means solutions with electrolytes may have abnormal colligative properties, as there are more solute particles than initially expected. This increase in particle number can significantly enhance or affect the noted colligative effects.

Degree of Ionization

The degree of ionization refers to the extent to which an electrolyte dissociates into ions in a solution. It is generally expressed as a percentage and indicates how much of the solute has converted into ions.
For instance, in the given example, \( \mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_{6}] \) is 60% ionized. This means that out of the full potential to dissociate into 5 ions, only 60% of that potential is realized.
Knowing the degree of ionization is crucial for calculating the actual number of particles in a solution, which is needed for determining the van’t Hoff factor. This factor aids in quantifying how the solution deviates from ideal behavior due to ionization.
Always consider this degree, as it dictates the real solute particle count affecting colligative properties.

Number of Solute Particles

Calculating the number of solute particles in a solution is vital for understanding colligative properties, especially in solutions with electrolytes. The total particle number can vary significantly due to the dissociation of electrolyte compounds.
To find the number of particles, one must consider both the dissociated and undissociated parts of the solute. The undissociated portion remains as formula units, while the dissociated part forms multiple ionic particles.
Using the previous example, if \( \mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_{6}] \) is 60% ionized, there would be 1 undissociated unit and 3 particles from dissociation. This results in a total of 4 solute particles.
Why is this crucial? Because the number of particles directly influences the solution's colligative properties and helps in calculating the van’t Hoff factor, which assesses the extent of dissociation. It's pivotal for understanding solutions that deviate from ideal conditions, notably in electrolyte solutions.