Understanding the van't Hoff Factor ($i$)

The van't Hoff factor, denoted by $i$, is a measure of the effect of a solute on colligative properties such as osmotic pressure, relative lowering of vapor pressure, elevation of boiling point, and depression of freezing point. For electrolytes, $i$ represents the number of particles (ions or molecules) that are formed when the solute dissolves in a solvent, relative to the number of formula units dissolved.

For a solute that undergoes complete dissociation in solution, the van't Hoff factor is equal to the number of ions or particles produced per formula unit of the substance.

Calculating van't Hoff Factor for $\text{K}_4[\text{Fe}(\text{CN})_6]$

The given substance is potassium ferrocyanide, $\text{K}_4[\text{Fe}(\text{CN})_6]$. This is a complex ionic compound. When it dissolves in water, it dissociates into potassium ions ($\text{K}^+$) and the ferrocyanide complex anion ($[\text{Fe}(\text{CN})_6]^{4-}$).

The dissociation reaction is:

\begin{equation*}\text{K}_4[\text{Fe}(\text{CN})_6](s) \rightarrow 4\text{K}^+(aq) + [\text{Fe}(\text{CN})_6]^{4-}(aq)\end{equation*}

From the dissociation equation, we can see that one formula unit of $\text{K}_4[\text{Fe}(\text{CN})_6]$ produces:

• 4 potassium ions ($\text{K}^+$)
• 1 ferrocyanide ion ($[\text{Fe}(\text{CN})_6]^{4-}$)

The total number of particles formed from one formula unit upon complete dissociation is $4 + 1 = 5$.

Therefore, the van't Hoff factor ($i$) for $\text{K}_4[\text{Fe}(\text{CN})_6]$, assuming complete dissociation, is 5.

Determining van't Hoff Factor for Given Options

Now, let's calculate the van't Hoff factor for each of the given options, assuming complete dissociation as is typical for calculating the theoretical van't Hoff factor for strong electrolytes.

Option 1: $\text{AlCl}_3$

Aluminium chloride ($\text{AlCl}_3$) is an ionic compound that dissociates in water into aluminium ions ($\text{Al}^{3+}$) and chloride ions ($\text{Cl}^-$).

The dissociation reaction is:

\begin{equation*}\text{AlCl}_3(s) \rightarrow \text{Al}^{3+}(aq) + 3\text{Cl}^-(aq)\end{equation*}

Number of particles formed = 1 $\text{Al}^{3+}$ ion + 3 $\text{Cl}^-$ ions = 4 particles.

van't Hoff factor for $\text{AlCl}_3$ ($i$) = 4.

Option 2: $\text{AlN}$

Aluminium nitride ($\text{AlN}$) is a covalent compound with a very strong lattice. It is largely insoluble in water and does not readily dissociate into ions in aqueous solution. If considered in the context of colligative properties in water, it would likely behave as a non-electrolyte or have minimal dissociation, leading to a van't Hoff factor close to 1. For typical purposes in this type of question involving ionic dissociation, it is not expected to produce multiple ions effectively.

van't Hoff factor for $\text{AlN}$ ($i$) $\approx$ 1 (as a non-electrolyte).

Option 3: $\text{AlF}_3$

Aluminium fluoride ($\text{AlF}_3$) is an ionic compound that dissociates in water into aluminium ions ($\text{Al}^{3+}$) and fluoride ions ($\text{F}^-$).

The dissociation reaction is:

\begin{equation*}\text{AlF}_3(s) \rightarrow \text{Al}^{3+}(aq) + 3\text{F}^-(aq)\end{equation*}

Number of particles formed = 1 $\text{Al}^{3+}$ ion + 3 $\text{F}^-$ ions = 4 particles.

van't Hoff factor for $\text{AlF}_3$ ($i$) = 4.

Option 4: $\text{Al}_2(\text{SO}_4)_3$

Aluminium sulfate ($\text{Al}_2(\text{SO}_4)_3$) is an ionic compound that dissociates in water into aluminium ions ($\text{Al}^{3+}$) and sulfate ions ($\text{SO}_4^{2-}$).

The dissociation reaction is:

\begin{equation*}\text{Al}_2(\text{SO}_4)_3(s) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{SO}_4^{2-}(aq)\end{equation*}

Number of particles formed = 2 $\text{Al}^{3+}$ ions + 3 $\text{SO}_4^{2-}$ ions = 5 particles.

van't Hoff factor for $\text{Al}_2(\text{SO}_4)_3$ ($i$) = 5.

Comparison of van't Hoff Factors

Let's compare the van't Hoff factor of $\text{K}_4[\text{Fe}(\text{CN})_6]$ with those of the given options:

The van't Hoff factor of $\text{K}_4[\text{Fe}(\text{CN})_6]$ is 5. The substance among the options that has the same theoretical van't Hoff factor is $\text{Al}_2(\text{SO}_4)_3$, which also has a theoretical van't Hoff factor of 5.

Conclusion

Based on the complete dissociation of the substances, $\text{Al}_2(\text{SO}_4)_3$ produces 5 particles (2 $\text{Al}^{3+}$ and 3 $\text{SO}_4^{2-}$) per formula unit, resulting in a van't Hoff factor of 5, which is the same as that of $\text{K}_4[\text{Fe}(\text{CN})_6]$.

Revision Table: Key Concepts for van't Hoff Factor

Additional Information on van't Hoff Factor and Colligative Properties

The van't Hoff factor is important because colligative properties depend on the number of solute particles in a solution, not their identity. For non-electrolytes, $i=1$ because they do not dissociate. For electrolytes, $i > 1$ due to dissociation into multiple ions. The higher the van't Hoff factor, the greater the effect on colligative properties for a given concentration.

The actual van't Hoff factor can be slightly less than the theoretical value (calculated assuming complete dissociation) due to ion association in concentrated solutions. However, for dilute solutions, the assumption of complete dissociation is often a good approximation.

Colligative properties include:

• Lowering of vapor pressure
• Elevation of boiling point
• Depression of freezing point
• Osmotic pressure

The formulas for colligative properties are modified by including the van't Hoff factor $i$. For example, the depression of freezing point ($\Delta T_f$) is given by $\Delta T_f = i \cdot K_f \cdot m$, where $K_f$ is the cryoscopic constant and $m$ is the molality of the solution. Similarly, osmotic pressure ($\Pi$) is given by $\Pi = i \cdot MRT$, where $M$ is molarity, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin.