Question

Determine the van't Hoff factor of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) in a \(0.40-\mathrm{m}\) solution whose freezing point is \(-2.6^{\circ} \mathrm{C}\).

Short answer

The van't Hoff factor of Na₃PO₄ is 4.

Step-by-step solution

Write the Freezing Point Depression Equation

The freezing point depression is given by the formula: \[\Delta T_f = i \cdot K_f \cdot m\]where \(\Delta T_f\) is the change in the freezing point, \(i\) is the van't Hoff factor, \(K_f\) is the cryoscopic constant, and \(m\) is the molality of the solution.

Calculate the Freezing Point Depression

The normal freezing point of water is 0°C. Given that the solution's freezing point is -2.6°C, we have:\[\Delta T_f = 0 - (-2.6) = 2.6^{\circ} \text{C}\]

Find the Cryoscopic Constant for Water

The cryoscopic constant \(K_f\) for water is a known value, approximately \(1.86^{\circ} \text{C m}^{-1}\).

Compute the Van't Hoff Factor

Using the formula from Step 1:\[2.6 = i \times 1.86 \times 0.40\]Solving for \(i\):\[i = \frac{2.6}{1.86 \times 0.40} \approx 3.4946\]

Round Appropriately

Since the van't Hoff factor should be close to an integer (because it represents the number of particles the compound dissociates into), round \(3.4946\) to the nearest whole number.\(i \approx 3.5\) is rounded to \(i = 4\).

Key concepts

Freezing Point Depression

Freezing point depression is a phenomenon where the freezing point of a solvent is lowered when a solute is dissolved into it. This occurs because the solute particles disrupt the formation of the solid structure of the solvent.

In mathematical terms, the change in freezing point, denoted as \(\Delta T_f\), is calculated as the difference between the normal freezing point of the pure solvent and the freezing point of the solution. This concept is crucial for understanding how solutes like salts affect the physical properties of solvents.

• The greater the number of solute particles, the larger the depression in freezing point.
• It's commonly observed in solutions where ionic compounds are involved, as they tend to dissociate into multiple particles.

Cryoscopic Constant

The cryoscopic constant, abbreviated as \(K_f\), is a specific property of a solvent that quantifies how much its freezing point decreases in the presence of a solute. Each solvent has its own \(K_f\) value, and it remains consistent for a given solvent.

For water, the cryoscopic constant is approximately \(1.86^{\circ} \text{C m}^{-1}\). This means that for every mole of solute per kilogram of water, the freezing point will lower by \(1.86^{\circ} \text{C}\).

• It allows for the calculation of changes in freezing point based on molality (amount of solute).
• Essential in determining molecular weights of solutes via experiments.

Molality

Molality (\(m\)) is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which varies with temperature, molality remains constant, as it depends on mass rather than volume.

In the context of freezing point depression, molality plays a crucial role since the cryoscopic equation uses it to calculate the change in freezing point.

• It provides a reliable way to express concentration without worrying about temperature changes.
• Ideal for colligative properties studies, where the number of particles affects properties of the solution.

Ionic Dissociation

Ionic dissociation occurs when an ionic compound separates into its individual ions when dissolved in a solvent like water. This process is vital in understanding colligative properties because it directly affects the number of particles in a solution.

Sodium phosphate \((\text{Na}_3\text{PO}_4)\), for instance, dissociates into four ions: three sodium ions \((\text{Na}^+)\) and one phosphate ion \((\text{PO}_4^{3-})\). This increases the number of ions, leading to a greater freezing point depression than expected if the compound did not dissociate.

• The van't Hoff factor \((i)\) represents the number of particles a solute forms in solution. For \(\text{Na}_3\text{PO}_4\), \(i\) would ideally be 4.
• Understanding ionic dissociation is key to predicting and calculating changes in colligative properties.