Question

Arrange the following solutions in order of decreasing freezing point: \(0.10 \mathrm{~m} \mathrm{Na}_{3} \mathrm{PO}_{4}, 0.35 \mathrm{~m} \mathrm{NaCl}, 0.20 \mathrm{~m}\) \(\mathrm{MgCl}_{2}, 0.15 \mathrm{~m} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}, 0.15 \mathrm{~m} \mathrm{CH}_{3} \mathrm{COOH}\)

Short answer

Order: C6H12O6 ≈ CH3COOH, Na3PO4, MgCl2, NaCl.

Step-by-step solution

Understand Freezing Point Depression

The freezing point of a solution is lowered compared to the pure solvent. The change in freezing point is given by the formula \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant, and \( m \) is the molality. The greater the \( \Delta T_f \), the lower the freezing point.

Calculate Van't Hoff Factors

For non-electrolytes like glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)), \( i = 1 \). For electrolytes, \( i \) is the number of ions the compound dissociates into: \( \text{Na}_3\text{PO}_4 \) dissociates into 4 ions (3 \( \text{Na}^+ \), 1 \( \text{PO}_4^{3-} \)), \( \text{NaCl} \) dissociates into 2 ions (\( \text{Na}^+ \), \( \text{Cl}^- \)), \( \text{MgCl}_2 \) dissociates into 3 ions (\( \text{Mg}^{2+} \), 2 \( \text{Cl}^- \)). \( \text{CH}_3\text{COOH} \) is a weak acid and partially ionizes, so \( i \) is slightly greater than 1, but closer to 1.

Compute Effective Molality

Use \( m_i = i \cdot m \) to find effective molality. For \( \text{Na}_3\text{PO}_4 \): \( 0.10 \cdot 4 = 0.40 \) m. For \( \text{NaCl} \): \( 0.35 \cdot 2 = 0.70 \) m. For \( \text{MgCl}_2 \): \( 0.20 \cdot 3 = 0.60 \) m. \( \text{C}_6\text{H}_{12}\text{O}_6 \) is 0.15 m. \( \text{CH}_3\text{COOH} \), being near 1, remains approximately 0.15 m.

Arrange by Decreasing Freezing Point

Solutions with higher effective molality have lower freezing points. Arrange ascending by effective molality: \( \text{C}_6\text{H}_{12}\text{O}_6 \approx \text{CH}_3\text{COOH} > \text{Na}_3\text{PO}_4 > \text{MgCl}_2 > \text{NaCl} \). They have decreasing freezing points in this order, reversed: \( \text{NaCl} < \text{MgCl}_2 < \text{Na}_3\text{PO}_4 < \text{C}_6\text{H}_{12}\text{O}_6 \approx \text{CH}_3\text{COOH} \).

Key concepts

The van't Hoff Factor

The van't Hoff factor, denoted as \( i \), is a crucial element in understanding freezing point depression in solutions. It represents the number of particles a solute dissociates into in solution. When a solute is added to a solvent, the solution's physical properties, like its freezing point, change.
For electrolytes, which dissociate into ions in solution, \( i \) is equal to the number of ions formed. For example:

• NaCl dissociates into two ions: \( \text{Na}^+ \) and \( \text{Cl}^- \), so \( i = 2 \).
• Na\(_3\)PO\(_4\) dissociates into four ions: three \( \text{Na}^+ \) and one \( \text{PO}_4^{3-} \), making \( i = 4 \).
• MgCl\(_2\) dissociates into three ions: one \( \text{Mg}^{2+} \) and two \( \text{Cl}^- \), so \( i = 3 \).

Electrolytes fully dissociate in solution, whereas non-electrolytes, such as glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)), do not, leading to \( i = 1 \) since they do not form any ions. This factor directly affects the extent to which the freezing point is depressed.

Effective Molality

Effective molality is a measure that takes into account both the concentration of the solute and the number of particles into which it dissociates. We calculate it by multiplying the van't Hoff factor \( i \) by the molality of the solution, symbolized by \( m \).
The formula is: \[ m_i = i \cdot m \] This helps determine the impact of a solute on the freezing point depression.
For example:

• For Na\(_3\)PO\(_4\) at \( 0.10 \) molality: \( 0.40 = 4 \times 0.10 \)
• For NaCl at \( 0.35 \) molality: \( 0.70 = 2 \times 0.35 \)
• For MgCl\(_2\) at \( 0.20 \) molality: \( 0.60 = 3 \times 0.20 \)

The effective molality indicates how many moles of particles are in solution per kilogram of solvent, playing a significant role in determining which solution will have a lower freezing point.

Electrolytes

Electrolytes are substances that ionize in solution, meaning they break down into ions. This process enables their solutions to conduct electricity. The van't Hoff factor for electrolytes is usually more than 1, reflecting the number of ions each compound produces.
Some common examples include:

• Sodium chloride (NaCl): Dissolves to form Na\(^+\) and Cl\(^-\), with \( i = 2 \).
• Sodium phosphate (Na\(_3\)PO\(_4\)): Forms three Na\(^+\) and one PO\(_4^{3-}\), so \( i = 4 \).
• Magnesium chloride (MgCl\(_2\)): Produces Mg\(^{2+}\) and two Cl\(^-\), resulting in \( i = 3 \).

Because they produce multiple ions, electrolytes significantly depress the freezing point compared to non-electrolytes of the same molality.

Non-Electrolytes

Non-electrolytes are substances that do not dissociate into ions in solution. They dissolve as whole molecules, and therefore, they do not conduct electricity in solutions. This results in a van't Hoff factor \( i = 1 \), indicating no increase in the number of particles.
Common examples of non-electrolytes are:

• Glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)): Stays intact as complete molecules in solution.
• Acetic acid (\( \text{CH}_3\text{COOH} \)): Although a weak acid, it only partially ionizes, so \( i \) is close to 1.

Non-electrolytes generally result in smaller freezing point depression effects due to the absence of ion formation, making them less effective in altering colligative properties like freezing point depression.