Problem 48 - Chemistry and Chemical Reactivity

Question

Arrange the following aqueous solutions in order of decreasing freezing point. (The last three are all assumed to dissociate completely into ions in water.) (a) \(0.20 m\) ethylene glycol (nonvolatile, nonelectrolyte) (b) \(0.12 m\) \(\mathrm{K}_{2} \mathrm{SO}_{4}\) (c) \(0.10 m\) \(\mathrm{MgCl}_{2}\) (d) \(0.12 m\) \(m \mathrm{KBr}\)

Step-by-Step Solution

Answer

(a) > (d) > (c) > (b)

Step 1: Understand Freezing Point Depression

The freezing point of a solution is lower than that of the pure solvent. The extent of this decrease depends on the molality of the solution and the van 't Hoff factor ( abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla ). The formula is abla abla abla abla abla: , where abla abla abla is the freezing point depression, abla abla abla is the van 't Hoff factor, abla abla abla is the freezing-point-depression constant, and abla abla is the molality.

Step 2: Determine the van 't Hoff factor (i)

Identify the van 't Hoff factor abla of each solution based on the number of particles each one produces in solution: (a) Ethylene glycol i = 1 (nonelectrolyte), (b) K2SO4 dissociates into 3 ions (2K+ and SO4^2-) so i = 3, (c) MgCl2 dissociates into 3 ions (Mg^2+ and 2 Cl-) so i = 3, (d) KBr dissociates into 2 ions (K+ and Br-) so i = 2.

Step 3: Calculate the effective concentration

For each solution, multiply the molality by the van 't Hoff factor to determine the concentration of particles affecting the freezing point. (a) Ethylene glycol: 0.20 m * 1 = 0.20, (b) K2SO4: 0.12 m * 3 = 0.36, (c) MgCl2: 0.10 m * 3 = 0.30, (d) KBr: 0.12 m * 2 = 0.24.

Step 4: Rank solutions by decreasing freezing point

The freezing point decreases as the effective concentration of solute particles increases. Thus, arrange the solutions from lowest effective concentration to highest: (a) Ethylene glycol (0.20) (d) KBr (0.24) (c) MgCl2 (0.30) (b) K2SO4 (0.36).

Key Concepts

van 't Hoff factor

The van 't Hoff factor, denoted by \( i \), is a measure that describes the number of particles a solute will form when dissolved in a solvent. This factor is crucial when understanding colligative properties like freezing point depression.
In simple terms, the van 't Hoff factor tells us how many ions or molecules are produced per formula unit of solute in a solution. For instance, a nonelectrolyte like ethylene glycol does not dissociate into ions, thus its \( i \) value is 1.

• For electrolytes, like salts, the factor depends on the number of particles formed. For example, potassium sulfate \((\mathrm{K}_2\mathrm{SO}_4)\) dissociates into three ions: two \( \mathrm{K}^+ \) and one \( \mathrm{SO}_4^{2-} \). Hence, its \( i \) value is 3.
• Similarly, magnesium chloride \((\mathrm{MgCl}_2)\) also separates into three ions: one \( \mathrm{Mg}^{2+} \) and two \( \mathrm{Cl}^- \), making \( i \) equal to 3 as well.
• For potassium bromide \((\mathrm{KBr})\), it splits into two ions: \( \mathrm{K}^+ \) and \( \mathrm{Br}^- \), resulting in \( i \) being 2.

The concept of the van 't Hoff factor is essential in determining how solutes affect the colligative properties of solutions.

Molality

Molality (abla molality abla molality) is a concentration unit used to describe the amount of solute in a solution. Unlike molarity, which depends on the volume of the solvent, molality considers the mass of the solvent, making it independent of temperature and pressure changes.
Mathematically, molality is defined as the amount of solute in moles divided by the mass of the solvent in kilograms. It is expressed as \( m \). To illustrate:

• If you have 0.12 moles of \( \mathrm{K}_2\mathrm{SO}_4 \) dissolved in 1 kg of water, the molality is simply 0.12 m.
• For \( \mathrm{MgCl}_2 \) with 0.10 moles in 1 kg of solvent, it's 0.10 m.
• For our nonelectrolyte, ethylene glycol, with 0.20 moles in that same amount, the molality is 0.20 m.

The use of molality in calculations like freezing point depression is beneficial because physical properties, such as pressure and temperature fluctuations, do not alter the mass of the solvent.

Electrolyte and Nonelectrolyte Solutions

Understanding the difference between electrolyte and nonelectrolyte solutions helps explain how various solutes affect freezing point depression.
Electrolytes are substances that, when dissolved in a solvent like water, dissociate into ions. These ions make the solution conductive. Examples include \( \mathrm{K}_2\mathrm{SO}_4 \), \( \mathrm{MgCl}_2 \), and \( \mathrm{KBr} \). When these solutes dissolve:

• \( \mathrm{K}_2\mathrm{SO}_4 \) releases three ions: \( 2 \mathrm{K}^+ \) and \( \mathrm{SO}_4^{2-} \).
• \( \mathrm{MgCl}_2 \) splits into \( \mathrm{Mg}^{2+} \) and two \( \mathrm{Cl}^- \) ions.
• \( \mathrm{KBr} \) separates into \( \mathrm{K}^+ \) and \( \mathrm{Br}^- \).

Nonelectrolytes like ethylene glycol do not dissociate into ions. When dissolved, they remain intact as molecules, resulting in a van 't Hoff factor of 1, significantly impacting the colligative properties of the solution differently than electrolytes.
This distinction is crucial in calculating how various types of solutes influence the freezing point of their solutions, with electrolytes generally having a larger effect due to their higher number of dissociated ions.