Problem 81 - Chemistry and Chemical Reactivity

Question

A \(2.00 \%\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in water freezes at \(-0.796^{\circ} \mathrm{C}\) (a) Calculate the van't Hoff factor, \(i\) (b) Which of the following best represents sulfuric acid in a dilute aqueous solution: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HSO}_{4}^{-},\) or \(2 \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{SO}_{4}^{2-2}\)

Step-by-Step Solution

Answer

(a) Van't Hoff factor, i = 2.06. (b) Best representation: H3O+ + HSO4^-.

Step 1: Determine Molality

First, we need to determine the molality (m) of the solution. A 2.00% w/w solution of \( \mathrm{H}_2\mathrm{SO}_4 \) means there are 2.00 grams of \( \mathrm{H}_2\mathrm{SO}_4 \) per 100 grams of solution. Thus, the mass of water is 98 grams or 0.098 kg.The molar mass of \( \mathrm{H}_2\mathrm{SO}_4 \) is 98.08 g/mol.The molality \( m \) is given by \( m = \frac{\text{moles of solute}}{\text{kg of solvent}} \).Using the values, we have:\[m = \frac{2.00 \, \text{g} \div 98.08 \, \text{g/mol}}{0.098 \, \text{kg}} = 0.208 \text{ mol/kg.}\]

Step 2: Use Freezing Point Depression Equation

Freezing point depression is calculated using the formula:\[\Delta T_f = i \cdot K_f \cdot m\]where \( \Delta T_f \) is the change in freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant for water (1.86 °C kg/mol), and \( m \) is the molality.Given that \( \Delta T_f = 0.796 ^\circ C \) and \( m = 0.208 \text{ mol/kg} \), we can solve for \( i \):\[0.796 = i \times 1.86 \times 0.208\]

Step 3: Solve for Van't Hoff Factor

Rearrange the formula to solve for \( i \):\[i = \frac{0.796}{1.86 \times 0.208} \approx 2.06\]The van't Hoff factor \( i \) is approximately 2.06.

Step 4: Explanation for Species in Solution

\( i \approx 2.06 \) suggests partial ionization. The initial complete dissociation of \( \mathrm{H}_2\mathrm{SO}_4 \) would result in three ions, which corresponds to \( i \approx 3 \), but since \( i \approx 2.06 \), it's more likely that \( \mathrm{H}_2\mathrm{SO}_4 \) dissociates into \( \mathrm{H}_3\mathrm{O}^{+} + \mathrm{HSO}_4^- \), which aligns closely with the observed van't Hoff factor.

Key Concepts

Freezing Point Depression

Freezing point depression is a colligative property, meaning it depends on the number of solute particles in a solution rather than the type of particles. When a solute is dissolved in a solvent like water, it disrupts the solvent's structure, causing the freezing point to lower. This is particularly important in understanding how solutions behave, such as when adding sulfuric acid to water. The decrease in temperature at which the water will freeze is proportional to the molality of the solution and the van’t Hoff factor, which represents ion dissociation.

• The formula used for freezing point depression is \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the change in freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant, and \( m \) is molality.
• In our sulfuric acid solution, the change in freezing point was observed to be \( -0.796 ^\circ \text{C} \).

Understanding freezing point depression helps us to grasp why solutions behave differently than pure solvents.

Sulfuric Acid Dissociation

When sulfuric acid (\( \text{H}_2\text{SO}_4 \)) is added to water, it dissociates into ions. However, this dissociation is not always complete. Typically, sulfuric acid can lose its first hydrogen ion (\( \text{H}^+ \)) relatively completely, forming \( \text{H}_3\text{O}^+ \) and \( \text{HSO}_4^- \). In more diluted solutions, the second ionization step may not occur as readily, demonstrating partial dissociation.
The van't Hoff factor, calculated in the exercise as approximately 2.06, suggests that the equilibrium lies more towards the first dissociation stage:

• \( \text{H}_2\text{SO}_4 \longrightarrow \text{H}_3\text{O}^+ + \text{HSO}_4^- \).
• This aligns with the observed value, as complete dissociation to \( 2 \text{H}_3\text{O}^+ \) and \( \text{SO}_4^{2-} \) would result in a factor closer to 3.

Ionization equilibrium is vital as it affects not just freezing points but also other colligative properties like boiling point elevation.

Molality

Molality (\( m \)) is a measure of solute concentration expressed in moles of solute per kg of solvent. It is particularly useful when dealing with colligative properties because it does not change with temperature fluctuations, unlike molarity, which is volume-based. In our sulfuric acid example, molality was used to calculate the changes in the freezing point.

• To find molality, first determine the number of moles of \( \text{H}_2\text{SO}_4 \). With 2 grams of \( \text{H}_2\text{SO}_4 \) and a molar mass of 98.08 g/mol, this yields approximately 0.0204 moles.
• The mass of the solvent (water) is 98 grams, which is 0.098 kg.

Thus, the molality of the sulfuric acid solution is about 0.208 mol/kg, which was crucial to calculating freezing point depression in the exercise.

Chemical Ionization Equilibria

Chemical ionization equilibria refer to the balance established between dissociated and undissociated forms of a compound in solution. This concept is especially relevant with strong acids, like sulfuric acid, which can dissociate into multiple ions. The van't Hoff factor provides insight into this equilibrium by indicating the effective number of solute particles.

• The partial dissociation reflected by a van't Hoff factor of 2.06 indicates that the equilibrium for sulfuric acid lies closer to the formation of \( \text{H}_3\text{O}^+ \) and \( \text{HSO}_4^- \).
• This means that not all molecules dissociate completely into ions, enabling the calculation of colligative properties to pinpoint approximate levels of dissociation.

Understanding ionization equilibria can help predict how substances influence various solution properties, such as electrical conductivity and freezing point depression, which are pivotal in chemical and industrial applications.