Problem 112 - Chemistry The Molecular Science

Question

A solution, prepared by dissolving \(9.41 \mathrm{~g} \mathrm{NaHSO}_{3}\) in \(1.00 \mathrm{~kg}\) water, freezes at \(-0.33{ }^{\circ} \mathrm{C}\). From these data, decide which of these equations is the correct expression for the dissociation of the salt. (a) \(\mathrm{NaHSO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq})\) (b) \(\mathrm{NaHSO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{SO}_{3}^{2-}(\mathrm{aq})\)

Step-by-Step Solution

Answer

The correct equation is (a): \( \text{NaHSO}_3 \rightarrow \text{Na}^+ + \text{HSO}_3^- \).

Step 1: Determine Moles of Solute

First, determine the molar mass of NaHSO₃. The molar mass is calculated by adding the atomic masses of its constituent atoms: \( \text{Na} = 22.99 \text{ g/mol}, \text{H} = 1.01 \text{ g/mol}, \text{S} = 32.07 \text{ g/mol}, \text{O} = 3 \times 16.00 \text{ g/mol} \). Thus, the molar mass of NaHSO₃ is \( 22.99 + 1.01 + 32.07 + 48.00 = 104.07 \text{ g/mol} \). Next, calculate the number of moles of NaHSO₃: \[ \text{moles of NaHSO₃} = \frac{9.41 \text{ g}}{104.07 \text{ g/mol}} = 0.0904 \text{ mol} \].

Step 2: Calculate Molality of Solution

Molality \( (m) \) is defined as the moles of solute per kilogram of solvent. Given that the solvent, water, has a mass of \( 1.00 \text{ kg} \), calculate the molality: \[ m = \frac{0.0904 \text{ mol}}{1.00 \text{ kg}} = 0.0904 \text{ mol/kg} \].

Step 3: Use Freezing Point Depression Formula

The freezing point depression formula is \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f = 0.33 ^{ ext{o}} \text{C} \) (the decrease in freezing point), \( K_f = 1.86 ^{ ext{o}} \text{C/molal} \) for water, and \( i \) is the van't Hoff factor that represents the number of particles the solute dissociates into. Solve for \( i \):\[ 0.33 = i \cdot 1.86 \cdot 0.0904 \].\[ i = \frac{0.33}{1.86 \times 0.0904} \approx 1.97 \].

Step 4: Determine the Correct Dissociation Equation

The calculated \( i \approx 1.97 \) suggests the solution dissociates into approximately 2 particles per formula unit.The first equation (a): \( \text{NaHSO}_3 \rightarrow \text{Na}^+ + \text{HSO}_3^- \) shows dissociation into 2 ions.The second equation (b): \( \text{NaHSO}_3 \rightarrow \text{Na}^+ + \text{H}^+ + \text{SO}_3^{2-} \) shows dissociation into 3 ions.Since \( i \approx 1.97 \), which is closer to 2, equation (a) is correct.

Key Concepts

Dissociation Equation

In chemistry, understanding how compounds dissociate in water is crucial, especially for electrolyte solutions. A **dissociation equation** describes how a compound separates into ions when dissolved in a solvent, like water. This is essential for calculating properties like the freezing point depression. Applying this to our problem, we are examining how sodium bisulfite, \( \text{NaHSO}_3 \), dissociates in water. The given possible equations are:

• \( \text{NaHSO}_3(\text{aq}) \rightarrow \text{Na}^+(\text{aq}) + \text{HSO}_3^-(\text{aq}) \)
• \( \text{NaHSO}_3(\text{aq}) \rightarrow \text{Na}^+(\text{aq}) + \text{H}^+(\text{aq}) + \text{SO}_3^{2-}(\text{aq}) \)

Equation (a) shows dissociation into two ions, sodium ion (\( \text{Na}^+ \)) and bisulfite ion (\( \text{HSO}_3^- \)), whereas equation (b) shows dissociation into three ions. By experimentally calculating the van’t Hoff factor, which indicates the number of ions a compound dissociates into, we can determine the correct equation to use. Thus, knowing dissociation helps us delve deeper into solute behavior in solutions.

Molality

**Molality** is a measure of the concentration of a solution in terms of the moles of solute per kilogram of solvent. This term is symbolized by \( m \). Unlike molarity, which is volume-dependent, molality remains unaffected by temperature changes because it uses mass rather than volume. In the freezing point depression formula, molality determines the extent to which the solute lowers the freezing point of the solvent.To calculate the molality of a solution containing \( 9.41 \text{ g} \) of sodium bisulfite in \( 1.00 \text{ kg} \) of water, we first find moles using the compound's molar mass, which is \( 104.07 \text{ g/mol} \). This gives us \( 0.0904 \text{ mol} \) of \( \text{NaHSO}_3 \). Thus, the molality \( m \) is:\[m = \frac{0.0904 \text{ mol}}{1.00 \text{ kg}} = 0.0904 \text{ mol/kg}\]This value is essential for calculating the freezing point depression in colligative property analysis, which can further inform us about the nature of solute dissociation.

Van't Hoff Factor

The **van’t Hoff factor**, represented by \( i \), is pivotal in understanding how compounds affect colligative properties like boiling point elevation and freezing point depression. It essentially signifies the number of particles a solute produces in a solution. For instance, if a substance completely dissociates into three ions, \( i \) would be approximately 3. In our exercise, we computed: \[\i = \frac{0.33}{1.86 \times 0.0904} \approx 1.97\]This near 2 value indicates that \( \text{NaHSO}_3 \) dissociates into roughly two particles per formula unit, aligning with dissociation equation (a) rather than equation (b). Using \( i \) effectively helps confirm the extent of dissociation and identify the correct dissociation equation based on the nature of ionic separation.