Question

The freezing point of \(0.10 \mathrm{M} \mathrm{KHSO}_{3}\) is \(-0.38^{\circ} \mathrm{C}\). Which of the following equations best represents what happens when \(\mathrm{KHSO}_{3}\) dissolves in water? (a) \(\mathrm{KHSO}_{3}(s) \longrightarrow \mathrm{KHSO}_{3}(a q)\) (b) \(\mathrm{KHSO}_{3}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{HSO}_{3}^{-}(a q)\) (c) \(\mathrm{KHSO}_{3}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{SO}_{3}^{2-}(a q)+\mathrm{H}^{+}(a q)\)

Short answer

Answer: The best equation representing the dissociation of KHSO3 in water is KHSO3(s) → K+(aq) + HSO3−(aq).

Step-by-step solution

Calculate the freezing point depression

The freezing point depression of the solution is the difference between the freezing point of pure water and the observed freezing point of the solution. In this case, the freezing point of pure water is \(0^{\circ}\mathrm{C}\) and the freezing point of the \(0.10 \mathrm{M}\) \(\mathrm{KHSO}_{3}\) solution is \(-0.38^{\circ}\mathrm{C}\). Therefore, the freezing point depression is: \(\Delta T_f = 0^{\circ}\mathrm{C} - (-0.38^{\circ}\mathrm{C}) = 0.38^{\circ}\mathrm{C}\)

Calculate the molality of the solution

The molality of the solution (in mol/kg) can be calculated with the following formula: \(molality = \frac{moles\,of\,solute}{mass\,of\,solvent \,in\,kg}\) We are given the molarity of the solution as \(0.10 \mathrm{M}\). Assuming 1 liter (1000 mL) of the solution, we can convert to molality. \(0.10\,M = \frac{0.10\,mol}{1\,L \times 1\,kg} = 0.10\,mol/kg\)

Determine the van't Hoff factor

Now we can use the freezing point depression formula to determine the van't Hoff factor, i, which indicates the number of particles the solute dissociates into when dissolved in water: \(\Delta T_{f}= K_{f} × molality × i\) We can rearrange the formula to find the van't Hoff factor. The constant \(K_{f}\) for water is 1.86 \(\frac{^{\circ}\mathrm{C}}{molal}\). \(i= \frac{\Delta T_{f}}{K_{f} \times molality}=\frac{0.38^{\circ}\mathrm{C}}{1.86\frac{^{\circ}\mathrm{C}}{molal}\times 0.10\,mol/kg} \approx 2\) The van't Hoff factor is approximately 2, which means that the \(\mathrm{KHSO}_{3}\) dissociates into 2 ions when it dissolves in water.

Determine the best equation

Based on the van't Hoff factor calculated above, the best equation that represents what happens when \(\mathrm{KHSO}_{3}\) dissolves in water is the one that shows the dissociation into 2 ions. Hence, the correct answer is: (b) $\mathrm{KHSO}_{3}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{HSO}_{3}^{-}(a q)$

Key concepts

Colligative Properties

Colligative properties are characteristics of a solution that depend solely on the number of solute particles present, rather than on the nature of the chemical species itself. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. Understanding freezing point depression is particularly useful in day-to-day life, from the antifreeze in cars to the salt on icy roads. Freezing point depression occurs when a solute is dissolved in a solvent; the solution’s freezing point is lower than that of the pure solvent. This is due to the solute particles disrupting the formation of a solid structure in the solvent, thereby requiring a lower temperature to freeze.

For example, adding salt to ice causes it to melt because the salt lowers the freezing point of water. This property is pivotal in scenarios such as the one found in our exercise, where the freezing point of a solution is measured to determine its properties.

Molality Calculation

Molality is a measure of the concentration of a solute in a solution and is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which is moles per liter of solution, molality is temperature-independent because it's based on mass, not volume, which can change with temperature. The calculation of molality is critical when dealing with colligative properties because these properties depend on the concentration of the solute particles.

To calculate molality, you use the formula:
\[molality = \frac{moles\,of\,solute}{mass\,of\,solvent\,in\,kg}\]

This value is essential to find other colligative properties of the solution, like the freezing point depression, as we saw in our exercise.

van't Hoff Factor

The van’t Hoff factor, denoted by the symbol 'i', is a dimensionless quantity used in the calculation of colligative properties. It represents the number of particles into which a compound dissociates in solution. For example, NaCl dissociates into two ions, Na+ and Cl-, so its van't Hoff factor is 2. In our exercise, we used the van't Hoff factor to determine the dissociation of potassium hydrogen sulfite (KHSO3).

You calculate the factor by rearranging the freezing point depression equation:
\[i = \frac{\Delta T_{f}}{K_{f} \times molality}\]

This calculation helps us understand how many particles a solute forms in a solution, which in turn provides insight into the solution's effects on the colligative properties.

Dissociation of Solutes

Dissociation refers to the separation of a molecule into two or more ions or simpler molecules when dissolved in a solvent. The extent to which a solute dissociates is important for understanding the behavior of solutions in various chemical and physical processes. Dissociation is essential in the context of colligative properties because the number of particles created upon dissolution of the solute influences properties like freezing point depression.

Our exercise questioned how KHSO3 dissociates in water. As we determined using the van't Hoff factor, it separates into two ions, K+ and HSO3-. Knowing how solutes dissociate into ions helps predict the outcome of certain reactions and explains observations like changes in conductivity or reactivity in the solution.