Problem 135 - Chemistry The Molecular Science

Question

Consider these data for aqueous solutions of ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\). $$\begin{array}{cc}\hline \text { Molality }(\mathrm{mol} / \mathrm{kg}) & \text { Freezing Point }\left({ }^{\circ} \mathrm{C}\right) \\\\\hline 0.0050 & -0.0158 \\\0.020 & -0.0709 \\\0.20 & -0.678 \\\1.0 & -3.33\end{array}$$ (a) Plot these data and from the graph determine the freezing point of a \(0.50 \mathrm{~mol} / \mathrm{kg}\) ammonium chloride solution. (b) Calculate the van't Hoff \(i\) factor for each concentration. Explain any trend that you see. (c) Calculate the percent dissociation of ammonium chloride in each solution.

Step-by-Step Solution

Answer

(a) -1.7°C; (b) i decreases with concentration; (c) Percent dissociation decreases with concentration.

Step 1: Plotting the Data

Start by plotting the given data with molality on the x-axis and freezing point on the y-axis. Use the points: (0.0050, -0.0158), (0.020, -0.0709), (0.20, -0.678), and (1.0, -3.33). After plotting, draw a smooth curve that best fits these points.

Step 2: Estimating the Freezing Point at 0.50 mol/kg

Using the plot from Step 1, locate the x-value of 0.50 mol/kg on the x-axis. From this point, draw a vertical line until it meets the curve. Then draw a horizontal line from this intersection point to the y-axis to determine the freezing point. The estimated freezing point is approximately -1.7°C.

Step 3: Calculating the van't Hoff Factor

The van't Hoff factor, \(i\), can be determined using the equation \(\Delta T_f = i \cdot K_f \cdot m\), where \(\Delta T_f\) is the change in freezing point, \(K_f\) is the freezing point depression constant for water (1.86°C kg/mol), and \(m\) is the molality. Calculate \(i\) for each concentration: - For 0.0050 mol/kg: \(i = \frac{0.0158}{1.86 \times 0.0050} \approx 1.70\)- For 0.020 mol/kg: \(i = \frac{0.0709}{1.86 \times 0.020} \approx 1.91\)- For 0.20 mol/kg: \(i = \frac{0.678}{1.86 \times 0.20} \approx 1.82\)- For 1.0 mol/kg: \(i = \frac{3.33}{1.86 \times 1.0} \approx 1.79\)

Step 4: Analyzing the Trend in van't Hoff Factor

The van't Hoff factor \(i\) should theoretically be \(2\) for complete dissociation of ammonium chloride to \(NH_4^+\) and \(Cl^-\). However, \(i\) decreases with increasing concentration, indicating ion pairing or incomplete dissociation at higher molalities.

Step 5: Calculating Percent Dissociation

Percent dissociation is calculated as:\[ \text{Percent dissociation} = \left( \frac{i - 1}{2 - 1} \right) \times 100 \%\]- For 0.0050 mol/kg: \( \frac{1.70 - 1}{2} \times 100 \approx 85\%\)- For 0.020 mol/kg: \( \frac{1.91 - 1}{2} \times 100 \approx 91\%\)- For 0.20 mol/kg: \( \frac{1.82 - 1}{2} \times 100 \approx 82\%\)- For 1.0 mol/kg: \( \frac{1.79 - 1}{2} \times 100 \approx 79\%\)The decrease in percent dissociation with increasing concentration reflects increased ion pairing.

Key Concepts

Molality

Molality is a way of expressing the concentration of a solution that is particularly useful in colligative properties like freezing point depression. It measures the number of moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of the solution, molality is concerned with the mass of the solvent, making it independent of temperature.Whenever you need to determine the freezing point depression of a solution, knowing the molality helps us understand how many particles are present to disrupt the solvent's freezing process:

• Definition: Molality (\(m\)) is calculated as: \[m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\]
• Usage: For the freezing point depression, the formula \(\Delta T_f = i \cdot K_f \cdot m\) is applied. Here \(\Delta T_f\) is the change in freezing point, \(K_f\) is the freezing point depression constant, and \(m\) is the molality.

Molality remains constant regardless of temperature changes. This makes it particularly handy in experiments where temperature control is challenging.

van't Hoff Factor

The van't Hoff factor (\(i\)) is an important concept when dealing with colligative properties like freezing point depression. It accounts for the dissociation of solute particles in a solution. The more a compound dissociates, the higher its van't Hoff factor becomes.When ammonium chloride (\(NH_4Cl\)) dissolves in water, it splits into two ions: ammonium (\(NH_4^+\)) and chloride (\(Cl^-\)). Ideally, the van't Hoff factor should be 2, because the solute is expected to dissociate fully into two ions in the solution.However, as the solution's concentration increases, you may notice:

• Reduced \(i\) values: These indicate incomplete dissociation or ion pairing at higher concentrations.
• Trend Analysis: With increasing concentration, \(i\) tends to be less than the theoretical value, reflecting the increased likelihood of ions remaining paired or grouped together instead of acting as free species.

Understanding the van't Hoff factor helps in predicting not just dissociation, but also the extent of colligative property changes like the lowering of freezing points.

Percent Dissociation

Percent dissociation is a measure of how much of a substance has dissociated into its ions in solution. For substances like ammonium chloride, which dissociate into ions when dissolved, this measure indicates the fraction of molecules that break apart.The formula for percent dissociation related to the van't Hoff factor, \(i\), becomes very useful:

• Formula: \[\text{Percent dissociation} = \left( \frac{i - 1}{n - 1} \right) \times 100 \%\]Where \(i\) is the van't Hoff factor, and \(n\) is the number of ions the solute dissociates into. Here, \(n\) would be 2 for ammonium chloride.
• Observation: With decreasing percent dissociation at higher molalities, one can infer increasing ion pairing. When ions form pairs, they do not contribute to the colligative properties as effectively as when they are free.

By evaluating the percent dissociation, one gains insight into the molecular behavior of solutes in varying concentrations, which can affect their overall chemical and physical behavior in solutions.