Problem 100 - Chemistry: The Central Science

Question

When \(10.0 \mathrm{~g}\) of mercuric nitrate, \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\), is dissolved in \(1.00 \mathrm{~kg}\) of water, the freezing point of the solution is \(-0.162^{\circ} \mathrm{C}\). When \(10.0 \mathrm{~g}\) of mercuric chloride \(\left(\mathrm{HgCl}_{2}\right)\) is dissolved in \(1.00 \mathrm{~kg}\) of water, the solution freezes at \(-0.0685^{\circ} \mathrm{C}\). Use these data to determine which is the stronger electrolyte, \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) or \(\mathrm{HgCl}_{2}\).

Step-by-Step Solution

Answer

The stronger electrolyte is mercuric nitrate \(\left(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\right)\) as it has a higher van 't Hoff factor (2.89) in comparison to mercuric chloride \(\left(\mathrm{HgCl}_{2}\right)\) (1.07).

Step 1: Calculate the molality of each solution

First, we need to find the molality of each solution, which is expressed as moles of solute per kilogram of solvent. To calculate molality, use the formula: \[m = \frac{moles\ of\ solute}{mass\ of\ solvent (in\ kg)}\] For both electrolytes, the mass of water used as a solvent is 1.00 kg. The molar masses of mercuric nitrate and mercuric chloride are 324.6 g/mol and 271.5 g/mol, respectively. Therefore, the molality of each solution can be calculated as follows: \[m_{Hg(NO_3)_2} = \frac{10.0 g}{324.6 g/mol \times 1.00 kg} = 0.0308 mol\ kg^{-1}\] \[m_{HgCl_2} = \frac{10.0 g}{271.5 g/mol \times 1.00 kg} = 0.0368 mol\ kg^{-1}\]

Step 2: Determine the Van 't Hoff factor (i) for each electrolyte

The formula for freezing point depression is given by: \[\Delta T_{f} = K_{f} \times m \times i\] We are given the freezing point depression (\(\Delta T_{f}\)) for each solution and we have calculated the molality, we can now find the van 't Hoff factor (i) for each electrolyte using the cryoscopic constant for water, which is \(K_{f} = 1.86\ K \cdot kg/mol \). Rearranging the formula above, we get: \[i = \frac{\Delta T_{f}}{K_{f} \times m}\] Let's determine the van 't Hoff factor for each electrolyte: \[i_{Hg(NO_3)_2} = \frac{-0.162^{\circ}C}{1.86\ K \cdot kg/mol \times 0.0308\ mol\ kg^{-1}} \approx 2.89\] \[i_{HgCl_2} = \frac{-0.0685^{\circ}C}{1.86\ K \cdot kg/mol \times 0.0368\ mol\ kg^{-1}} \approx 1.07\]

Step 3: Compare the Van 't Hoff factors to determine the stronger electrolyte

We have calculated the van 't Hoff factor for each electrolyte, and now we must compare the factors to determine the stronger electrolyte. The larger the van 't Hoff factor, the more ions are present in the solution, making the electrolyte stronger. Comparing the calculated van 't Hoff factors, we find that: \[i_{Hg(NO_3)_2} > i_{HgCl_2}\] So, the stronger electrolyte is mercuric nitrate \(\left(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\right)\) as it has a higher van 't Hoff factor in comparison to mercuric chloride \(\left(\mathrm{HgCl}_{2}\right)\).

Key Concepts

Molality

Molality is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which is affected by changes in volume due to temperature and pressure, molality remains constant, making it extremely useful in situations involving physical changes like freezing or boiling.

To understand it with a practical example, if you dissolve a substance in water, you'd measure the moles of substance you have, and then divide it by the mass of the water in kilograms. In the textbook problem, molality helps determine how much the solute (mercuric nitrate or mercuric chloride) affects the freezing point of water.

Van 't Hoff factor

The Van 't Hoff factor, often represented as 'i', reflects the number of particles into which a compound dissociates in solution. It indicates whether a solute is an electrolyte and whether it's strong or weak. For nonelectrolytes that don’t dissociate, 'i' is typically equal to 1. For strong electrolytes that dissociate completely, 'i' corresponds to the number of ions formed per formula unit. The Van 't Hoff factor plays a pivotal role when calculating properties such as freezing point depression and boiling point elevation.

In the example given, 'i' is used to evaluate the strength of the electrolytes. By comparing the calculated values of 'i' for both mercuric nitrate and mercuric chloride, it becomes possible to determine which substance is the stronger electrolyte.

Freezing point depression

Freezing point depression is a colligative property, which means it depends on the number of particles dissolved in the solvent, not on their identity. This phenomenon occurs when a solute is added to a solvent, causing the freezing point of the solution to lower compared to the pure solvent. The more particles (or higher molality) you have in the solution, the more significant the depression of the freezing point.

The physical explanation behind this effect lies in the disruption of the solvent's structure by the solute particles, which requires a lower temperature to achieve the ordered arrangement necessary for the solvent to solidify. In our textbook problem, freezing point depression is used to infer the strength of electrolytes by observing how much the addition of each solute lowers water's freezing point.

Cryoscopic constant

The cryoscopic constant, denoted as 'Kf', is a proportionality constant that varies with each solvent. It's defined for a particular solvent and quantifies how much the freezing point will decrease per molal concentration of the solute. Generally, it is measured in degrees Celsius per molal (°C/m).

In the case of water, the cryoscopic constant is 1.86 K kg/mol, a value commonly used when calculating the freezing point depression of aqueous solutions. By incorporating the specific cryoscopic constant into the freezing point depression formula, you can calculate the degree to which the solute affects the freezing point. This was adeptly used in the example given to help find the stronger electrolyte based on the observed freezing point depression.