Problem 59 - Chemistry: Principles and Reactions

Question

An aqueous solution of \(L \mathrm{iX}\) is prepared by dissolving \(3.58 \mathrm{~g}\) of the electrolyte in \(283 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}(d=1.00 \mathrm{~g} / \mathrm{mL})\). The solution freezes at \(-1.81^{\circ} \mathrm{C}\). What is \(\mathrm{X}^{-}\) ? (Assume complete dissociation of \(\mathrm{LiX}\) to \(\mathrm{Li}^{+}\) and \(\mathrm{X}^{-}\).)

Step-by-Step Solution

Answer

Answer: The identity of the anion X- is F-. Therefore, the electrolyte is LiF.

Step 1: Calculate the molality of the solution

First, we have to find the molality of the solution using the mass of LiX and the mass of solvent. Molality (m) is defined as the moles of solute per kilogram of solvent. Given the mass of LiX is 3.58 g and the mass of H2O (283 mL) is 283 g (since the density of water is 1.00 g/mL). We can represent the moles of LiX as n_LiX and the molar mass of LiX as M_LiX. To calculate the molality, we need to determine the moles of LiX. Thus, we have the equation: molality (m) = n_LiX / (mass of H2O / 1000) However, we don't know the molar mass of LiX since we don't know the identity of X-. Let's represent the molality in terms of the mass of LiX and its molar mass: m = (mass of LiX / M_LiX) / (mass of H2O / 1000)

Step 2: Apply the freezing point depression equation

The change in freezing point of a solution (∆Tf) is related to the molality of the solution as follows: ∆Tf = Kf * m * i Here, Kf is the cryoscopic constant of water (1.86 K kg/mol), i is the van't Hoff factor, which is 2 for LiX since it dissociates into 2 ions (Li+ and X-). The freezing point depression is given as 1.81°C. Plugging this information into the equation, we get: 1.81 = 1.86 * m * 2 Solving for m, we have: m = 1.81 / (1.86 * 2) m ≈ 0.486 mol/kg

Step 3: Determine the identity of X-

Now, we can substitute the molality value back into the equation from Step 1: 0.486 = (3.58 / M_LiX) / (283 / 1000) Rearrange the equation to solve for M_LiX: M_LiX ≈ 3.58 / (0.486 * 0.283) M_LiX ≈ 27.542 g/mol Since the molar mass of Li is 6.939 g/mol, the molar mass of X- can be determined by subtracting the molar mass of Li from the molar mass of LiX: M_X- ≈ 27.542 - 6.939 M_X- ≈ 20.603 g/mol Looking at the periodic table, the element that has a molar mass close to 20.603 g/mol is Fluorine (F) with a molar mass of 18.998 g/mol. Therefore, X- is F-, and the electrolyte is LiF.

Key Concepts

Understanding Molality

Molality is a measure of the concentration of a solute in a solution, and it's defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which is another concentration unit involving volume, molality is not affected by temperature changes because mass remains constant regardless of temperature.

In the context of freezing point depression, understanding molality is crucial. The problem provided may seem complex, but the concept of molality simplifies the process to find the concentration of the dissolved substance, in this case, LiX. Once the mass of the solute and the mass of the solvent are known, the molality (m) can be calculated using the formula:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kilograms}} \].

Therefore, when you're given the mass of lithium halide (LiX) and the mass of water, you can determine the molality, which is instrumental for further calculations in the freezing point depression equation.

Colligative Properties and Freezing Point Depression

Colligative properties are characteristics of solutions that depend solely on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present. One of the most important colligative properties is freezing point depression, which is the decrease in the freezing point of a solvent when a solute is dissolved in it.

The connection between freezing point depression and molality can be seen in the formula used to calculate the change in freezing point:
\[ \Delta T_f = K_f \times m \times i \].

Here, \(\Delta T_f\) represents the change in freezing point, \(K_f\) is the cryoscopic constant (specific to the solvent), m is the molality of the solution, and i is the van't Hoff factor. By recognizing the relationship between these variables, students can understand how the presence of a solute lowers the freezing point of a solvent.

The Cryoscopic Constant (Kf)

The cryoscopic constant, represented as \(K_f\), is a proportionality factor that links the molality of a solution to the magnitude of the freezing point depression. It is a unique value for each solvent and is measured in degrees Celsius per molal (°C/m).

The cryoscopic constant takes into account how a specific solvent's properties, such as entropy and enthalpy of fusion, affect the freezing point when a solute is added. In the textbook exercise, \(K_f\) for water is given as 1.86 K kg/mol, which is used to calculate the freezing point depression. This value allows us to predict how much the freezing point will drop for each mole of solute particles dissolved in a kilogram of water. Therefore, understanding the cryoscopic constant is essential when solving problems related to the freezing point depression of solutions.

The Role of the van't Hoff factor (i)

The van't Hoff factor, notated as \(i\), plays a critical role in freezing point depression calculations. It represents the number of particles into which a compound dissociates in solution. For non-electrolytes, which don’t dissociate, the van't Hoff factor is 1. However, for electrolytes like LiX, which dissociate into ions, the factor is equal to the number of ions produced.

In the given problem, LiX dissociates into Li+ and X-, resulting in a van't Hoff factor of 2. This means that each formula unit of LiX produces two particles in solution. It is vital to remember that the van't Hoff factor must be considered when calculating the freezing point depression, otherwise the extent of the freezing point drop might be underestimated if we erroneously assume the substance does not dissociate.