Problem 82 - Chemistry and Chemical Reactivity

Question

A compound is known to be a potassium halide, KX. If 4.00 g of the salt is dissolved in exactly \(100 \mathrm{g}\) of water, the solution freezes at \(-1.28^{\circ} \mathrm{C}\) Identify the halide ion in this formula.

Step-by-Step Solution

Answer

The halide ion is fluoride (\(\text{F}^-\)).

Step 1: Understanding the Freezing Point Depression

The freezing point depression (abla T_f) can be calculated using the formula: \(abla T_f = K_f \cdot m\) where \(K_f\) is the cryoscopic constant of water (\(1.86 \degree C/m\)), and \(m\) is the molality of the solution.

Step 2: Calculate Molality

Given the freezing point depression \(-1.28 \degree C\), the molality \(m\) can be found: \( m = \frac{-1.28}{1.86} = 0.688 \ m\). Molality \(m\) is defined as moles of solute per kilogram of solvent. Since the solvent is 100 grams, or 0.1 kg, \(m = \frac{\text{moles of } KX}{0.1}\).

Step 3: Calculate Moles of Solute

From the molality, moles of \(KX\) \(= m \times 0.1 = 0.0688 \text{ moles}\).

Step 4: Determine Molar Mass of "KX"

Since 4 grams of \(KX\) give 0.0688 moles, the molar mass \(M\) is \(M = \frac{4}{0.0688} \approx 58.14 \text{ g/mol}\).

Step 5: Identify the Halide Ion

The potassium ion \(K^+\) has an atomic mass of approximately 39 g/mol. Therefore, the halide ion \(X^-\) will have an atomic mass of \(58.14 - 39 = 19.14 \text{ g/mol}\). This corresponds to the atomic mass of fluorine (19 g/mol), so the halide ion is \(\text{F}^-\).

Key Concepts

Potassium Halide

A potassium halide is a type of compound that consists of potassium ions (\( K^+ \)) and halide ions (\( X^- \)).These halides include fluoride, chloride, bromide, and iodide.When a potassium halide dissolves in water, it dissociates into its constituent ions.This process affects the colligative properties of the solution, such as freezing point depression.Understanding the behavior of potassium halides in solutions is crucial.This knowledge helps us explore various applications in chemistry and related scientific fields.

Molality Calculation

Molality is a measure of the concentration of a solute in a solution.It is defined as the moles of solute per kilogram of solvent.This concept is especially useful in calculating changes in colligative properties, such as freezing point depression.For example, in the given exercise, the molality \( m \) of the solution is calculated as follows:* Freezing point depression: \(-1.28 \degree C\)* Cryoscopic constant of water: \(1.86 \degree C/m\)* Formula: \( \Delta T_f = K_f \cdot m \)Solving for \( m \):\[ m = \frac{-1.28}{1.86} = 0.688 \ \text{mol/kg} \]This reveals the concentration of potassium halide in the solution, which helps in identifying the solute’s molar characteristics.

Cryoscopic Constant

The cryoscopic constant, \( K_f \), is a property of the solvent that affects depression in the freezing point.For water, \( K_f \) is approximately \(1.86 \degree C/m\).This constant is used in the formula for calculating the freezing point depression:\[ \Delta T_f = K_f \cdot m \]Where:

• \( \Delta T_f \) is the change in the freezing point.
• \( m \) is the molality of the solution.

Knowing the cryoscopic constant allows us to predict how much the freezing point of water will drop when a known concentration of solute is added.This is a key factor in understanding the behavior of solutions.

Halide Ion Identification

Identifying the halide ion involves calculating the molar mass of the entire compound and comparing it to the known atomic masses of halogens.In the exercise, knowing that potassium's atomic mass is about 39 g/mol helps.Steps to identify the halide:

• Calculate the moles of \( KX \) based on the weight of the solute and the molality of the solution.
• Determine the molar mass of \( KX \): \[ M = \frac{4 \text{ g}}{0.0688 \text{ moles}} = 58.14 \text{ g/mol} \]
• Subtract the atomic mass of potassium: \[ 58.14 - 39 = 19.14 \text{ g/mol} \]

The value obtained approximates the atomic mass of Fluorine.Therefore, the halide ion is identified as \( \text{F}^- \), solving the mystery of the compound's composition.