Problem 35 - Chemistry in 30 days

Question

An electrolyte \(\mathrm{XY}_{2}\) is \(40 \%\) ionized. Calculate the van't Hoff factor. (a) \(1.4\) (b) \(0.4\) (c) \(1.8\) (d) \(1.6\)

Step-by-Step Solution

Answer

The correct answer is (c) 1.8.

Step 1: Understand Ionization

The electrolyte \( \mathrm{XY}_{2} \) dissociates in water to form ions \( \mathrm{X}^{2+} \) and \( 2 \mathrm{Y}^{-} \). The chemical equation for this ionization is: \[ \mathrm{XY}_{2} \rightarrow \mathrm{X}^{2+} + 2 \mathrm{Y}^{-} \].

Step 2: Calculate Number of Ions

From the equation, one molecule of \( \mathrm{XY}_{2} \) splits into three ions (1 \( \mathrm{X}^{2+} \) and 2 \( \mathrm{Y}^{-} \)). This means the degree of ionization, \( \alpha \), affects the count of these ions.

Step 3: Apply Degree of Ionization

The problem states \( \mathrm{XY}_{2} \) is \( 40 \% \) ionized, so \( \alpha = 0.4 \). Meaning, 40% of the \( \mathrm{XY}_{2} \) molecules dissociate into ions.

Step 4: Use Formula for Van't Hoff Factor

The formula relating the degree of ionization to the van't Hoff factor \( i \) is given by: \[ i = 1 + \alpha (n - 1) \] where \( n \) is the total number of particles (ions) produced from one formula unit. Here \( n = 3 \).

Step 5: Solve for Van't Hoff Factor

Substitute \( \alpha = 0.4 \) and \( n = 3 \) into the formula: \[ i = 1 + 0.4 \times (3 - 1) = 1 + 0.4 \times 2 = 1 + 0.8 = 1.8 \]. Thus, the van't Hoff factor is \( 1.8 \).

Key Concepts

Understanding the Degree of Ionization

The degree of ionization, represented by the symbol \( \alpha \), is a way to describe how much an electrolyte dissociates into its ions when dissolved in a solvent like water. In simpler terms, it's a measure of how well the compound breaks apart into its ionic components. For instance, if a compound has a degree of ionization of \( 0.4 \) (or 40%), this means that 40% of the original molecules have dissociated into ions, while 60% remain intact.
\( \alpha = 0 \) indicates no ionization at all, while \( \alpha = 1 \) indicates complete ionization. The degree of ionization depends on several factors such as the nature of the electrolyte, concentration, and temperature.

What is Electrolyte Ionization?

Electrolyte ionization is the process by which an electrolyte splits into its positive and negative ions when dissolved in a solvent. This process is crucial for the conductivity of the solution, enabling it to carry an electric current. When a strong electrolyte ionizes, it dissociates almost completely, leading to a high concentration of ions in the solution. Conversely, a weak electrolyte partially ionizes and thus conducts less electricity due to the lower concentration of ions.
Consider the compound \( \mathrm{XY}_2 \). When dissolved in water, it ionizes into one positive ion \( \mathrm{X}^{2+} \) and two negative ions \( \mathrm{Y}^{-} \), forming a total of three ions. This transformation from molecular to ionic form is what allows the solution to become conductive.

Calculation of Ions and their Contribution

When dealing with electrolyte ionization, it becomes essential to calculate the number of ions produced from a single molecule to understand the solution's properties better. From the chemical dissociation of \( \mathrm{XY}_2 \), three ions are formed: one \( \mathrm{X}^{2+} \) and two \( \mathrm{Y}^{-} \). Calculating this correctly is vital for understanding how the solution behaves.
Knowing the degree of ionization \( \alpha \), one can determine the actual count of ions in a solution. Suppose an electrolyte is 40% ionized; then 40% of the original molecules convert into ions. This understanding is essential for various practical applications such as determining the resultant van't Hoff factor in a solution.

Fundamentals of Chemical Dissociation

Chemical dissociation refers to the process where a compound breaks down into its constituent ions. It's a central concept in understanding electrolyte behaviors. In our example with \( \mathrm{XY}_{2} \), dissociation involves the formula \( \mathrm{XY}_2 \rightarrow \mathrm{X}^{2+} + 2 \mathrm{Y}^{-} \). When this dissociation occurs in water, it results in three ions from a single molecule.
It's crucial to recognize that dissociation affects properties like the van't Hoff factor, which tells us about the number of particles in a solution compared to the original compound. Through understanding dissociation fully, one can apply other related concepts such as the degree of ionization and its implications in calculating properties of solutions, enhancing academic and practical comprehension in chemistry.