Question

Which of the following salt will have the same volume of vant Hoff factor 'i'as that of \(\mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})] ?\) (a) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (b) \(\mathrm{NaCl}\) (c) \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) (d) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)

Short answer

The salt \( \mathrm{Al}_{2} (\mathrm{SO}_{4})_{3} \) has the same Vant Hoff factor of 5.

Step-by-step solution

Understanding Vant Hoff Factor

The Vant Hoff factor, denoted as 'i', is the number of particles a compound dissociates into in solution. For ionic compounds, it relates to the ions produced when dissolved in water.

Dissociation of K4[Fe(CN)6]

The compound \( \mathrm{K}_{4} [ \mathrm{Fe} ( \mathrm{CN} ) ] \) dissociates into 4 \( \mathrm{K}^+ \) ions and one \( [\mathrm{Fe} (\mathrm{CN})]_6^{4-} \) ion, resulting in a total of 5 particles. Thus, \( i = 5 \) for \( \mathrm{K}_{4} [\mathrm{Fe} (\mathrm{CN}) ] \).

Option a: Al2(SO4)3

\( \mathrm{Al}_{2} (\mathrm{SO}_{4})_{3} \) dissociates into 2 \( \mathrm{Al}^{3+} \) ions and 3 \( \mathrm{SO}_{4}^{2-} \) ions, totaling 5 ions. Hence, \( i = 5 \).

Option b: NaCl

\( \mathrm{NaCl} \) dissociates into 1 \( \mathrm{Na}^+ \) ion and 1 \( \mathrm{Cl}^- \) ion, totaling 2 ions. Thus, \( i = 2 \).

Option c: Al(NO3)3

\( \mathrm{Al}(\mathrm{NO}_3)_3 \) dissociates into 1 \( \mathrm{Al}^{3+} \) ion and 3 \( \mathrm{NO}_3^- \) ions, totaling 4 ions. Therefore, \( i = 4 \).

Option d: Na2SO4

\( \mathrm{Na}_{2} \mathrm{SO}_{4} \) dissociates into 2 \( \mathrm{Na}^+ \) ions and 1 \( \mathrm{SO}_4^{2-} \) ion, totaling 3 ions. Thus, \( i = 3 \).

Identifying the Salt with Matching Vant Hoff Factor

The compound \( \mathrm{Al}_2(\mathrm{SO}_4)_3 \) has a Vant Hoff factor of 5, which matches that of \( \mathrm{K}_{4} [\mathrm{Fe} (\mathrm{CN}) ] \).

Key concepts

Ionic Dissociation

When ionic compounds dissolve in water, they break apart into individual ions in a process known as ionic dissociation. This is fundamental to understanding the Vant Hoff factor because it describes the number of particles that are formed.

For example, consider the dissociation of potassium ferrocyanide, \( \mathrm{K}_{4} [ \mathrm{Fe} (\mathrm{CN}) ] \), into 4 \( \mathrm{K}^+ \) ions and 1 \([\mathrm{Fe} (\mathrm{CN})]_6^{4-}\) ion. The result is a total of 5 ions, making the Vant Hoff factor, \( i \), equal to 5.

The degree of dissociation is incredibly important in determining the properties of solutions. The more particles that result from the dissociation, the greater the impact on properties like boiling point elevation and freezing point depression.

Ionic Compounds

Ionic compounds are composed of cations and anions held together by the electrostatic forces between oppositely charged particles. These compounds are typically formed when metals react with nonmetals.

Consider \( \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\), which consists of aluminum cations \( \mathrm{(Al^{3+})} \) and sulfate anions \( \mathrm{(SO_4^{2-})} \). In solution, this compound dissociates into 2 aluminum ions and 3 sulfate ions, totaling 5 ions.

The dissociation number of ionic compounds is crucial to understanding their impact on solution properties and calculating the Vant Hoff factor. This factor informs us how many effective particles are present, which directly affects colligative properties.

Solution Chemistry

Solution chemistry explains how substances dissolve and interact in a solvent, usually water. In the case of ionic compounds, when these dissolve in water, they separate into their constituent ions.

Each dissolved ion affects solution properties like conductivity, boiling point, and osmotic pressure. For instance, when \( \mathrm{NaCl} \) dissolves, it separates into \( \mathrm{Na}^+ \) and \( \mathrm{Cl}^- \), resulting in 2 ions, providing a basic Vant Hoff factor of 2. This simple dissociation illustrates how ionic compounds modify the behavior of solutions.

Understanding the mechanics of solution chemistry helps grasp how ionic dissociation affects various properties and why different compounds have different Vant Hoff factors. Through this understanding, we can predict the behavior of solutions and manipulate them for desired outcomes, from creating saline solutions to refining antifreeze mixtures.