Problem 24 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(\mathrm{AlF}_{3}\) is insoluble in anhydrous HF but it becomes soluble in presence of little amount of KF. Addition of boron trifluoride to the resulting solution causes reprecipitation of \(\mathrm{AlF}_{3}\). Explain with balanced chemical equations.

Step-by-Step Solution

Answer

\(\text{KF}\) solubilizes \(\text{AlF}_3\) by forming \(\text{K}_3\text{AlF}_6\); \(\text{BF}_3\) addition re-precipitates \(\text{AlF}_3\) by forming \(\text{KBF}_4\).

Step 1: Understand the system with anhydrous HF

Aluminum fluoride (\(\text{AlF}_3\)) is insoluble in anhydrous hydrogen fluoride (HF) due to the ionic nature of \(\text{AlF}_3\), which does not dissolve in nonpolar solvents like HF.

Step 2: Effect of KF addition

When a small amount of potassium fluoride (KF) is added to the system, it dissociates into \(\text{K}^+\) and \(\text{F}^-\) ions. The presence of these ions enables the formation of a complex ion \(\text{K}_3\text{AlF}_6\), which is soluble. This process can be represented by the equation: \[ \text{AlF}_3 + 3\text{KF} \rightarrow \text{K}_3\text{AlF}_6 \].

Step 3: Addition of Boron Trifluoride causes reprecipitation

Boron trifluoride (\(\text{BF}_3\)) reacts with the complex \(\text{K}_3\text{AlF}_6\) to form an insoluble compound of \(\text{AlF}_3\) again. This process can be described by the equation: \[ \text{K}_3\text{AlF}_6 + 3\text{BF}_3 \rightarrow 3\text{KF} + 3\text{KBF}_4 + \text{AlF}_3 \]. The aluminum fluoride thus precipitates back out of the solution.

Key Concepts

Solubility

Solubility is a key concept in understanding how substances dissolve in a solvent to form a solution. Solubility depends on the nature of the solute and the solvent, and often involves the interaction between their respective molecules. Typically, polar compounds dissolve in polar solvents, while nonpolar compounds dissolve in nonpolar solvents.
\(\text{AlF}_3\) is an ionic compound, which means it has a specific solubility behavior due to its charged ions. In anhydrous hydrogen fluoride (HF), \(\text{AlF}_3\) is insoluble because HF is a nonpolar solvent and does not effectively stabilize the separated ions of \(\text{AlF}_3\).
This explains why we do not observe \(\text{AlF}_3\) dissolving initially in HF, despite the fundamental aspects of its ionic solubility.

Complex Ions

Complex ions are entities formed from a central metal atom or ion bonded chemically to a surrounding array of molecules or ions, known as ligands. They play a crucial role in altering the solubility of certain compounds.
When potassium fluoride (KF) is introduced to the \(\text{AlF}_3\) and HF mixture, KF dissociates into \(\text{K}^+\) and \(\text{F}^-\) ions. The fluoride ions form a complex with \(\text{AlF}_3\), resulting in \(\text{K}_3\text{AlF}_6\), a soluble complex ion.
Creating such a complex ion involves coordination chemistry, where the aluminum atom serves as the central ion, coordinated by the fluorine atoms from the fluoride ions. By forming this soluble complex, the previously insoluble \(\text{AlF}_3\) is effectively dissolved in the solution.

Precipitation Reactions

Precipitation reactions involve the formation of a solid in a solution during a chemical reaction. The resulting solid, known as a precipitate, falls out of the solution.
In the context of this exercise, after \(\text{K}_3\text{AlF}_6\) has been formed, the addition of boron trifluoride (\(\text{BF}_3\)) reverts the complex back into an insoluble form of \(\text{AlF}_3\). This happens because \(\text{BF}_3\) interacts with the fluoride ions, forming \(\text{KBF}_4\), which frees up \(\text{AlF}_3\) without its coordinating fluoride ions.
Thus, \(\text{AlF}_3\) precipitates out again, showcasing the dynamic equilibrium involved in complex ion formation and dissolution. Precipitation effectively shows the reactions leading to the initial dissolved state being undone, leaving a visible solid in the solution.