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Chapter 12: Problem 102

Iodine \(\left(\mathrm{I}_{2}\right)\) is only sparingly soluble in water (left photo). Yet upon the addition of iodide ions (for example, from KI), iodine is converted to the triiodide ion, which readily dissolves (right photo): $$\mathrm{I}_{2}(s)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{I}_{3}^{-}(a q) $$ Describe the change in solubility of \(\mathrm{I}_{2}\) in terms of the change in intermolecular forces.

Short Answer

Expert verified
The solubility of iodine increases when it reacts with an iodide ion to form a triiodide ion, which is charged. This charged ion can interact more strongly with water molecules, through ion-dipole interactions, than the uncharged iodine molecules could, thus the triiodide ion is more soluble in water than iodine.

Step by step solution

01

Understanding the Chemical Equation

First, look at the chemical equation provided. It shows that iodine \(\mathrm{I}_{2}\) and iodide ions \(\mathrm{I}^{-}\) undergoes an equilibrium reaction to form triiodide ions \(\mathrm{I}_{3}^{-}\). This explains the change in solubility of iodine upon the addition of iodine ions.
02

Identify the Change in Intermolecular Forces

Before the reaction, there's weak dispersion forces between iodine molecules and water molecules, which make iodine only sparingly soluble in water. Now, triiodide is an ion (charge -1), it's formed by iodine, which is nonpolar, and iodide ions, which are polar.
03

Describing how the Intermolecular Forces Affect Solubility

When \(\mathrm{I}_{2}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{I}_{3}^{-}\), it becomes charged. In water, a polar solvent, it can now interact with water molecules through ion-dipole forces. Ion-dipole interactions are significantly stronger than dispersion forces, therefore \(\mathrm{I}_{3}^{-}\) is more soluble in water compared to \(\mathrm{I}_{2}\). This accounts for the increase in solubility of iodine in water in the presence of iodide ions.

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