Problem 77 - General Chemistry: Principles and Modern Applications

Question

The molecular solid \(\mathrm{I}_{2}(\mathrm{s})\) is only slightly soluble in water but will dissolve to a much greater extent in an aqueous solution of \(\mathrm{KI}\), because the \(\mathrm{I}_{3}^{-}\) anion forms. Write an equation for the formation of the \(I_{3}^{-}\) anion, and indicate the Lewis acid and Lewis base.

Step-by-Step Solution

Answer

The equation for formation of the I₃⁻ ion is \( I_{2(s)} + I_{-(aq)} \rightarrow I_{3^{-}(aq)} \). In this reaction, I- acts as a Lewis base while I₂ acts as a Lewis acid.

Step 1: Formulate Equation

You start with the compounds provided: I₂ and KI. When I₂ dissolves in KI, it reacts with I⁻ (from KI) to form the I₃⁻ ion. The reaction can be written as follows: \[ I_{2(s)} + I_{-(aq)} \rightarrow I_{3^{-}(aq)} \]

Step 2: Identify Lewis Acids and Bases

Next we identify the Lewis acid and Lewis base in the reaction. The Lewis base is a species that has an electron pair available for bonding, while the Lewis acid is a species that can accept this pair. In this reaction, I- (from KI) acts as a Lewis base as it donates an electron pair. I₂ acts as a Lewis acid, as it accepts the electron pair.

Key Concepts

Solubility

Solubility refers to the ability of a substance to dissolve in a solvent, forming a homogeneous mixture at the molecular or ionic level. When \( \mathrm{I}_2(\mathrm{s}) \) is mentioned as slightly soluble in water, it means that only a small amount of iodine can dissolve in water. However, in an aqueous solution of \( \mathrm{KI} \), the solubility of \( \mathrm{I}_2 \) increases significantly. This increase is due to the formation of the \( \mathrm{I}_3^- \) anion.Ul

Solubility can be affected by temperature, pressure, and the presence of other substances.

In our context, molecular interactions with \( \mathrm{KI} \) increase solubility.

This is due to the creation of new chemical species.

When \( \mathrm{I}_2 \), a non-polar molecule, encounters \( \mathrm{KI} \), iodine gains the ability to interact with polar water molecules by forming charged species (\( \mathrm{I}_3^- \)). This makes it dissolve more easily in water, showing how chemical reactions can change solubility dynamics.

Lewis Acids and Bases

Understanding Lewis acids and bases helps in visualizing how electron exchanges occur in reactions. A Lewis acid is a molecule or ion that can accept an electron pair, while a Lewis base donates an electron pair.In the reaction \[ I_{2(s)} + I_{-(aq)} \rightarrow I_{3^{-}(aq)} \]

• \( I^- \) (from \( \mathrm{KI} \)) acts as a Lewis base.
• It donates its electron pair to \( \mathrm{I}_2 \).
• \( \mathrm{I}_2 \) is the Lewis acid as it accepts this electron pair.

This interaction leads to the formation of \( \mathrm{I}_3^- \). Identifying these roles is important because it shows how molecules interact at a fundamental level, beyond traditional acid-base concepts. This broadened perspective includes reactions that don't involve \( \mathrm{H}^+ \) ions, expanding our understanding of chemical reactivity.

Chemical Equations

Chemical equations represent chemical reactions, showing how reactants transform into products. In this reaction, the equation is:\[ I_{2(s)} + I_{-(aq)} \rightarrow I_{3^{-}(aq)} \]This equation shows the process of \( \mathrm{I}_2 \) interacting with \( \mathrm{I}^- \) to form \( \mathrm{I}_3^- \).

• The left side lists the reactants: \( \mathrm{I}_2 \) and \( \mathrm{I}^- \).
• The right side lists the product: \( \mathrm{I}_3^- \).
• The arrow indicates the direction of the reaction.

Writing and balancing chemical equations is crucial because it describes the quantitative aspects of chemistry. It ensures the conservation of mass and charge, illustrating that the same number of each type of atom appears in both reactants and products, obeying the law of conservation of mass. This balanced breakdown aids in predicting how substances will interact under set conditions.