Problem 20 - Chemistry in 30 days

Question

A mixture of \(\mathrm{Nal}\) and \(\mathrm{NaIO}_{3}\) is treated with hot concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The iodine-containing product formed is (a) \(\mathrm{HIO}_{3}\) (b) \(\mathrm{NalO}_{4}\) (o) \(\mathrm{I}_{2}\) (d) \(\mathrm{HIO}_{4}\)

Step-by-Step Solution

Answer

The iodine-containing product formed is \(\mathrm{I}_2\).

Step 1: Analyze the Reaction

When a mixture of sodium iodide (\(\mathrm{NaI}\)) and sodium iodate (\(\mathrm{NaIO}_3\)) is treated with hot concentrated sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)), several reactions can occur. The goal is to determine the iodine-containing product formed by this reaction.

Step 2: Consider Oxidation States

In \(\mathrm{NaI}\), iodine is in the -1 oxidation state, while in \(\mathrm{NaIO}_3\), iodine is in the +5 oxidation state. These tend to form products where iodine is in a different oxidation state.

Step 3: Predict the Likely Product

Hot concentrated sulfuric acid can oxidize iodide ions. Given the choices, sulfuric acid's oxidizing ability typically leads to the formation of diatomic iodine \(\mathrm{I}_2\), especially when considering the redox behavior of iodide and iodate under the presence of sulfuric acid.

Step 4: Eliminate Other Options

The other options \(\mathrm{HIO}_3\), \(\mathrm{NalO}_4\), and \(\mathrm{HIO}_4\) either form under different, less likely conditions or have more complex formation processes not typical for this reaction.

Key Concepts

Oxidation States

Oxidation states, also referred to as oxidation numbers, help in understanding the transfer of electrons in a chemical reaction. They indicate the degree of oxidation or reduction an atom has undergone within a compound. For iodine in sodium iodide (\(\mathrm{NaI}\)), the oxidation state is -1 because iodine gains an electron to form an iodide ion. In sodium iodate (\(\mathrm{NaIO}_3\)), iodine has an oxidation state of +5 since it is bonded to oxygen atoms, which are more electronegative. This shift in oxidation states is essential to predict the products in a reaction, as atoms tend to move towards more stable oxidation states. Understanding these transitions helps you recognize how electrons are transferred or shared, which is vital for predicting reaction outcomes.

Sulfuric Acid Reactions

Sulfuric acid (\(\mathrm{H}_2o\mathrm{SO}_4}\)) is a powerful acid and an oxidizing agent, often used in chemical reactions to drive changes in oxidation states. When hot and concentrated, sulfuric acid has the potential to oxidize other substances. In the reaction involving sodium iodide and sodium iodate, sulfuric acid acts on the iodide ions (\(\mathrm{I}^-\)) and potentially oxidizes them to form diatomic iodine (\(\mathrm{I}_2\)). This happens because sulfuric acid can remove electrons from iodide ions, leading to the formation of iodine molecules. Knowing how sulfuric acid interacts with other compounds can help predict reaction products more accurately.

Redox Reactions

Redox reactions are chemical processes where oxidation and reduction occur simultaneously. In the described scenario, the interaction of sulfuric acid with sodium iodide and sodium iodate is a classic redox reaction. Here, the oxidation state of iodine in \(\mathrm{NaI}\)) is increased from -1 to 0 in \(\mathrm{I}_2\)), indicating oxidation. On the other hand, sulfur from \(\mathrm{H}_2o\mathrm{SO}_4}\)) can change as it reduces, showing the dual nature of these reactions. In redox reactions, recognizing which substances are oxidized and which are reduced can help determine the final products. This understanding is crucial when analyzing chemical equilibria involving oxidizing agents like sulfuric acid.

Iodine Compounds

Iodine compounds exhibit diverse chemistry because of iodine's ability to exist in different oxidation states, such as -1, +1, +5, and +7. In the exercise, when sodium iodide and sodium iodate are mixed with hot concentrated sulfuric acid, simplifying the variety of possible iodine compounds is key. The reaction typically results in the formation of diatomic iodine (\(\mathrm{I}_2\)), which is the molecular form of iodine often encountered in chemistry laboratories. This is because the conditions favor the release of iodine from iodide ions as \(\mathrm{I}_2\)) gas, which is a well-documented product under specific strong oxidizing conditions. Understanding iodine's versatile chemistry aids in predicting the structure and reactivity of different iodine-containing compounds.