Question

Identify the spectator ions in the complete ionic equation. $$ \begin{aligned} \mathrm{Ba}^{2+}(a q)+2 \mathrm{I}^{-}(a q)+2 \mathrm{Na}^{+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow & \longrightarrow \\ & \mathrm{BaSO}_{4}(s)+2 \mathrm{I}^{-}(a q)+2 \mathrm{Na}^{+}(a q) \end{aligned} $$

Short answer

The spectator ions in the reaction are \(\mathrm{I}^{-}(aq)\) and \(\mathrm{Na}^{+}(aq)\).

Step-by-step solution

Identify the Reactants and Products

Identify all the ions present as reactants and products in the equation. Reactants: \(\mathrm{Ba}^{2+}(aq)\), \(\mathrm{2I}^{-}(aq)\), \(\mathrm{2Na}^{+}(aq)\), and \(\mathrm{SO}_{4}^{2-}(aq)\). Products: \(\mathrm{BaSO}_{4}(s)\), \(\mathrm{2I}^{-}(aq)\), \(\mathrm{2Na}^{+}(aq)\).

Determine the Products Formed in the Reaction

Note that \(\mathrm{BaSO}_{4}\) is the only product that forms a solid precipitate, indicating that it is not in the aqueous (aq) state but in the solid (s) state. This compound is the result of the reaction and is not a spectator ion.

Identify the Spectator Ions

Spectator ions are ions that do not change their oxidation state and are present in the same form on both the reactant and product side of the equation. Comparing both sides, \(\mathrm{I}^{-}(aq)\) and \(\mathrm{Na}^{+}(aq)\) are present as reactants and as products in the same state and therefore are spectator ions.

Key concepts

Complete Ionic Equation

In chemistry, a complete ionic equation is a way of displaying the chemical species involved in a reaction as distinct ions, assuming they exist in an aqueous solution. This form of the equation provides a clearer understanding of the role each ion plays in the reaction.

Unlike a traditional balanced chemical equation that shows the reactants and products as compounds, the complete ionic equation breaks down soluble ionic compounds into their individual ions. For example, in the given exercise where barium (Ba²⁺), iodide (I^-), sodium (Na⁺), and sulfate (SO₄^2-) ions react, the complete ionic equation is used to illustrate their reactivity in water.

When writing complete ionic equations, we follow these steps: firstly, separate the ionic compounds into their constituent ions, and secondly, show the state of each substance to distinguish between the soluble ions and the insoluble precipitate. This approach is critical for identifying the spectator ions—ions that remain unchanged throughout the reaction.

Precipitation Reaction

A precipitation reaction occurs when two soluble ionic compounds react in an aqueous solution to form at least one insoluble product, known as a precipitate. This type of reaction is essential for understanding how ionic compounds interact in solutions.

In the exercise provided, the reaction between barium and sulfate ions forms barium sulfate (BaSO₄), which is insoluble in water and precipitates out of the solution as a solid. This creates a visual cue that a reaction has taken place—it's the formation of a solid that distinguishes precipitation reactions from others.

To predict whether a precipitation reaction will occur, one must be familiar with solubility rules. Since barium sulfate is known to be insoluble in water, upon mixing barium and sulfate ion-containing solutions, a precipitation reaction is guaranteed, resulting in its formation as a solid precipitate.

Ionic Compounds in Reactions

Reactions involving ionic compounds are fundamental in chemistry, especially when discussing solubility, electrolytes, and reaction types like precipitation. In aqueous solutions, ionic compounds dissociate into cations and anions, which can partake in various reactions.

In our original exercise, we see the interaction of barium, iodide, sodium, and sulfate ions—all of which come from ionic compounds dissolved in water. Understanding how these ions interact in solution allows chemists to predict the formation of new compounds. For instance, the provided reaction shows that when these ions meet, barium and sulfate ions form BaSO₄, while the sodium and iodide ions remain in the solution as they are not part of the precipitate.

An important aspect to consider is that not all ionic compounds participate actively in the reaction. Some, like sodium (Na⁺) and iodide (I^-) ions in this case, are spectator ions, which means they don't contribute to the formation of the precipitate and remain unaltered in the solution. Recognizing these ions is crucial for simplifying chemical equations and understanding the actual changes occurring during the reaction.