Question

Write a balanced net ionic equation for each of the following solution reactions. Refer to Table 5 for electrolyte information (a) \(\mathrm{AgNO}_{3}(a q)+\mathrm{KI}(a q) \longrightarrow \mathrm{AgI}(s)+\mathrm{KNO}_{3}(a q)\) (b) \(\mathrm{BaCl}_{2}(a q)+\mathrm{K}_{2} \mathrm{CrO}_{4}(a q) \longrightarrow \mathrm{BaCrO}_{4}(\mathrm{~s})+\mathrm{KCl}(a q)\)

Short answer

(a) \( \mathrm{Ag}^+(aq) + \mathrm{I}^-(aq) \rightarrow \mathrm{AgI}(s) \) (b) \( \mathrm{Ba}^{2+}(aq) + \mathrm{CrO}_4^{2-}(aq) \rightarrow \mathrm{BaCrO}_4(s) \)

Step-by-step solution

Identify Reactants and Products

For each reaction, list the reactants and products provided. For Reaction (a), the reactants are \( \mathrm{AgNO}_3(aq) \) and \( \mathrm{KI}(aq) \), while the products are \( \mathrm{AgI}(s) \) and \( \mathrm{KNO}_3(aq) \). For Reaction (b), the reactants are \( \mathrm{BaCl}_2(aq) \) and \( \mathrm{K}_2\mathrm{CrO}_4(aq) \), and the products are \( \mathrm{BaCrO}_4(s) \) and \( \mathrm{KCl}(aq) \).

Write the Complete Ionic Equations

Dissociate all soluble strong electrolytes into their respective ions. For Reaction (a): \( \mathrm{Ag}^+(aq) + \mathrm{NO}_3^-(aq) + \mathrm{K}^+(aq) + \mathrm{I}^-(aq) \rightarrow \mathrm{AgI}(s) + \mathrm{K}^+(aq) + \mathrm{NO}_3^-(aq) \). For Reaction (b): \( \mathrm{Ba}^{2+}(aq) + 2\mathrm{Cl}^-(aq) + 2\mathrm{K}^+(aq) + \mathrm{CrO}_4^{2-}(aq) \rightarrow \mathrm{BaCrO}_4(s) + 2\mathrm{K}^+(aq) + 2\mathrm{Cl}^-(aq) \).

Identify and Cancel Spectator Ions

In these ionic equations, identify spectator ions that appear on both sides without undergoing change. For Reaction (a), \( \mathrm{K}^+ \) and \( \mathrm{NO}_3^- \) ions are spectators. For Reaction (b), \( \mathrm{K}^+ \) and \( \mathrm{Cl}^- \) ions are spectators.

Write the Net Ionic Equations

Remove the spectator ions from the complete ionic equation to form the net ionic equation. For Reaction (a), the net ionic equation is \( \mathrm{Ag}^+(aq) + \mathrm{I}^-(aq) \rightarrow \mathrm{AgI}(s) \). For Reaction (b), the net ionic equation is \( \mathrm{Ba}^{2+}(aq) + \mathrm{CrO}_4^{2-}(aq) \rightarrow \mathrm{BaCrO}_4(s) \).

Key concepts

Balanced Equations

In chemistry, balanced equations are a fundamental aspect of accurately representing chemical reactions. They ensure that the number of each type of atom is the same on the reactants side as on the products side of the equation. This aligns with the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.
To balance a chemical equation, we adjust the coefficients—the numbers in front of molecules or atoms—to ensure an equal number of each atom type on both sides of the equation. For instance, in the reaction involving silver nitrate and potassium iodide:

• We start with the compounds: \( \mathrm{AgNO}_3 \) and \( \mathrm{KI} \) as reactants and \( \mathrm{AgI} \) and \( \mathrm{KNO}_3 \) as products.
• By writing this as a balanced chemical equation, we ensure that the silver (Ag), potassium (K), iodine (I), and nitrate (NO₃) all appear with the same frequencies on both sides.

This balance is crucial for calculating reaction stoichiometry and for moving toward writing net ionic equations.

Reaction Stoichiometry

Stoichiometry involves quantifying the relationships between reactants and products in a chemical reaction. It relies on balanced equations to determine how much of a substance is consumed or produced.
In our example reactions, stoichiometry helps us predict how much barium chromate will precipitate when barium chloride reacts with potassium chromate.

• Consider the balanced equation: \( \mathrm{BaCl}_2(aq) + \mathrm{K}_2\mathrm{CrO}_4(aq) \longrightarrow \mathrm{BaCrO}_4(s) + \mathrm{KCl}(aq) \).
• We can deduce the mole-to-mole relationships, such as one mole of barium chloride reacting with one mole of potassium chromate to yield one mole of barium chromate.

This molar ratio provides critical insights when calculating the amounts in larger-scale reactions. Stoichiometry bridges the balanced equation with practical, quantitative lab work, allowing chemists to accurately mix reactants and predict product yields.

Spectator Ions

In aqueous reactions, some ions do not participate in the chemical change, known as spectator ions. These ions appear unchanged on both the reactant and product sides of the complete ionic equation.
Knowing how to identify and exclude them is essential for writing accurate net ionic equations.

• In Reaction (a): \( \mathrm{AgNO}_3(aq) + \mathrm{KI}(aq) \rightarrow \mathrm{AgI}(s) + \mathrm{KNO}_3(aq) \), potassium ions \( \mathrm{K}^+ \) and nitrate ions \( \mathrm{NO}_3^- \) are spectators.
• For Reaction (b): \( \mathrm{BaCl}_2(aq) + \mathrm{K}_2 \mathrm{CrO}_4(aq) \rightarrow \mathrm{BaCrO}_4(s) + \mathrm{KCl}(aq) \), potassium \( \mathrm{K}^+ \) and chloride \( \mathrm{Cl}^- \) ions do not participate actively.

By removing these ions, we derive net ionic equations, focusing on the essence of the reaction—the formation of solid precipitates. Recognizing and removing spectator ions simplifies reactions and highlights the true transformative processes in solution chemistry.