Problem 88 - Chemistry The Molecular Nature of Matter

Question

Complete and balance the following equations. For each, write the molecular, ionic, and net ionic equations. (All of the products are soluble in water.) (a) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow\) (b) \(\mathrm{HClO}_{4}(a q)+\mathrm{NH}_{3}(a q) \longrightarrow\) (c) \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{NH}_{3}(a q) \longrightarrow\)

Step-by-Step Solution

Answer

For reaction (a): Net ionic: \(\mathrm{Mg}(\mathrm{OH})_{2}(s) + 2\mathrm{H}^{+}(aq) \rightarrow \mathrm{Mg}^{2+}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l)\). For reaction (b): Net ionic: \(\mathrm{NH}_{3}(aq) + \mathrm{H}^{+}(aq) \rightarrow \mathrm{NH}_{4}^{+}(aq)\). For reaction (c): Net ionic: \(2\mathrm{NH}_{3}(aq) + 2\mathrm{H}^{+}(aq) \rightarrow 2\mathrm{NH}_{4}^{+}(aq)\).

Step 1: Balance the Equation for Reaction (a)

First, write the balanced molecular equation for the reaction between acetic acid, \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\), and magnesium hydroxide, \(\mathrm{Mg}(\mathrm{OH})_{2}\). The products will be magnesium acetate, \(\mathrm{Mg}(C_{2}H_{3}O_{2})_{2}\), and water, \(\mathrm{H}_{2}\mathrm{O}\). The balanced molecular equation is: \[(2\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}(aq) + \mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Mg}(C_{2}H_{3}O_{2})_{2}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l))\].

Step 2: Write the Ionic and Net Ionic Equations for Reaction (a)

Next, split the soluble compounds into their ions for the ionic equation. Since \(\mathrm{Mg}(\mathrm{OH})_{2}\) is a solid, it remains unchanged, and water is a liquid, so it also remains unchanged in the ionic equation. The ionic equation is: \[2\mathrm{H}^{+}(aq) + 2\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}(aq) + \mathrm{Mg}^{2+}(aq) + 2\mathrm{OH}^{-}(s) \rightarrow \mathrm{Mg}^{2+}(aq) + 2\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l)\]. Cancel out identical ions on both sides of the equation to find the net ionic equation: \[\mathrm{Mg}(\mathrm{OH})_{2}(s) + 2\mathrm{H}^{+}(aq) \rightarrow \mathrm{Mg}^{2+}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l)\].

Step 3: Balance the Equation for Reaction (b)

For the reaction between perchloric acid, \(\mathrm{HClO}_{4}(aq)\), and ammonia, \(\mathrm{NH}_{3}(aq)\), the products are ammonium perchlorate, \(\mathrm{NH}_{4}ClO_{4}(aq)\). The balanced molecular equation is: \[\mathrm{HClO}_{4}(aq) + \mathrm{NH}_{3}(aq) \rightarrow \mathrm{NH}_{4}ClO_{4}(aq)\].

Step 4: Write the Ionic and Net Ionic Equations for Reaction (b)

The ionic equation separates the soluble compounds into their ions: \[\mathrm{H}^{+}(aq) + \mathrm{ClO}_{4}^{-}(aq) + \mathrm{NH}_{3}(aq) \rightarrow \mathrm{NH}_{4}^{+}(aq) + \mathrm{ClO}_{4}^{-}(aq)\]. The net ionic equation cancels out common ions on both sides, leading to: \[\mathrm{NH}_{3}(aq) + \mathrm{H}^{+}(aq) \rightarrow \mathrm{NH}_{4}^{+}(aq)\].

Step 5: Balance the Equation for Reaction (c)

When carbonic acid, \(\mathrm{H}_{2}\mathrm{CO}_{3}(aq)\), reacts with ammonia, \(\mathrm{NH}_{3}(aq)\), the products are ammonium carbonate, \(\mathrm{(NH}_{4})_{2}\mathrm{CO}_{3}(aq)\). The balanced molecular equation is: \[\mathrm{H}_{2}\mathrm{CO}_{3}(aq) + 2\mathrm{NH}_{3}(aq) \rightarrow \mathrm{(NH}_{4})_{2}\mathrm{CO}_{3}(aq)\].

Step 6: Write the Ionic and Net Ionic Equations for Reaction (c)

Write the ionic equation by dissociating the soluble compounds into ions: \[2\mathrm{H}^{+}(aq) + \mathrm{CO}_{3}^{2-}(aq) + 2\mathrm{NH}_{3}(aq) \rightarrow 2\mathrm{NH}_{4}^{+}(aq) + \mathrm{CO}_{3}^{2-}(aq)\]. The net ionic equation is found by canceling out the common ions, leaving: \[2\mathrm{NH}_{3}(aq) + 2\mathrm{H}^{+}(aq) \rightarrow 2\mathrm{NH}_{4}^{+}(aq)\].

Key Concepts

Molecular Equations

Molecular equations are representations of chemical reactions where the compounds are expressed as molecules, not distinguishing between ionic and molecular species. These equations show the exact number of reactant and product molecules involved in a reaction. For students to fully grasp molecular equations, let's break down the fundamental concept.

A key aspect of understanding molecular equations lies in balancing them. Balancing ensures that the number of atoms for each element is the same on both sides of the reaction. This adheres to the Law of Conservation of Mass, stating that matter cannot be created or destroyed in a chemical reaction.

To improve on the exercise, involve the concept of stoichiometry, which is the calculation of reactants and products in chemical reactions. For the balanced reaction between acetic acid and magnesium hydroxide, stoichiometry would help students understand the ratio of reacting molecules to the products formed, which is vital in practical applications like mixing correct amounts in a lab setting.

Ionic Equations

Ionic equations provide a more nuanced view of chemical reactions involving ionic compounds. These compounds, when in an aqueous solution, dissociate into ions. An ionic equation breaks down the soluble ionic compounds into their respective cations and anions.

In an ionic equation, the focus shifts from whole compounds to the ions they form in solution. This is particularly useful for predicting the outcomes of reactions and understanding the movement of individual ions. It's important to note that solid salts, liquids, and gases are represented as complete formulas and not dissociated ions.

Identifying Spectator Ions

In improving the exercise, emphasize identifying spectator ions. These are ions that remain unchanged on both sides of a reaction. They can be omitted when writing net ionic equations, which simplifies and highlights the actual chemical change.

Net Ionic Equations

Net ionic equations streamline the chemical reaction to its bare essence, showcasing only the species that undergo change. They are derived from the full ionic equation by removing spectator ions. For students, mastering this concept is key to focusing on the heart of the reaction.

To arrive at a net ionic equation, one should cancel out ions appearing identically on both reactant and product sides, as they do not participate in the actual chemical reaction. This process highlights the driving force behind the reaction, which is particularly important in understanding precipitation, acid-base, and redox reactions.

Importance in Laboratory and Industrial Applications

Including the importance of net ionic equations in laboratory and industrial contexts can further enrich the learning experience. These equations can simplify the analysis of complex mixtures, and predict reaction outcomes, thereby aiding in the synthesis of desired compounds with increased efficiency.

Acid-Base Reactions

Acid-base reactions are a fundamental concept in chemistry, often represented through neutralization reactions where an acid reacts with a base to produce water and a salt. Recognizing and balancing these reactions is invaluable for students as they are pervasive in both laboratory and real-world situations.

An acid is a substance that donates protons (hydrogen ions, H+) in an aqueous solution, while a base provides hydroxide ions (OH-) or accepts protons. When they react, the H+ from the acid and the OH- from the base combine to form water (H2O), and the remaining ions form a salt, which is usually soluble in water.

Understanding pH Changes

In line with the exercise, explaining the concept of pH changes during acid-base reactions would be an excellent addition. pH is a measure of the acidity or basicity of a solution, and acid-base reactions usually entail a significant change in pH. Illustrating how titrations work to determine concentrations of acid or base in a solution could also deepen students' understanding of this chemical process.