Question

Complete and balance each of the following molecular equations (in aqueous solution); include phase labels. Then, for each, write the net ionic equation. (a) $$ \mathrm{Al}(\mathrm{OH})_{3}+\mathrm{HCl} \longrightarrow $$ (b) $$ \mathrm{HClO}+\mathrm{Sr}(\mathrm{OH})_{2} \longrightarrow $$ (c) $$ \mathrm{Ba}(\mathrm{OH})_{2}+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \longrightarrow $$ (d) $$ \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{KOH} \longrightarrow $$

Short answer

All reactions produce water as part of their net ionic equations.

Step-by-step solution

Completing and balancing equation for reaction (a)

The initial reaction combines \(\mathrm{Al}(\mathrm{OH})_{3}\) with \(\mathrm{HCl}\). The balanced molecular equation is:\[\mathrm{Al}(\mathrm{OH})_{3}(s) + 3 \mathrm{HCl}(aq) \longrightarrow \mathrm{AlCl}_3(aq) + 3 \mathrm{H}_2\mathrm{O}(l)\]

Writing net ionic equation for reaction (a)

Dissociate the soluble compounds and cancel out the spectator ions. The \(\mathrm{H}_2\mathrm{O}\) remains intact as it is a liquid:\[\mathrm{Al}(\mathrm{OH})_3(s) + 3\mathrm{H}^+(aq) \longrightarrow \mathrm{Al}^{3+}(aq) + 3\mathrm{H}_2\mathrm{O}(l)\]

Completing and balancing equation for reaction (b)

The reaction between \(\mathrm{HClO}\) and \(\mathrm{Sr}(\mathrm{OH})_{2}\) produces water and \(\mathrm{Sr}(\mathrm{ClO})_2\):\[2\mathrm{HClO}(aq) + \mathrm{Sr}(\mathrm{OH})_2(aq) \longrightarrow \mathrm{Sr}(\mathrm{ClO})_2(aq) + 2\mathrm{H}_2\mathrm{O}(l)\]

Writing net ionic equation for reaction (b)

Remove the spectator ions from the dissociated ionic compounds and write the net reaction as:\[2\mathrm{H}^+(aq) + 2\mathrm{ClO}^-(aq) + \mathrm{Sr}^{2+}(aq) + 2\mathrm{OH}^-(aq) \longrightarrow \mathrm{Sr}^{2+}(aq) + 2\mathrm{ClO}^-(aq) + 2\mathrm{H}_2\mathrm{O}(l)\]The net ionic equation simplifies to:\[2\mathrm{H}^+(aq) + 2\mathrm{OH}^-(aq) \longrightarrow 2\mathrm{H}_2\mathrm{O}(l)\]

Completing and balancing equation for reaction (c)

The reaction between \(\mathrm{Ba}(\mathrm{OH})_2\) and \(\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\) results in formation of water and barium acetate:\[\mathrm{Ba}(\mathrm{OH})_2(aq) + 2\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2(aq) \longrightarrow \mathrm{Ba}(\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2)_2(aq) + 2\mathrm{H}_2\mathrm{O}(l)\]

Writing net ionic equation for reaction (c)

Since all relevant products and reactants dissociate, the net ionic equation is:\[2\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-(aq) + \mathrm{Ba}^{2+}(aq) + 2\mathrm{H}^+(aq) + 2\mathrm{OH}^-(aq) \longrightarrow \mathrm{Ba}^{2+}(aq) + 2\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-(aq) + 2\mathrm{H}_2\mathrm{O}(l)\]Simplifies to:\[2\mathrm{H}^+(aq) + 2\mathrm{OH}^-(aq) \longrightarrow 2\mathrm{H}_2\mathrm{O}(l)\]

Completing and balancing equation for reaction (d)

Combining \(\mathrm{H}_2\mathrm{SO}_4\) with \(\mathrm{KOH}\) yields water and potassium sulfate:\[\mathrm{H}_2\mathrm{SO}_4(aq) + 2\mathrm{KOH}(aq) \longrightarrow \mathrm{K}_2\mathrm{SO}_4(aq) + 2\mathrm{H}_2\mathrm{O}(l)\]

Writing net ionic equation for reaction (d)

For the net ionic equation, only the ions of interest react to form water:\[2\mathrm{H}^+(aq) + 2\mathrm{OH}^-(aq) \longrightarrow 2\mathrm{H}_2\mathrm{O}(l)\]

Key concepts

Net Ionic Equation

In chemistry, reactions can be represented in various ways, and one of the most simplified forms is the net ionic equation. The net ionic equation showcases only the ions and molecules directly involved in the chemical change, removing the spectator ions that do not participate directly. This simplification helps chemists and students focus on the actual transformation happening during the reaction.

To write the net ionic equation, one must first start with the balanced molecular equation. Then, identify all the compounds that are soluble in water and dissociate them into their respective ions. Finally, cancel out the ions that appear on both sides of the equation—these are the spectator ions. For instance, when aluminum hydroxide reacts with hydrochloric acid, the aluminum ions and hydroxide ions react to form water, leaving out the chloride ions as spectators. Hence, the net ionic equation for Reaction (a) becomes:

• \[ \mathrm{Al}(\mathrm{OH})_3(s) + 3\mathrm{H}^+(aq) \longrightarrow \mathrm{Al}^{3+}(aq) + 3\mathrm{H}_2\mathrm{O}(l) \]

This equation emphasizes the essential chemical change while offering a clearer, more concise understanding of the reaction dynamics.

Aqueous Solutions

A key part of understanding chemical reactions involves aqueous solutions—solutions where water is the solvent. Many reactions occur in aqueous solutions due to water's ability to dissolve a wide variety of substances. In such solutions, ionic compounds tend to dissociate into ions, allowing them to interact more freely.

Consider the reaction between hydrochloric acid (\( \mathrm{HCl} \)) and aluminum hydroxide (\( \mathrm{Al(OH)}_3 \)). When these substances are dissolved in water, \( \mathrm{HCl} \) dissociates into \( \mathrm{H}^+ \) and \( \mathrm{Cl}^- \) ions. Similarly, any soluble substance in an aqueous solution will ionize based on its solubility rules and the ions present in the solution. This ionization is crucial for reactions such as neutralization, precipitation, and redox reactions.

Therefore, understanding the behavior of substances in aqueous environments is critical for predicting and explaining reaction processes. Water’s role as a solvent facilitates these reactions through its effect on the dissociation and mobility of ions, ultimately affecting the entire reaction pathway.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is obeyed. Every chemical reaction must have equal numbers of each type of atom on both sides of the equation. This ensures that matter is neither created nor destroyed during a reaction.

To balance a chemical equation, you adjust the coefficients before each compound to obtain the same number of each type of atom on both sides of the equation. Let's look at Reaction (a) involving \( \mathrm{Al(OH)}_3 \) and \( \mathrm{HCl} \). The initial unbalanced equation might have unequal numbers of atoms, but by adjusting coefficients, we achieve a balanced state:

• \[ \mathrm{Al}(\mathrm{OH})_3(s) + 3 \mathrm{HCl}(aq) \longrightarrow \mathrm{AlCl}_3(aq) + 3 \mathrm{H}_2\mathrm{O}(l) \]

With these coefficients, the equation now reflects equal amounts of each atom involved: 1 aluminum atom, 3 chlorine atoms, 3 oxygen atoms, and 6 hydrogen atoms on each side.

Balancing equations is critical in providing the correct stoichiometric proportions of reactants and products. It also aids in determining the amounts of substances used and produced, making this skill indispensable for practical laboratory work and further chemical analyses.