Question

Write the balanced molecular and net ionic equations for each of the following neutralization reactions: (a) Aqueous acetic acid is neutralized by aqueous potassium hydroxide. (b) Solid chromium(III) hydroxide reacts with nitric acid. (c) Aqueous hypochlorous acid and aqueous calcium hydroxide react.

Short answer

(a) Balanced molecular equation: ${\text{CH}}_3\text{COOH (aq)}+\text{KOH (aq)}\to {\text{KCH}}_3\text{COO (aq)}+{\text{H}}_2\text{O (l)}$. Net ionic equation: ${\text{H}}^{+}(aq)+{\text{OH}}^{-}(aq)\to {\text{H}}_2\text{O (l)}$. (b) Balanced molecular equation: $\text{Cr}(\text{OH}{)}_3\text{(s)}+3{\text{HNO}}_3\text{(aq)}\to \text{Cr}({\text{NO}}_3{)}_3(\text{aq)}+3{\text{H}}_2\text{O (l)}$. Net ionic equation: $3{\text{OH}}^{-}(\text{s})+3{\text{H}}^{+}(\text{aq})\to 3{\text{H}}_2\text{O (l)}$. (c) Balanced molecular equation: $2\text{HClO (aq)}+{\text{Ca(OH)}}_2\text{(aq)}\to {\text{Ca(ClO)}}_2(\text{aq)}+2{\text{H}}_2\text{O (l)}$. Net ionic equation: $2{\text{H}}^{+}(\text{aq})+2{\text{OH}}^{-}(\text{aq})\to 2{\text{H}}_2\text{O (l)}$.

Step-by-step solution

Write the reactants

The reactants for this reaction are aqueous acetic acid and aqueous potassium hydroxide. Thus, we have: ${\text{CH}}_3\text{COOH (aq)}+\text{KOH (aq)}$

Predict the products

When an acid reacts with a base, it forms a salt and water. The salt in this case will be potassium acetate. So, our products will be: ${\text{KCH}}_3\text{COO (aq)}+{\text{H}}_2\text{O (l)}$

Balance the molecular equation

The equation is already balanced, so the balanced molecular equation is: ${\text{CH}}_3\text{COOH (aq)}+\text{KOH (aq)}\to {\text{KCH}}_3\text{COO (aq)}+{\text{H}}_2\text{O (l)}$

Write the net ionic equation

First, we need to write the total ionic equation: ${\text{CH}}_3{\text{COO}}^{-}(aq)+{\text{H}}^{+}(aq)+{\text{K}}^{+}(aq)+{\text{OH}}^{-}(aq)\to {\text{K}}^{+}(aq)+{\text{CH}}_3{\text{COO}}^{-}(aq)+{\text{H}}_2\text{O (l)}$ Now eliminate the spectator ions (K^+ and CH3COO^-): ${\text{H}}^{+}(aq)+{\text{OH}}^{-}(aq)\to {\text{H}}_2\text{O (l)}$ This is our net ionic equation for the neutralization of acetic acid by potassium hydroxide. (b)

Write the reactants

The reactants for this reaction are solid chromium(III) hydroxide and nitric acid. Thus, we have: $\text{Cr}(\text{OH}{)}_3\text{(s)}+{\text{HNO}}_3\text{(aq)}$

Predict the products

When an acid reacts with a metal hydroxide, it forms a salt and water. The salt in this case will be chromium(III) nitrate. So, our products will be: $\text{Cr}({\text{NO}}_3{)}_3(\text{aq)}+{\text{H}}_2\text{O (l)}$

Balance the molecular equation

To balance this equation, we need to have 3 nitrate ions for each chromium ion and 3 water molecules for every 3 hydroxide ions: $\text{Cr}(\text{OH}{)}_3\text{(s)}+3{\text{HNO}}_3\text{(aq)}\to \text{Cr}({\text{NO}}_3{)}_3(\text{aq)}+3{\text{H}}_2\text{O (l)}$

Write the net ionic equation

First, we need to write the total ionic equation: ${\text{Cr}}^{3+}(\text{s})+3{\text{OH}}^{-}(\text{s})+3{\text{H}}^{+}(\text{aq})+3{\text{NO}}_3^{-}(\text{aq})\to {\text{Cr}}^{3+}(\text{aq})+3{\text{NO}}_3^{-}(\text{aq})+3{\text{H}}_2\text{O (l)}$ Now eliminate the spectator ions (Cr^3+ and NO3^-): $3{\text{OH}}^{-}(\text{s})+3{\text{H}}^{+}(\text{aq})\to 3{\text{H}}_2\text{O (l)}$ This is our net ionic equation for the reaction between chromium(III) hydroxide and nitric acid. (c)

Write the reactants

The reactants for this reaction are aqueous hypochlorous acid and aqueous calcium hydroxide. Thus, we have: $\text{HClO (aq)}+{\text{Ca(OH)}}_2\text{(aq)}$

Predict the products

When an acid reacts with a base, it forms a salt and water. The salt in this case will be calcium hypochlorite. So, our products will be: ${\text{Ca(ClO)}}_2(\text{aq)}+{\text{H}}_2\text{O (l)}$

Balance the molecular equation

To balance this equation, we need 2 hypochlorous acid molecules for every calcium ion: $2\text{HClO (aq)}+{\text{Ca(OH)}}_2\text{(aq)}\to {\text{Ca(ClO)}}_2(\text{aq)}+2{\text{H}}_2\text{O (l)}$

Write the net ionic equation

First, we need to write the total ionic equation: $2{\text{H}}^{+}(\text{aq})+2{\text{ClO}}^{-}(\text{aq})+{\text{Ca}}^{2+}(\text{aq})+2{\text{OH}}^{-}(\text{aq})\to {\text{Ca}}^{2+}(\text{aq})+2{\text{ClO}}^{-}(\text{aq})+2{\text{H}}_2\text{O (l)}$ Now eliminate the spectator ions (Ca^2+ and ClO^-): $2{\text{H}}^{+}(\text{aq})+2{\text{OH}}^{-}(\text{aq})\to 2{\text{H}}_2\text{O (l)}$ This is our net ionic equation for the reaction between hypochlorous acid and calcium hydroxide.

Key concepts

Acid-Base Reactions

Acid-base reactions are a fundamental type of chemical reaction where an acid reacts with a base to produce water and a salt. These reactions are central to the field of chemistry due to their prevalence and importance in various natural and industrial processes. When an acid and a base come together, they neutralize each other. The acid typically provides a proton ($extrm{H}^{+}$), while the base provides a hydroxide ion ($extrm{OH}^{-}$). When these ions meet, they form water ($extrm{H}_{2}extrmO$), a neutral compound.

• Acids like acetic acid ($extrm{CH}_{3}extrmCOOH$) or nitric acid ($extrm{HNO}_3$) donate protons.
• Bases such as potassium hydroxide ($extrmKOH$) and calcium hydroxide ($extrm{Ca(OH)}_2$) provide hydroxide ions.

In a neutralization reaction, the completion of the reaction is often signaled by the absence of either the acid or the base reactants, resulting in the formation of water and a salt.

Balanced Equations

Balanced chemical equations are vital because they ensure that the Law of Conservation of Mass is upheld in a reaction. This law states that matter cannot be created or destroyed. Thus, a balanced equation ensures that the number of each type of atom on the reactant side equals the number on the product side. For instance, when balancing the reaction between acetic acid and potassium hydroxide, you must have equal moles of reactants and products:$${\text{CH}}_3\text{COOH (aq)}+\text{KOH (aq)}\to {\text{KCH}}_3\text{COO (aq)}+{\text{H}}_2\text{O (l)}$$Key points in balancing:

• Count the number of atoms for each element in reactants and products.
• Adjust coefficients to get the same number of each type of atom on both sides.
• Start balancing more complex molecules first, then move to simpler ones.

Balanced equations not only provide the correct stoichiometry but also ensure that reactions are properly analyzed for both theoretical and practical purposes.

Net Ionic Equations

Net ionic equations focus specifically on the components of the reaction that undergo a chemical change. This type of equation highlights the substance(s) that are actually involved in the chemical change, thus removing the spectator ions. For the neutralization of acetic acid by potassium hydroxide, the net ionic equation simplifies everything to the basic acid-base interaction:$${\text{H}}^{+}(\text{aq})+{\text{OH}}^{-}(\text{aq})\to {\text{H}}_2\text{O (l)}$$This equation shows the hydrogen ion from the acid reacting with the hydroxide ion from the base to form water. Other products, such as potassium acetate in this reaction, are not involved in the key chemical transformation involved in neutralization.Net ionic equations are useful because:

• They simplify the equation to the essence of the chemical reaction.
• Help concentrate on the reaction mechanism.
• Remove unnecessary components, focusing attention on what chemically changes.

Spectator Ions

Spectator ions are ions that do not directly participate in the chemical reaction. They remain in solution unchanged and can be found on both the reactants and products side of the equation. These ions are relevant in balancing a complete ionic equation but disappear in the net ionic equation. For example, in the reaction between acetic acid and potassium hydroxide, the potassium ion (${\text{K}}^{+}$) and the acetate ion (${\text{CH}}_3{\text{COO}}^{-}$) do not participate in the neutralization process directly. Thus, they are eliminated when writing the net ionic equation.Understanding spectator ions helps because:

• They demonstrate that not all ions in a reaction are reactive.
• Assist in cleaning up the equation to focus on the active components.
• Provide insights into solubility and ion behavior in aqueous solutions.

Recognizing these ions can clarify both the mechanism and stoichiometry involved in a reaction, leaving behind the core chemistry of interest.