Problem 43 - Chemistry The Molecular Science

Question

Identify the acid and base used to form these salts, and write the overall neutralization reaction in both complete and net ionic form. (a) \(\mathrm{NaCH}_{3} \mathrm{COO}\) (b) \(\mathrm{CaCl}_{2}\) (c) \(\mathrm{LiBr}\) (d) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\)

Step-by-Step Solution

Answer

Acid-base pairs are identified, and reactions are written for each salt: \( \mathrm{NaOH} + \mathrm{CH}_3 \mathrm{COOH} \), \( \mathrm{Ca(OH)_2} + \mathrm{HCl} \), \( \mathrm{LiOH} + \mathrm{HBr} \), and \( \mathrm{Ba(OH)_2} + \mathrm{HNO}_3 \). Common net ionic reactions involve water formation.

Step 1: Identify the Ions in the Salt

For each salt, identify the cation and the anion. This step will help determine the acid and base used to form the salt.- (a) For \( \mathrm{NaCH}_3 \mathrm{COO} \), the cation is \( \mathrm{Na}^+ \) and the anion is \( \mathrm{CH}_3 \mathrm{COO}^- \) (acetate ion).- (b) For \( \mathrm{CaCl}_2 \), the cation is \( \mathrm{Ca}^{2+} \) and the anions are \( \mathrm{Cl}^- \).- (c) For \( \mathrm{LiBr} \), the cation is \( \mathrm{Li}^+ \) and the anion is \( \mathrm{Br}^- \).- (d) For \( \mathrm{Ba(NO}_3)_2 \), the cation is \( \mathrm{Ba}^{2+} \) and the anion is \( \mathrm{NO}_3^- \).

Step 2: Associate Ions with Acid and Base

Determine which acids and bases contribute to the formation of the salt by neutralization.- (a) \( \mathrm{NaCH}_3 \mathrm{COO} \): Sodium cation from \( \mathrm{NaOH} \) (base) and acetate anion from \( \mathrm{CH}_3 \mathrm{COOH} \) (acid).- (b) \( \mathrm{CaCl}_2 \): Calcium cation from \( \mathrm{Ca(OH)_2} \) (base) and chloride anions from \( \mathrm{HCl} \) (acid).- (c) \( \mathrm{LiBr} \): Lithium cation from \( \mathrm{LiOH} \) (base) and bromide anion from \( \mathrm{HBr} \) (acid).- (d) \( \mathrm{Ba(NO}_3)_2 \): Barium cation from \( \mathrm{Ba(OH)_2} \) (base) and nitrate anions from \( \mathrm{HNO}_3 \) (acid).

Step 3: Write the Complete Neutralization Reaction

Write the balanced chemical reaction for the neutralization where the acid and base react to form the given salt and water.- (a) \( \mathrm{NaOH} + \mathrm{CH}_3 \mathrm{COOH} \rightarrow \mathrm{NaCH}_3 \mathrm{COO} + \mathrm{H_2O} \)- (b) \( \mathrm{Ca(OH)_2} + 2\mathrm{HCl} \rightarrow \mathrm{CaCl}_2 + 2\mathrm{H_2O} \)- (c) \( \mathrm{LiOH} + \mathrm{HBr} \rightarrow \mathrm{LiBr} + \mathrm{H_2O} \)- (d) \( \mathrm{Ba(OH)_2} + 2\mathrm{HNO_3} \rightarrow \mathrm{Ba(NO}_3)_2 + 2\mathrm{H_2O} \)

Step 4: Write the Net Ionic Equation

Identify the spectator ions and eliminate them to write the net ionic equation for each neutralization reaction.- (a) \( \mathrm{OH}^- + \mathrm{CH}_3 \mathrm{COOH} \rightarrow \mathrm{CH}_3 \mathrm{COO}^- + \mathrm{H_2O} \)- (b) \( 2\mathrm{OH}^- + 2\mathrm{H}^+ \rightarrow 2\mathrm{H_2O} \) (Note: The \( \mathrm{Ca}^{2+} \) and \( \mathrm{Cl}^- \) ions are spectators)- (c) \( \mathrm{OH}^- + \mathrm{H}^+ \rightarrow \mathrm{H_2O} \) (Note: The \( \mathrm{Li}^+ \) and \( \mathrm{Br}^- \) ions are spectators)- (d) \( 2\mathrm{OH}^- + 2\mathrm{H}^+ \rightarrow 2\mathrm{H_2O} \) (Note: The \( \mathrm{Ba}^{2+} \) and \( \mathrm{NO}_3^- \) ions are spectators)

Key Concepts

Acid-Base Chemistry

Acid-base chemistry is fundamental in understanding how substances neutralize each other through reactions. An acid is typically a substance that can donate a proton (\(\text{H}^+\)) which, when dissolved in water, yields an oppositely charged anion. Conversely, a base can accept a proton and often donates a hydroxide ion (\(\text{OH}^-\)) in aqueous solution. These reactions are called neutralizations because they result in the formation of water and a salt, making the solution less acidic or basic. In simple terms, when an acid and a base react, they "cancel out" the effects of each other by forming water.

Taking sodium acetate (\( \text{NaCH}_3\text{COO} \)) as an example, it forms from the neutralization of acetic acid (\( \text{CH}_3\text{COOH} \)) and sodium hydroxide (\( \text{NaOH} \)). Acetic acid provides the anion acetate, while sodium hydroxide provides the sodium cation. Understanding these components helps in predicting how acids and bases will interact.

Common terms associated with these reactions include:

• **Acid:** Proton (\( \text{H}^+ \)) donor.
• **Base:** Proton (\( \text{H}^+ \)) acceptor.
• **Neutralization:** Formation of water and a salt from acids and bases.

This basic concept is powerful in driving not only chemical reactions but also biological processes and industrial applications.

Ions Identification

To successfully understand neutralization reactions, ions identification is a crucial first step. Each salt is comprised of different ions, typically a positively charged cation and a negatively charged anion. By identifying these ions, we can trace back to the acids and bases involved in forming the salt via a neutralization reaction.

Let's look at the salt calcium chloride (\( \text{CaCl}_2 \)):

• The cation is calcium (\( \text{Ca}^{2+} \)), typically coming from a base like calcium hydroxide (\( \text{Ca(OH)}_2 \)).
• The anion is chloride (\( \text{Cl}^- \)), which originates from an acid like hydrochloric acid (\( \text{HCl} \)).

By breaking down these components, we can better understand the reaction pathway and the related chemistry.

This process is not only limited to theoretical exercises but is also applied in industries for applications like water treatment, where specific ions need to be identified and managed. Mastering ions identification is an invaluable tool in a chemist's toolkit.

Net Ionic Equations

Net ionic equations are crucial for simplifying acid-base reactions to focus on the actual chemical changes. These equations show only the ions and molecules directly involved in the chemical reaction, excluding spectator ions that do not participate in the reaction. Simplifying reactions in this way helps clarify the chemical processes occurring.

For example, in the neutralization of lithium bromide (\( \text{LiBr} \)), spectating ions such as lithium (\( \text{Li}^+ \)) and bromide (\( \text{Br}^- \)) are excluded to focus on what truly changes. The net ionic equation simplifies to:

• \( \text{OH}^- + \text{H}^+ \rightarrow \text{H}_2\text{O} \).

This equation highlights the essential process of water formation from hydroxide and hydrogen ions, which is the core of neutralization reactions.

Understanding net ionic equations enables learners and professionals alike to streamline their focus on active components in a reaction, which is especially beneficial in complex systems or when predicting reaction outcomes.