Problem 4 - Chemistry and Chemical Reactivity

Question

What are the products of each of the following acid-base reactions? Indicate the acid and its conjugate base, and the base and its conjugate acid. (a) \(\mathrm{HClO}_{4}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) (b) \(\mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) (c) \(\mathrm{HCO}_{3}^{-}+\mathrm{OH}^{-} \longrightarrow\)

Step-by-Step Solution

Answer

(a) \( \mathrm{ClO}_{4}^{-} + \mathrm{H}_{3} \mathrm{O}^{+} \); (b) \( \mathrm{NH}_{3} + \mathrm{H}_{3} \mathrm{O}^{+} \); (c) \( \mathrm{CO}_{3}^{2-} + \mathrm{H}_{2} \mathrm{O} \).

Step 1: Identify the Reactants

In each reaction, identify the acid and the base. For part (a), \( \mathrm{HClO}_{4} \) is the acid and \( \mathrm{H}_{2} \mathrm{O} \) acts as the base. For part (b), \( \mathrm{NH}_{4}^{+} \) is the acid and \( \mathrm{H}_{2} \mathrm{O} \) acts as the base. For part (c), \( \mathrm{HCO}_{3}^{-} \) acts as the acid and \( \mathrm{OH}^{-} \) acts as the base.

Step 2: Determine the Products

Find out what products each reactant pair will form. For part (a), the proton from \( \mathrm{HClO}_{4} \) is transferred to \( \mathrm{H}_{2} \mathrm{O} \), forming \( \mathrm{ClO}_{4}^{-} \) and \( \mathrm{H}_{3} \mathrm{O}^{+} \). For part (b), \( \mathrm{NH}_{4}^{+} \) donates a proton to water, forming \( \mathrm{NH}_{3} \) and \( \mathrm{H}_{3} \mathrm{O}^{+} \). For part (c), \( \mathrm{HCO}_{3}^{-} \) donates a proton to \( \mathrm{OH}^{-} \), forming \( \mathrm{CO}_{3}^{2-} \) and \( \mathrm{H}_{2} \mathrm{O} \).

Step 3: Identify Conjugate Acids and Bases

In part (a), \( \mathrm{HClO}_{4} \) is the acid and its conjugate base is \( \mathrm{ClO}_{4}^{-} \); \( \mathrm{H}_{2} \mathrm{O} \) is the base and its conjugate acid is \( \mathrm{H}_{3} \mathrm{O}^{+} \). In part (b), \( \mathrm{NH}_{4}^{+} \) is the acid and its conjugate base is \( \mathrm{NH}_{3} \); \( \mathrm{H}_{2} \mathrm{O} \) is the base and its conjugate acid is \( \mathrm{H}_{3} \mathrm{O}^{+} \). In part (c), \( \mathrm{HCO}_{3}^{-} \) is the acid and its conjugate base is \( \mathrm{CO}_{3}^{2-} \); \( \mathrm{OH}^{-} \) is the base and its conjugate acid is \( \mathrm{H}_{2} \mathrm{O} \).

Step 4: Write the Completed Reactions

For part (a), the reaction is \( \mathrm{HClO}_{4} + \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ClO}_{4}^{-} + \mathrm{H}_{3} \mathrm{O}^{+} \). For part (b), it is \( \mathrm{NH}_{4}^{+} + \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NH}_{3} + \mathrm{H}_{3} \mathrm{O}^{+} \). For part (c), it is \( \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-} \longrightarrow \mathrm{CO}_{3}^{2-} + \mathrm{H}_{2} \mathrm{O} \).

Key Concepts

Conjugate Acid-Base Pairs

In acid-base chemistry, a key concept is the idea of conjugate acid-base pairs. These pairs consist of two species that transform into each other by the gain or loss of a proton (\(\text{H}^+\)). Let's break it down with examples from the reactions provided.

• In reaction (a), \(\text{HClO}_4\) is the acid because it donates a proton to \(\text{H}_2\text{O}\), forming the conjugate base \(\text{ClO}_4^-\).
• \(\text{H}_2\text{O}\) acts as the base and turns into the conjugate acid \(\text{H}_3\text{O}^+\).
• In reaction (b), \(\text{NH}_4^+\) donates a proton and becomes \(\text{NH}_3\), its conjugate base.
• In reaction (c), \(\text{HCO}_3^-\) gives a proton to \(\text{OH}^-\), resulting in \(\text{CO}_3^{2-}\), its conjugate base.

Understanding conjugate pairs helps in predicting the direction of proton transfer reactions, as they maintain a balance between acids and bases.

Proton Transfer

Proton transfer is the core of acid-base reactions. It's what differentiates acids from bases in the Brønsted-Lowry theory.
For each reaction:

• In part (a), the proton (...\(\text{H}^+\)...) moves from \(\text{HClO}_4\) to \(\text{H}_2\text{O}\), creating hydronium ion \(\text{H}_3\text{O}^+\) and perchlorate ion \(\text{ClO}_4^-\).
• In reaction (b), \(\text{NH}_4^+\) loses a proton to \(\text{H}_2\text{O}\), forming ammonia \(\text{NH}_3\).
• In part (c), \(\text{HCO}_3^-\) transfers its proton to \(\text{OH}^-\), generating water \(\text{H}_2\text{O}\) and carbonate ions \(\text{CO}_3^{2-}\).

Proton transfer reactions highlight the roles of reactants as either proton donors (acids) or proton acceptors (bases). This step is crucial to understanding the nature of products formed in acid-base reactions.

Reaction Products

The final products of acid-base reactions are determined by proton transfer and the resulting conjugate pairs. Let’s look at what forms in each case.

• For reaction (a), the products are hydronium ion \(\text{H}_3\text{O}^+\) and perchlorate ion \(\text{ClO}_4^-\). The initial acid \(\text{HClO}_4\) gives away its proton to \(\text{H}_2\text{O}\).
• In reaction (b), ammonia \(\text{NH}_3\) and hydronium ion \(\text{H}_3\text{O}^+\) emerge as products. Here, \(\text{NH}_4^+\), as the acid, loses a proton.
• The products for reaction (c) are water \(\text{H}_2\text{O}\) and carbonate ion \(\text{CO}_3^{2-}\). The bicarbonate ion \(\text{HCO}_3^-\) gives a proton to hydroxide \(\text{OH}^-\).

Knowing the products of these reactions allows us to understand the changes that occur during acid-base interactions. This understanding helps predict outcomes in a variety of chemical environments.