Problem 38 - Chemistry An Atoms First Approach

Question

For each of the following aqueous reactions, identify the acid, the base, the conjugate base, and the conjugate acid. a. \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}\) b. \(\mathrm{H}_{2} \mathrm{O}+\mathrm{HONH}_{3}^{+} \rightleftharpoons \mathrm{HONH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+}\) c. HOCl \(+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \rightleftharpoons \mathrm{OCl}^{-}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)

Step-by-Step Solution

Answer

a. Acid: \(\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\), Base: \(\mathrm{H}_{2}\mathrm{O}\), Conjugate base: \(\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}(\mathrm{OH})^{2+}\), Conjugate acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\) b. Acid: \(\mathrm{HONH}_{3}^{+}\), Base: \(\mathrm{H}_{2}\mathrm{O}\), Conjugate base: \(\mathrm{HONH}_{2}\), Conjugate acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\) c. Acid: HOCl, Base: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), Conjugate base: \(\mathrm{OCl}^{-}\), Conjugate acid: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)

Step 1: Identify the acid and the base

In this reaction, the acid donates a proton to the base, forming the products. We see that \(\mathrm{H}_{3}\mathrm{O}^{+}\) has an extra proton in comparison to \(\mathrm{H}_{2}\mathrm{O}\), so we can identify the substances as follows: - Acid: \(\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\) - Base: \(\mathrm{H}_{2}\mathrm{O}\)

Step 2: Identify the conjugate base and the conjugate acid

The conjugate base is formed from the acid after it loses a proton, and the conjugate acid is formed from the base after it gains a proton. Therefore, we have: - Conjugate base: \(\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}(\mathrm{OH})^{2+}\) - Conjugate acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\) b. $\mathrm{H}_{2} \mathrm{O}+\mathrm{HONH}_{3}^{+} \rightleftharpoons \mathrm{HONH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+}$

Step 1: Identify the acid and the base

Again, we look for the substances that donate and accept protons in this reaction. Here, \(\mathrm{HONH}_{3}^{+}\) donates the proton to \(\mathrm{H}_{2}\mathrm{O}\), leading to the formation of the products. Therefore, we can identify the substances as: - Acid: \(\mathrm{HONH}_{3}^{+}\) - Base: \(\mathrm{H}_{2}\mathrm{O}\)

Step 2: Identify the conjugate base and the conjugate acid

As before, the conjugate base is formed from the acid after it loses a proton, and the conjugate acid is formed from the base after it gains a proton. In this case: - Conjugate base: \(\mathrm{HONH}_{2}\) - Conjugate acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\) c. HOCl $+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \rightleftharpoons \mathrm{OCl}^{-}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}$

Step 1: Identify the acid and the base

In this reaction, we can see that \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) has an extra proton in comparison to \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), so it appears that HOCl donates a proton to \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\). This allows us to identify the substances as: - Acid: HOCl - Base: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\)

Step 2: Identify the conjugate base and the conjugate acid

Following our definitions, the conjugate base is formed from the acid after it loses a proton, and the conjugate acid is formed from the base after it gains a proton. Thus, we have: - Conjugate base: \(\mathrm{OCl}^{-}\) - Conjugate acid: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)

Key Concepts

Conjugate Acid

In every acid-base reaction, the conjugate acid is a crucial component. A conjugate acid is the species that forms when a base gains a proton (H⁺). This means it is the counterpart of the base in the reaction pair. Let's look at these reactions for better understanding:

• In reaction (a), water (\(\mathrm{H}_{2} \mathrm{O}\)) gains a proton from \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\), forming \(\mathrm{H}_{3} \mathrm{O}^{+}\), which acts as the conjugate acid.
• For reaction (b), water again acts as a base and gains a proton from \(\mathrm{HONH}_{3}^{+}\), forming \(\mathrm{H}_{3} \mathrm{O}^{+}\). Thus, \(\mathrm{H}_{3} \mathrm{O}^{+}\) is the conjugate acid again.
• In reaction (c), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) gains a proton from HOCl, turning into \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\), which is the conjugate acid here.

Conjugate acids and their reactions are essential for maintaining chemical balance and understanding how molecules interact.

Conjugate Base

Opposite to conjugate acids, conjugate bases are the species formed when an acid loses a proton. They play a vital role in the equilibrium of acid-base reactions. Here’s how they appear in the given reactions:

• In reaction (a), \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) donates a proton and transforms into \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}\), the conjugate base.
• During reaction (b), \(\mathrm{HONH}_{3}^{+}\) donates a proton to \(\mathrm{H}_{2} \mathrm{O}\), creating \(\mathrm{HONH}_{2}\) as the conjugate base.
• For reaction (c), HOCl loses a proton, forming \(\mathrm{OCl}^{-}\), which is the conjugate base here.

Recognizing the conjugate base helps us understand the transformations that occur during the reaction, showing how acids are converted to their base forms through proton loss.

Proton Transfer

In the realm of acid-base reactions, proton transfer is the central process. It involves the movement of protons from acids to bases. This exchange is what defines these reactions and leads to the formation of conjugate acid-base pairs. Here's how this operates in our examples:

• In reaction (a), \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) transfers a proton to \(\mathrm{H}_{2} \mathrm{O}\)
• In reaction (b), \(\mathrm{HONH}_{3}^{+}\) gives a proton to water, \(\mathrm{H}_{2} \mathrm{O}\)
• And in reaction (c), HOCl transfers its proton to \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\).

Each reaction showcases one molecule donating a proton and another accepting it. This defines the roles of acids and bases in these reactions. Understanding proton transfer helps students see how molecular changes occur and how equilibrium is reached in chemical reactions.