Question

In each of the following chemical equations, identify the conjugate acid-base pairs. a. \(\mathrm{CH}_{3} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-}\) b. \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{NH}_{4}^{+}\) c. \(\mathrm{HF}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{F}^{-}+\mathrm{NH}_{4}^{+}\)

Short answer

The conjugate acid-base pairs for each equation are as follows: a. \((\mathrm{CH}_{3} \mathrm{NH}_{2}, \mathrm{CH}_{3} \mathrm{NH}_{3}^{+})\) and \((\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-})\) b. \((\mathrm{CH}_{3} \mathrm{COOH}, \mathrm{CH}_{3} \mathrm{COO}^{-})\) and \((\mathrm{NH}_{3}, \mathrm{NH}_{4}^{+})\) c. \((\mathrm{HF}, \mathrm{F}^{-})\) and \((\mathrm{NH}_{3}, \mathrm{NH}_{4}^{+})\)

Step-by-step solution

Identify the acid and base

In the given equation, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) is donating a proton (H+) to \(\mathrm{H}_{2} \mathrm{O}\), so \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) is the acid and \(\mathrm{H}_{2} \mathrm{O}\) is the base.

Identify conjugate acid and conjugate base

After donating the proton, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) forms \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\), which is the conjugate acid, and \(\mathrm{H}_{2} \mathrm{O}\) forms \(\mathrm{OH}^{-}\), which is the conjugate base.

Final conjugate acid-base pairs

Thus, the conjugate acid-base pairs for Equation 1 are: \((\mathrm{CH}_{3} \mathrm{NH}_{2}, \mathrm{CH}_{3} \mathrm{NH}_{3}^{+})\) and \((\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-})\). #b. Equation 2#

Identify the acid and base

In the given equation, \(\mathrm{CH}_{3} \mathrm{COOH}\) is donating a proton (H+) to \(\mathrm{NH}_{3}\), so \(\mathrm{CH}_{3} \mathrm{COOH}\) is the acid and \(\mathrm{NH}_{3}\) is the base.

Identify conjugate acid and conjugate base

After donating the proton, \(\mathrm{CH}_{3} \mathrm{COOH}\) forms \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), which is the conjugate base, and \(\mathrm{NH}_{3}\) forms \(\mathrm{NH}_{4}^{+}\), which is the conjugate acid.

Final conjugate acid-base pairs

Thus, the conjugate acid-base pairs for Equation 2 are: \((\mathrm{CH}_{3} \mathrm{COOH}, \mathrm{CH}_{3} \mathrm{COO}^{-})\) and \((\mathrm{NH}_{3}, \mathrm{NH}_{4}^{+})\). #c. Equation 3#

Identify the acid and base

In the given equation, \(\mathrm{HF}\) is donating a proton (H+) to \(\mathrm{NH}_{3}\), so \(\mathrm{HF}\) is the acid and \(\mathrm{NH}_{3}\) is the base.

Identify conjugate acid and conjugate base

After donating the proton, \(\mathrm{HF}\) forms \(\mathrm{F}^{-}\), which is the conjugate base, and \(\mathrm{NH}_{3}\) forms \(\mathrm{NH}_{4}^{+}\), which is the conjugate acid.

Final conjugate acid-base pairs

Thus, the conjugate acid-base pairs for Equation 3 are: \((\mathrm{HF}, \mathrm{F}^{-})\) and \((\mathrm{NH}_{3}, \mathrm{NH}_{4}^{+})\).

Key concepts

Acid-Base Reactions

Acid-base reactions are fascinating processes where acids and bases interact, often resulting in the transfer of protons (H⁺ ions). In simple terms, an acid is a substance that can donate a proton, while a base is a substance that can accept a proton. These reactions occur because molecules tend to move towards lower energy states, and proton transfer can help them achieve that.

• Acid: Donates a proton (H⁺)
• Base: Accepts a proton (H⁺)

When an acid loses a proton, it forms its conjugate base, which can potentially accept a proton back, if the conditions allow it. Similarly, when a base gains a proton, it forms its conjugate acid, which can release that proton under certain conditions. This dual relationship creates what are called conjugate acid-base pairs. These pairs are crucial in understanding the mechanism of acid-base reactions and are key to predicting the behavior of substances in a chemical reaction.
Conjugate acid-base pairs elegantly explain the reversible nature of many acid-base reactions, where the products can revert back to the original reactants under different conditions.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show how reactants are transformed into products. In the context of acid-base reactions, chemical equations specifically illustrate the process of proton transfer between molecules.
Each chemical equation should be balanced, meaning the number of atoms for each element should be equal on both sides of the equation. This balancing act also includes ensuring that the charges are balanced. In acid-base reactions, it's crucial to account for the transfer of protons, which also affects the balance of charge.
Let's look at an example:
In the equation \[\mathrm{CH}_{3}\mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CH}_{3}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-},\]\(\mathrm{CH}_{3}\mathrm{NH}_{2}\) is the acid donating a proton to \(\mathrm{H}_{2}\mathrm{O}\), resulting in \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) and \(\mathrm{OH}^{-}\).

• The left side contains reactants: \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\).
• The right side contains products: \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) and \(\mathrm{OH}^{-}\).

Understanding these transformations helps students grasp how chemical equations represent real-world chemical interactions. Being able to interpret these equations is crucial for predicting the outcomes of chemical reactions.

Proton Transfer

Proton transfer is the core mechanism driving acid-base reactions. This process involves the movement of a proton from an acid to a base. It is a fundamental concept in chemistry, especially in the study of reactions involving acids and bases.
In a typical proton transfer, the acid donates a hydrogen ion to the base. This ion is essentially just a proton, as it lacks any electrons. The receiving base accepts this proton, completing the transfer.

• Proton donor: Typically the acid in the reaction.
• Proton acceptor: Typically the base in the reaction.

Taking the example from the original problem, in the equation \[\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{NH}_{3} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{NH}_{4}^{+},\]\(\mathrm{CH}_{3}\mathrm{COOH}\) donates a proton to \(\mathrm{NH}_{3}\). As a result, \(\mathrm{CH}_{3}\mathrm{COOH}\) transforms into \(\mathrm{CH}_{3}\mathrm{COO}^{-}\), and \(\mathrm{NH}_{3}\) becomes \(\mathrm{NH}_{4}^{+}\).
This transfer is critical in many physiological processes and industrial applications. Recognizing how proton transfer works allows students not only to understand chemical equations but also to predict how molecules will behave in different environments.