Question

For each of the following reactions, identify the Bronsted-Lowry acids and bases and the conjugate acid-base pairs: (a) \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons\) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(a q)\) (b) \(\mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{HCO}_{3}^{-}(a q)\) (c) \(\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{SO}_{3}^{2-}(a q)\) (d) \(\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{OH}(a q)\)

Short answer

(a) Acids: CH3CO2H, NH4+; Bases: NH3, CH3CO2-. (b) Acids: H3O+, HCO3-; Bases: CO3^2-, H2O. (c) Acids: HSO3-, H3O+; Bases: H2O, SO3^2-. (d) Acids: H2O, H2SO3; Bases: HSO3-, OH-.

Step-by-step solution

Analyze Reaction (a)

In the reaction \( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} + \mathrm{NH}_{3} \rightleftharpoons \mathrm{NH}_{4}^{+} + \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \), the acetic acid \( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \) donates a proton to become \( \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \), making it the Bronsted-Lowry acid. \( \mathrm{NH}_{3} \) accepts a proton to form \( \mathrm{NH}_{4}^{+} \), making it the Bronsted-Lowry base. The conjugate acid-base pairs are \((\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{CH}_{3} \mathrm{CO}_{2}^{-})\) and \((\mathrm{NH}_{3}, \mathrm{NH}_{4}^{+})\).

Analyze Reaction (b)

For the reaction \( \mathrm{CO}_{3}^{2-} + \mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{O} + \mathrm{HCO}_{3}^{-} \), \(\mathrm{H}_{3} \mathrm{O}^{+}\) donates a proton to become \( \mathrm{H}_{2} \mathrm{O} \), making it the Bronsted-Lowry acid, and \( \mathrm{CO}_{3}^{2-} \) accepts a proton to form \( \mathrm{HCO}_{3}^{-} \), making it the Bronsted-Lowry base. The conjugate acid-base pairs are \((\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{H}_{2} \mathrm{O})\) and \((\mathrm{CO}_{3}^{2-}, \mathrm{HCO}_{3}^{-})\).

Analyze Reaction (c)

In the reaction \( \mathrm{HSO}_{3}^{-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+} + \mathrm{SO}_{3}^{2-} \), \( \mathrm{HSO}_{3}^{-} \) donates a proton to \( \mathrm{H}_{2} \mathrm{O} \), thus \( \mathrm{HSO}_{3}^{-} \) is the Bronsted-Lowry acid and \( \mathrm{H}_{2} \mathrm{O} \) acts as the base. The conjugate pairs are \((\mathrm{HSO}_{3}^{-}, \mathrm{SO}_{3}^{2-})\) and \((\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{3} \mathrm{O}^{+})\).

Analyze Reaction (d)

In the reaction \( \mathrm{HSO}_{3}^{-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3} + \mathrm{OH}^{-} \), \( \mathrm{H}_{2} \mathrm{O} \) donates a proton to form \( \mathrm{OH}^{-} \), making it the Bronsted-Lowry acid, and \( \mathrm{HSO}_{3}^{-} \) accepts the proton to form \( \mathrm{H}_{2} \mathrm{SO}_{3} \), making it the base. The conjugate pairs are \((\mathrm{HSO}_{3}^{-}, \mathrm{H}_{2} \mathrm{SO}_{3})\) and \((\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-})\).

Key concepts

Acid-Base Reactions

Acid-base reactions are processes where acids and bases interact to form a chemical equilibrium. According to the Bronsted-Lowry theory, an acid is a substance that donates a proton ( H^+ ), and a base is a substance that accepts a proton. This concept of proton transfer is the crux of acid-base reactions.
In a reversible reaction, such as the ones in the exercises provided, one molecule will act as an acid by donating a proton, while the other will act as a base by accepting that proton. This exchange ultimately leads to the formation of a conjugate acid and a conjugate base, creating conjugate acid-base pairs. In this dynamic equilibrium, both forward and reverse reactions occur simultaneously, and the roles of acids and bases can switch based on the direction of the reaction. Understanding the delicate balance of shifting protons in acid-base reactions is essential for grasping chemical equilibrium.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are directly linked to the Bronsted-Lowry definition. When an acid donates a proton, it transforms into its conjugate base. Similarly, when a base accepts a proton, it becomes its conjugate acid. To illustrate this, consider an acid-base reaction:

• In Reaction (a), the acetic acid (\( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \)) donates a proton, forming its conjugate base \( \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \). \( \mathrm{NH}_{3} \) accepts a proton, becoming its conjugate acid \( \mathrm{NH}_{4}^{+} \).
• In Reaction (b), \( \mathrm{H}_{3} \mathrm{O}^{+} \) loses a proton to become \( \mathrm{H}_{2} \mathrm{O} \), its conjugate base, while \( \mathrm{CO}_{3}^{2-} \) becomes \( \mathrm{HCO}_{3}^{-} \), its conjugate acid.

These relationships are always observed in pairs, and recognizing them can help you track the flow of protons throughout the reaction. Understanding conjugate acid-base pairs is essential for predicting the behavior of substances in different conditions.

Proton Transfer

Proton transfer is the key mechanism at play in Bronsted-Lowry acid-base reactions. It involves the movement of a proton ( H^+ ) from an acid to a base. The ability of a substance to donate or accept a proton is what classifies it as either an acid or a base. Then it's all about what happens after that crucial proton has moved.
In this process, we see that the acid becomes its conjugate base after losing the proton, and the base transforms into its conjugate acid after gaining the proton. This transfer can go backwards and forwards, especially in reversible reactions, and it's the heart of how these reactions work. The dynamic exchange of protons is what allows the chemical reaction to be reversible, and this process helps maintain chemical balance and stability in various systems.

Acids and Bases Identification

Identifying acids and bases in chemical reactions is crucial, and the Bronsted-Lowry theory offers a clear approach to this. You begin by looking for the substance that donates a proton; this is your acid. Conversely, the substance that receives the proton is your base. In the given exercises, identifying these components helps us understand the reactions:

• In Reaction (c), \( \mathrm{HSO}_{3}^{-} \) is identified as the acid as it donates a proton to \( \mathrm{H}_{2} \mathrm{O} \).
• In Reaction (d), \( \mathrm{H}_{2} \mathrm{O} \) acts as the acid, while \( \mathrm{HSO}_{3}^{-} \) serves as the base.

With practice, identifying acids and bases becomes intuitive, making it easier to predict reaction outcomes. This skill is fundamental for any chemistry student to develop, ensuring you can manage and predict chemical interactions effectively.