Problem 19 - Chemistry Matter and Change

Question

Identify the conjugate acid-base pairs in each equation. a. \(H C O O H(a q)+H_{2} O(1) \rightleftharpoons H C O O^{-}(a q)+H_{3} O^{+}(a q)\) b. \(\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{NH}_{4}+(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)

Step-by-Step Solution

Answer

In each given equation, the conjugate acid-base pairs are: a. \(HCOOH(aq)\) and \(HCOO^{-}(aq)\); \(H_{2}O(l)\) and \(H_{3}O^{+}(aq)\) b. \(NH_{3}(aq)\) and \(NH_{4}^{+}(aq)\); \(H_{2}O(l)\) and \(OH^{-}(aq)\)

Step 1: Identify the Acids and Bases in the Equations

Firstly, we need to identify the acids and bases present in each equation. Keep in mind that acids donate protons and bases accept protons. For Equation a: Acid: \(HCOOH(aq)\) Base: \(H_2O(l)\) For Equation b: Acid: \(NH_{3}(aq)\) Base: \(H_{2}O(l)\)

Step 2: Identify the Conjugate Acids and Bases

In the next step, we will determine the conjugate acids and bases formed after the proton transfer occurs. For Equation a: After the acid \(HCOOH(aq)\) donates a proton to \(H_{2}O(l)\), it forms the conjugate base \(HCOO^{-}(aq)\), and water forms its conjugate acid \(H_{3}O^{+}(aq)\) by accepting a proton. For Equation b: When the base \(NH_{3}(aq)\) accepts a proton from \(H_{2}O(l)\), it forms its conjugate acid \(NH_{4}^{+}(aq)\) while water forms its conjugate base \(OH^{-}(aq)\) by donating a proton.

Step 3: Identify Conjugate Acid-Base Pairs

Now, we can identify the conjugate acid-base pairs for both equations. For Equation a: The conjugate acid-base pairs are 1. \(HCOOH(aq)\) and \(HCOO^{-}(aq)\) 2. \(H_{2}O(l)\) and \(H_{3}O^{+}(aq)\) For Equation b: The conjugate acid-base pairs are 1. \(NH_{3}(aq)\) and \(NH_{4}^{+}(aq)\) 2. \(H_{2}O(l)\) and \(OH^{-}(aq)\)

Key Concepts

Proton Transfer

In an acid-base reaction, proton transfer is a fundamental process. A proton, which is a hydrogen ion ( ext{H}^+), is moved from one molecule to another. This transfer is what defines substances as acids or bases.

According to the Bronsted-Lowry theory, acids are proton donors and bases are proton acceptors. For example, in equation a, the acid

• \(HCOOH(aq)\) donates a proton to water, leading to \(HCOO^{-}(aq)\) and \(H_3O^{+}(aq)\).
• In equation b, \(NH_3(aq)\) accepts a proton from \(H_2O(l)\), forming \(NH_4^{+}(aq)\) and \(OH^{-}(aq)\).

This simple proton movement is the essence of these reactions.

Acid-Base Reactions

Acid-base reactions involve acids and bases reacting to form their conjugates. Each acid-base pair is linked through the transfer of a proton.

For instance, in equation a:

• The acid \(HCOOH(aq)\) transfers a proton to water, forming \(H_3O^{+}(aq)\) and the conjugate base \(HCOO^{-}(aq)\).
• In equation b, \(NH_3(aq)\) acts as a base, receiving a proton and creating \(NH_4^{+}(aq)\) and \(OH^{-}(aq)\).

The essence of these reactions is the transformation of acid to base and vice versa, driven by proton exchange.

Bronsted-Lowry Theory

The Bronsted-Lowry theory offers a way to understand acids and bases beyond traditional definitions. It classifies:

• Acids as substances that donate protons and
• Bases as substances that accept protons.

In application, this theory explains:

• How \(HCOOH(aq)\) (a Bronsted-Lowry acid) gives a proton to \(H_2O(l)\), generating \(HCOO^{-}(aq)\) and \(H_3O^{+}(aq)\).
• How \(NH_3(aq)\) (a Bronsted-Lowry base) accepts a proton from water, resulting in \(NH_4^{+}(aq)\) and \(OH^{-}(aq)\).

This theory enhances our understanding by focusing on proton exchange rather than other chemical characteristics.

Equilibrium Reactions

Equilibrium in acid-base reactions indicates the state where forward and reverse reactions occur at equal rates.

Take equation a for example:

• The reaction \(HCOOH(aq) + H_2O(l) ightleftharpoons HCOO^{-}(aq) + H_3O^{+}(aq)\) involves a balance between the donating and accepting of protons.
• Similarly, equation b achieves equilibrium with \(NH_3(aq) + H_2O(l) ightleftharpoons NH_4^{+}(aq) + OH^{-}(aq)\)

In equilibrium, the concentration of reactants and products remains stable, showing a dynamic balance in the system. It's a key concept for understanding how reactions stabilize.