Problem 20 - Chemistry: The Central Science

Question

(a) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l)\). (b) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l)\). (c) What is the conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) ? What is its conjugate base?

Step-by-Step Solution

Answer

(a) When \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base, the equation for the reaction is: A\(^-\) + H\(_2\)O (l) \(\rightleftharpoons\) HA (aq) + OH\(^-\) (aq). (b) When \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid, the equation for the reaction is: A\(^-\) + H\(_2\)O (l) \(\rightleftharpoons\) A\(^{2-}\) (aq) + H\(_3\)O\(^+\) (aq). (c) The conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) is \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}(a q)\) and its conjugate base is \(\mathrm{H}_{1} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{2-}(a q)\).

Step 1: (a) Act as a Base

When the molecule acts as a base, it will accept a proton from water. The given molecule, \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\), can be written as A\(^-\) for simplicity. The equation for the reaction will be: A\(^-\) + H\(_2\)O (l) \(\rightleftharpoons\) HA (aq) + OH\(^-\) (aq)

Step 2: (b) Act as an Acid

When the molecule acts as an acid, it will donate a proton to water. Using the same notation of A\(^-\) for the given molecule, the equation for the reaction will be: A\(^-\) + H\(_2\)O (l) \(\rightleftharpoons\) A\(^{2-}\) (aq) + H\(_3\)O\(^+\) (aq)

Step 3: (c) Conjugate Acid and Conjugate Base

The conjugate acid of the given molecule is the species formed after the molecule accepts a proton, which is A (or HA) where A\(^-\) represents the given molecule, \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\). So the conjugate acid is \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}(a q)\) (with the added proton). The conjugate base of the given molecule is the species formed after the molecule donates a proton, which is A\(^{2-}\). So the conjugate base is \(\mathrm{H}_{1} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{2-}(a q)\) (with one less proton).

Key Concepts

Conjugate Acid-Base Pairs

Understanding conjugate acid-base pairs is crucial in the context of acid-base reactions. A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton. This is a fundamental concept of the Brønsted-Lowry theory, which emphasizes the role of protons in acid-base reactions.

To illustrate, when an acid donates a proton during a reaction, it forms its conjugate base, whereas when a base accepts a proton, it forms its conjugate acid. It's important to recognize that these transformations are reversible, showing the dynamic relationship in acid-base chemistry.

In the exercise provided, \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(aq)\), acts as both a base and an acid. As a base, it pairs with \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}(aq)\) as its conjugate acid. Conversely, as an acid, it forms \(\mathrm{H}_{1} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{2-}(aq)\) as its conjugate base.

Brønsted-Lowry Acid-Base Theory

The Brønsted-Lowry acid-base theory provides a broader definition of acids and bases compared to its predecessors. This theory, introduced by Johannes Nicolaus Brønsted and Thomas Martin Lowry, proposes that an acid is a substance that can donate a proton, while a base is a substance that can accept a proton.

Every acid-base reaction according to this theory involves a transfer of protons between the reacting species. This transfer results in the creation of conjugate acid-base pairs, as seen in the exercise, where the roles of acids and bases are reversible depending on the direction of the reaction. The theory fundamentally changed the understanding of acid-base chemistry, emphasizing the role of proton exchange rather than the production of hydrogen or hydroxide ions.

Chemical Equilibrium

In an acid-base reaction, the formation of reactants and products often reaches a state of chemical equilibrium. This is when the forward and reverse reactions occur at the same rate, resulting in no net change in the concentration of the substances involved over time.

At equilibrium, the concentrations of all reactants and products remain constant but are not necessarily equal. It's essential to realize that reaching equilibrium does not mean the reaction has stopped but that it has reached a dynamic balance.

The provided equations in the step-by-step solutions exemplify a chemical equilibrium, denoted by the \( \rightleftharpoons \) symbol. Knowing how to write and interpret these equations helps students understand that reactions are reversible and that equilibrium is a critical concept in studying chemical reactions.

Proton Transfer Reactions

Acid-base reactions are often classified as proton transfer reactions because they involve the transfer of hydrogen ions (\(H^+\) ions, also known as protons) between the reactant molecules. This is a useful way to view acids and bases, as it focuses on the movement of protons from one substance to another.

Proton transfers can be visualized easily when interpreting chemical equations. For instance, in the exercise solutions, when \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(aq)\) acts as a base, it accepts a proton, becoming \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}(aq)\). Conversely, when it acts as an acid, it donates a proton, producing \(\mathrm{H}_{1} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{2-}(aq)\). The ability to track proton movement is a key aspect of understanding chemical reactions at a molecular level.