Question

Write balanced equations showing how the hydrogen oxalate ion, \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-},\) can be both a Bronsted acid and a Bronsted base.

Short answer

\(\mathrm{HC}_{2}\mathrm{O}_{4}^{-}\) can donate a proton to become \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) or accept a proton to become \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\).

Step-by-step solution

Understanding the Problem

We need to write two separate balanced chemical equations where hydrogen oxalate ion \(\mathrm{HC}_{2}\mathrm{O}_{4}^{-}\) acts as both a Bronsted acid and a Bronsted base. A Bronsted acid donates a proton (\(\mathrm{H}^{+}\)), while a Bronsted base accepts a proton.

Equilibrium Equation as a Bronsted Acid

As a Bronsted acid, \(\mathrm{HC}_{2}\mathrm{O}_{4}^{-}\) will donate a proton. The reaction can be written as follows: \[ \mathrm{HC}_{2}\mathrm{O}_{4}^{-} \rightarrow \mathrm{C}_{2}\mathrm{O}_{4}^{2-} + \mathrm{H}^{+} \] This equation is balanced, showing the ion losing a proton to become \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\).

Equilibrium Equation as a Bronsted Base

As a Bronsted base, \(\mathrm{HC}_{2}\mathrm{O}_{4}^{-}\) will accept a proton. The reaction can be written as follows: \[ \mathrm{HC}_{2}\mathrm{O}_{4}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \] This balanced equation shows the ion gaining a proton to become \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\).

Key concepts

Hydrogen Oxalate Ion

The hydrogen oxalate ion, represented as \( \mathrm{HC}_{2}\mathrm{O}_{4}^{-} \), is a fascinating ion because of its ability to act in different ways in chemical reactions. It is derived from oxalic acid, and it possesses both the ability to donate and accept a proton, which makes it an amphiprotic species.

To visualize this, imagine the ion as being in a seesaw with its proton: it can either give it away or capture another. This dual capability is central to its behavior once it participates in Bronsted acid-base reactions. By understanding this ion's role, we can see how chemistry embodies balance and flexibility, allowing for numerous reaction pathways crucial in both nature and synthetic processes.

Proton Donation

Proton donation is a key characteristic of acids in Bronsted-Lowry acid-base theory. When an acid donates a proton (\( \mathrm{H}^{+} \)), it transforms into its conjugate base. For the hydrogen oxalate ion \( \mathrm{HC}_{2}\mathrm{O}_{4}^{-} \), donating a proton is a natural reaction path.

The balanced chemical equation for this process is written as:

• \[ \mathrm{HC}_{2}\mathrm{O}_{4}^{-} \rightarrow \mathrm{C}_{2}\mathrm{O}_{4}^{2-} + \mathrm{H}^{+} \]

Here, the ion donates a proton to become \( \mathrm{C}_{2}\mathrm{O}_{4}^{2-} \), its conjugate base. This ability to donate a proton emphasizes its role as a Bronsted acid. This concept not only helps in understanding chemical processes but also lays the groundwork for predicting the outcomes in various reactions, such as titrations or buffer solutions.

Proton Acceptance

Proton acceptance is equally important as donation in the Bronsted-Lowry framework. A base, in this context, accepts a proton and becomes its conjugate acid.

The hydrogen oxalate ion \( \mathrm{HC}_{2}\mathrm{O}_{4}^{-} \), when acting as a base, accepts a proton as follows:

• \[ \mathrm{HC}_{2}\mathrm{O}_{4}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \]

In this equilibrium, the ion captures an additional proton resulting in the formation of \( \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \), its conjugate acid. This illustrates its dual nature in that it can shift roles based on its environment, which is vital in biological systems and industrial applications, including pharmaceuticals and agriculture.

Acid-Base Equilibrium

An understanding of acid-base equilibrium is crucial for grasping how reactions reach a state of balance. In a Bronsted acid-base reaction, equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction.

For the hydrogen oxalate ion \( \mathrm{HC}_{2}\mathrm{O}_{4}^{-} \) reactions, the equilibrium allows us to see its transforming roles, transitioning seamlessly between an acid and a base. This dynamic balance is portrayed in these equations:

• \[ \mathrm{HC}_{2}\mathrm{O}_{4}^{-} \rightarrow \mathrm{C}_{2}\mathrm{O}_{4}^{2-} + \mathrm{H}^{+} \]
• \[ \mathrm{HC}_{2}\mathrm{O}_{4}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \]

Equilibrium is vital not only in comprehending reaction pathways and their directionality but also in practical scenarios like the formulation of buffers. Buffers are systems that maintain stable pH levels which is crucial in many biological and chemical processes. This principle is foundational to many concepts in chemistry and highlights the intricate symphony of reactions occurring in a system at equilibrium.