Question

The conjugate acid of \(\mathrm{HPO}_{4}^{2-}\) is (a) \(\mathrm{PO}_{4}^{3-}\) (b) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} ;(\mathrm{c}) \mathrm{H}_{3} \mathrm{PO}_{4} ;(\mathrm{d}) \mathrm{H}_{3} \mathrm{O}^{+} ;\) (e) none of these.

Short answer

(b) \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\)

Step-by-step solution

Checking the Correct Option

\(\mathrm{HPO}_{4}^{2-}\) forms \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) by gaining a proton (H+), so \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) is the conjugate acid of \(\mathrm{HPO}_{4}^{2-}\). Utilizing the concept of conjugate acid-base pairs, one can conclude that when a base gains a hydrogen ion (H+), it becomes its conjugate acid. Thus, each option can be checked according to this principle. By adding a proton (H+) to \(\mathrm{HPO}_{4}^{2-}\), we get \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\), which makes option (b) the correct answer.

Key concepts

Acid-Base Pairs

Understanding acid-base pairs is fundamental in grasping various chemical reactions and processes. An acid-base pair consists of two substances that differ by only one proton (H+). The acid can donate a proton, while the base can accept one. This proton transfer changes the acid into its conjugate base and the base into its conjugate acid, as seen in the relationship between \( \mathrm{HPO}_{4}^{2-} \) and \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \).

In the given exercise, the correct answer identifies \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \) as the conjugate acid of \( \mathrm{HPO}_{4}^{2-} \). It highlights that for every base, there is an associated acid that forms when the base gains a proton (H+). This connection is not only crucial for solving exercises but also for understanding chemical behaviors in different environments, such as biological systems or industrial processes.

Proton (H+) Transfer

The transfer of a proton (H+) between substances is a central theme in acid-base chemistry. A proton is essentially a hydrogen atom that has lost its electron, leaving behind a positively charged ion, \( \text{H}^{+} \). When we refer to acids and bases, we often talk about their ability to donate or accept protons, respectively.

When an acid donates a proton to a base, the acid becomes its conjugate base, and the base becomes its conjugate acid - forming an acid-base pair. The exercise illustrates this point by showing how \( \mathrm{HPO}_{4}^{2-} \) accepts a proton to become \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \). This is an essential concept in explaining how substances react with each other and helps in predicting the outcomes of chemical reactions.

Significance of Proton Transfer

Proton transfer reactions are vital in many biological processes, like enzyme function and energy production in cells. They are also the basis for the pH scale, which measures the acidity or basicity of a solution.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate. As a result, the concentrations of the reactants and products remain unchanged over time. Equilibrium can be applied to acid-base reactions, including the proton transfer between conjugate acid-base pairs.

Once the \( \mathrm{HPO}_{4}^{2-} \) gains a proton to become \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \), an equilibrium is established between these two forms. This concept is important because it dictates the concentrations of different species in a solution at equilibrium. In the human body, for example, blood pH is maintained through equilibrium mechanisms involving buffer systems of conjugate acid-base pairs.

Relevance to Real-world Applications

Understanding how chemical equilibrium works is crucial for the development of industrial processes, pharmaceuticals, and more. It informs how certain conditions, like temperature and pH, can affect the position of equilibrium and alter the concentrations of substances in a system.