Question

Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each equation, and also designate the conjugate acid and conjugate base of each on the right side. (a) $\mathrm{HBrO}(aq)+{\mathrm{H}}_2\mathrm{O}(l)\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}(aq)+{\mathrm{BrO}}^{-}(aq)$ (b) ${\mathrm{HSO}}_4^{-}(aq)+{\mathrm{HCO}}_3^{-}(aq)\rightleftharpoons {\mathrm{SO}}_4^{2-}(aq)+{\mathrm{H}}_2{\mathrm{CO}}_3(aq)$ (c) ${\mathrm{HSO}}_3^{-}(aq)+{\mathrm{H}}_3{\mathrm{O}}^{+}(aq)\rightleftharpoons {\mathrm{H}}_2{\mathrm{SO}}_3(aq)+{\mathrm{H}}_2\mathrm{O}(l)$

Short answer

(a) Acid: $\mathrm{HBrO}$, Base: ${\mathrm{H}}_2\mathrm{O}$, Conjugate acid: ${\mathrm{H}}_3{\mathrm{O}}^{+}$, Conjugate base: ${\mathrm{BrO}}^{-}$ (b) Acid: ${\mathrm{HSO}}_4^{-}$, Base: ${\mathrm{HCO}}_3^{-}$, Conjugate acid: ${\mathrm{H}}_2{\mathrm{CO}}_3$, Conjugate base: ${\mathrm{SO}}_4^{2-}$ (c) Acid: ${\mathrm{H}}_3{\mathrm{O}}^{+}$, Base: ${\mathrm{HSO}}_3^{-}$, Conjugate acid: ${\mathrm{H}}_2{\mathrm{SO}}_3$, Conjugate base: ${\mathrm{H}}_2\mathrm{O}$

Step-by-step solution

Identify the acid and base in the reactants

First, we're looking for the species that donates a proton and the species that accepts a proton. In this case, $\mathrm{HBrO}$ will donate a proton to ${\mathrm{H}}_2\mathrm{O}$, which will accept it.

Identify the conjugate acid and conjugate base in the products

After donating a proton, $\mathrm{HBrO}$ will become ${\mathrm{BrO}}^{-}$, which is the conjugate base. On the other hand, after accepting a proton, ${\mathrm{H}}_2\mathrm{O}$ will become ${\mathrm{H}}_3{\mathrm{O}}^{+}$, which is the conjugate acid. (b) ${\mathrm{HSO}}_4^{-}(aq)+{\mathrm{HCO}}_3^{-}(aq)\rightleftharpoons {\mathrm{SO}}_4^{2-}(aq)+{\mathrm{H}}_2{\mathrm{CO}}_3(aq)$

Identify the acid and base in the reactants

In this reaction, we can see that ${\mathrm{HSO}}_4^{-}$ will donate a proton to ${\mathrm{HCO}}_3^{-}$, which will accept it.

Identify the conjugate acid and conjugate base in the products

After donating a proton, ${\mathrm{HSO}}_4^{-}$ will become ${\mathrm{SO}}_4^{2-}$, which is the conjugate base. After accepting a proton, ${\mathrm{HCO}}_3^{-}$ will become ${\mathrm{H}}_2{\mathrm{CO}}_3$, which is the conjugate acid. (c) ${\mathrm{HSO}}_3^{-}(aq)+{\mathrm{H}}_3{\mathrm{O}}^{+}(aq)\rightleftharpoons {\mathrm{H}}_2{\mathrm{SO}}_3(aq)+{\mathrm{H}}_2\mathrm{O}(l)$

Identify the acid and base in the reactants

Here, we can readily see that ${\mathrm{HSO}}_3^{-}$ will accept a proton from ${\mathrm{H}}_3{\mathrm{O}}^{+}$, which will donate it.

Identify the conjugate acid and conjugate base in the products

After accepting a proton, ${\mathrm{HSO}}_3^{-}$ will become ${\mathrm{H}}_2{\mathrm{SO}}_3$, which is the conjugate acid. Conversely, after donating a proton, ${\mathrm{H}}_3{\mathrm{O}}^{+}$ will become ${\mathrm{H}}_2\mathrm{O}$, which is the conjugate base.

Key concepts

Acid-Base Reactions

An acid-base reaction is a fundamental chemical reaction, central to chemistry, where an acid and a base interact with each other. In the context of Bronsted-Lowry theory, which we are focusing on here, the reaction involves the transfer of protons between the molecules involved. When we refer to a 'proton' in chemistry, we're usually talking about a hydrogen ion, or H⁺.
In these reactions:

• The acid donates a proton to the base.
• The base accepts that proton from the acid.
• The reactions are typically reversible, reaching an equilibrium over time.

In any given Bronsted-Lowry acid-base reaction, identifying the species that acts as the acid and the one that acts as the base is the first step. For example, in the reaction $\mathrm{HBrO}(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}(aq)+{\mathrm{BrO}}^{-}(aq)$, $\mathrm{HBrO}$ is the acid because it donates a proton to water, ${\mathrm{H}}_2\mathrm{O}$, which acts as the base by accepting the proton.

Conjugate Acid-Base Pairs

After the proton has been transferred, the original acid becomes a new species known as its conjugate base, and the original base becomes its conjugate acid. These new species form what's called a conjugate acid-base pair.
Here is how it works step by step using our examples:

• When $\mathrm{HBrO}$ donates a proton, it becomes ${\mathrm{BrO}}^{-}$, which is its conjugate base.
• When ${\mathrm{H}}_2\mathrm{O}$ accepts a proton, it becomes ${\mathrm{H}}_3{\mathrm{O}}^{+}$, its conjugate acid.
• These pairs highlight the reversible nature of acid-base reactions.

Similarly, for other given equations, you can link each acid with its conjugate base and each base with its conjugate acid based on how protons are transferred. Understanding these pairs is crucial; it frames our grasp of chemical equilibriums and buffer mechanisms.

Proton Donation and Acceptance

At the heart of the Bronsted-Lowry acid-base theory is the donation and acceptance of protons. This characterizes the role of acids and bases respectively:
- An acid is defined by its ability to donate a proton. - A base is characterized by its ability to accept a proton.The donation and acceptance process is more than just labels; it’s a dynamic interaction that explains how substances behave in a solution. For instance, ${\mathrm{HSO}}_4^{-}$ donates a proton to ${\mathrm{HCO}}_3^{-}$, transforming into ${\mathrm{SO}}_4^{2-}$, its conjugate base, while ${\mathrm{HCO}}_3^{-}$, after accepting the proton, becomes ${\mathrm{H}}_2{\mathrm{CO}}_3$, its conjugate acid.
This interaction is key to many processes, including natural phenomena and industrial applications, such as biological enzyme functions and pharmaceutical formulations.