Question

Identify the Bronsted-Lowry acid and base in each of the following neutralization reactions: (a) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{HClO}_{4}(a q) \longrightarrow \mathrm{HNO}_{2}(a q)+\mathrm{ClO}_{4}^{-}(a q)\)

Short answer

(a) \(\mathrm{HCO}_{3}^{-}\) is the acid, \(\mathrm{OH}^{-}\) is the base. (b) \(\mathrm{HClO}_{4}\) is the acid, \(\mathrm{NO}_{2}^{-}\) is the base.

Step-by-step solution

Understanding the Reaction Components - Part (a)

In the given reaction \(\mathrm{HCO}_{3}^{-}(aq)+\mathrm{OH}^{-}(aq) \longrightarrow \mathrm{CO}_{3}^{2-}(aq)+\mathrm{H}_{2}\mathrm{O}(l)\), we have to identify the acid and base according to the Bronsted-Lowry theory. In this context, a Bronsted-Lowry acid is a species that donates a proton (\(\text{H}^+\)) and a Bronsted-Lowry base is a species that accepts a proton.

Identify Bronsted-Lowry Acid in Reaction (a)

In the reaction, \(\mathrm{HCO}_{3}^{-}\) donates a proton to become \(\mathrm{CO}_{3}^{2-}\). Therefore, \(\mathrm{HCO}_{3}^{-}\) is acting as the Bronsted-Lowry acid.

Identify Bronsted-Lowry Base in Reaction (a)

The \(\mathrm{OH}^{-}\) ion accepts a proton in the process of forming water, \(\mathrm{H}_{2}\mathrm{O}\). Consequently, \(\mathrm{OH}^{-}\) is acting as the Bronsted-Lowry base.

Understanding the Reaction Components - Part (b)

In the given reaction \(\mathrm{NO}_{2}^{-}(aq)+\mathrm{HClO}_{4}(aq) \longrightarrow \mathrm{HNO}_{2}(aq)+\mathrm{ClO}_{4}^{-}(aq)\), we again need to use the Bronsted-Lowry concept to identify the acid and the base. The acid will donate a proton, and the base will accept a proton.

Identify Bronsted-Lowry Acid in Reaction (b)

In this reaction, \(\mathrm{HClO}_{4}\) donates a proton to form \(\mathrm{ClO}_{4}^{-}\). Thus, \(\mathrm{HClO}_{4}\) is the Bronsted-Lowry acid.

Identify Bronsted-Lowry Base in Reaction (b)

Here, \(\mathrm{NO}_{2}^{-}\) accepts the proton and forms \(\mathrm{HNO}_{2}\). Therefore, \(\mathrm{NO}_{2}^{-}\) is the Bronsted-Lowry base.

Key concepts

Neutralization Reactions

Neutralization reactions are fascinating and fundamental concepts in chemistry where an acid and a base react to form water and a salt. These reactions are key examples of the Bronsted-Lowry theory in action.
When an acid and a base come together, they essentially perform a dance of protons. The acid is a proton donor, handing over an \( ext{H}^+\) ion, while the base is a proton acceptor, ready to receive that proton.
The result of this exchange is the creation of water from the added protons and hydroxide ions, and often a salt from the remaining ions of the reactants. Such reactions are not only theoretical; they occur regularly in daily life.

• When baking soda is mixed with vinegar, a classic kitchen experiment.
• Anti-acids neutralizing stomach acid.

Acid-Base Identification

Acid-base identification is about recognizing who plays which role in a chemical reaction.
According to the Bronsted-Lowry theory:

• Acids are proton donors.
• Bases are proton acceptors.

This appears in various chemical equations where different substances donate and accept protons.
Consider the equation \( \\mathrm{HCO}_{3}^{-}(aq)+\mathrm{OH}^{-}(aq) \longrightarrow \mathrm{CO}_{3}^{2-}(aq)+\mathrm{H}_{2} \mathrm{O}(l) \) :
\(\mathrm{HCO}_{3}^{-}\) is the acid because it donates a proton to become \(\text{CO}_{3}^{2-}\), while \(\mathrm{OH}^{-}\) is the base because it accepts the proton to create water.
Recognizing these roles helps in balancing equations, predicting reaction outcomes, and understanding reaction mechanisms.

Proton Transfer

Proton transfer is the heart of the chemical universe when it comes to the Bronsted-Lowry theory.
This transfer is a movement of protons (\(\text{H}^+\) ions) from acids to bases.
Envision a proton packed with potential energy transferred from one molecular structure to another, initiating a transformation.
For instance, in the reaction \( \\mathrm{NO}_{2}^{-}(aq)+\mathrm{HClO}_{4}(aq) \longrightarrow \mathrm{HNO}_{2}(aq)+\mathrm{ClO}_{4}^{-}(aq) \):

• \(\mathrm{HClO}_{4}\) generously donates a proton to \(\mathrm{NO}_{2}^{-}\). Here, a proton isn't just transferred; it's a transaction that strengthens bonds in new molecules.
• \(\mathrm{HNO}_{2}\) and \(\text{ClO}_{4}^{-}\) are born from this proton exchange as new products.

Witnessing this handoff gives insight into the chemical behavior and properties of substances, which is vital for students and professionals in chemistry.