Problem 42 - Chemistry and Chemical Reactivity

Question

Write two chemical equations, one in which \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is a Bronsted acid (in reaction with the carbonate ion, \(\mathrm{CO}_{3}^{2-}\) ), and a second in which \(\mathrm{HPO}_{4}^{2-}\) is a Bronsted base (in reaction with acetic acid, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) ).

Step-by-Step Solution

Answer

1. \(\mathrm{H}_2\mathrm{PO}_4^- + \mathrm{CO}_3^{2-} \rightarrow \mathrm{HPO}_4^{2-} + \mathrm{HCO}_3^-\) 2. \(\mathrm{HPO}_4^{2-} + \mathrm{CH}_3\mathrm{CO}_2\mathrm{H} \rightarrow \mathrm{H}_2\mathrm{PO}_4^- + \mathrm{CH}_3\mathrm{CO}_2^-\)

Step 1: Identify the reactions

First, understand which species act as a Bronsted acid or base in the given reactions. 1. Di-hydrogen phosphate ion \(\mathrm{H}_2\mathrm{PO}_4^-\) acts as a Bronsted acid, meaning it donates a proton. In this reaction, it will donate a proton to the carbonate ion \(\mathrm{CO}_3^{2-}\). 2. Hydrogen phosphate ion \(\mathrm{HPO}_4^{2-}\) acts as a Bronsted base, meaning it accepts a proton. In this reaction, it will accept a proton from acetic acid \(\mathrm{CH}_3\mathrm{CO}_2\mathrm{H}\).

Step 2: Write the first equation with \(\mathrm{H}_2\mathrm{PO}_4^-\) as a Bronsted acid

To show the reaction where \(\mathrm{H}_2\mathrm{PO}_4^-\) donates a proton to \(\mathrm{CO}_3^{2-}\), write the following equation:\[\mathrm{H}_2\mathrm{PO}_4^- + \mathrm{CO}_3^{2-} \rightarrow \mathrm{HPO}_4^{2-} + \mathrm{HCO}_3^- \]Here, \(\mathrm{H}_2\mathrm{PO}_4^-\) donates a proton to \(\mathrm{CO}_3^{2-}\), forming \(\mathrm{HPO}_4^{2-}\) and \(\mathrm{HCO}_3^-\).

Step 3: Write the second equation with \(\mathrm{HPO}_4^{2-}\) as a Bronsted base

To show the reaction where \(\mathrm{HPO}_4^{2-}\) accepts a proton from \(\mathrm{CH}_3\mathrm{CO}_2\mathrm{H}\), write the following equation:\[\mathrm{HPO}_4^{2-} + \mathrm{CH}_3\mathrm{CO}_2\mathrm{H} \rightarrow \mathrm{H}_2\mathrm{PO}_4^- + \mathrm{CH}_3\mathrm{CO}_2^- \]Here, \(\mathrm{HPO}_4^{2-}\) accepts a proton from \(\mathrm{CH}_3\mathrm{CO}_2\mathrm{H}\), forming \(\mathrm{H}_2\mathrm{PO}_4^-\) and \(\mathrm{CH}_3\mathrm{CO}_2^-\).

Key Concepts

Chemical Equations

In chemistry, a chemical equation is a symbolic representation of a chemical reaction. It shows the reactants and products, along with their respective quantities. To write a chemical equation correctly:

• Identify the reactants involved in the reaction.
• Determine what products are formed after the reaction.
• Balance the equation to ensure the conservation of mass, meaning the number of atoms for each element must be the same on both sides of the equation.

The given exercise asks us to write two chemical equations involving phosphate ions. The first equation involves a di-hydrogen phosphate ion, \(\mathrm{H}_2\mathrm{PO}_4^{-}\), reacting with a carbonate ion, \(\mathrm{CO}_3^{2-}\). Here, \(\mathrm{H}_2\mathrm{PO}_4^{-}\) acts as a proton donor.
The second equation involves the hydrogen phosphate ion, \(\mathrm{HPO}_4^{2-}\), which accepts a proton from acetic acid. Chemical equations help visualize these interactions and the transfer of a proton (\(\mathrm{H}^+\)) that is characteristic of Bronsted-Lowry reactions.

Proton Transfer

The concept of proton transfer is central to Bronsted-Lowry acid-base reactions. In these reactions, a proton, represented as \(\mathrm{H}^+\), is transferred from the acid to the base.

• When an acid donates a \(\mathrm{H}^+\), it becomes its corresponding base.
• Simultaneously, when a base accepts a \(\mathrm{H}^+\), it becomes its corresponding acid.

In our first reaction, \(\mathrm{H}_2\mathrm{PO}_4^-\) donates a proton to \(\mathrm{CO}_3^{2-}\), resulting in the formation of \(\mathrm{HPO}_4^{2-}\) and \(\mathrm{HCO}_3^-\). This is a classic example of proton transfer. In our second reaction, \(\mathrm{HPO}_4^{2-}\) accepts a proton from \(\mathrm{CH}_3\mathrm{CO}_2\mathrm{H}\), illustrating the reverse process. Proton transfer reactions are fundamental to understanding how acids and bases interact in solutions.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are essential in the study of Bronsted-Lowry reactions. These pairs consist of two species that transform into each other by gain or loss of a proton:

• The species that loses the proton is the acid, and the result after the proton is donated is the conjugate base.
• The species that gains the proton is the base, and the result after the proton is accepted is the conjugate acid.

For the first reaction in our exercise, \(\mathrm{H}_2\mathrm{PO}_4^-\) is the acid, and its conjugate pair is \(\mathrm{HPO}_4^{2-}\), the conjugate base. Meanwhile, \(\mathrm{CO}_3^{2-}\) acts as the base, and \(\mathrm{HCO}_3^-\) is its conjugate acid.
In the second reaction, \(\mathrm{HPO}_4^{2-}\) serves as the base, forming the conjugate acid \(\mathrm{H}_2\mathrm{PO}_4^-\) after the proton transfer. \(\mathrm{CH}_3\mathrm{CO}_2\mathrm{H}\) donates a proton to become \(\mathrm{CH}_3\mathrm{CO}_2^-\), forming its conjugate base. Understanding these pairs helps in predicting the direction and extent of acid-base reactions.