Question

(a) Give the conjugate base of the following BronstedLowry acids: (i) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\). (ii) \(\mathrm{HPO}_{4}{ }^{2-}\). (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{CO}_{3}^{2-}\) (ii) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\)

Short answer

a) (i) Conjugate base: \(\mathrm{C}_6\mathrm{H}_5\mathrm{COO}^{-}\); (ii) Conjugate base: \(\mathrm{PO}_4^{3-}\) b) (i) Conjugate acid: \(\mathrm{HCO}_3^{-}\); (ii) Conjugate acid: \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^{+}\)

Step-by-step solution

(i) Conjugate base of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\)

To find the conjugate base of the acid \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), we must remove a proton (H+) from the molecule. This occurs at the carboxyl group (-COOH), leaving us with \(\mathrm{C}_6\mathrm{H}_5\mathrm{COO}^{-}\) as the conjugate base.

(ii) Conjugate base of \(\mathrm{HPO}_{4}^{2-}\)

To find the conjugate base of the acid \(\mathrm{HPO}_{4}^{2-}\), we must remove a proton (H+) from the molecule. This results in the formation of \(\mathrm{PO}_4^{3-}\) as the conjugate base. b) Conjugate acids of given Bronsted-Lowry bases

(i) Conjugate acid of \(\mathrm{CO}_{3}^{2-}\)

To find the conjugate acid of the base \(\mathrm{CO}_{3}^{2-}\), we must add a proton (H+) to the molecule. This results in the formation of \(\mathrm{HCO}_3^{-}\) as the conjugate acid.

(ii) Conjugate acid of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\)

To find the conjugate acid of the base \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\), we must add a proton (H+) to the molecule. This results in the formation of \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^{+}\) as the conjugate acid.

Key concepts

Conjugate Base

In the realm of acid-base chemistry, the concept of a conjugate base is pivotal. A conjugate base is formed when an acid donates a proton (H+) to another molecule. Once the proton is lost, what remains of the original acid is the conjugate base.

For example, in the case of \(\mathrm{C}_6\mathrm{H}_5\mathrm{COOH}\), when it relinquishes a proton, it transforms into its conjugate base \(\mathrm{C}_6\mathrm{H}_5\mathrm{COO}^{-}\). This process highlights the Bronsted-Lowry theory in action, where acids and bases are defined by their ability to donate or accept protons, respectively.

Another example can be seen with \(\mathrm{HPO}_4^{2-}\). When this acid loses a proton, it becomes \(\mathrm{PO}_4^{3-}\). The removal of a proton from an acid thus results in the formation of its conjugate base, which is essential in understanding the dynamic relationship between acids and bases in a solution.

Conjugate Acid

The term "conjugate acid" refers to the species formed when a base gains a proton in a chemical reaction. This is the exact opposite of forming a conjugate base from an acid. In each reaction, a conjugate acid-base pair is formed.

Taking the Bronsted-Lowry base \(\mathrm{CO}_3^{2-}\) as an example, we see that when it accepts a proton, it becomes the conjugate acid \(\mathrm{HCO}_3^{-}\). Similarly, when \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\) accepts a proton, it transforms into \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^{+}\), its conjugate acid.

Understanding these transformations is crucial in predicting the behavior of substances in acid-base reactions. By learning how a base becomes a conjugate acid, students can better grasp the reversible nature of these reactions and their extensive applications in chemistry.

Acid-Base Chemistry

Acid-base chemistry revolves around the interactions between acids and bases as they exchange protons. According to the Bronsted-Lowry theory, acids are proton donors and bases are proton acceptors. This theory broadens the concept beyond hydrogen ions and expands to various molecular species.

In an acid-base reaction, the acid will donate a proton to the base. Consequently, the acid becomes its conjugate base, while the base turns into its conjugate acid. This illustrates the reversible and dynamic nature of these reactions.

Practical examples, like the transformation from \(\mathrm{C}_6\mathrm{H}_5\mathrm{COOH}\) to \(\mathrm{C}_6\mathrm{H}_5\mathrm{COO}^{-}\) or from \(\mathrm{CO}_3^{2-}\) to \(\mathrm{HCO}_3^{-}\), show the tangible applications of the theory. This foundational understanding is essential in many fields, including biochemistry, environmental science, and industrial chemistry.