Problem 17 - Chemistry: The Central Science

Question

(a) Give the conjugate base of the following BronstedLowry acids: (i) \(\mathrm{HIO}_{3}\), (ii) \(\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}\). (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{O}^{2-}\), (ii) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\).

Step-by-Step Solution

Answer

The conjugate bases of the given Bronsted-Lowry acids are: (i) \(\mathrm{IO}_{3}^{-}\) and (ii) \(\mathrm{NH}_{3}\). The conjugate acids of the given Bronsted-Lowry bases are: (i) \(\mathrm{OH}^{-}\) and (ii) \(\mathrm{H}_{3}\mathrm{PO}_{4}\).

Step 1: (i) Conjugate base of \(\mathrm{HIO}_{3}\)

Remove one proton (H+) from the acid to form the conjugate base: \(\mathrm{HIO}_{3} - \mathrm{H}^{+} = \mathrm{IO}_{3}^{-}\) The conjugate base of \(\mathrm{HIO}_{3}\) is: \(\mathrm{IO}_{3}^{-}\).

Step 2: (ii) Conjugate base of \(\mathrm{NH}_{4}{ }^{+}\)

Remove one proton (H+) from the acid to form the conjugate base: \(\mathrm{NH}_{4}^{+} - \mathrm{H}^{+} = \mathrm{NH}_{3}\) The conjugate base of \(\mathrm{NH}_{4}^{+}\) is: \(\mathrm{NH}_{3}\). (b) Give the conjugate acid of the following Bronsted-Lowry bases:

Step 3: (i) Conjugate acid of \(\mathrm{O}^{2-}\)

Add one proton (H+) to the base to form the conjugate acid: \(\mathrm{O}^{2-} + \mathrm{H}^{+} = \mathrm{OH}^{-}\) The conjugate acid of \(\mathrm{O}^{2-}\) is: \(\mathrm{OH}^{-}\).

Step 4: (ii) Conjugate acid of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

Add one proton (H+) to the base to form the conjugate acid: \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-} + \mathrm{H}^{+} = \mathrm{H}_{3}\mathrm{PO}_{4}\) The conjugate acid of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is: \(\mathrm{H}_{3}\mathrm{PO}_{4}\).

Key Concepts

Conjugate Acid

In the Bronsted-Lowry acid-base theory, a conjugate acid is formed when a base receives a proton (H⁺). Imagine pairing protons like a dance—where whenever a base steps up and grabs an extra proton, it turns into its conjugate acid.

For example, consider the base \(\mathrm{O}^{2-}\). When it knabs one proton, it transitions into \(\mathrm{OH}^{-}\).
Thus \(\mathrm{OH}^{-}\) is the conjugate acid of \(\mathrm{O}^{2-}\).

• Base gains a proton: \( X + \mathrm{H}^{+} \rightarrow \text{Conjugate Acid}\)
• More protons = more positive charge after the transformation.

This concept emphasizes how bases become acids by accepting protons, forming a dynamic duo with their original self—one as the original base and the other as the newly minted conjugate acid.

Conjugate Base

A conjugate base is created when an acid sheds or donates a proton. Picture an acid like a generous friend, giving away a proton. Once it has done so, what's left is known as the conjugate base.

Consider \(\mathrm{HIO}_3\). When it gives away a proton, it turns into \(\mathrm{IO}_3^-\), its conjugate base.

• Acid loses a proton: \( \text{Acid} - \mathrm{H}^{+} \rightarrow \text{Conjugate Base} \)
• Less protons mean a more negative charge after losing the proton.

Understanding conjugate bases helps grasp how acids transition in a reaction, showing their pivotal role in the acid-base balance.

Proton Transfer Reactions

In the realm of chemistry, proton transfer reactions are the heart of Bronsted-Lowry theory. These reactions involve the movement of protons between molecules.

Imagine molecules exchanging dance partners—a proton jumps from one molecule (the acid) to another (the base). The intricacies of proton transfer can be seen when:

• An acid donates a proton to create a conjugate base.
• A base accepts a proton to form a conjugate acid.

For example, \(\mathrm{NH}_{4}^+\) gives a proton to become \(\mathrm{NH}_{3}\), while \(\mathrm{OH}^{-}\) accepts a proton from water to become \(\mathrm{H}_2\mathrm{O}\). The speed and ease of these reactions can vary, but the essence is the tangible movement of that little proton, making these reactions pivotal in chemical processes. This transferring of protons fosters equilibrium in both biological and chemical systems.