Problem 15 - Chemistry The Central Science

Question

(a) Give the conjugate base of the following Bronsted- Lowry acids: (i) \(\mathrm{HIO}_{3}\), (ii) \(\mathrm{NH}_{4}^{+}\). (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

(a) The conjugate bases are: (i) \( IO_{3}^{-} \) for \( HIO_{3} \) (ii) \( NH_{3} \) for \( NH_{4}^{+} \) (b) The conjugate acids are: (i) \( OH^{-} \) for \( O^{2-} \) (ii) \( H_{3}PO_{4} \) for \( H_{2}PO_{4}^{-} \)

Step 1: Find the conjugate base of HIO3 (i)

To find the conjugate base of the Bronsted-Lowry acid HIO3, we need to remove one proton (H+) from its formula. So, HIO3 loses one proton to give the conjugate base IO3-. The balanced equation can be written as: HIO3 ⇆ H+ + IO3^-

Step 2: Find the conjugate base of NH4+ (ii)

To find the conjugate base of the Bronsted-Lowry acid NH4+, we need to remove one proton (H+) from its formula. So, NH4+ loses one proton to give the conjugate base NH3. The balanced equation can be written as: NH4+ ⇆ H+ + NH3

Step 3: Find the conjugate acid of O^2- (i)

To find the conjugate acid of the Bronsted-Lowry base O^2-, we need to add one proton (H+) to its formula. So, O^2- gains one proton to give the conjugate acid OH^-. The balanced equation can be written as: O^2- + H+ ⇆ OH^-

Step 4: Find the conjugate acid of H2PO4^- (ii)

To find the conjugate acid of the Bronsted-Lowry base H2PO4^-, we need to add one proton (H+) to its formula. So, H2PO4^- gains one proton to give the conjugate acid H3PO4. The balanced equation can be written as: H2PO4^- + H+ ⇆ H3PO4 So, the conjugate bases are IO3^- for HIO3 and NH3 for NH4+, while the conjugate acids are OH^- for O^2- and H3PO4 for H2PO4^-.

Key Concepts

Conjugate Acid-Base Pair

In the realm of chemistry, the Bronsted-Lowry theory elegantly describes the dynamic nature of acids and bases. Central to this theory is the concept of conjugate acid-base pairs. Every acid has a corresponding conjugate base, and similarly, every base has a conjugate acid.

A conjugate acid-base pair differs by a single proton ( ext{H}^+ ext{ion}). For example:

• When an acid donates a proton, it transforms into its conjugate base. Consider ext{HIO}_3, when it loses a proton, the conjugate base ext{IO}_3^- is formed.
• When a base accepts a proton, it becomes its conjugate acid. For instance, ext{O}^{2-} gains a proton to become ext{OH}^-.

Therefore, the conjugate pairs help us understand how acids and bases exist in equilibrium with their conjugates, serving as the foundation for more complex reactions.

Proton Transfer

Proton transfer is the heart of the Bronsted-Lowry acid-base reactions. This mechanism involves the movement of ext{H}^+ ext{ions} from acids to bases, which is intrinsic to the definition of acids and bases in this theory.

When an acid donates a proton, it triggers a change in its chemical structure. For example:

• ext{NH}_4^+ donates a proton, transforming into its conjugate base ext{NH}_3. This process illustrates how the ext{NH}_4^+ ion reacts as a donor in a solution.

Similarly, bases receive a proton. This process is crucial as it balances the reaction:

• ext{H}_2 ext{PO}_4^- accepts a proton becoming ext{H}_3 ext{PO}_4.

Understanding proton transfer is essential, as it explains the reversible interactions of acid-base reactions observed in chemical systems.

Chemical Equilibrium

Chemical equilibrium occurs when the forward and reverse reactions in a system happen at the same rate, leading to stable concentrations of products and reactants.

In the context of Bronsted-Lowry acid-base reactions, equilibrium is reached when the rate at which the acid donates protons equals the rate at which the conjugate base accepts them. This equilibrium can be represented as:

• ext{HIO}_3 ightleftarrows ext{H}^+ + ext{IO}_3^-, illustrating how ext{HIO}_3 reaches equilibrium with its conjugate base.
• ext{NH}_4^+ ightleftarrows ext{H}^+ + ext{NH}_3.

The equilibrium state means that while the components of our reaction are dynamically interconverting, their overall amounts remain constant. Studying this balance helps in understanding how reactions will proceed under different conditions and how they can be manipulated in practical applications.