Problem 18 - Chemistry The Central Science In Si Units

Question

Identify the Brønsted-Lowry acid and the BrønstedLowry base on the left side of each equation, and also identify the conjugate acid and conjugate base of each on the right side. $$ \begin{array}{l} \text { (a) } \mathrm{HBrO}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{BrO}^{-}(a q) \\\ \text { (b) } \mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \\ \text { (c) } \mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{array} $$

Step-by-Step Solution

Answer

(a) Acid: \(\mathrm{HBrO}\), Base: \(\mathrm{H}_2\mathrm{O}\); Conjugate Acid: \(\mathrm{H}_3\mathrm{O}^+\), Conjugate Base: \(\mathrm{BrO}^-\). (b) Acid: \(\mathrm{HSO}_4^-\), Base: \(\mathrm{HCO}_3^-\); Conjugate Acid: \(\mathrm{H}_2\mathrm{CO}_3\), Conjugate Base: \(\mathrm{SO}_4^{2-}\). (c) Acid: \(\mathrm{H}_3\mathrm{O}^+\), Base: \(\mathrm{HSO}_3^-\); Conjugate Acid: \(\mathrm{H}_2\mathrm{SO}_3\), Conjugate Base: \(\mathrm{H}_2\mathrm{O}\).

Step 1: Identify the Brønsted-Lowry Acid in Equation (a)

In the reaction \( \mathrm{HBrO}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{BrO}^{-}(a q) \), \( \mathrm{HBrO} \) is the Brønsted-Lowry acid because it donates a proton (H\(^+\)) to water.

Step 2: Identify the Brønsted-Lowry Base in Equation (a)

In the same reaction, \( \mathrm{H}_{2} \mathrm{O} \) acts as the Brønsted-Lowry base, as it accepts a proton from \( \mathrm{HBrO} \), forming \( \mathrm{H}_{3} \mathrm{O}^{+} \).

Step 3: Identify the Conjugate Acid and Base in Equation (a)

After \( \mathrm{HBrO} \) donates a proton, it becomes \( \mathrm{BrO}^{-} \), which is the conjugate base. \( \mathrm{H}_{3} \mathrm{O}^{+} \) is the conjugate acid formed from \( \mathrm{H}_{2} \mathrm{O} \).

Step 4: Identify the Brønsted-Lowry Acid in Equation (b)

In the reaction \( \mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \), \( \mathrm{HSO}_{4}^{-} \) is the Brønsted-Lowry acid as it donates a proton to \( \mathrm{HCO}_{3}^{-} \).

Step 5: Identify the Brønsted-Lowry Base in Equation (b)

\( \mathrm{HCO}_{3}^{-} \) is the Brønsted-Lowry base in this reaction, because it accepts a proton from \( \mathrm{HSO}_{4}^{-} \).

Step 6: Identify the Conjugate Acid and Base in Equation (b)

The conjugate base in this reaction is \( \mathrm{SO}_{4}^{2-} \), which forms after \( \mathrm{HSO}_{4}^{-} \) donates a proton. \( \mathrm{H}_{2} \mathrm{CO}_{3} \) is the conjugate acid of \( \mathrm{HCO}_{3}^{-} \).

Step 7: Identify the Brønsted-Lowry Acid in Equation (c)

In the reaction \( \mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \), \( \mathrm{H}_{3} \mathrm{O}^{+} \) is the Brønsted-Lowry acid because it donates a proton.

Step 8: Identify the Brønsted-Lowry Base in Equation (c)

\( \mathrm{HSO}_{3}^{-} \) acts as the Brønsted-Lowry base as it accepts a proton from \( \mathrm{H}_{3} \mathrm{O}^{+} \).

Step 9: Identify the Conjugate Acid and Base in Equation (c)

The conjugate acid is \( \mathrm{H}_{2} \mathrm{SO}_{3} \), formed when \( \mathrm{HSO}_{3}^{-} \) accepts a proton. The conjugate base is \( \mathrm{H}_{2} \mathrm{O} \), formed as \( \mathrm{H}_{3} \mathrm{O}^{+} \) donates a proton.

Key Concepts

Conjugate Acid-Base Pairs

In the Brønsted-Lowry acid-base theory, acids and bases always work together. When an acid donates a proton, it becomes what is known as a conjugate base. Similarly, when a base accepts a proton, it turns into a conjugate acid. This relationship forms what we call conjugate acid-base pairs.
For example, in equation (a) of the exercise, we see this clearly:

• The acid, HBrO, donates a proton, transforming itself into BrO\(^-\), which is the conjugate base.
• The base, water (H\(_2\)O), accepts a proton, forming H\(_3\)O\(^+\) as its conjugate acid.

Recognizing these pairs in reactions helps us understand the dual role these compounds can play. Many substances can behave both as acids and bases, depending on the environment, showcasing the versatility of conjugate acid-base pairs.

Proton Transfer Reactions

Proton transfer is at the heart of Brønsted-Lowry acid-base reactions. It's the process where a proton (H\(^+\)) shifts from the acid to the base. This shift is what defines the roles of acids and bases in a chemical reaction.
Consider equation (b) from our exercise:

• The bisulfate ion (HSO\(_4^-\)) acts as an acid because it donates a proton to bicarbonate (HCO\(_3^-\)), which accepts it.
• Bicarbonate, accepting the proton, acts as a base. This causes the production of sulfate (SO\(_4^{2-}\)) and carbonic acid (H\(_2\)CO\(_3\)).

This proton transfer is essential for the transformation process. By analyzing the flow of protons, chemists can decipher the behavior of substances in different reactions, helping them understand reaction mechanisms more thoroughly.

Chemical Equilibria

Chemical equilibria involve the balance of reactions and their reversibility, a critical aspect of many chemical processes. When reactions reach equilibrium, the rate of the forward reaction matches the rate of the reverse, leading to a steady state of components.
For instance, in equation (c) from the exercise, observe how:

• The reaction between hydronium (H\(_3\)O\(^+\)) and bisulfite (HSO\(_3^-\)) reaches a point where the rate of forward and reverse reactions balance each other.
• This results in the production of sulfurous acid (H\(_2\)SO\(_3\)) and water (H\(_2\)O) in a dynamic yet stable state.

Understanding chemical equilibria is fundamental for controlling reaction conditions and yields in practical applications. By knowing how to influence this balance, chemists can optimize industrial processes and laboratory experiments alike.