Problem 18 - Chemistry The Central Science

Question

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

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 Acid and Base

On the left side, \(\mathrm{HBrO}\) donates \(\mathrm{H}^{+}\), so it is a Bronsted-Lowry acid, and \(\mathrm{H}_{2} \mathrm{O}\) accepts \(\mathrm{H}^{+}\), so it is a Bronsted-Lowry base.

Step 2: Identify Conjugate Acid and Conjugate Base

On the right side, \(\mathrm{H}_{3} \mathrm{O}^{+}\) is the conjugate acid formed by the acceptance of \(\mathrm{H}^{+}\) by \(\mathrm{H}_{2} \mathrm{O}\), and \(\mathrm{BrO}^{-}\) is the conjugate base formed by donating \(\mathrm{H}^{+}\) from \(\mathrm{HBrO}\). (b) \(\mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\)

Step 1: Identify Acid and Base

On the left side, \(\mathrm{HSO}_{4}^{-}\) donates \(\mathrm{H}^{+}\), so it is a Bronsted-Lowry acid, and \(\mathrm{HCO}_{3}^{-}\) accepts \(\mathrm{H}^{+}\), so it is a Bronsted-Lowry base.

Step 2: Identify Conjugate Acid and Conjugate Base

On the right side, \(\mathrm{H}_{2} \mathrm{CO}_{3}\) is the conjugate acid formed by the acceptance of \(\mathrm{H}^{+}\) by \(\mathrm{HCO}_{3}^{-}\), and \(\mathrm{SO}_{4}^{2-}\) is the conjugate base formed by donating \(\mathrm{H}^{+}\) from \(\mathrm{HSO}_{4}^{-}\). (c) \(\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Step 1: Identify Acid and Base

On the left side, \(\mathrm{HSO}_{3}^{-}\) accepts \(\mathrm{H}^{+}\), so it is a Bronsted-Lowry base, and \(\mathrm{H}_{3} \mathrm{O}^{+}\) donates \(\mathrm{H}^{+}\), so it is a Bronsted-Lowry acid.

Step 2: Identify Conjugate Acid and Conjugate Base

On the right side, \(\mathrm{H}_{2} \mathrm{SO}_{3}\) is the conjugate acid formed by the acceptance of \(\mathrm{H}^{+}\) by \(\mathrm{HSO}_{3}^{-}\), and \(\mathrm{H}_{2} \mathrm{O}\) is the conjugate base formed by donating \(\mathrm{H}^{+}\) from \(\mathrm{H}_{3} \mathrm{O}^{+}\).

Key Concepts

Conjugate Acid

In the fascinating world of chemistry, a conjugate acid is the species formed when a base gains a proton (\(\text{H}^+\)). Imagine you start with a base, like water (\(\text{H}_2\text{O}\)), and add a proton to it. That transformed water into hydronium (\(\text{H}_3\text{O}^+\)), which is the conjugate acid.

A conjugate acid can donate a proton, reversing the process. This interchange between donating and accepting protons is at the heart of the Bronsted-Lowry acid-base theory.

• For example, in the reaction where \(\text{H}_2\text{O}\) accepts a proton from \(\text{HBrO}\), \(\text{H}_3\text{O}^+\) is the resulting conjugate acid.
• Another instance shows \(\text{HCO}_3^-\) accepting a proton and becoming \(\text{H}_2\text{CO}_3\), the conjugate acid.

Once you've grasped this, identifying the conjugate acids in reactions becomes easier. Every proton gain conversion gives birth to a conjugate acid, helping complete the cycle of acid-base interactions.

Conjugate Base

A conjugate base is something you'll often come across when studying chemical reactions. It forms when an acid donates a proton (\(\text{H}^+\)). Picture an acid molecule that gives away a proton, what remains is its partner-in-crime, the conjugate base.

Think of the acid \(\text{HBrO}\), which after losing a proton becomes \(\text{BrO}^-\), its conjugate base.

• This leftover particle is crucial as it can acquire a proton again, reforming the original acid.
• In a different scenario, \(\text{HSO}_4^-\) gives up its proton turning into \(\text{SO}_4^{2-}\), its conjugate base.

Conjugate bases often have negative charges, thanks to the lost proton. They play a central role in acid-base balance and reactivity, especially when a base receives and an acid donates a proton.

Proton Transfer

One might wonder, what exactly happens when acids and bases interact? The magic behind this interaction is proton transfer, an exchange of protons between the reacting substances.

The Bronsted-Lowry theory elegantly describes this: Acids are proton donors and bases are proton acceptors. This transfer of \(\text{H}^+\) ions makes the entire process dynamic and fascinating.

• For instance, in the reaction \(\text{H}_2\text{O}\) acts as a base accepting a proton from \(\text{HBrO}\).
• Meanwhile, in the interplay between \(\text{HSO}_3^-\) and \(\text{H}_3\text{O}^+\), we see the latter donate a proton, facilitating transfer.

Each side of the reaction has its Bronsted-Lowry acid and base. Understanding the concept of proton transfer is vital, as it's the key to predicting and explaining the behavior of acid-base reactions in chemical equations.