Problem 16 - Chemistry: An Atoms-Focused Approach

Question

Identify the acids and bases in the following reactions: a. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) b. \(\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) c. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}_{2}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)\)

Step-by-Step Solution

Answer

Question: Identify the acids and bases in each of the following reactions: a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(a q)+\mathrm{H}_{2}\mathrm{O}(\ell) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3}\mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) b) \(\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}(a q)+\mathrm{H}_{3}\mathrm{O}^{+}(a q) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}_{2}^{+}(a q)+\mathrm{H}_{2}\mathrm{O}(\ell)\) Answer: a) Base: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\), Acid: \(\mathrm{H}_{2}\mathrm{O}\) b) Base: \(\mathrm{CO}_{2}\), Acid: \(\mathrm{H}_{2} \mathrm{O}\) c) Base: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\), Acid: \(\mathrm{H}_{3} \mathrm{O}^{+}\)

Step 1: 1. Reaction a:

In this reaction, we have: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(a q)+\mathrm{H}_{2}\mathrm{O}(\ell) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3}\mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) By comparing the reactants and products, we can see that \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\) gains a proton (\(\mathrm{H}^{+}\)) and turns into \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{NH}^{+}\). So, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\) is acting as a base in this reaction. On the other hand, \(\mathrm{H}_{2} \mathrm{O}\) loses a proton to become \(\mathrm{OH}^{-}\). Thus, water is acting as an acid in this reaction.

Step 2: 2. Reaction b:

In this reaction, we have: \(\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) By comparing the reactants and products, we can see that \(\mathrm{CO}_{2}\) gains a proton (\(\mathrm{H}^{+}\)) and turns into \(\mathrm{HCO}_{3}^{-}\). So, \(\mathrm{CO}_{2}\) is acting as a base in this reaction. On the other hand, \(\mathrm{H}_{2} \mathrm{O}\) donates a proton to become \(\mathrm{H}_{3} \mathrm{O}^{+}\). Thus, water is acting as an acid in this reaction.

Step 3: 3. Reaction c:

In this reaction, we have: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}(a q)+\mathrm{H}_{3}\mathrm{O}^{+}(a q) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}_{2}^{+}(a q)+\mathrm{H}_{2}\mathrm{O}(\ell)\) By comparing the reactants and products, we can see that \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\) gains a proton (\(\mathrm{H}^{+}\)) to become \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}_{2}^{+}\). So, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\) is acting as a base in this reaction. On the other hand, \(\mathrm{H}_{3}\mathrm{O}^{+}\) donates a proton and turns into \(\mathrm{H}_{2}\mathrm{O}\). Thus, \(\mathrm{H}_{3} \mathrm{O}^{+}\) is acting as an acid in this reaction.