Answer :
Let's analyze each reaction and identify the highlighted reactant for each category:
1. Reaction: \( I (aq) + H_2O (l) \rightarrow HI (aq) + OH^- (aq) \)
- Bronsted-Lowry acid: A Bronsted-Lowry acid is a proton (H+) donor.
- Bronsted-Lowry base: A Bronsted-Lowry base is a proton (H+) acceptor.
- Neither: The reactant doesn't fit the criteria of either a Bronsted-Lowry acid or base.
Here, \(I (aq)\) reacts with \(H_2O (l)\) to form \(HI (aq)\) and \(OH^-(aq)\). To identify if the \(I (aq)\) is acting as a proton donor (Bronsted-Lowry acid) or proton acceptor (Bronsted-Lowry base), note that \(H_2O (l)\) is a proton donor resulting in the production of \(HI (aq)\) which contains the proton. So, \(I (aq)\) in this case doesn't actively participate in the proton-exchange process but aids in forming an acidic product.
Therefore, in this reaction:
- Bronsted-Lowry acid: 9
- Bronsted-Lowry base: 0
- Neither: 0
2. Reaction: \( I^- (aq) + H_2O (l) \rightarrow HI (aq) + OH^- (aq) \)
Here, the iodide ion \(I^- (aq)\) interacts with water to produce \(HI (aq)\) and \(OH^- (aq)\). The \(I^- (aq)\) doesn't donate or accept a proton itself, but facilitates the exchange in the medium, thus it doesn't strictly behave as a Bronsted-Lowry acid or base.
Therefore, in this reaction:
- Bronsted-Lowry acid: 0
- Bronsted-Lowry base: 0
- Neither: 0
3. Reaction: \( HI (aq) + OH^- (aq) \rightarrow I^- (aq) + H_2O (l) \)
In this reaction, \(HI (aq)\) donates a proton to \(OH^- (aq)\), producing \(I^- (aq)\) and \(H_2O (l)\). Here, \(HI (aq)\) acts as a proton donor (Bronsted-Lowry acid), and \(OH^- (aq)\) acts as a proton acceptor (Bronsted-Lowry base).
Therefore, in this reaction:
- Bronsted-Lowry acid: 0
- Bronsted-Lowry base: 0
- Neither: 0
4. Reaction: \( HI (aq) + OH^- (aq) \rightarrow I^- (aq) + H_2O (g) \)
Similar to the third reaction, \(HI (aq)\) again donates a proton to \(OH^- (aq)\), resulting in \(I^- (aq)\) and \(H_2O\) (gaseous form). Thus, the categorization for this reaction remains:
- Bronsted-Lowry acid: 0
- Bronsted-Lowry base: 0
- Neither: 0
Putting these details together in the tabular format:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline \multirow{2}{*}{ reaction } & \multicolumn{3}{|c|}{ highlighted reactant } \\ \hline & \begin{tabular}{l} Bronsted-Lowry \\ acid \end{tabular} & \begin{tabular}{l} Bronsted-Lowry \\ base \end{tabular} & neither \\ \hline [tex]$I (aq) + H_2O (l) \rightarrow HI (aq) + OH^-(aq)$[/tex] & 9 & 0 & 0 \\
\hline [tex]$I^-(aq) + H_2O (l) \rightarrow HI (aq) + OH^-(aq)$[/tex] & 0 & 0 & 0 \\
\hline [tex]$HI (aq) + OH^-(aq) \rightarrow I^-(aq) + H_2O (l)$[/tex] & 0 & 0 & 0 \\
\hline [tex]$HI (aq) + OH^-(aq) \rightarrow I^-(aq) + H_2O (g)$[/tex] & 0 & 0 & 0 \\
\hline
\end{tabular}
\][/tex]
Thus, the table provided in the problem statement is correct and reflects the categorization of the reactants in each reaction accurately.
1. Reaction: \( I (aq) + H_2O (l) \rightarrow HI (aq) + OH^- (aq) \)
- Bronsted-Lowry acid: A Bronsted-Lowry acid is a proton (H+) donor.
- Bronsted-Lowry base: A Bronsted-Lowry base is a proton (H+) acceptor.
- Neither: The reactant doesn't fit the criteria of either a Bronsted-Lowry acid or base.
Here, \(I (aq)\) reacts with \(H_2O (l)\) to form \(HI (aq)\) and \(OH^-(aq)\). To identify if the \(I (aq)\) is acting as a proton donor (Bronsted-Lowry acid) or proton acceptor (Bronsted-Lowry base), note that \(H_2O (l)\) is a proton donor resulting in the production of \(HI (aq)\) which contains the proton. So, \(I (aq)\) in this case doesn't actively participate in the proton-exchange process but aids in forming an acidic product.
Therefore, in this reaction:
- Bronsted-Lowry acid: 9
- Bronsted-Lowry base: 0
- Neither: 0
2. Reaction: \( I^- (aq) + H_2O (l) \rightarrow HI (aq) + OH^- (aq) \)
Here, the iodide ion \(I^- (aq)\) interacts with water to produce \(HI (aq)\) and \(OH^- (aq)\). The \(I^- (aq)\) doesn't donate or accept a proton itself, but facilitates the exchange in the medium, thus it doesn't strictly behave as a Bronsted-Lowry acid or base.
Therefore, in this reaction:
- Bronsted-Lowry acid: 0
- Bronsted-Lowry base: 0
- Neither: 0
3. Reaction: \( HI (aq) + OH^- (aq) \rightarrow I^- (aq) + H_2O (l) \)
In this reaction, \(HI (aq)\) donates a proton to \(OH^- (aq)\), producing \(I^- (aq)\) and \(H_2O (l)\). Here, \(HI (aq)\) acts as a proton donor (Bronsted-Lowry acid), and \(OH^- (aq)\) acts as a proton acceptor (Bronsted-Lowry base).
Therefore, in this reaction:
- Bronsted-Lowry acid: 0
- Bronsted-Lowry base: 0
- Neither: 0
4. Reaction: \( HI (aq) + OH^- (aq) \rightarrow I^- (aq) + H_2O (g) \)
Similar to the third reaction, \(HI (aq)\) again donates a proton to \(OH^- (aq)\), resulting in \(I^- (aq)\) and \(H_2O\) (gaseous form). Thus, the categorization for this reaction remains:
- Bronsted-Lowry acid: 0
- Bronsted-Lowry base: 0
- Neither: 0
Putting these details together in the tabular format:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline \multirow{2}{*}{ reaction } & \multicolumn{3}{|c|}{ highlighted reactant } \\ \hline & \begin{tabular}{l} Bronsted-Lowry \\ acid \end{tabular} & \begin{tabular}{l} Bronsted-Lowry \\ base \end{tabular} & neither \\ \hline [tex]$I (aq) + H_2O (l) \rightarrow HI (aq) + OH^-(aq)$[/tex] & 9 & 0 & 0 \\
\hline [tex]$I^-(aq) + H_2O (l) \rightarrow HI (aq) + OH^-(aq)$[/tex] & 0 & 0 & 0 \\
\hline [tex]$HI (aq) + OH^-(aq) \rightarrow I^-(aq) + H_2O (l)$[/tex] & 0 & 0 & 0 \\
\hline [tex]$HI (aq) + OH^-(aq) \rightarrow I^-(aq) + H_2O (g)$[/tex] & 0 & 0 & 0 \\
\hline
\end{tabular}
\][/tex]
Thus, the table provided in the problem statement is correct and reflects the categorization of the reactants in each reaction accurately.
