Answer :
Let's look at the chemical equation:
[tex]$ \text{HCl} + \text{C}_5 \text{H}_5 \text{N} \rightleftharpoons \text{Cl}^- + \text{HC}_5 \text{H}_5 \text{N}^+ $[/tex]
To identify the acids and bases:
1. Identify the initial reactants and products:
- HCl (Hydrochloric acid)
- C_5 H_5 N (Pyridine)
- Cl^- (Chloride ion)
- HC_5 H_5 N^+ (Pyridinium ion)
2. Determine the role of each species:
- HCl is a well-known acid because it dissociates in solution to release protons (H⁺).
- C_5 H_5 N (Pyridine) is a base, as it can accept a proton.
- Cl^- is the conjugate base of HCl after it has donated a proton.
- HC_5 H_5 N^+ is the conjugate acid of C_5 H_5 N after it has accepted a proton.
Next, we assign these roles to the corresponding positions in the equation:
[tex]$ \text{HCl} (\text{acid}) + \text{C}_5 \text{H}_5 \text{N} (\text{base}) \rightleftharpoons \text{Cl}^- (\text{base}) + \text{HC}_5 \text{H}_5 \text{N}^+ (\text{acid}) $[/tex]
So, the final labeled equation would be:
[tex]$ \text{HCl} (\boxed{\text{acid}}) + \text{C}_5 \text{H}_5 \text{N} (\boxed{\text{base}}) \rightleftharpoons \text{Cl}^- (\boxed{\text{base}}) + \text{HC}_5 \text{H}_5 \text{N}^+ (\boxed{\text{acid}}) $[/tex]
[tex]$ \text{HCl} + \text{C}_5 \text{H}_5 \text{N} \rightleftharpoons \text{Cl}^- + \text{HC}_5 \text{H}_5 \text{N}^+ $[/tex]
To identify the acids and bases:
1. Identify the initial reactants and products:
- HCl (Hydrochloric acid)
- C_5 H_5 N (Pyridine)
- Cl^- (Chloride ion)
- HC_5 H_5 N^+ (Pyridinium ion)
2. Determine the role of each species:
- HCl is a well-known acid because it dissociates in solution to release protons (H⁺).
- C_5 H_5 N (Pyridine) is a base, as it can accept a proton.
- Cl^- is the conjugate base of HCl after it has donated a proton.
- HC_5 H_5 N^+ is the conjugate acid of C_5 H_5 N after it has accepted a proton.
Next, we assign these roles to the corresponding positions in the equation:
[tex]$ \text{HCl} (\text{acid}) + \text{C}_5 \text{H}_5 \text{N} (\text{base}) \rightleftharpoons \text{Cl}^- (\text{base}) + \text{HC}_5 \text{H}_5 \text{N}^+ (\text{acid}) $[/tex]
So, the final labeled equation would be:
[tex]$ \text{HCl} (\boxed{\text{acid}}) + \text{C}_5 \text{H}_5 \text{N} (\boxed{\text{base}}) \rightleftharpoons \text{Cl}^- (\boxed{\text{base}}) + \text{HC}_5 \text{H}_5 \text{N}^+ (\boxed{\text{acid}}) $[/tex]