Write the balanced equation for the ionization of the weak base pyridine, C, H, N, in water, H₂O. Phases optional.

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**Title: Ionization of Weak Base Pyridine in Water**

**Objective:**
To write the balanced equation for the ionization of the weak base pyridine, \( \text{C}_5\text{H}_5\text{N} \), in water, \( \text{H}_2\text{O} \). Phases are optional.

**Task:**
* Write the balanced equation for the ionization of pyridine in water.

**Equation Template:**

\[
\text{Equation:} \quad \underline{\hspace{3in}}
\]

In the context of chemistry, a base like pyridine accepts a proton (\( \text{H}^+ \)) from water, forming its conjugate acid and hydroxide ions. The general equation for a weak base \( \text{B} \) in water is:

\[ 
\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^- 
\]

For pyridine, the equation is:

\[ 
\text{C}_5\text{H}_5\text{N} + \text{H}_2\text{O} \rightleftharpoons \text{C}_5\text{H}_5\text{NH}^+ + \text{OH}^- 
\]

**Note:**
The arrow (\( \rightleftharpoons \)) indicates that the reaction is at equilibrium, typical of weak bases in water.

**Discussion:**
* Pyridine acts as a proton acceptor in water.
* The ionization process forms the pyridinium ion (\( \text{C}_5\text{H}_5\text{NH}^+ \)) and hydroxide ion (\( \text{OH}^- \)).
* Understanding these processes is crucial in various applications, including buffer solutions and pharmaceutical formulations.
Transcribed Image Text:**Title: Ionization of Weak Base Pyridine in Water** **Objective:** To write the balanced equation for the ionization of the weak base pyridine, \( \text{C}_5\text{H}_5\text{N} \), in water, \( \text{H}_2\text{O} \). Phases are optional. **Task:** * Write the balanced equation for the ionization of pyridine in water. **Equation Template:** \[ \text{Equation:} \quad \underline{\hspace{3in}} \] In the context of chemistry, a base like pyridine accepts a proton (\( \text{H}^+ \)) from water, forming its conjugate acid and hydroxide ions. The general equation for a weak base \( \text{B} \) in water is: \[ \text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^- \] For pyridine, the equation is: \[ \text{C}_5\text{H}_5\text{N} + \text{H}_2\text{O} \rightleftharpoons \text{C}_5\text{H}_5\text{NH}^+ + \text{OH}^- \] **Note:** The arrow (\( \rightleftharpoons \)) indicates that the reaction is at equilibrium, typical of weak bases in water. **Discussion:** * Pyridine acts as a proton acceptor in water. * The ionization process forms the pyridinium ion (\( \text{C}_5\text{H}_5\text{NH}^+ \)) and hydroxide ion (\( \text{OH}^- \)). * Understanding these processes is crucial in various applications, including buffer solutions and pharmaceutical formulations.
Expert Solution
Step 1

Answer:

In the given question base mentioned is pyridine and we have to write down the Ionization reaction that takes place when pyridine is dissolved in water. Chemical formula of pyridine is C5H5N.

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