28. When electrodes connected to a lightbulb are inserted into an aqueous solution of acetic acid, the bulb glows dimly. Will the bulb become brighter, main the same, or turn off after one equivalent of aqueous NaOH is added to the solution? Explain and write a balanced net ionic equation that supports your answer.

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**Question 28:** When electrodes connected to a lightbulb are inserted into an aqueous solution of acetic acid, the bulb glows dimly. Will the bulb become brighter, remain the same, or turn off after one equivalent of aqueous NaOH is added to the solution? Explain and write a balanced net ionic equation that supports your answer.

**Explanation:** 

When acetic acid (CH₃COOH) is in solution, it partially ionizes, producing ions that allow the bulb to glow dimly. Acetic acid is a weak acid, so its ionization is limited, resulting in the dim glow.

Upon adding an equivalent of NaOH, which is a strong base, the acetic acid reacts to form sodium acetate (CH₃COONa) and water (H₂O). The reaction is:

\[ \text{CH}_3\text{COOH (aq)} + \text{NaOH (aq)} \rightarrow \text{CH}_3\text{COONa (aq)} + \text{H}_2\text{O (l)} \]

The net ionic equation focuses on the principal ions involved in the reaction:

\[ \text{CH}_3\text{COOH (aq)} + \text{OH}^- \text{(aq)} \rightarrow \text{CH}_3\text{COO}^- \text{(aq)} + \text{H}_2\text{O (l)} \]

In this net ionic equation, hydrogen ions from acetic acid react with hydroxide ions from NaOH to produce water. The acetate ions (\( \text{CH}_3\text{COO}^- \)) remain in solution, leading to a decrease in free hydrogen ions and thus lowering the conductivity. This would cause the bulb to glow even more dimly or possibly turn off, as sodium acetate is a weak electrolyte compared to the initial acetic acid solution.
Transcribed Image Text:**Question 28:** When electrodes connected to a lightbulb are inserted into an aqueous solution of acetic acid, the bulb glows dimly. Will the bulb become brighter, remain the same, or turn off after one equivalent of aqueous NaOH is added to the solution? Explain and write a balanced net ionic equation that supports your answer. **Explanation:** When acetic acid (CH₃COOH) is in solution, it partially ionizes, producing ions that allow the bulb to glow dimly. Acetic acid is a weak acid, so its ionization is limited, resulting in the dim glow. Upon adding an equivalent of NaOH, which is a strong base, the acetic acid reacts to form sodium acetate (CH₃COONa) and water (H₂O). The reaction is: \[ \text{CH}_3\text{COOH (aq)} + \text{NaOH (aq)} \rightarrow \text{CH}_3\text{COONa (aq)} + \text{H}_2\text{O (l)} \] The net ionic equation focuses on the principal ions involved in the reaction: \[ \text{CH}_3\text{COOH (aq)} + \text{OH}^- \text{(aq)} \rightarrow \text{CH}_3\text{COO}^- \text{(aq)} + \text{H}_2\text{O (l)} \] In this net ionic equation, hydrogen ions from acetic acid react with hydroxide ions from NaOH to produce water. The acetate ions (\( \text{CH}_3\text{COO}^- \)) remain in solution, leading to a decrease in free hydrogen ions and thus lowering the conductivity. This would cause the bulb to glow even more dimly or possibly turn off, as sodium acetate is a weak electrolyte compared to the initial acetic acid solution.
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