The conjugate base of `H_2SO_4` in the following reaction is
`H_2SO_4 + H_2O iff H_3O^+ HSO_4^-`

Options

A. `H_2O`
B. `H_3O^+`
C. `HSO_4^-`
D. `SO_4^(2-)`

Answer

To identify the conjugate base of \( H_2SO_4 \) in the given reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: The reaction given is:
\[ H_2SO_4 + H_2O \rightleftharpoons H_3O^+ + HSO_4^- \]
Here, sulfuric acid (\( H_2SO_4 \)) reacts with water (\( H_2O \)).

2. **Identify Acid and Base**: In this reaction, \( H_2SO_4 \) donates a proton (\( H^+ \)) to water. Therefore, \( H_2SO_4 \) acts as an acid, and water acts as a base.

3. **Determine the Products**: When \( H_2SO_4 \) donates a proton, it forms \( HSO_4^- \) (the bisulfate ion) and \( H_3O^+ \) (the hydronium ion).

4. **Identify the Conjugate Base**: The conjugate base of an acid is what remains after the acid has donated a proton. Since \( H_2SO_4 \) donates a proton to become \( HSO_4^- \), the conjugate base of \( H_2SO_4 \) is:
\[ HSO_4^- \]

5. **Conclusion**: Therefore, the conjugate base of \( H_2SO_4 \) in the reaction is \( HSO_4^- \).

### Final Answer:
The conjugate base of \( H_2SO_4 \) is \( HSO_4^- \).

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