The hydrolysis of an ester was carried out separately with `0.05 N HCl and 0.05 NH_(2)SO_(4)`. Which of the following will be true?

Options

A. `k_(Hl)gtk_(H_(2)SO_(4))`
B. `k_(H_(2)SO_(4)) gt k_(HCl)`
C. `k_(H_(2)SO_(4))=2k_(HCl)`
D. `k_(H_(2)SO_(4))=k_(HCl)`

Answer

To solve the problem regarding the hydrolysis of an ester with `0.05 N HCl` and `0.05 N H2SO4`, we need to analyze the dissociation of these acids and their effect on the rate of the reaction.

### Step-by-Step Solution:

1. **Understanding Normality**:
- Normality (N) is a measure of concentration equivalent to molarity multiplied by the number of equivalents. For acids, the number of equivalents is determined by the number of protons (H⁺ ions) that can be donated.
- `0.05 N HCl` means it can donate 0.05 equivalents of H⁺ ions per liter.
- `0.05 N H2SO4` can donate 2 equivalents of H⁺ ions since sulfuric acid (H2SO4) can dissociate to give 2 H⁺ ions.

2. **Calculating H⁺ Ion Concentration**:
- For `0.05 N HCl`: This gives 0.05 moles of H⁺ ions per liter.
- For `0.05 N H2SO4`: Since it can donate 2 H⁺ ions, the effective concentration of H⁺ ions will be:
\[ 0.05 \, \text{N} \times 2 = 0.10 \, \text{N} \]
- Therefore, H2SO4 provides a higher concentration of H⁺ ions compared to HCl.

3. **Rate of Hydrolysis**:
- The rate of hydrolysis of the ester is directly proportional to the concentration of H⁺ ions present in the solution.
- Since H2SO4 provides a higher concentration of H⁺ ions (0.10 N) compared to HCl (0.05 N), the rate of hydrolysis will be greater in the presence of H2SO4.

4. **Conclusion**:
- Given that H2SO4 produces more H⁺ ions than HCl at the same normality, the rate of hydrolysis of the ester will be faster in the presence of `0.05 N H2SO4` than in `0.05 N HCl`.

### Final Answer:
The correct statement is that the rate of hydrolysis of the ester will be greater with `0.05 N H2SO4` than with `0.05 N HCl`.

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