When `10 ml` of `0.1 M` acetic acid `(pk_(a)=5.0)` is titrated against `10 ml` of `0.1M` ammonia solution `(pk_(b)=5.0)`,the equivalence point occurs at `pH`

Options

A. `5.0`
B. `6.0`
C. `7.0`
D. `9.0`

Explanation

To solve the problem of determining the pH at the equivalence point when titrating 10 mL of 0.1 M acetic acid with 10 mL of 0.1 M ammonia, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between acetic acid (CH₃COOH) and ammonia (NH₃) can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NH}_3 \rightarrow \text{CH}_3\text{COONH}_4 \]
At the equivalence point, all acetic acid will have reacted with ammonia to form ammonium acetate (CH₃COONH₄).

2. **Determine the pKa and pKb**:
The given values are:
- \( pK_a \) of acetic acid = 5.0
- \( pK_b \) of ammonia = 5.0

3. **Use the Formula for pH at Equivalence Point**:
The formula to calculate the pH at the equivalence point when a weak acid is titrated with a weak base is:
\[ \text{pH} = 7 + \frac{1}{2} (pK_a - pK_b) \]

4. **Substitute the Values**:
Plugging in the values of \( pK_a \) and \( pK_b \):
\[ \text{pH} = 7 + \frac{1}{2} (5.0 - 5.0) \]
This simplifies to:
\[ \text{pH} = 7 + \frac{1}{2} \times 0 = 7 \]

5. **Conclusion**:
Therefore, the pH at the equivalence point is 7.

### Final Answer:
The pH at the equivalence point is **7**.