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

Options

A. 5
B. 6
C. 7
D. 9

Answer

To solve the problem, we need to determine the pH at the equivalence point when 10 ml of 0.1 M acetic acid is titrated against 10 ml of 0.1 M ammonia solution.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between acetic acid (CH₃COOH) and ammonia (NH₃) can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NH}_3 \rightarrow \text{CH}_3\text{COONH}_4 + \text{H}_2\text{O} \]
At the equivalence point, all the acetic acid will have reacted with ammonia to form ammonium acetate (CH₃COONH₄).

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

3. **Use the Formula for pH at Equivalence Point**:
At the equivalence point of a weak acid-strong base titration, the pH can be calculated using the formula:
\[ \text{pH} = 7 + \frac{1}{2} (pK_a - pK_b) \]

4. **Substitute the Values**:
Since both \( pK_a \) and \( pK_b \) are equal to 5.0:
\[ \text{pH} = 7 + \frac{1}{2} (5.0 - 5.0) = 7 + \frac{1}{2} (0) = 7 \]

5. **Final Answer**:
Therefore, the pH at the equivalence point is:
\[ \text{pH} = 7 \]

### Conclusion:
The pH at the equivalence point when titrating 10 ml of 0.1 M acetic acid with 10 ml of 0.1 M ammonia is **7**.

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