Calculate the `pH` of the following mixture
`50mL` of `0.05M CH_(3)COOH+50mL` of `0.05M NH_(4) OH`
Given : `pK_(a)=pK_(b)=4.74`

Explanation

To calculate the pH of the mixture of 50 mL of 0.05 M acetic acid (CH₃COOH) and 50 mL of 0.05 M ammonium hydroxide (NH₄OH), we can follow these steps:

### Step 1: Identify the Components
- **Acetic Acid (CH₃COOH)**: This is a weak acid.
- **Ammonium Hydroxide (NH₄OH)**: This is a weak base.

### Step 2: Determine the Reaction
When a weak acid reacts with a weak base, they form a salt and water. The reaction can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NH}_4\text{OH} \rightarrow \text{CH}_3\text{COONH}_4 + \text{H}_2\text{O} \]
Here, CH₃COONH₄ is the salt formed from the weak acid and weak base.

### Step 3: Use the pH Formula for Weak Acid and Weak Base
For a solution containing a salt of a weak acid and a weak base, the pH can be calculated using the formula:
\[ \text{pH} = 7 + \frac{1}{2}(\text{pK}_a - \text{pK}_b) \]

### Step 4: Substitute the Given Values
Given that \( \text{pK}_a = \text{pK}_b = 4.74 \):
\[ \text{pH} = 7 + \frac{1}{2}(4.74 - 4.74) \]

### Step 5: Simplify the Equation
Since \( 4.74 - 4.74 = 0 \):
\[ \text{pH} = 7 + \frac{1}{2}(0) \]
\[ \text{pH} = 7 + 0 \]
\[ \text{pH} = 7 \]

### Final Answer
The pH of the mixture is **7**.

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