The `pH` of a solution containing `0.1mol` of `CH_(3)COOH, 0.2 mol` of `CH_(3)COONa`,and `0.05 mol` of `NaOH` in `1L. (pK_(a) of CH_(3)COOH = 4.74)` is:

Options

A. `5.44`
B. `5.20`
C. `5.04`
D. `4.74`

Explanation

To find the pH of the solution containing acetic acid (CH₃COOH), sodium acetate (CH₃COONa), and sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Identify the components and their initial concentrations
We have:
- 0.1 mol of CH₃COOH (weak acid)
- 0.2 mol of CH₃COONa (salt of the weak acid)
- 0.05 mol of NaOH (strong base)

### Step 2: Determine the reaction between CH₃COOH and NaOH
The reaction between acetic acid and sodium hydroxide can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

### Step 3: Calculate the moles after the reaction
- Initial moles of CH₃COOH = 0.1 mol
- Moles of NaOH = 0.05 mol

After the reaction:
- Moles of CH₃COOH remaining = 0.1 - 0.05 = 0.05 mol
- Moles of CH₃COONa formed = 0.2 + 0.05 = 0.25 mol

### Step 4: Calculate the concentrations of CH₃COOH and CH₃COONa
Since the total volume of the solution is 1 L:
- Concentration of CH₃COOH = 0.05 mol / 1 L = 0.05 M
- Concentration of CH₃COONa = 0.25 mol / 1 L = 0.25 M

### Step 5: Use the Henderson-Hasselbalch equation
For an acidic buffer, the pH can be calculated using the formula:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

Where:
- \(\text{pK}_a\) of CH₃COOH = 4.74
- \([\text{Salt}] = [\text{CH}_3\text{COONa}] = 0.25 \, \text{M}\)
- \([\text{Acid}] = [\text{CH}_3\text{COOH}] = 0.05 \, \text{M}\)

### Step 6: Substitute the values into the equation
\[ \text{pH} = 4.74 + \log\left(\frac{0.25}{0.05}\right) \]
\[ \text{pH} = 4.74 + \log(5) \]

### Step 7: Calculate \(\log(5)\)
Using a calculator, we find:
\[ \log(5) \approx 0.699 \]

### Step 8: Final pH calculation
\[ \text{pH} = 4.74 + 0.699 \]
\[ \text{pH} \approx 5.44 \]

Thus, the pH of the solution is approximately **5.44**.