100 mL of a 0.1 M `CH_(3)COOH` is titrated with 0.1 M `NaOH` solution .The pH of the solution in the titration flask at the titre value of 50 is `[pK_(a)(CH_(3)COOH)=4.74]`

Options

A. 2.37
B. 4.74
C. 1. 34
D. 5. 74

Explanation

To solve the problem of finding the pH of the solution in the titration flask at the titrate value of 50 mL, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- We have 100 mL of 0.1 M acetic acid (CH₃COOH).
- The concentration of acetic acid (CH₃COOH) is 0.1 M.
- We are titrating it with 0.1 M sodium hydroxide (NaOH).

2. **Determine the Amount of Acid and Base**:
- The number of moles of CH₃COOH in 100 mL (0.1 L) is:
\[ \text{Moles of CH₃COOH} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{moles} \]
- At the titrate value of 50 mL, we have added 50 mL of 0.1 M NaOH:
\[ \text{Moles of NaOH} = 0.1 \, \text{M} \times 0.05 \, \text{L} = 0.005 \, \text{moles} \]

3. **Reaction Between Acid and Base**:
- The reaction is:
\[ \text{CH₃COOH} + \text{NaOH} \rightarrow \text{CH₃COONa} + \text{H₂O} \]
- At the point of adding 50 mL of NaOH, half of the acetic acid has reacted:
- Moles of CH₃COOH remaining = 0.01 - 0.005 = 0.005 moles
- Moles of CH₃COONa formed = 0.005 moles

4. **Calculate Concentrations**:
- The total volume of the solution after titration is 100 mL + 50 mL = 150 mL (0.15 L).
- Concentration of remaining CH₃COOH:
\[ [\text{CH₃COOH}] = \frac{0.005 \, \text{moles}}{0.15 \, \text{L}} = \frac{1}{30} \approx 0.0333 \, \text{M} \]
- Concentration of CH₃COONa:
\[ [\text{CH₃COONa}] = \frac{0.005 \, \text{moles}}{0.15 \, \text{L}} = \frac{1}{30} \approx 0.0333 \, \text{M} \]

5. **Use the Henderson-Hasselbalch Equation**:
- The pH can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
- Given that both concentrations are equal:
\[ \text{pH} = 4.74 + \log\left(\frac{0.0333}{0.0333}\right) \]
- Since \(\log(1) = 0\):
\[ \text{pH} = 4.74 + 0 = 4.74 \]

### Final Answer:
The pH of the solution in the titration flask at the titrate value of 50 mL is **4.74**.