When `0.2M` solution of acetic acid is neutralised with `0.2M NaOH` in `500 mL` of water, the `pH` of the resulting solution will be: `[pK_(a)` of acetic acid `= 4.74]`

Options

A. `12.67`
B. `7.87`
C. `8.87`
D. `7`

Explanation

To find the pH of the resulting solution when a 0.2 M solution of acetic acid is neutralized with 0.2 M NaOH in 500 mL of water, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the moles of NaOH and acetic acid:**
- The concentration of NaOH is 0.2 M and the volume is 500 mL (0.5 L).
- Moles of NaOH = Molarity × Volume = 0.2 mol/L × 0.5 L = 0.1 moles (or 100 millimoles).
- The concentration of acetic acid is also 0.2 M and the volume is 500 mL (0.5 L).
- Moles of acetic acid = 0.2 mol/L × 0.5 L = 0.1 moles (or 100 millimoles).

2. **Neutralization Reaction:**
- The neutralization reaction between acetic acid (CH₃COOH) and NaOH produces sodium acetate (CH₃COONa) and water:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]
- Since both reactants are present in equal moles (100 millimoles), they will completely neutralize each other.

3. **Calculate the total volume of the solution:**
- Total volume after mixing = Volume of acetic acid + Volume of NaOH = 500 mL + 500 mL = 1000 mL (or 1 L).

4. **Determine the concentration of the salt (sodium acetate):**
- After neutralization, we have 100 millimoles of sodium acetate in a total volume of 1000 mL.
- Concentration of sodium acetate = Moles / Volume = 0.1 moles / 1 L = 0.1 M.

5. **Calculate the pH using the hydrolysis of the salt:**
- Sodium acetate is a salt of a weak acid (acetic acid) and a strong base (NaOH). The pH can be calculated using the formula:
\[ \text{pH} = 7 + \frac{1}{2} \text{pK}_a + \log [\text{Salt concentration}] \]
- Given pKₐ of acetic acid = 4.74, we can substitute the values:
\[ \text{pH} = 7 + \frac{1}{2} \times 4.74 + \log(0.1) \]
- Calculate:
\[ \text{pH} = 7 + 2.37 - 1 = 8.37 \]

6. **Final pH Calculation:**
- The final pH of the resulting solution after neutralization is approximately **8.37**.