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

### Step 1: Write the neutralization reaction
The neutralization reaction between acetic acid (CH₃COOH) and sodium hydroxide (NaOH) can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

### Step 2: Calculate the number of moles of acetic acid and NaOH
Given that both solutions are 0.2 M and the volume is 500 mL (which is 0.5 L), we can calculate the moles of each reactant:
- Moles of acetic acid:
\[ \text{Moles of CH}_3\text{COOH} = 0.2 \, \text{M} \times 0.5 \, \text{L} = 0.1 \, \text{mol} \]
- Moles of NaOH:
\[ \text{Moles of NaOH} = 0.2 \, \text{M} \times 0.5 \, \text{L} = 0.1 \, \text{mol} \]

### Step 3: Determine the limiting reactant
Since both acetic acid and NaOH are present in equal amounts (0.1 mol each), they will completely neutralize each other, producing 0.1 mol of sodium acetate (CH₃COONa).

### Step 4: Calculate the concentration of sodium acetate
After the neutralization, the total volume of the solution is still approximately 500 mL (since the volume change is negligible). Thus, the concentration of sodium acetate in the solution is:
\[ \text{Concentration of CH}_3\text{COONa} = \frac{0.1 \, \text{mol}}{0.5 \, \text{L}} = 0.2 \, \text{M} \]

### Step 5: Use the Henderson-Hasselbalch equation
The pH of the solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \(\text{pK}_a\) of acetic acid = 4.74
- \([\text{A}^-]\) is the concentration of the conjugate base (sodium acetate) = 0.2 M
- \([\text{HA}]\) is the concentration of acetic acid after neutralization = 0 M (since it is fully neutralized)

Since acetic acid is fully neutralized, we can consider the pH to be determined by the sodium acetate alone. The formula simplifies to:
\[ \text{pH} = 7 + \frac{1}{2} \text{pK}_a + \log \left( \text{C} \right) \]
Where \(C\) is the concentration of the acetate ion (0.2 M).

### Step 6: Substitute the values into the equation
Substituting the values:
\[ \text{pH} = 7 + \frac{1}{2} \times 4.74 + \log(0.2) \]
Calculating \(\log(0.2)\):
\[ \log(0.2) = -0.699 \]
Now substituting:
\[ \text{pH} = 7 + 2.37 - 0.699 \]
\[ \text{pH} = 8.071 \]

### Final Calculation:
The pH of the resulting solution is approximately **8.07**.