The pH of the solution obtained on neutralisation of 40 mL 0.1 M NaOH with 40 ml 0.1 M` CH_(3)COOH` is

Options

A. 7
B. 8
C. 6
D. 3

Explanation

To solve the problem of finding the pH of the solution obtained from the neutralization of 40 mL of 0.1 M NaOH with 40 mL of 0.1 M acetic acid (CH₃COOH), we can follow these steps:

### Step 1: Determine the moles of NaOH and acetic acid
- **Moles of NaOH** = Volume (L) × Molarity (mol/L)
- Volume of NaOH = 40 mL = 0.040 L
- Molarity of NaOH = 0.1 M
- Moles of NaOH = 0.040 L × 0.1 mol/L = 0.004 moles

- **Moles of acetic acid (CH₃COOH)** = Volume (L) × Molarity (mol/L)
- Volume of acetic acid = 40 mL = 0.040 L
- Molarity of acetic acid = 0.1 M
- Moles of acetic acid = 0.040 L × 0.1 mol/L = 0.004 moles

### Step 2: Determine the neutralization reaction
The neutralization reaction between NaOH and acetic acid is:
\[ \text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

Since the moles of NaOH and acetic acid are equal (0.004 moles each), they will completely neutralize each other.

### Step 3: Identify the resulting solution
After neutralization, the solution contains:
- 0.004 moles of sodium acetate (CH₃COONa), which is the salt formed.
- The solution is now a buffer solution because it contains the conjugate base (acetate ion, CH₃COO⁻) from acetic acid.

### Step 4: Calculate the pH of the resulting solution
To find the pH of the resulting solution, we can use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \(\text{pK}_a\) of acetic acid is approximately 4.76.
- \([\text{A}^-]\) is the concentration of acetate ion (CH₃COO⁻).
- \([\text{HA}]\) is the concentration of acetic acid (CH₃COOH).

Since all acetic acid has been neutralized, \([\text{HA}] = 0\) and we only have the acetate ion. However, the acetate ion will hydrolyze in water, which will affect the pH.

### Step 5: Calculate the concentration of acetate ion
The total volume of the solution after mixing is:
- Total volume = 40 mL + 40 mL = 80 mL = 0.080 L

The concentration of acetate ion (\([\text{A}^-]\)) is:
\[ [\text{A}^-] = \frac{0.004 \text{ moles}}{0.080 \text{ L}} = 0.05 \text{ M} \]

### Step 6: Calculate the pH using the hydrolysis of acetate ion
The acetate ion will hydrolyze according to the reaction:
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

Using the Kb of acetate ion:
\[ K_b = \frac{K_w}{K_a} \]

Where \( K_w = 1.0 \times 10^{-14} \) and \( K_a \) for acetic acid is \( 1.8 \times 10^{-5} \):
\[ K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]

Using the Kb expression:
\[ K_b = \frac{[OH^-]^2}{[\text{CH}_3\text{COO}^-]} \]

Assuming \( x \) is the concentration of \( OH^- \):
\[ 5.56 \times 10^{-10} = \frac{x^2}{0.05} \]
\[ x^2 = 5.56 \times 10^{-10} \times 0.05 \]
\[ x^2 = 2.78 \times 10^{-11} \]
\[ x = \sqrt{2.78 \times 10^{-11}} \approx 5.27 \times 10^{-6} \]

### Step 7: Calculate pOH and then pH
\[ pOH = -\log(5.27 \times 10^{-6}) \approx 5.28 \]
\[ pH = 14 - pOH = 14 - 5.28 = 8.72 \]

### Final Answer
The pH of the solution obtained on neutralization of 40 mL of 0.1 M NaOH with 40 mL of 0.1 M acetic acid is approximately **8.72**.

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