Find the change in pH when `0.01` mole `CH_(3)COONa` is added to one litre of `0.01 M CH_(3)COOH` soluton `(pK_(a) = 4.74)`

Options

A. `3.27`
B. `4.74`
C. `1.37`
D. `2.74`

Explanation

To find the change in pH when `0.01` mole of `CH₃COONa` (sodium acetate) is added to one liter of `0.01 M CH₃COOH` (acetic acid) solution, we can follow these steps:

### Step 1: Calculate the initial pH of the acetic acid solution

1. **Determine the dissociation constant (Ka) of acetic acid**:
The dissociation constant \( K_a \) can be calculated from the \( pK_a \):
\[ K_a = 10^{-pK_a} = 10^{-4.74} \]

2. **Calculate the concentration of \( H^+ \)**:
For a weak acid, the concentration of \( H^+ \) ions can be approximated using:
\[ [H^+] = \sqrt{K_a \cdot C} \]
where \( C \) is the concentration of the acid (0.01 M).

3. **Calculate the pH**:
The pH can be calculated as:
\[ pH = -\log[H^+] \]

### Step 2: Calculate the pH after adding sodium acetate

1. **Understand the buffer system**:
When sodium acetate is added to the acetic acid solution, it forms a buffer system. The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
where \( [A^-] \) is the concentration of the salt (sodium acetate) and \( [HA] \) is the concentration of the weak acid (acetic acid).

2. **Calculate the concentrations**:
- Moles of acetic acid: \( 0.01 \, \text{mol} \) in \( 1 \, \text{L} \) gives \( [HA] = 0.01 \, \text{M} \).
- Moles of sodium acetate: \( 0.01 \, \text{mol} \) in \( 1 \, \text{L} \) gives \( [A^-] = 0.01 \, \text{M} \).

3. **Substitute into the Henderson-Hasselbalch equation**:
\[ pH = 4.74 + \log\left(\frac{0.01}{0.01}\right) = 4.74 + \log(1) = 4.74 \]

### Step 3: Calculate the change in pH

1. **Find the change in pH**:
\[ \Delta pH = pH_{\text{final}} - pH_{\text{initial}} = 4.74 - 3.37 \]

2. **Calculate**:
\[ \Delta pH = 1.37 \]

### Final Answer:
The change in pH when `0.01` mole of `CH₃COONa` is added to one liter of `0.01 M CH₃COOH` solution is **1.37**.

---