What would be the pH of 0.1 molar sodium acetate solution, given that the dissociation constant of acetic acid is `1.8xx10^(-5)`.

Explanation

To find the pH of a 0.1 molar sodium acetate solution, we can use the relationship between the pH of a salt solution and the dissociation constant of its corresponding weak acid. Sodium acetate is the salt of acetic acid, which is a weak acid.

### Step-by-Step Solution:

1. **Identify the Weak Acid and Its Dissociation Constant**:
- The weak acid here is acetic acid (CH₃COOH).
- The dissociation constant (Kₐ) for acetic acid is given as \(1.8 \times 10^{-5}\).

2. **Calculate the pKₐ**:
- pKₐ is calculated using the formula:
\[ pK_a = -\log(K_a) \]
- Substituting the value of Kₐ:
\[ pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

3. **Use the Henderson-Hasselbalch Equation**:
- For a salt solution of a weak acid, the pH can be calculated using the formula:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
- In this case, since sodium acetate completely dissociates in solution, we can assume that the concentration of acetate ion \([A^-]\) is equal to the concentration of the sodium acetate solution, which is 0.1 M, and the concentration of the weak acid \([HA]\) is negligible (approaching 0).

4. **Substituting Values into the Equation**:
- Since \([HA]\) is negligible, we can simplify the equation:
\[ pH = pK_a + \log\left(\frac{[0.1]}{[0]}\right) \]
- However, we need to consider that we can use a modified version of the equation for a basic solution:
\[ pH = 7 + \frac{1}{2}pK_a + \log[C] \]
- Here, \(C\) is the concentration of the salt, which is 0.1 M.

5. **Calculating the pH**:
- Substitute \(pK_a\) and \(C\) into the modified equation:
\[ pH = 7 + \frac{1}{2}(4.74) + \log(0.1) \]
- Calculate \(\log(0.1)\):
\[ \log(0.1) = -1 \]
- Now substitute:
\[ pH = 7 + 2.37 - 1 \]
- Finally, calculate:
\[ pH = 8.37 \]

### Final Answer:
The pH of the 0.1 molar sodium acetate solution is approximately **8.37**.