The dissociation constant of acetic acid is `8 xx 10^(-5)` ta `25^(@)C`. Find the `pH` of
i. `M//10` ii. `M//100` solution of acetic acid.

Explanation

To find the pH of acetic acid solutions with concentrations of \( \frac{M}{10} \) and \( \frac{M}{100} \), we will follow these steps:

### Given:
- Dissociation constant of acetic acid, \( K_a = 8 \times 10^{-5} \)
- Temperature = \( 25^\circ C \)

### Step 1: Calculate \( pK_a \)
The \( pK_a \) is calculated using the formula:
\[ pK_a = -\log(K_a) \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(8 \times 10^{-5}) = -(\log(8) + \log(10^{-5})) = -\log(8) + 5 \]

Using \( \log(8) \approx 0.903 \):
\[ pK_a \approx -0.903 + 5 = 4.097 \]

### Step 2: Calculate pH for \( \frac{M}{10} \) solution
For a \( \frac{M}{10} \) solution:
\[ C = \frac{1}{10} M = 0.1 \, M \]
Using the formula for pH of a weak acid:
\[ pH = \frac{1}{2} pK_a - \log C \]
Substituting the values:
\[ pH = \frac{1}{2} \times 4.097 - \log(0.1) \]
Calculating \( \log(0.1) = -1 \):
\[ pH = 2.0485 + 1 = 3.0485 \]

### Step 3: Calculate pH for \( \frac{M}{100} \) solution
For a \( \frac{M}{100} \) solution:
\[ C = \frac{1}{100} M = 0.01 \, M \]
Using the same formula:
\[ pH = \frac{1}{2} pK_a - \log C \]
Substituting the values:
\[ pH = \frac{1}{2} \times 4.097 - \log(0.01) \]
Calculating \( \log(0.01) = -2 \):
\[ pH = 2.0485 + 2 = 4.0485 \]

### Final Answers:
- pH of \( \frac{M}{10} \) solution = 3.05 (approximately)
- pH of \( \frac{M}{100} \) solution = 4.05 (approximately)