The `pH` of `0.1 M` solution of `CH_(3)COOH` if it ionizes to an extent of `1%` is .

Options

A. `1`
B. `2`
C. `3`
D. `4`

Explanation

To find the pH of a 0.1 M solution of acetic acid (CH₃COOH) that ionizes to an extent of 1%, we can follow these steps:

### Step 1: Determine the concentration of acetic acid
The initial concentration of acetic acid (C) is given as 0.1 M.

### Step 2: Calculate the degree of dissociation (α)
The degree of dissociation (α) is given as 1%. To convert this percentage into a fraction, we can express it as:
\[ \alpha = \frac{1}{100} = 0.01 \]

### Step 3: Calculate the concentration of H⁺ ions
The concentration of H⁺ ions produced from the dissociation of acetic acid can be calculated using the formula:
\[ [\text{H}^+] = C \times \alpha \]
Substituting the values:
\[ [\text{H}^+] = 0.1 \, \text{M} \times 0.01 = 0.001 \, \text{M} \]

### Step 4: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the concentration of H⁺ ions:
\[ \text{pH} = -\log(0.001) = -\log(10^{-3}) = 3 \]

### Final Answer
Thus, the pH of the 0.1 M solution of acetic acid that ionizes to an extent of 1% is **3**.

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