`CH_3COOH` is 1% ionised in its aqueous solution of 0.1 M strength . Its pOH will be

Options

A. 11
B. 3
C. 13
D. 2

Answer

To find the pOH of a 0.1 M acetic acid solution that is 1% ionized, we can follow these steps:

### Step 1: Determine the ionization of acetic acid
Acetic acid (CH₃COOH) ionizes in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Calculate the concentration of ionized species
Given that the concentration of acetic acid is 0.1 M and it is 1% ionized:
- The degree of ionization (α) is given as 1%, which can be expressed as:
\[ \alpha = \frac{1}{100} = 0.01 \]

### Step 3: Calculate the concentration of H⁺ ions
The concentration of H⁺ ions produced from the ionization can be calculated using the formula:
\[ [\text{H}^+] = C \cdot \alpha \]
Where:
- C = initial concentration of acetic acid = 0.1 M
- α = degree of ionization = 0.01

Substituting the values:
\[ [\text{H}^+] = 0.1 \times 0.01 = 0.001 \, \text{M} \]

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

### Step 5: Calculate the pOH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
We can rearrange this to find pOH:
\[ \text{pOH} = 14 - \text{pH} \]
Substituting the value of pH:
\[ \text{pOH} = 14 - 3 = 11 \]

### Final Answer
Thus, the pOH of the solution is **11**.

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