0.01M `CH_(3),COOH` is 4.24% ionised. What will be the pereentage ionisation of 0.1 `M CH_(3),COOH.`

Options

A. `1.34%`
B. `4.23%`
C. `5.24%`
D. `0.33%`

Answer

To find the percentage ionization of 0.1 M acetic acid (CH₃COOH) given that 0.01 M acetic acid is 4.24% ionized, we can use the relationship between the degree of ionization (α) and concentration (C) for a weak acid.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Concentration of acetic acid (C₁) = 0.01 M
- Percentage ionization of 0.01 M acetic acid = 4.24%
- Concentration of acetic acid (C₂) = 0.1 M

2. **Convert Percentage Ionization to Degree of Ionization:**
- Degree of ionization (α₁) for 0.01 M acetic acid:
\[ \alpha_1 = \frac{4.24}{100} = 0.0424 \]

3. **Use the Relationship Between Degrees of Ionization and Concentrations:**
- The formula relating the degrees of ionization and concentrations is:
\[ \frac{\alpha_1}{\alpha_2} = \sqrt{\frac{C_2}{C_1}} \]
- Here, α₂ is the degree of ionization for 0.1 M acetic acid.

4. **Substitute the Values into the Formula:**
- Substitute C₁ = 0.01 M and C₂ = 0.1 M into the formula:
\[ \frac{0.0424}{\alpha_2} = \sqrt{\frac{0.1}{0.01}} \]
- Simplifying the right side:
\[ \sqrt{\frac{0.1}{0.01}} = \sqrt{10} \approx 3.1623 \]

5. **Rearranging the Equation to Solve for α₂:**
- Rearranging gives:
\[ \alpha_2 = \frac{0.0424}{3.1623} \]

6. **Calculate α₂:**
- Performing the calculation:
\[ \alpha_2 \approx \frac{0.0424}{3.1623} \approx 0.0134 \]

7. **Convert Degree of Ionization Back to Percentage:**
- Convert α₂ to percentage:
\[ \text{Percentage Ionization} = \alpha_2 \times 100 = 0.0134 \times 100 \approx 1.34\% \]

### Final Answer:
The percentage ionization of 0.1 M acetic acid is approximately **1.34%**.

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