0.1 M `CH_(3)`COOH solution is 1.0% ionized. In another diluted solution acetic acid is 10% ionised. In other solution concentration of acetic acid is

Options

A. 0.001 M
B. 0.01 M
C. 0.0001 M
D. 0.1 M

Answer

To solve the problem, we need to determine the concentration of acetic acid in a second solution where it is 10% ionized, given that a 0.1 M solution is 1% ionized. We will use the concept of ionization constant (K_a) for acetic acid, which remains the same for both solutions since they are the same acid.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- For the first solution:
- Concentration (C₁) = 0.1 M
- Degree of ionization (α₁) = 1% = 0.01
- For the second solution:
- Degree of ionization (α₂) = 10% = 0.10
- Concentration (C₂) = ?

2. **Write the Expression for Ionization Constant (K_a):**
The ionization of acetic acid (CH₃COOH) can be represented as:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
The expression for the ionization constant (K_a) is given by:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]

3. **Calculate K_a for the First Solution:**
- Initial concentration of acetic acid = C₁ = 0.1 M
- Change in concentration due to ionization = C₁ * α₁ = 0.1 * 0.01 = 0.001 M
- At equilibrium:
- [CH₃COO⁻] = 0.001 M
- [H⁺] = 0.001 M
- [CH₃COOH] = C₁ - C₁ * α₁ = 0.1 - 0.001 = 0.099 M

Now, substituting these values into the K_a expression:
\[ K_a = \frac{(0.001)(0.001)}{0.099} \approx \frac{0.000001}{0.099} \approx 1.01 \times 10^{-5} \]

4. **Use K_a to Find Concentration for the Second Solution:**
- For the second solution, we know:
- K_a = 1.01 × 10⁻⁵ (from the first solution)
- α₂ = 0.10

Using the K_a expression again:
\[ K_a = \frac{C_2 \cdot \alpha_2^2}{1 - \alpha_2} \approx \frac{C_2 \cdot (0.10)^2}{1 - 0.10} = \frac{C_2 \cdot 0.01}{0.90} \]
Setting this equal to K_a:
\[ 1.01 \times 10^{-5} = \frac{C_2 \cdot 0.01}{0.90} \]

5. **Solve for C₂:**
Rearranging the equation to solve for C₂:
\[ C_2 = \frac{1.01 \times 10^{-5} \cdot 0.90}{0.01} = 0.000909 \approx 0.001 \text{ M} \]

### Final Answer:
The concentration of acetic acid in the second solution is approximately **0.001 M**.