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 solution where it is 10% ionized, given that a 0.1 M solution is 1% ionized. We'll use the concept of ionization and the equilibrium constant (K) for acetic acid.

### Step-by-Step Solution:

1. **Understanding Ionization**:
- The ionization of acetic acid (CH₃COOH) can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
- The degree of ionization (α) is the fraction of the acid that ionizes.

2. **Calculate K for the First Solution**:
- For the first solution:
- Concentration (C) = 0.1 M
- Ionization (α) = 1% = 0.01
- At equilibrium:
- Concentration of CH₃COOH = C(1 - α) = 0.1(1 - 0.01) = 0.1 × 0.99 = 0.099 M
- Concentration of CH₃COO⁻ = Concentration of H⁺ = Cα = 0.1 × 0.01 = 0.001 M
- The equilibrium constant (K) is given by:
\[ K = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} = \frac{(0.001)(0.001)}{0.099} = \frac{10^{-6}}{0.099} \approx 10^{-5} \]

3. **Using K for the Second Solution**:
- For the second solution:
- Ionization (α) = 10% = 0.10
- We need to find the concentration (C).
- Using the same expression for K:
\[ K = Cα^2 \implies 10^{-5} = C(0.10)^2 = C(0.01) \]
- Rearranging gives:
\[ C = \frac{10^{-5}}{0.01} = 10^{-3} \text{ M} \]

4. **Final Answer**:
- The concentration of acetic acid in the second solution is:
\[ C = 0.001 \text{ M} \]