The hydrogen ion concentration in 0.2 M ethanoic acid
`(K_a = 2 xx 10^(-5) mol dm^(-3))` is

Options

A. `2 xx 10^(-2)`
B. `2 xx 10^(-4)`
C. `2 xx 10^(-3)`
D. `2 xx 10^(-5)`

Answer

To find the hydrogen ion concentration in a 0.2 M solution of ethanoic acid (acetic acid), we can use the dissociation constant \( K_a \) and the degree of dissociation.

### Step-by-Step Solution:

1. **Write the dissociation equation**:
Ethanoic acid (CH₃COOH) dissociates in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^- \]

2. **Define the initial concentration**:
Let the initial concentration of ethanoic acid be \( C = 0.2 \, \text{M} \).

3. **Define the degree of dissociation**:
Let \( \alpha \) be the degree of dissociation. At equilibrium, the concentration of hydrogen ions \( [\text{H}^+] \) and acetate ions \( [\text{CH}_3\text{COO}^-] \) will be \( C\alpha \).

4. **Set up the expression for \( K_a \)**:
The expression for the dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]
Substituting the concentrations at equilibrium, we have:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C^2\alpha^2}{C(1 - \alpha)} \]

5. **Simplify the equation**:
This simplifies to:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]

6. **Assume \( 1 - \alpha \approx 1 \)**:
For weak acids, we can assume that \( \alpha \) is small, so \( 1 - \alpha \approx 1 \). Thus:
\[ K_a \approx C\alpha^2 \]

7. **Rearranging for \( \alpha \)**:
We can rearrange this to find \( \alpha \):
\[ \alpha = \sqrt{\frac{K_a}{C}} \]

8. **Substituting the values**:
Given \( K_a = 2 \times 10^{-5} \, \text{M} \) and \( C = 0.2 \, \text{M} \):
\[ \alpha = \sqrt{\frac{2 \times 10^{-5}}{0.2}} = \sqrt{1 \times 10^{-4}} = 0.01 \]

9. **Calculate hydrogen ion concentration**:
The concentration of hydrogen ions \( [\text{H}^+] \) is given by:
\[ [\text{H}^+] = C\alpha = 0.2 \times 0.01 = 0.002 \, \text{M} = 2 \times 10^{-3} \, \text{M} \]

### Final Answer:
The hydrogen ion concentration in 0.2 M ethanoic acid is \( 2 \times 10^{-3} \, \text{M} \).

---