Calculate the concentration of `H^+` ions in 0.2 M solution of `CH_3COOH`. `K_a` for `CH_3COOH=1.8xx10^(-5)`

Explanation

To calculate the concentration of \( H^+ \) ions in a 0.2 M solution of acetic acid (\( CH_3COOH \)), we will use the dissociation constant \( K_a \) and the equilibrium expression for the weak acid.

### Step-by-Step Solution:

1. **Write the dissociation equation:**
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]

2. **Set up the initial concentration and change in concentration:**
- Let the initial concentration of \( CH_3COOH \) be \( C = 0.2 \, M \).
- At equilibrium, let \( \alpha \) be the degree of dissociation. Thus, the concentrations at equilibrium will be:
- \( [CH_3COOH] = C - C\alpha = 0.2 - 0.2\alpha \)
- \( [CH_3COO^-] = C\alpha = 0.2\alpha \)
- \( [H^+] = C\alpha = 0.2\alpha \)

3. **Write the expression for \( K_a \):**
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} = \frac{(0.2\alpha)(0.2\alpha)}{0.2 - 0.2\alpha} \]

4. **Substitute \( K_a \) value:**
Given \( K_a = 1.8 \times 10^{-5} \):
\[ 1.8 \times 10^{-5} = \frac{(0.2\alpha)(0.2\alpha)}{0.2 - 0.2\alpha} \]

5. **Simplify the equation:**
Assuming \( \alpha \) is small, we can approximate \( 0.2 - 0.2\alpha \approx 0.2 \):
\[ 1.8 \times 10^{-5} = \frac{(0.2\alpha)^2}{0.2} \]
\[ 1.8 \times 10^{-5} = 0.2\alpha^2 \]

6. **Solve for \( \alpha^2 \):**
\[ \alpha^2 = \frac{1.8 \times 10^{-5}}{0.2} = 9.0 \times 10^{-5} \]

7. **Calculate \( \alpha \):**
\[ \alpha = \sqrt{9.0 \times 10^{-5}} \approx 0.00949 \]

8. **Calculate the concentration of \( H^+ \):**
\[ [H^+] = C\alpha = 0.2 \times 0.00949 \approx 0.001898 \, M \]

9. **Final result:**
The concentration of \( H^+ \) ions in the solution is approximately:
\[ [H^+] \approx 1.898 \times 10^{-3} \, M \]