Understanding the Problem
To find the percentage dissociation of 0.2 M acetic acid (CH₃COOH) in 0.1 M HCl, we need to consider the common ion effect due to the presence of HCl, which will suppress the dissociation of acetic acid.

Given Data
- \( K_a \) (for CH₃COOH) = \( 1.8 \times 10^{-5} \)
- Initial concentration of CH₃COOH = 0.2 M
- Concentration of HCl = 0.1 M

Initial Setup
1. **Dissociation of Acetic Acid**:
\( CH₃COOH \rightleftharpoons CH₃COO^- + H^+ \)
2. **Equilibrium Concentrations**:
- Let \( x \) be the amount dissociated.
- At equilibrium:
- \( [CH₃COOH] = 0.2 - x \)
- \( [H^+] = 0.1 + x \)
- \( [CH₃COO^-] = x \)

Applying the Equilibrium Expression
Using the equilibrium expression for acetic acid:
\[ K_a = \frac{[CH₃COO^-][H^+]}{[CH₃COOH]} \]
Substituting the equilibrium concentrations:
\[ 1.8 \times 10^{-5} = \frac{x(0.1 + x)}{0.2 - x} \]
Assuming \( x \) is small compared to 0.1:
\[ 1.8 \times 10^{-5} = \frac{x(0.1)}{0.2} \]

Solving for x
1. Rearranging gives:
\[ x = \frac{1.8 \times 10^{-5} \times 0.2}{0.1} \]
\[ x = 3.6 \times 10^{-5} \]

Calculating Percentage Dissociation
\[ \text{Percentage Dissociation} = \left( \frac{x}{0.2} \right) \times 100 \]
Substituting \( x \):
\[ \text{Percentage Dissociation} = \left( \frac{3.6 \times 10^{-5}}{0.2} \right) \times 100 = 0.018\% \]

Final Answer
The percentage dissociation of 0.2 M CH₃COOH in 0.1 M HCl is **0.018%** (Option A).

(A)