The dissociation of acetic acid can be represented as:

HCl dissociated completely.
Now, let the concentration of acetate ion be x. Therefore, hydrogen ion contribution from acetic acid = x.
Since HCl dissociates completely, hydrogen ion contribution from HCl =0.1M
Total [H^+] concentration = x + 0.1
Now, Ka for acetic acid = 1.8 x 10^-5
Hence,

Introduction:
To find the percent dissociation of CH3COOH in an HCl solution, we need to consider the equilibrium reaction between CH3COOH and HCl. The dissociation constant (Ka) of CH3COOH will help us determine the extent of dissociation.

Equilibrium Reaction:
The equilibrium reaction between CH3COOH and HCl can be represented as follows:
CH3COOH + HCl ⇌ CH3COO- + H3O+

Expressing Equilibrium Concentrations:
Let's assume the initial concentration of CH3COOH is 0.2 M. Since HCl is a strong acid and dissociates completely, its concentration is 0.1 M. At equilibrium, let's assume x M of CH3COOH dissociates.

The equilibrium concentrations can be expressed as:
[CH3COOH] = 0.2 - x
[HCl] = 0.1
[CH3COO-] = x
[H3O+] = x

Writing the Expression for Ka:
The expression for Ka is given by:
Ka = ([CH3COO-] * [H3O+]) / [CH3COOH]

Substituting the equilibrium concentrations, the expression becomes:
Ka = (x * x) / (0.2 - x)

Using the Value of Ka:
Given that the Ka for CH3COOH is 1.8x10^-5, we can substitute this value into the equation:
1.8x10^-5 = (x * x) / (0.2 - x)

Solving for x:
Rearranging the equation, we get:
x^2 = 1.8x10^-5 * (0.2 - x)
x^2 = 3.6x10^-6 - 1.8x10^-5x
x^2 + 1.8x10^-5x - 3.6x10^-6 = 0

Solving this quadratic equation, we find that x ≈ 1.8x10^-3 M.

Calculating the Percent Dissociation:
The percent dissociation can be calculated using the formula:
% dissociation = (x / initial concentration) * 100

Substituting the values, we get:
% dissociation = (1.8x10^-3 / 0.2) * 100
% dissociation ≈ 0.9%

Conclusion:
The percent dissociation of 0.2 M CH3COOH in a 0.1 M HCl solution is approximately 0.9%. This means that only a small fraction of CH3COOH molecules dissociate into CH3COO- and H3O+ ions in the presence of HCl.

0.0175