The Ka is given by the expression:

Ka = [H+][CH3COO-] / [CH3COOH]

In this case, the initial concentration of acetic acid is 0.2 M, and initially, there is no reaction, so the concentration of CH3COO- and H+ ions is 0 M.

Let's assume that x is the concentration of H+ ions at equilibrium. Then, the concentration of CH3COO- ions will also be x, and the concentration of CH3COOH will be 0.2 - x.

Now we can substitute these values into the Ka expression:

1.8 x 10^-5 = x * x / (0.2 - x)

Simplifying the equation:

1.8 x 10^-5 = x^2 / (0.2 - x)
x^2 = 1.8 x 10^-5 * (0.2 - x)
x^2 = 3.6 x 10^-6 - 1.8 x 10^-5x

Since the value of x is expected to be very small compared to 0.2, we can neglect the -1.8 x 10^-5x term compared to 3.6 x 10^-6 to simplify the equation further.

x^2 = 3.6 x 10^-6

Taking the square root of both sides:

x = √(3.6 x 10^-6)

x ≈ 1.897 x 10^-3 M

Now, we can find the pOH of the solution using the equation:

pOH = -log10 [OH-]

Since [OH-] = x (since the concentration of H+ ions and OH- ions in water is the same), we find:

pOH = -log10 (1.897 x 10^-3)
pOH ≈ 2.72

Finally, we can calculate the pH using the equation:

pH = 14 - pOH
pH ≈ 14 - 2.72
pH ≈ 11.28

Therefore, the pH of the 0.2 M acetic acid solution is approximately 11.28.

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