To calculate the pH of a 0.1 M solution of acetic acid, we can use the formula for the dissociation constant Ka and the definition of pH. The dissociation of acetic acid CH3COOH in water can be represented as:CH3COOH CH3COO- + H+The Ka expression for this reaction is:Ka = [CH3COO-][H+]/[CH3COOH]Let x be the concentration of H+ ions at equilibrium. Since the reaction is 1:1, the concentration of CH3COO- ions will also be x, and the concentration of CH3COOH will be 0.1 - x. We can now write the Ka expression as:1.8 x 10^-5 = x x / 0.1 - x Since Ka is very small, we can assume that x is much smaller than 0.1, so we can approximate the equation as:1.8 x 10^-5 x^2/0.1Now, we can solve for x:x^2 1.8 x 10^-5 * 0.1x^2 1.8 x 10^-6x 1.8 x 10^-6 x 1.34 x 10^-3Now that we have the concentration of H+ ions x , we can calculate the pH using the formula:pH = -log[H+]pH = -log 1.34 x 10^-3 pH 2.87So, the pH of a 0.1 M solution of acetic acid with a dissociation constant Ka of 1.8 x 10^-5 is approximately 2.87.