To calculate the pH of a 0.1 M solution of acetic acid CH3COOH , we need to know the acid dissociation constant Ka for acetic acid. The Ka for acetic acid at 25C is 1.8 x 10^-5.The dissociation of acetic acid in water can be represented by the following equilibrium equation:CH3COOH aq CH3COO- aq + H+ aq We can set up an equilibrium expression using the Ka value:Ka = [CH3COO-][H+]/[CH3COOH]Let x represent the concentration of H+ ions and CH3COO- ions at equilibrium. Since the initial concentration of acetic acid is 0.1 M, the equilibrium concentrations can be represented as:[CH3COOH] = 0.1 - x[CH3COO-] = x[H+] = xNow, we can substitute these values into the Ka expression: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 simplify the equation: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:pH = -log[H+]pH = -log 1.34 x 10^-3 pH 2.87The pH of a 0.1 M solution of acetic acid at 25C is approximately 2.87.