The dissociation of acetic acid CH3COOH in water can be represented by the following chemical equation:CH3COOH aq CH3COO- aq + H+ aq The equilibrium constant expression for this reaction is given by:K_a = [CH3COO-][H+]/[CH3COOH]Given that the pH of the solution is 4.5, we can find the concentration of H+ ions:pH = -log[H+]4.5 = -log[H+][H+] = 10^-4.5 = 3.16 x 10^-5 MLet x be the change in concentration of CH3COOH, CH3COO-, and H+ at equilibrium. Then:[CH3COOH] = 0.1 - x[CH3COO-] = x[H+] = 3.16 x 10^-5 + xSince the initial concentration of acetic acid is much larger than the concentration of H+ ions, we can assume that x is small compared to 0.1 M. Therefore, we can approximate [CH3COOH] 0.1 M. Now, we can substitute these values into the K_a expression:K_a = x 3.16 x 10^-5 + x / 0.1 Since x is small, we can also assume that 3.16 x 10^-5 + x 3.16 x 10^-5 . Now, we can solve for K_a:K_a = x 3.16 x 10^-5 / 0.1 x = [CH3COO-] = [H+] - 3.16 x 10^-5 = 3.16 x 10^-5 M since x is small K_a = 3.16 x 10^-5 3.16 x 10^-5 / 0.1 = 1.00 x 10^-9 The equilibrium constant value K_a for the dissociation of acetic acid in water is approximately 1.00 x 10^-9 .