First, we need to recognize that the reaction is actually a buffer system with the following equilibrium reaction:CH3COOH acetic acid + H2O CH3COO- acetate ion + H3O+ hydronium ion The equilibrium constant for this reaction is Ka, the acid dissociation constant, which is given as 1.8 x 10^-5.To find the concentration of CH3COONa sodium acetate at equilibrium, we can use the Henderson-Hasselbalch equation:pH = pKa + log [A-]/[HA] where pH is the negative logarithm of the H3O+ concentration, pKa is the negative logarithm of Ka, [A-] is the concentration of the conjugate base acetate ion , and [HA] is the concentration of the weak acid acetic acid .First, we need to find the pKa:pKa = -log Ka = -log 1.8 x 10^-5 = 4.74Next, we need to find the initial concentrations of acetic acid and acetate ion. Since sodium acetate dissociates completely in water, the initial concentration of acetate ion is 0.40 mol/L. The initial concentration of acetic acid is 0.25 mol/L.Now we can use the Henderson-Hasselbalch equation to find the pH of the solution:pH = 4.74 + log [0.40]/[0.25] = 4.74 + log 1.6 4.94Since the pH is less than 7, the solution is acidic, and there will be some dissociation of acetic acid. However, because the solution is a buffer, the concentrations of acetic acid and acetate ion will not change significantly. Therefore, the concentration of CH3COONa sodium acetate at equilibrium will still be approximately 0.40 mol/L.