To solve this problem, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa and the ratio of the concentrations of the conjugate base A- and the weak acid HA :pH = pKa + log [A-]/[HA] First, we need to find the pKa of acetic acid. The pKa is the negative logarithm of the Ka:pKa = -log Ka = -log 1.8 x 10^-5 4.74Now, we have a 0.1 M solution of both acetic acid HA and its corresponding acetate ion A- . Since the concentrations of the acid and its conjugate base are equal, the ratio [A-]/[HA] is 1:[A-]/[HA] = [CH3COO-]/[CH3COOH] = 0.1 M / 0.1 M = 1Now we can plug these values into the Henderson-Hasselbalch equation:pH = pKa + log [A-]/[HA] = 4.74 + log 1 = 4.74 + 0 = 4.74Therefore, the pH of the 0.1 M solution of acetic acid and its corresponding acetate ion is 4.74.