To solve this problem, we can use the Henderson-Hasselbalch equation:pH = pKa + log [A-]/[HA] where pH is the solution's pH, pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base acetate ion , and [HA] is the concentration of the weak acid acetic acid .Initially, the pH is 4.5, and the pKa of acetic acid is 4.76. We can plug these values into the equation to find the initial ratio of [A-]/[HA]:4.5 = 4.76 + log [A-]/[HA] Rearranging the equation, we get:log [A-]/[HA] = -0.26Taking the antilog of both sides:[A-]/[HA] = 10^-0.26 0.55Now, let's consider the addition of 0.02 moles of HCl. The HCl will react with the sodium acetate the conjugate base to form acetic acid and sodium chloride:NaCH3COO + HCl CH3COOH + NaClSince both acetic acid and sodium acetate have initial concentrations of 0.1 M, we can calculate the new concentrations after the reaction:[A-] = 0.1 - 0.02 = 0.08 M sodium acetate [HA] = 0.1 + 0.02 = 0.12 M acetic acid Now we can plug these new concentrations back into the Henderson-Hasselbalch equation:pH = 4.76 + log [A-]/[HA] pH = 4.76 + log 0.08/0.12 pH = 4.76 - 0.41pH 4.35So, after adding 0.02 moles of HCl, the pH of the solution will be approximately 4.35.