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 of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.First, we need to find the pKa of acetic acid. The pKa of acetic acid is 4.76.The initial pH of the solution is 4.7, which is close to the pKa of acetic acid. This means that the ratio of [A-] to [HA] is approximately 1:1. Since the solution is 0.1 M in both acetic acid and sodium acetate, we can assume that [A-] = 0.1 M and [HA] = 0.1 M.Now, we will add 0.01 mol of hydrochloric acid HCl to the solution. HCl is a strong acid, so it will react completely with the sodium acetate the conjugate base to form acetic acid and sodium chloride:HCl + NaCH3COO -> CH3COOH + NaClSince we are adding 0.01 mol of HCl to a 500 mL solution, the concentration of HCl is:[HCl] = 0.01 mol / 0.5 L = 0.02 MThe reaction will consume 0.02 M of sodium acetate and produce 0.02 M of acetic acid. The new concentrations of acetic acid and sodium acetate are:[CH3COOH] = 0.1 M + 0.02 M = 0.12 M[NaCH3COO] = 0.1 M - 0.02 M = 0.08 MNow we can use the Henderson-Hasselbalch equation to find the new pH:pH = pKa + log [A-]/[HA] pH = 4.76 + log 0.08/0.12 pH 4.76 - 0.4pH 4.36The new pH of the solution after adding 0.01 mol of hydrochloric acid is approximately 4.36.