First, we need to determine the amount of HCl and NaOH that will react with each other. Since HCl is a strong acid and NaOH is a strong base, they will react completely with each other to form water and NaCl. The limiting reactant in this case is NaOH, as it has a lower concentration.0.05 M NaOH will react with 0.05 M HCl, leaving 0.05 M HCl unreacted.Now, we have a solution containing 0.05 M HCl, 0.1 M CH3COOH, and 0.1 M CH3COONa. The unreacted HCl will react with the CH3COONa, which is the conjugate base of the weak acid CH3COOH. This reaction will produce CH3COOH and NaCl.CH3COONa + HCl -> CH3COOH + NaClSince the initial concentration of CH3COONa and CH3COOH are equal 0.1 M , and the concentration of HCl is 0.05 M, the reaction will consume 0.05 M of CH3COONa and produce 0.05 M of CH3COOH. The new concentrations will be:CH3COOH: 0.1 M + 0.05 M = 0.15 MCH3COONa: 0.1 M - 0.05 M = 0.05 MNow we can use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution:pH = pKa + log [A-]/[HA] where pKa = -log Ka , [A-] is the concentration of the conjugate base CH3COONa , and [HA] is the concentration of the weak acid CH3COOH .pKa = -log 1.8x10^-5 = 4.74pH = 4.74 + log 0.05/0.15 pH = 4.74 - 0.48pH 4.26The equilibrium position of the system after adding the buffer solution will have a pH of approximately 4.26.