To calculate the new pH of the solution, we first need to determine the moles of acetic acid CH3COOH , sodium acetate CH3COONa , and hydrochloric acid HCl present in the solution.Initial moles of acetic acid CH3COOH = 0.1 M * 0.5 L = 0.05 molesInitial moles of sodium acetate CH3COONa = 0.1 M * 0.5 L = 0.05 molesMoles of hydrochloric acid HCl added = 1 M * 0.002 L = 0.002 molesWhen HCl is added to the buffer solution, it reacts with the sodium acetate to form acetic acid and sodium chloride NaCl :CH3COONa + HCl CH3COOH + NaClThe moles of sodium acetate and HCl will decrease by 0.002 moles, while the moles of acetic acid will increase by 0.002 moles:Final moles of acetic acid CH3COOH = 0.05 moles + 0.002 moles = 0.052 molesFinal moles of sodium acetate CH3COONa = 0.05 moles - 0.002 moles = 0.048 molesNow, we can use the Henderson-Hasselbalch equation to calculate the new pH of the solution:pH = pKa + log [A-]/[HA] The pKa of acetic acid is 4.74. The concentrations of acetic acid and sodium acetate can be calculated by dividing the moles by the total volume of the solution 500 mL + 2 mL = 502 mL = 0.502 L :[CH3COOH] = 0.052 moles / 0.502 L = 0.1036 M[CH3COONa] = 0.048 moles / 0.502 L = 0.0956 MNow, we can plug these values into the Henderson-Hasselbalch equation:pH = 4.74 + log 0.0956 / 0.1036 pH 4.74 - 0.036pH 4.704The new pH of the solution is approximately 4.704.The buffer system resists changes in pH because it contains both a weak acid acetic acid and its conjugate base sodium acetate . When a strong acid like HCl is added, it reacts with the conjugate base sodium acetate to form more of the weak acid acetic acid , which has a limited ability to donate protons and lower the pH. This reaction helps to neutralize the added acid and minimize the change in pH. Similarly, if a strong base were added, it would react with the weak acid to form more of the conjugate base, again minimizing the change in pH.