To solve this problem, we will use the Henderson-Hasselbalch equation:pH = pKa + log [A-]/[HA] First, we need to find the pKa of the acetic acid CH3COOH . The Ka of acetic acid is 1.8 x 10^-5. To find the pKa, we use the following equation:pKa = -log Ka pKa = -log 1.8 x 10^-5 = 4.74Now, let's find the initial concentrations of CH3COO- A- and CH3COOH HA in the 0.2 M sodium acetate solution. Since sodium acetate dissociates completely, the initial concentration of CH3COO- is 0.2 M. The initial concentration of CH3COOH is 0 since it is not present initially.When 0.1 mol of NaOH is added, it will react with the CH3COOH to form more CH3COO- and water. Let's calculate the new concentrations of CH3COO- and CH3COOH after the reaction:Moles of CH3COO- = 0.2 M * volume of the solution in L Moles of NaOH = 0.1 molSince NaOH reacts with CH3COOH in a 1:1 ratio, the moles of CH3COOH formed will be equal to the moles of NaOH added:Moles of CH3COOH = 0.1 molThe moles of CH3COO- will increase by the moles of NaOH added:Moles of CH3COO- new = Moles of CH3COO- initial + Moles of NaOHNow, we need to find the new volume of the solution after adding NaOH. Let's assume the volume of NaOH added is V_NaOH in L . The new volume of the solution will be:New volume = Initial volume + V_NaOHNow, we can calculate the new concentrations of CH3COO- and CH3COOH:[CH3COO-] = Moles of CH3COO- new / New volume[CH3COOH] = Moles of CH3COOH / New volumeNow, we can plug these values into the Henderson-Hasselbalch equation:pH = 4.74 + log [CH3COO-]/[CH3COOH] Since we don't have the exact volume of the solution or the volume of NaOH added, we cannot calculate the exact pH value. However, we can say that the pH will increase due to the addition of NaOH, as it will shift the equilibrium towards the formation of more CH3COO- and OH-.