To solve this problem, we will use the Henderson-Hasselbalch equation, which relates the pH, pKa, and the ratio of the concentrations of the conjugate base [A-] and the weak acid [HA] :pH = pKa + log [A-]/[HA] First, we need to find the pKa of acetic acid. The pKa of acetic acid is 4.74.Now, we can use the initial pH and pKa values to find the initial ratio of the concentrations of the conjugate base and the weak acid:4.5 = 4.74 + log [CH3COO-]/[CH3COOH] Rearranging the equation, we get:log [CH3COO-]/[CH3COOH] = 4.5 - 4.74 = -0.24Taking the antilog of both sides:[CH3COO-]/[CH3COOH] = 10^-0.24 0.58Now, let's find the initial concentration of CH3COO-:[CH3COO-] = 0.58 * [CH3COOH] = 0.58 * 0.2 M = 0.116 MWhen the student adds the buffer solution containing equal amounts of acetic acid and sodium acetate, the concentrations of CH3COOH and CH3COO- will both increase. Let's say the student adds x moles of the buffer solution to the initial solution. The new concentrations will be:[CH3COOH] = 0.2 + x M[CH3COO-] = 0.116 + x MSince the buffer solution contains equal amounts of acetic acid and sodium acetate, the ratio of the concentrations of the conjugate base and the weak acid will remain the same: [CH3COO-] + x / [CH3COOH] + x = 0.58Now, we can use the Henderson-Hasselbalch equation again to find the new pH of the solution:pH = pKa + log [CH3COO-]/[CH3COOH] Since the ratio of the concentrations remains the same, the new pH will be:pH = 4.74 + log 0.58 4.5Thus, the pH of the solution remains the same at 4.5 after adding the buffer solution.As for the extent of the shift in the equilibrium position, since the pH remains the same, the equilibrium position does not shift significantly. The addition of the buffer solution helps to maintain the pH of the solution by resisting changes in the concentrations of the weak acid and its conjugate base.