To solve this problem, we can use the Henderson-Hasselbalch equation:pH = pKa + log [A-]/[HA] where pH is the desired pH of the buffer solution, pKa is the acid dissociation constant of the weak acid, [A-] is the concentration of the conjugate base sodium acetate , and [HA] is the concentration of the weak acid acetic acid .We are given the pH 4.5 , pKa 4.76 , and the concentration of acetic acid 0.1 M . We need to find the concentration of sodium acetate [A-] .4.5 = 4.76 + log [A-]/0.1 Rearranging the equation to solve for [A-]:log [A-]/0.1 = 4.5 - 4.76log [A-]/0.1 = -0.26Now, we can find the ratio of [A-] to 0.1:[A-]/0.1 = 10^-0.26 [A-] = 0.1 * 10^-0.26 Calculating the concentration of sodium acetate:[A-] 0.1 * 0.55[A-] 0.055 MNow, we need to find the mass of sodium acetate needed to achieve this concentration in 100 mL of solution:moles of sodium acetate = 0.055 moles/L * 0.100 Lmoles of sodium acetate 0.0055 molesThe molar mass of sodium acetate CH3COONa is approximately 82.03 g/mol. Therefore, the mass of sodium acetate needed is:mass of sodium acetate = 0.0055 moles * 82.03 g/molmass of sodium acetate 0.451 gThe student should add approximately 0.451 grams of sodium acetate to the solution to achieve the desired pH of 4.5.