To solve this problem, we will 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 for acetic acid, [A-] is the concentration of the conjugate base sodium acetate , and [HA] is the concentration of the weak acid acetic acid .The pKa of acetic acid is 4.74. We are given the desired pH 7.10 and the concentration of acetic acid 0.20 M . We can now solve for the concentration of sodium acetate:7.10 = 4.74 + log [NaCH3COO]/[CH3COOH] 2.36 = log [NaCH3COO]/0.20 10^2.36 = [NaCH3COO]/0.20229.09 = [NaCH3COO]/0.20[NaCH3COO] = 45.82 MNow, we need to find the number of moles of sodium acetate needed for the 250 mL buffer solution:moles of NaCH3COO = Molarity Volume in L moles of NaCH3COO = 45.82 0.250moles of NaCH3COO = 11.455Finally, we need to convert moles of sodium acetate to grams. The molar mass of NaCH3COO is approximately 82.03 g/mol:grams of NaCH3COO = moles molar massgrams of NaCH3COO = 11.455 82.03grams of NaCH3COO 939.56 gTherefore, the laboratory assistant will need approximately 939.56 grams of sodium acetate to make the buffer solution with a pH of 7.10.