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 negative logarithm of the acid dissociation constant Ka , [A-] is the concentration of the conjugate base sodium acetate , and [HA] is the concentration of the weak acid acetic acid .First, we need to find the pKa of acetic acid:pKa = -log Ka = -log 1.8x10^-5 = 4.74Now, we can plug the values into the Henderson-Hasselbalch equation:7.20 = 4.74 + log [NaCH3COO]/[CH3COOH] Rearrange the equation to solve for the ratio of [NaCH3COO] to [CH3COOH]:log [NaCH3COO]/[CH3COOH] = 7.20 - 4.74 = 2.46Now, we can find the ratio of the concentrations:[NaCH3COO]/[CH3COOH] = 10^2.46 288.4We know that the total buffer concentration is 0.20 M, and the volume of the solution is 200 mL 0.2 L . Therefore, the total moles of buffer components in the solution are:0.20 M * 0.2 L = 0.04 molesSince we already have 200 mL of 0.10 M NaCH3COO, we have:0.10 M * 0.2 L = 0.02 moles of NaCH3COONow, we can use the ratio we found earlier to determine the moles of CH3COOH:0.02 moles NaCH3COO / 288.4 = x moles CH3COOH / 1x moles CH3COOH = 0.02 moles / 288.4 6.94x10^-5 molesFinally, we can find the mass of CH3COOH needed by multiplying the moles by the molar mass of CH3COOH 60.05 g/mol :mass CH3COOH = 6.94x10^-5 moles * 60.05 g/mol 0.00417 gSo, the student needs to add approximately 0.00417 grams of CH3COOH to the 200 mL of 0.10 M NaCH3COO to obtain the desired pH of 7.20.