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 of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.For acetic acid, the pKa is 4.74. We are given the desired pH of 7.2. Plugging these values into the equation, we get:7.2 = 4.74 + log [A-] / [HA] Now, we need to solve for the ratio of [A-] to [HA]. Rearranging the equation:log [A-] / [HA] = 7.2 - 4.74log [A-] / [HA] = 2.46Taking the antilog base 10 of both sides:[A-] / [HA] = 10^2.46 288.4Now, let's set up the equation for the volumes of the two solutions. Let x be the volume of the acetic acid solution and y be the volume of the sodium acetate solution:x + y = 500 mL total volume Since we have 100 mL of 0.1 M acetic acid and 100 mL of 0.1 M sodium acetate, we can calculate the moles of each:moles of acetic acid HA = 0.1 M * x mLmoles of sodium acetate A- = 0.1 M * y mLNow, we can set up the ratio of [A-] to [HA]: y * 0.1 / x * 0.1 = 288.4Simplifying the equation:y / x = 288.4Now, we have a system of two equations with two variables:x + y = 500y = 288.4xWe can solve for x:x = 500 / 1 + 288.4 1.73 mLNow, we can solve for y:y = 288.4 * x 498.27 mLTherefore, the student needs to mix approximately 1.73 mL of the 0.1 M acetic acid solution and 498.27 mL of the 0.1 M sodium acetate solution to prepare the desired buffer solution with a total volume of 500 mL and a pH of 7.2.