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 .The pKa of acetic acid is 4.74. We can plug in the given values and solve for the ratio of [A-] to [HA]:5.2 = 4.74 + log [A-]/[HA] 0.46 = log [A-]/[HA] To remove the logarithm, we can use the inverse function, which is the exponentiation with base 10:10^0.46 = [A-]/[HA]2.885 = [A-]/[HA]Now, we need to find the moles of acetic acid and sodium acetate:moles of acetic acid HA = 0.15 M * 10 mL = 1.5 mmolmoles of sodium acetate A- = 0.20 M * 15 mL = 3 mmolWe can now plug these values into the equation we derived earlier:2.885 = 3 mmol / 1.5 mmol The ratio is satisfied, so the current mixture will give us the desired pH of 5.2. Now we need to find the volume of water to add.Since adding water does not change the number of moles of acetic acid and sodium acetate, the ratio of their concentrations will remain the same. Therefore, we can add any volume of water to the solution, and the pH will still be 5.2.In conclusion, the student can add any volume of water to the mixture of 10 mL of 0.15 M acetic acid and 15 mL of 0.20 M sodium acetate to achieve a buffer solution with a pH of 5.2.