Final answer:
To obtain a pH of 6.993 in a solution with 1.80 g of imidazole, one would need to add approximately 24.73 mL of 1.07 M HClO4, assuming a 1:1 stoichiometry and an imidazole to imidazolium ion ratio close to 1:1 due to the proximity of the desired pH to the pKa value.
Step-by-step explanation:
To calculate the amount of 1.07 M HClO4 needed to be added to 1.80 g of imidazole to achieve a pH of 6.993, we first need to determine the number of moles of imidazole present. The molar mass of imidazole (C3H4N2) is approximately 68.08 g/mol. The amount of imidazole in moles is therefore 1.80 g / 68.08 g/mol = 0.02644 moles.
Since imidazole is a weak base, we need to consider the pKa of its conjugate acid (imidazolium ion), which we may assume to be around 7.0 (as it's not given, this is a typical pKa value for such compounds). To achieve a pH of 6.993, which is slightly less than the pKa, we would need a slightly greater amount of HClO4 than imidazole to form the imidazolium ion and have a small excess of H3O+ ions to lower the pH.
The Henderson-Hasselbalch equation could usually be used here to determine the exact ratio of base to acid required, but since the pH is very close to the pKa, we can infer that the ratio of imidazole to imidazolium ion will be close to 1:1. Thus we would need approximately the same number of moles of HClO4 as we have moles of imidazole, which is 0.02644 moles. Since the concentration of HClO4 is 1.07 M, the volume of HClO4 needed is 0.02644 moles / 1.07 moles/L = 0.02473 L, or 24.73 mL.