Final answer:
To reach the equivalence point in the titration of 100 ml of 0.250 M aniline with 0.5 M HCl, 50 ml of HCl is required, considering a 1:1 mole ratio between aniline and HCl.
Step-by-step explanation:
To determine the volume of HCl required to reach the equivalence point in the titration of 100 ml of 0.250 M aniline with 0.5 M HCl, we use the concept of molarity and the stoichiometry of the titration reaction. Aniline (C6H5NH2) reacts with HCl in a 1:1 mole ratio.
First, calculate the moles of aniline in 100 ml:
Moles of aniline = Molarity (M) × Volume (L)
Moles of aniline = 0.250 M × 0.100 L = 0.025 moles.
Since the reaction is in a 1:1 mole ratio, the moles of HCl required will also be 0.025 moles. Now, to find the volume of 0.5 M HCl needed, use the moles of HCl and its molarity:
Volume of HCl = Moles of HCl / Molarity of HCl
Volume of HCl = 0.025 moles / 0.5 M = 0.050 L or 50 ml.
Therefore, 50 ml of 0.5 M HCl is required to reach the equivalence point.