A 25.0 mL sample of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalence point. (Kb for aniline = 4.0 x 10-10)
C6H5NH2(aq) + H3O+(aq) ↔ C6H5NH3+(aq) + H2O(l)
What was the concentration of aniline in the original solution?