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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.

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)

a) What was the concentration of aniline in the original solution?


b) What is the pH at the equivalence point of the titration?


c) What is the pH at the half equivalence point?

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