Explain the ph at the equivalence point of the titration

Reference no: EM13233053

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 is the pH at the equivalence point of the titration and at the half equivalence point?

Reference no: EM13233053

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