Question

A 100.0-mL sample of 0.250 M aniline (C6H5NH2) is titrated with 0.500 M HCl. What is the pH at the
equivalence point?

Answer 1

The pH at the equivalence point is determined by the hydrolysis of the salt, aniline hydrochloride.
....C6H5NH3^+ + H2O ==> H3O^+ + C6H5NH2
I.....0.167M.............0........0
C........-x..............x........x
E.....0.167-x............x........x

Ka for the salt - (Kw/Kb for aniline)=(H3O^+)(C6H5NH2)/(0.167-x)
salt concn determined as follows:
100 mL x 0.250M = 25.0 millimoles.
M = mmol/mL = 25/150 = 0.1667M
Solve for x = (H3O^+) and convert to pH.

Answer 2

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