Question

How do I do this??
Calculate the pH at the equivalence point for the titration of 0.20 M aniline (C6H5NH2, Kb = 3.8 x 10-10) with 0.20 M HCl. [Hint: remember the dilution factor caused by addition of titrant to the aniline solution.]

Answer 1

The equation is
C6H5NH2 + HCl ==> C6H5NH3^+ + Cl^-
So at the equivalence point, the solution is C6H5NH3^+ (the amine salt).
What is its concentration.
It will be You don't specify how much of the aniline is being titrated BUT since it is 0.2 M and the HCl is the same, then the concn of the salt will be 0.1 M at the equivalence point. The pH of the solution will be determined by the hydrolysis of the salt.
C6H5NH3^+ + HOH ==> C6H5NH2 + H3O^+
Do an ICE procedure for starting with 0.1 M of the salt.
Ka = Kw/Kb = (C6H5NH2)(H3O^+)/(C6H5NH3^+)
and solve for (H3O^+).
Then pH = - log (H3O^+)
Post your work if you get stuck.