What is the ph at the equivalence point for a titration of 250 mL of 0.17M piperidine with 0.17 M nitric acid.
(piperidine: Kb = 1.3 x 10-3)
Call piperidine BH; therefore, the titration will be
HB + HCl ==> BH^+ + Cl^- and hydrolysis at the end point will determine the pH at the equivalence point.(salt, BH^+, will be 0.085M (that's 0.17M x 1/2 = 0.085M).
........BH^+ + H2O ==> H3O^+ + BH
I......0.085............0.......0
C........-x.............x.......x
E......0.095-x..........x.......x
Ka for BH^+ = (Kw/Kb for piperidine) = (x)(x)/(0.085-x)
Solve for x = (H3O^+) and convert to pH.