Question

Consider the titration of 25.00 mL of 0.08364 M pyridine (see structure below) with 0.1067 M HCl. Find the pH at the following acid volumes (Va): 4.63 mL, Ve, and 25.0 mL. pKa = 11.3

I got 11.81, 6.63, and 1.74 respectively. Yet, the last two are incorrect and I am unsure what I did incorrectly

Answer 1

11.81 is ok for Va.

Where is the equivalence point? That will be at
25.00 x 0.08364 = mLHCl x 0.1067
Solve for mL HCl and I get approximately 19 mL. Therefore, the concn of the salt at the equivalence point will be mols/L or
(25.00*0.08364/(25+19) = about 0.05.
Calling pyridine PN, the hydolysis of the salt will be
.......PNH^+ + H2O ==> H3O^+ + PN
I.......0.05............0.......0
C........-x..............x......x
E.......0.05-x..........x.......x

Ka for PNH^+ = (Kw/Kb) for pyridine = (x)(x)/(0.05-x) and solve for x = (H3O^+) and covert to pH. Answer should be about 6.3

For 25 mL, that should be just the excess HCl (diluted of course which is what most students forget).
mmols PN = 25.00*0.08364 = about 2.091
mmols HCl = 25.00*0.1067 = about 2.67 which is an excess of about 0.576 mmols in (25+25 mL) = about 0.0115 or pH about 1.94.