Question

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.210 M pyridine, C5H5N(aq) with 0.210 M HBr(aq):

Kb= 1.7 x 10^-9

(d) after addition of 25.0 mL of HBr

(e) after addition of 32.0 mL of HBr

Answer 1

Pyridine (C5H5N) is a weak base, at the equivalence point its salt, pyridinium (C5H5NH+) gives an acidic pH (that is pH < 7).
The pH of the solution, when none of the HCl has been added, is determined from the concentration of OH- from the dissociation of pyridine and from the Kb (1.7 X 10^-9).
The rxn:
C5H5N + H2O --- - - - -> C5H5NH+ + OH-
kb = [OH-][C5H5NH+]/[C5H5N]
Since kb x Cb (analytical base conc.) = 1.7x10^-10 >> Kw
then, the main source is only from pyridine.

kb = 1.7x10^-9 = [OH-]^2 / 0.210
[OH-] = 1.8894 x 10^-5 M
pH = -log[7.669652718 x 10 ^-10] = 9.115

Second instance, after the addition of 25 mL of 0.210 M HCl:
The concentration of H+ is determined from the dissociation of remaining pyridine and hydrolysis of pyridinium salt.
The rxns to be considered are:
C5H5N + H+ - - - - - > C5H5NH+ + Cl-
C5H5NH+ + H2O - - - - -> C5H5N + H3O+

All the acid is consumed to produce the same amount of the salt, and the excess pyridine is equal to:
C C5H5N = (25.0mL x 0.21M) - (25.0mL x 0.21M) / 25+25 mL
C C5H5N = 0. M

C C5H5NH+ = 25.0mL x 0.21 M / 25+25 mL = 0.105M

Kb = 1.7x10^-9 = [OH-]^2 / 0M
[OH-] from pyridine= 4.123 x 10^-5 M

Since, from the very 1st equation, [OH-] = [C5H5NH+]
[C5H5NH+] total = 0.105M + 4.123x10^-5 M
= 0.1050 M

We get Ka = Kw / Kb = 5.88x10^-6 for pyridine, and from the 3rd equation: Ka = [C5H5N][H3O+] / [C5H5NH+] = 5.88x10^-6
Substituting the values of [C5H5NH+]total and [C5H5N], and rearranging gives,
[H3O+] = (5.88x10^-6)(4.444834611x10^-2) / (0.011111M)
= 2.35223 x10^-5
pH = 4.629