Calculate the pH for each of the following cases in the titration of 25.0 mL of
ID: 791375 • Letter: C
Question
Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 M HBr(aq):
a. Before and addition of HBr
b. After addition of 12.5ml HBr
c. After addition of 15ml HBr
d. After addition of 25ml HBr
e. After addition of 33ml HBr
Explanation / Answer
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 HBr 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 / 0.100
[OH-] = 1.30384 x 10^-5 M
pH = -log[7.669652718 x 10 ^-10] = 9.115
Second instance, after the addition of 12.5 mL of 0.100 M HBr:
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 0.1M) - (12.5.0mL x 0.1M) / 12.5+25 mL
C C5H5N = 0.033333 M
C C5H5NH+ = 12.5.0mL x 0.1 M / 12.5+25 mL = 0.033333M
Kb = 1.7x10^-9 = [OH-]^2 / 0.033333M
[OH-] from pyridine= 4.3460 x 10^-6 M
Since, from the very 1st equation, [OH-] = [C5H5NH+]
[C5H5NH+] total = 0.033333M + 4.3460x10^-6 M
= 3.333734x10^-2M
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)(.033333x10^-2) / (0.033333M)
= 5.88x10^-6
pH = 5.23
continue in the same manner with the next steps
That is all! Hope this help you...