During a titration of a weak base with a strong acid, you are slowing converting

Question

During a titration of a weak base with a strong acid, you are slowing converting molecules of the weak base into molecules of its conjugate acid. For the hypothetical weak base, B we see the following:

In the problem below you will be adding some strong acid, but not enough to reach the endpoint of the titration.

You are titrating 22.00 mL of a 0.80 M solution of C5H5N (pyridine) with a strong acid. If you add 1.20 mL of a 1.80 M solution of hydrochloric acid, what is the final pH of the remaining weak base solution? The Kb for C5H5N (pyridine) is 1.7 X 10-9.

1.20 mL of hydrochloric acid added to the weak base


1.80M hydrochloric acid
in the buret as the titrant



22.0 mL of original 0.80 MC5H5N (pyridine)sample

Explanation / Answer

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 = (22.0mL x 0.8M) - (1.2mL x 1.8M) / 22+1.2 mL
C C5H5N = 0.71

C C5H5NH+ = 1.2.0mL x 1.8 M / 1.2+22 mL = 0.093M

Kb = 1.7x10^-9 = [OH-]^2 / 0.71M
[OH-] from pyridine= 0.346 x 10 ^ (-4)M

[OH-] = [C5H5NH+]
[C5H5NH+] total = 0.093M + 0.34611x10^-4 M

= 0.093 M

We get Ka = Kw / Kb = 5.88x10^-6 for pyridine

Ka = [C5H5N][H3O+] / [C5H5NH+] = 5.88x10^-6
Substituting the values of [C5H5NH+]total and [C5H5N], and rearranging gives,

[H3O+] = (5.88x10^-6)(9.93x10^-2) / (0.71M)

= 0.76 x 10 ^ (-8)

pH = 8.11

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