107.0 mL of a solution of quinine contains 224.7 x 10-3 mol of quinine.?

Question

107.0 mL of a solution of quinine contains 224.7 x 10-3 mol of quinine.
If this is titrated with 0.200 M HCl, what is the pH half-way to the stoichiometric point ?

Quinine is monobasic with Kb = 7.90 x 10-6

Answer 1

First, figure out the molarity of the solution.
Use proportion.
224.7x10^-3 /0.107 = M / 1 (denominators in L). For arguement, say M = 2.4x10^-4. At equivalence point, for the dissociation of Q,
[QH+] [OH-] /[Q] = Kb and [QH+]=[Q]
To "get here" from the start of the titration, the addition of H+ from hydrochloric acid removes [OH-]. Thus, more quinine dissociates. At the halfway point, [QH+]/[Q] = 1/3.
Then [OH-]/3 = Kb or [OH-] = 3Kb.
Roughly, 3Kb = 2.4 x 10-5 = [OH-]
In water, [H+][OH-] = 10x10-15
Then [H+]= 4.2 x 10-10 (appx)
pH= 10-log 4.2 = 9.35 appx