[umpa]

20ml of .800M pyridine is titrated with .800M HBr.  Assuming this reaction goes to completion, calculate the mmol of each reactant and product present after the the following amounts of titrant...
0mL, 5mL, 19mL, 20mL, 20.5mL, and 32mL.  This is what I've done:

Net ionic eqn:  C5H5N + H+ --> C5H5NH+

I know that  for 0mL, 20mL * .800M = 16mmol C5H5N and H+
I know that for 5mL, (5mL *.800M) = 4mmol = C5H5NH+, and therefore, 16mmol - 4mmol = 12mmol C5H5N and H+
19mL:  19mL * .800M = 15.2mmol C5H5NH+, and therefore, 16mmol - 15.2mmol = 0.8mmol C5H5N and H+

But when I get to the equivalence pt and beyond, I'm a little confused.

At 20mL, I know C5H5NH+  <--> C5H5N + H+

at this point, I am stuck.  Please *delete me*

[sdekivit]

20ml of .800M pyridine is titrated with .800M HBr.  Assuming this reaction goes to completion, calculate the mmol of each reactant and product present after the the following amounts of titrant...
0mL, 5mL, 19mL, 20mL, 20.5mL, and 32mL.  This is what

this exactly the same question that i answered on another forum, only there is titrated with HCl, but it makes no difference. Here are my replies there:

reply 1:

the titration goes to completion.

pyridine is a base that can take up 1 proton. if 0,00 mL HCl is added, we have 20 * 0,8 = 16 mmol pyridine.

When we add 5,00 mL hydrochlorous acid we add 5 * 0,8 = 4 mmol H+. This reacts with the 16 mmol pyridine leaving 12 mmol pyridine. Now you calculate the rest.

The pH can be calculated with:

Kb = [C5H5NH+][OH-] / [C5H5N] = x2 / [C5H5N] = 2,0 x 10-9

when no HCl is added we get: 2,0 x 10-9 = x2 / 0,8 --> x = 4,0 x 10-5 --> pOH = 4,397940009 --> pH = 9,60

when we add 5,00 mL HCl we have to consider that the concentration of pyridine changes. we now have 12 mmol pyridine in 25 mL --> [pyridine] = 12 / 25 = 0,48 M

--> x = 3,098386677 x 10-5 --> pOH = 4,508864383 --> pH = 9,49.

and reply 2:

when u go beyonfd the equivalnece point you get a solution containing a strong acid. There will be an excess of H+-ions and because the equilibrium for pyridine lies strongly to the acid side, there is no equilibrium anymore.

--> you can calculate the pH by doing as you have a HCl-solution. So when u ad 17 mmol HCl, we have a 1 mmol H+ solution

--> [H+] = 1 / 41,25 = 0,024242424242 --> pH = 1,62

At the equivalence pint whe have the equilibrium C5H5NH+ <--> C5H5N + H+

Ka = x2 / [C5H5NH+] = 1,0 x 1014 / 2,0 x 10-9 = 5,0 x 10-6 (--> Ka * Kb = Kw)

--> [basic pyridine] = 16 / 40 = 0,4

--> x = 0,0014142136 --> pH = 2,85.

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=titration-curves-calculation

You will be probably mostly interested in the paragraph before the last.

To check your results try BATE.