# Chemistry

1. Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.240 M pyridine, C5H5N(aq) with 0.240 M HBr(aq):

(a) before addition of any HBr
(b) after addition of 12.5 mL of HBr
(c) after addition of 22.0 mL of HBr
(d) after addition of 25.0 mL of HBr
(e) after addition of 37.0 mL of HBr

I believe I know how to do part a, I got pH=9.31. After that I don't understand what to do if HBr is added. Any help is really appreciated! Thanks!

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1. py + HBr ==> PyH + Br^-
You know how many mols pyridine you started with. That's M x L = ?

Now take M x L of the titrant to give yo mols HBr added. Since the reaction is 1:1 you know 1 mol HBr neutralizes 1 mol pyridine.
Subtract mols py-mol HBr to see which is in excess. If py is in excess use the Henderson-Hasselbalch equation to solve for pH. If HBr is in excess, see how much in excess, determine molarity, and use pH = -log(HBr)

Those two procedures will work for all points AFTER the beginning and AFTER the equivalence point. AT THE equivalence point you must use the hydrolysis of the salt formed.
That is
.......pyH^+ + H2O==>H3O^+ + py
I.......x..............0.......0
C.......-y.............y.......y
E......x-y.............y........y

Ka for salt = (Kw/Kb for py) = (y)(y)/(x-y) and solve for y = (H3O^) and convert to pH.
Don't be confused by the two unknowns here. x is just the salt concn at the eq. pt. whatever that happens to be. I just didn't work through to see what it was.

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