Calculate the pH of a solution....?

Question

Calculate the pH of a solution that is 1.50M C5H5N and 2.50 M C5H5NHCl.

This one is killing me! Please help. Thanks in advance!

Answer 1

Let C5H5N (pyridine) be called Py

PyH+ ===> Py + H+

Ka = [Py][H+]/[PyH+] = 6.46 x 10^-6

(1.50)[H+]/(2.50) = 6.46 x 10^-6

[H+] = (2.50)(6.46x10^-6/(1.50) = 1.08x10^-5

pH=4.97

Answer 2

You would use the Henderson-Hasselbalch equation to solve this problem.

http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch

Note that this solution is of a weak base, so read on a bit in the above page and look for the pOH form of the H-H Eq. It's a bit hard to understand, the B+ thing should be HB+ (the conjugate acid) and the BOH should be just B, standing for the base. This equation deals with weak acids and bases, which tend not to have any hydroxide in their formula. That's why I like B rather than BOH

You will need the pKb of pyridine; the Kb is 1.78 x 10^-9.

The C5H5N is the base and the C5H5NH+ is the conjugate acid.

I did the calculation in pieces and hopefuly there is no faulty memory, but it seems like the pOH is right around 9, making the pH about 5. That has the right feel to it since there is more acid (C5H5NH+) than base in the solution.

HTH