What is the pH of a 0.41 M solution of pyridinium chloride (C5H5NH+Cl−)? (Kb for pyridine is 1.5×10−9.)

Question

Answer in units of pH

Answer 1

.....C5H5NH^+ + HOH ==> C5H5N + H3O^+
I.....0.41M...............0.......0
C......-x.................x.......x
E.....0.41-x..............x.......x

Ka for C4H5NH^+ = Kw/Kb for pyradine = (C5H5N)(H3O^+)/(C5H5NH^+)

Plug in Kw/Kb to find Ka, then solve for x = (H3O^+). Then use pH = -log(H3O^+) for find (H3O^+).
Post your work if you get stuck.

Answer 2

9.39443755788771

Answer 3

Ignore J's response. Just plug into DrBob222's explanation