# AP Chemistry

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

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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.

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2. 9.39443755788771

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posted by J
3. Ignore J's response. Just plug into DrBob222's explanation

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posted by FeMan

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