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A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

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PyH+Cl- → PyH+ + Cl-

PyH+ undergoes hydrolysis in aqueous solution

PyH+ + H2O ⇌ Py + H3O+ 0.02 M

pH = 3.44

Therefore, [H+] = Antilog (-3.44)

= 3.63 x 10-4 M

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