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Question
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Solution
pH = 3.44
We know that,
pH = – log [H+]
`therefore ["H"^+] = 3.63 xx 10^(-4)`
Then `"K"_"h" = (3.63 xx 10^(-4))^2/0.02` (∵ concentration = 0.02 M)
`=> "K"_"h" = 6.6 xx 10^(-6)`
Now `"K"_"h" = "K"_"w"/"K"_"a"`
`=> "K"_"a" = "K"_"w"/"K"_"h" = 10^(-14)/(6.6 xx 10^(-6))`
`= 1.51 xx 10^(-9)`
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