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Question
The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5NH) in a 0.10 M aqueous pyridine solution (Kb for C5H5N = 1.7 × 10−9) is ____________.
Options
0.006%
0.013%
0.77%
1.6%
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Solution
The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5NH) in a 0.10 M aqueous pyridine solution (Kb for C5H5N = 1.7 × 10−9) is 0.013%.
Explanation:
\[\ce{C5H5N + H - OH ⇌ C5H5\overset{+}{N}H + OH^-}\]
`(α^2"C")/(1 - α)` = Kb
α2C ≂ Kb
α = `sqrt("K"_"b")/"C" = sqrt(1.7 xx 10^-9)/0.1`
= `sqrt1.7 xx 10^-4`
Percentage of dissociation = `sqrt1.7 xx 10^-4 xx 100`
= 1.3 × 10−2
= 0.013%
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