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Ionic Equilibrum question

2016 · Q108
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Ionic Equilibrum question

2016 · Q108

NEETChemistryIonic EquilibrumMCQ+4 / −1
The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5N+H) ina 0.10 M aqueous pyridine solution (Kb for C5H5N = 1.7 ×\times× 10−-−9) is
  1. A
    0.0060%
  2. B
    0.013%
  3. C
    0.77%
  4. D
    1.6%
View written solutionFree

Correct answer: B

C5H5N + H2O → C5H5N+H + OH–

So, the amount of pyridine that forms pyridinium ion is α\alpha α.

Now, α\alpha α = Kbconc. of pyridine\sqrt {{{{K_b}} \over {conc.\,of\,pyridine}}} conc.ofpyridineKb​​​

= 1.7×1090.10\sqrt {{{1.7 \times {{10}^9}} \over {0.10}}} 0.101.7×109​​

= 1.30 ×\times× 10-4

So, percentage of pyridine that forms pyridinium ion

= 1.30 × 10–4 × 100

= 1.30 × 10–2 = 0.013%

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