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NEET 2016
Exam
NEET
Year
2016
Subject
Chemistry
Chapter
Ionic Equilibrium

NEET 2016 · Chemistry · Ionic Equilibrium

The percentage of pyridine ( C _ 5 H _ 5 N) that forms pyridinium...

The percentage of pyridine (C5H5N)\left({C}_{5}{H}_{5}N\right) that forms pyridinium ion \left({C}_{5}{H}_{5}{N}^{+}H\right)inain a0 . 10Maqueouspyridinesolution(KbforC5H5N=1.7×10−9)aqueous pyridine solution \left({K}_{b}for{C}_{5}{H}_{5}N = 1 .7 \times {10}^{- 9}\right) is

Options

  1. A0.00600 . 0060 %
  2. B0.0130 . 013 %
  3. C0.770 . 77 %
  4. D1.61 . 6 %

Correct answer

B. 0.0130 . 013 %

Step-by-step solution

{C5H5N+H2O⇌C5H5N⊕H+OH⊖t=0Ct=eqC(1−α)CαCα\begin{cases} & {\text{C}}_{5}{\text{H}}_{5}\text{N} + {\text{H}}_{2}\text{O} & ⇌ & {\text{C}}_{5}{\text{H}}_{5}{\text{N}}^{⊕}\text{H} & + & {\text{OH}}^{⊖} \\ \text{t} = 0 & \text{C} & & & & \\ \text{t} = \text{eq} & \text{C}\left(1 - \alpha\right) & & \text{C}\alpha & & \text{C}\alpha\end{cases}

∴Kb=Cα21−α≈Cα2∴ {K}_{b} = \dfrac{C{\alpha}^{2}}{1 - \alpha} \approx C{\alpha}^{2}

∴α=KbC=1.7×10−90.1∴ \alpha = \sqrt{\dfrac{{K}_{b}}{C}} = \sqrt{\dfrac{1.7 \times {10}^{- 9}}{0.1}}

=1.3×10−4= \text{1.3} \times 1 {0}^{- 4}

∴∴ %\alpha = 1.3 \times {10}^{- 4} \times 100 = 0.013%
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