The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5N*H) in a 0.10 M aqueous pyridine solution (Kb for C5H5N = 1.7 x 10-9) is

  • Option 1)

    0.0060%

  • Option 2)

    0.013%

  • Option 3)

    0.77%

  • Option 4)

    1.6%

 

Answers (8)

 

Relation between Ka and Kb -

K_{a}\times K_{b}=K_{W}

or\:P(K_{a})+P(K_{b})=P(K_{W})=14   (at\:298K)

- wherein

K_{a}=dissociation \:constant\:of\:acid

K_{b}=dissociation \:constant\:of\:base

K_{W}=ionic \:product\:of\:water

 

K_{a} =\frac{K_{w}}{K_{e}}  K_{a} =\frac{K_{w}}{K_{e}}\: = \frac{10^{-14}}{1.7\times 10^{-9}} = 5.8\times 10^{-6}

C_{5}H_{5}N + H_{2}O  \Leftrightarrow C_{5}H_{5}NH+C_{5}H_{5}NH ^{\left ( + \right )} + OH^{\left ( - \right )}

1-x                                                                  x                         x

5.8\times 10^{-6} = x^{2}

x=7.6\times 10^{-4}

% of pyridie = \frac{7.6\times 10^{-4}\times 100}{0.1}      = 0.76

Ulnser  is 3 Chanage eusur  to 3


Option 1)

0.0060%

Incorrect

Option 2)

0.013%

Correct

Option 3)

0.77%

Incorrect

Option 4)

1.6%

Incorrect

Answer 1

 

0.77

Option 2

 

Option 3

0.013 %

 

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