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# Need explanation for: 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

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%

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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

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