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The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅N*H) in a 0.10 M aqueous pyridine solution (K_b 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 (10)

best_answer

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

Posted by

prateek

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Answer 1

Posted by

S. Moulika

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0.77

Posted by

Ranjith

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Option 2

Posted by

Angel Anna Jiby

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3

Posted by

Dheerendra Pratap Singh Chauhan

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3

Posted by

Khan vahidunnisha

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Option 3

Posted by

talhask2002

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0.013 %

Posted by

Shubhrajit Nayak

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3

Posted by

lakshita malvannan

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0.013%

Posted by

www.jananieswaran2004

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