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At 25^(@)C the dissociation constant of ...

At `25^(@)C` the dissociation constant of HCN is `4.9 xx 10^(-10) M.` Calculate the degree of dissociation of HCN if the concentration is 0.1 M.

A

`7 xx 10 ^(-5)`

B

`5 xx 10 ^(-5)`

C

`6 xx 10 ^(-5)`

D

`8 xx 10 ^(-5)`

Text Solution

Verified by Experts

The correct Answer is:
D
The reaction can be represented as
`HCN + H _(2)O hArr H _(3) O ^() + CN ^(-)`
`{:(C, 0, 0), (C (1-alpha), alpha C, alpha C):}`
Therefore, `K _(C) = (Calpha ^(2))/( 1 - alpha ) , 1 - alpha =1, alpha = sqrt (( K_(a))/( c )) , alpha = sqrt ((4.9 xx 10 ^(-1))/( 0.1)) = 7 xx 10 ^(-5)`
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