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K(a) of HA at 25^(@) is 10^(-5). If 0.1m...

`K_(a)` of `HA` at `25^(@)` is `10^(-5)`. If `0.1mol` of this acid is dissolved in `1L` of aqueous solution, the percent dissociation at equilibrium will be closer to

A

`0.1%`

B

`1.0%`

C

`99.0%`

D

`99.9%`

Text Solution

Verified by Experts

The correct Answer is:
B

`alpha = sqrt((K_(a))/(c)) = sqrt((10^(-5))/(0.1)) = 10^(-2)`
`alpha% 10^(-2) xx 100 = 1%`
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