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0.2 M solution of a weak acid HA is 1 % ...

`0.2 M` solution of a weak acid `HA` is `1 %` ionised `25^(@)C`. `K_(a)` for the acid is equal to

A

`(0.002xx0.002)/(0.198)`

B

`(0.02xx0.02)/(0.18)`

C

`(0.01xx0.01)/(0.19)`

D

`(0.19)/(0.01xx0.01)`

Text Solution

Verified by Experts

The correct Answer is:
A

`HA+H_(2)OhArr H_(3)O^(-)+A^(-)`
`[HA] =0.2-(0.02xx1)/(100)=0.198 M`
`[A^(-1)]=0.002 M`
`[H_(3)O^(+)]=0.002 M`
`K_(a)=([H_(3)O^(+)][A-])/([HA])=(0.002xx0.002)/(0.198)`
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