`K_(a)` for an acid HA is `4.9 xx 10^(-8)`. Calculate percentage dissocitation `H^(+)` ion concentration for its 0.1 M aqueous solution.

Answer

`alpha = 7xx 10^(-4) ,[H^(+)]=7 xx 10^(-5) "mol " L^(-1)`

Explanation

Degree of dissociation of acid `(alpha) = sqrt((K_(a))/(C))= ((4.9xx 10^(-8))/(0.1))^(1//2) =7xx 10^(-4)`
`H^(+)" ion concentration i.e.," [H^(+)] =Calpha =0.1 xx7xx10^(-4) =7xx10^(5) " mol "L^(-1)`