At 298 K, 0.1 M solution of acetic acid is `1.34 %` ionised . What is the ionisation constant `(K_(a))` for the acid ?

Answer

`1.82 xx 10^(-5)`

Explanation

The degree of ionisation `(alpha) =1.34% =(1.34)/(100) =0.0134`
The ionisation of acetic may be represented as :
`{:(CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),(C(1-alpha),,Calpha,,Calpha):}`
`[CH_(3)COO^(-)] =Calpha =0.1 xx 0.0134 =0.00134 M , [H^(+)] =Calpha =0.1 xx 0.0134 =0.00134 M`
`[CH_(3)COOH]=C(1-alpha)=0.1(1-0.0134) =0.09866 M`
`K_(a) =[[CH_(3)COO^(-)][H^(+)]]/[[CH_(3)COOH]] =((0.00134)xx(0.00134))/(0.09866) =1.82 xx 10^(-5)`