In a 0.1 solution, acetic acid is 1.34% ionized. Calculate `[H^(+)], [CH_(3)COO^(-)]` and `[CH_(3)COOH]` in the solution and calculate `K_(a)` of acetic acid.

Explanation

`CH_(3)COOH hArr CH_(3)COO^(-)+H^(+)`
Ionisation of `CH_(3)COOH=1.34%`
So for 0.1 M the ionisation `=(1.34xx0.1)/100`
`=1.34xx10^(-3)M`
`:.[H^(+)]=[CH_(3)COOH]=1.34xx10^(-3)M`
`[CH_(3)COOH]=1-1.34xx10^(-3)=0.09866M`
`K_(a)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])`
`=(1.34xx10^(-3)xx1.34xx10^(-3))/0.1=1.8xx10^(-5)`