A solution contains `0.09 HC1, 0.09 M CHC1_(2)COOH`, and `0.1M CH_(3)COOH`. The `pH` of this solution is one. Calculate `K_(a)`for `CHC1_(2)COOH`. (Given `K_(a)CH_(3)COOH = 10^(-5))`

Explanation

`pH` will be decided by `[H^(o+)]` furnished by `HCI` and `CHCI_(2)COOH`.
`{:(,CHCI_(2)COOH hArr,CHCI_(2)COO^(Theta)+,H^(o+),),("Initial conce",0.09,0,0.09("from"HCI),),("Final conce",(0.03-x),x,(0.09+x),):}`
`[H^(o+)] = 0.09 +x`,
but `pH = 1, :. [H^(o+)] = 10^(-1) = 0.1`
`0.09 +x = 0.1, :. x = 0.01`
`K_(a)` for `CHCI_(2)COOH` can be given as
`K_(a) = ([H^(o+)][CHCI_(2)COO^(-)])/([CHCI_(2)COOH]) = (0.1xx0.01)/((0.09-0.01)) = 1.25 xx 10^(-2)`