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^(+)]` furnished by HCl and `CHCl_(2)COOH`
` {:(,CH_(2)COOH,hArr,CHl_(2)COO^(-),+,H.,),("Intial conc.",0.09,,0,,0.09,("from HCl")),("Final conc",(0.09 -x),,x,,(0.09 +x),):}`
` :. [ H^(+)] = 0.09 +x,`
but pH = 1
` :. " " [H^(+)] = 10^(-1)= 0.1 `
` :. 0.09 + x = 0.1 `
` :. x = 0.01 `
`K_(a)` for `CHCl_(2)COOH ` can be given as
` K_(a) ` for `CHCl_(2)COOH ` can be given as
` K_(a) = ([H^(+)][CHCl_(2)COO^(-)] )/([CHCl_(2)COOH]) = (0.1 xx 0.01 )/((0.09 - 0.01 )) = 1.25 xx10^(-2) ` .