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Calculate the [CH(2)FCOOH] (fluoroacetic...

Calculate the `[CH_(2)FCOOH]` (fluoroacetic acid) which is required to get `[H^(o+)] = 1.5 xx 10^(-3)M. K_(a)` of acid `=2.6 xx 10^(-3)`.

Text Solution

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`{:(,CH_(2)FCOOH hArr,CH_(2)FCOO^(Theta)+,H^(o+),),("Mole before dissociation",1,0,0,),("Mole after dissociation",(1-alpha),alpha,alpha,):}`
Given, `[H^(o+)] = C.alpha = 1.5 xx 10^(-3) "mol litre"^(-1)`
`:. K_(a) = ((Calpha)(Calpha))/(C(1-alpha))=(Calpha^(2))/((1-alpha))`
`2.6 xx 10^(-3) = (1.5 xx 10^(-3) xx alpha)/((1-alpha)) :. alpha =0.634`
Now, `C.alpha = 1.50 xx 10^(-3)`
`:. C = (1.50 xx 10^(-3))/(0.634) = 2.37 xx 10^(-3)M`
Note: Since `K_(a)` is of the order of `10^(-3)M` and this it is not advisable to use `K_(a) = C alpha^(2)`. Because `(1-alpha)` is not equal to `1` since `alpha` is not small.
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