Calculate the amount of acetic acid presnt in `1L` of solution having `alpha = 1%` and `K_(a) = 1.8 xx 10^(-5)`.

Answer

A

Explanation

`{:(CH_(3)COOH hArr,CH_(3)COO^(Theta)+,H^(o+),,),(1,0,0,,),(1-alpha,alpha,alpha,,):}`
where `alpha` is degree of dissociation of acid, if `C mol L^(-1)` is concentration of acid, then
`[H^(o+)] = C alpha, [CH_(3)COO^(Theta)] = C alpha, [CH_(3)COOH] = C(1-alpha)`
Also `K_(a) = ([H^(o+)][CH_(3)COO^(Theta)])/([CH_(3)COOH]) = (C alpha.C alpha)/(C(1-alpha)) = C alpha^(2)`
[`:. K_(a)` is small `:. alpha` will also be small and thus `1-alpha =1]`
or `1.8 xx 10^(-5) = X xx ((1)/(100))^(2) :. C = 0.18`
`:. 1L` solution contains = 0.18 mole of `CH_(3)COOH`
`:. 1L` solution contains `= 0.18 xx 60 = 10.8 gCH_(3) COOH`