InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Calculate the amount of acetic acid presnt in `1L` of solution having `alpha = 1%` and `K_(a) = 1.8 xx 10^(-5)`. |
|
Answer» Correct Answer - A `{:(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` |
|