InterviewSolution
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InterviewSolution
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Calculate the degree of ionisation of `0.05 M` acetic acid if its `pK_(a)` value is `4.74`. How is the degree of dissociation affected when its solution also contains a. `0.01 M`, b. `0.1 M` in `HCl`? |
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Answer» `{:(CH_(3)COOHhArr,CH_(3)COO^(-), +H^(+)), (1,0,0), (1-alpha,alpha,alpha):}` `(pK_(a)= - Iog K_(a)= 4.74, :. K_(a)= 1.82xx10^(-5))` `K_(a)= (C alpha^(2))/((1-alpha))=C alpha^(2)` , `(.: 1-alpha ~~1)` `:. alpha= sqrt((K_(a))/(C ))= sqrt((1.82xx10^(-5))/(0.05))` `= 0.019` or `1.9%` always calculate `alpha` first by `K_(b)= C alpha^(2)`, if `alphagt 5%` then use again `K= (C alpha^(2))/((1-alpha))` (a) If `H^(+)` are already present (due to HCI). `{:(CH_(3)COOHhArr,CH_(3)COO^(-), +H^(+)), (1,0,0.01), (C(1-alpha),Calpha,[0.01+Calpha]):}` `K_(b)= ([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH]) = (C alphaxx(0.01+C alpha))/(C(1-alpha))` Since presence of `H^(+)` will favour the reverse reaction or `alpha` will decrease, i.e., `0.01+Calpha=0.01` and `1-alpha=1` (due to common ion effect) `:. 1.82xx10^(-5) = (0.05xxalphaxx0.01)/(0.05)` `:. alpha=1.82xx10^(-3)= 0.0018` (b) Similary solve for `0.1 MHCL` `alpha= 0.00018` |
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