Question

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`?

Answer 1

`{:(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`