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`?

Explanation

Here we have to consider common ion effect (i.e., contribution of `H^(+)` ion from HCl).
Let contribution of `H^(+)`ions from dissociation of `CH_(3)COOH" in " 0.01 " M HCl be x mol " L^(-1) `, then,
`CH_(3) COOH hArr H^(+) + CH_(3)COO^(-), " but " K_(a) = ([H^(+)]_("Total") [CH_(3)COO^(-)])/([CH_(3)COOH^(-)]) ` ....(i)
`{:("Initially",c," "-," "-),("At equil.",(c-calpha)=c-x,x=calpha,x=calpha):}`
Now, `[H^(+)]_("Total") = x ("from"CH_(3)COOH)+0.01("from HCl") = (x + 0.01)`
`[CH_(3)COO^(-)] = x ("only from " CH_(3)COOH),` therefore , from equation (i) we get
`1.8 xx10^(-5) = ((x + 0.01)x)/c `, as x is very small we get
`1.8 xx10^(-5 ) = (0.01 x)/c " and " x = (1.8 xx10^(-5)c)/(0.01) = (1.8 xx 10^(-5)xx0.05)/(0.01) = 9.0 xx 10^(-5)`
but . `x = c alpha`, hence , `alpha = x/c = (9.0 xx 10^(-5))/(0.05) = 0.0018`