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

a) `0.0018` , b) `0.00018`

Explanation

`c = 0.05 M`
`pK_(a) = 4.74`
`pK_(a) = -"log" (K_(a))`
`K_(a) = 1.82 xx 10^(-5)`
`K_(a) = calpha^(2) = calpha^(2) , alpha = sqrt((K_(a))/(c))`
`alpha = sqrt((1.82 xx 10^(-5))/(5 xx 10^(-2))) = 1.908 xx 10^(-2)`
When HCl is added to the solution, the concentration of `H^(+)` ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.
Case I: When 0.01 M HCl is taken.
Let x be the amount of acetic acid dissociated after the addition of HCl.
`{:(,CH_(3)COOH,harr,H^(+),+,CH_(3)COO^(-)),("Initial conc.",0.05M,,0,,0),("After dissociation",0.05-x,,0.01+x,,x):}`
As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 - x and 0.01 + x can be taken as 0.05 and 0.01 respectively.
`K_(a) = ([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])`
`:. K_(a) = ((0.01)x)/(0.05)`
`x = (1.82 xx 10^(-5) xx 0.05)/(0.01)`
`x = 1.82 xx 10^(-3) xx 0.05 M`
Now
`alpha = ("Amount of acid dissociated")/("Amount of acid teken")`
`= (1.82 xx 10^(-4) xx 0.05)/(0.05)`
`= 1.82 xx 10^(-4)`
Case II: When 0.1 M HCl is taken.
Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:
`[CH_(3)COOH] = 0.05 - X , 0.05 M`
`[CH_(3)COO^(-)] = X`
`[H^(+)] = 0.1 + X , 0.1M`
`K_(a) = ([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])`
`:. K_(a) = ((0.1)X)/(0.05)`
`x = (1.82 xx 10^(-5) xx 0.05)/(0.1)`
`x = 1.82 xx 10^(-4) xx 0.05 M`
Now,
`alpha = ("Amount of acid dissociated")/("Amount of acid taken")`
`= (1.82 xx 10^(-4) xx 0.05)/(0.05)`
`= 1.82 xx 10^(-4)`