What is the `pH` of the solution when `0.20 mol` of `HCI` is added to `1L` of a solution containing
a. `1M` each of acetic acid and acetate ion.
b. `0.1M`each of aceta acid and acetate ion.
Assume the total volume is `1L. K_(a)` for acetic acid is `1.8 xx 10^(-5)`.

Explanation

(i) Addition of 0.2 mole of HCI increases the concentration of `CH_3COOH` by 0.2 and reduces the concentration of `CH_3COO^-` by 0.2 mole/L.
`[CH_(3)COOH]=1+0.2=1.2M`
`[CH_(3)COO^-]=1-0.2=0.8M =[CH_(3)COONa]`
Using `pH =pK_(a)+log (["salt"])/(["acid"])`
`=-log (1.8 xx 10^(-5))+log ""(0.8)/(1.2)`
=4.57
(ii) In this case 0.1 M of `CH_3COO^-` will combine with 0.1 M of HCI forming 0.1 M `CH_3COOH.`
Total `[CH_3COOH]= 0.1 +0.1 =0-.2 M.`
But in the presence of the remaining 0:1 M HCl, the dissociation of 0.2 M `CH_3COOH` will be further suppressed by the common-ion effect. Hence neglecting [H] produced by `CH_3COOH,` the pH may be calculated for just 0.1 M HCl.
`pH =-log [H^+]=-log (0.1)=1.0`