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) 0.20 mole of HCl will neutralizes 0.20 mole `CH_(3)COONa`, producing 0.20 mol `CH_(3)COOH`. Therefore, in the solution moles of `CH_(3)COOH` = 1.20.
Moles of `CH_(3)COONa` = 0.80
`pH = pK_(a) + log [Sal t]/[Acid] = -log(1.8 xx 10^(-5)) + log(0.80)/(1.20) = 4.56`
`CH_(3)COONa + HCl rarr CH_(3)COOH + NaCl`
Initial 0.10 0.20 0 0
Final 0 0.10 0.10 0.10
Now, the solution has 0.2 mole acetic acid and 0.1 mole HCl. Due to presence of HCl, ionisation of `CH_(3)COOH` can be ignored (common ion effect) and `H^(+)` in solution is mainly due to HCl.
`[H^(+)] = 0.10`
pH = -log(0.10) = 10