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)`.

Answer

A, B, D

Explanation

a. A mixture of acetic acid and acetate ions forms an acidic buffer.
`pH = pK_(a) +"log" (["Salt"])/(["Acid"])`
Rule `A A A`: When `HCI` is added, the amount of acid increases because acetate ions react with hydorgen ions to form undissociated salt. Therefore, the amount of acetate decreases.
New `["Acid"] = 1 + 0.2 = 1.2`
New `["Salt"] = 1- 0.2 = 0.8`
`pH = pK_(a) + "log" (["Salt"])/(["Acid"])`
`=- log (1.8 xx 10^(-5)) +"log" (0.8)/(1.2)`
`=4.47 + "log" (2)/(3) =4.568`
b. When `0.2M` hydrochloric acid is added in the second case, `0.1M` aceate is consumed totally in the reaction with hydrogen ion. only `0.1M HCI` is left. Also there is acetic acid. So there is no buffer, `pH` is calculated according to strong acid `HCI` only.
`pH =- log [H^(o+)] =- log 0.1 = 1`