A-level Chemistry buffer question

Hi, could someone please clear up a misconception I have about weak acid equilibriums and buffers?

If you have weak acid HA ⇌ A- + H+ and you add A-/H+, what will happen? I thought nothing would happen because there is virtually no H+/A- to react to form the HA acid due to it only slightly dissociating due to it being a weak acid? I know to make it a buffer solution you would add a salt, e.g. NaA --> Na+ + A- fully dissociating, which provides the A- for added H+ to react with. For previous buffer solutions I have just assumed that when one of the products (H+ or A-) is added by itself the concentration of HA remains constant and this has always worked for me. Sorry if this is rambling but I would really appreciate some help. Thanks.

Reply 1

JA03

Volunteer Team

To answer the main question, it shifts the position of equilibrium so it favours the reverse reaction.

Reply 2

Har6547

Original post by JA03

To answer the main question, it shifts the position of equilibrium so it favours the reverse reaction.

But how can it form HA if there is virtually 0 of the other reactant?

Reply 3

charco

Study Forum Helper

Original post by Har6547

But how can it form HA if there is virtually 0 of the other reactant?

In a weak acid there is usually about 1% of the dissociated side (broad assumption, based on ethanoic acid)

Just plug in some approx values to convince yourself.

0.1 mol dm^-3 ethanoic acid will be about 0.001 mol dm^-3 in the hydrogen and conjugate base ions.

ka = (0.001 x 0.001)/0.1 = 1 x 10^-5

If you add 0.001 mol of conjugate base ...

Q*(disturbed) = (0.001 x 0.002)/0.1 = 2 x 10^-5

The system must respond by making more ethanoic acid to restore the value of ka