Physics, Chemistry, Maths Forum

Solving this implies IONIC EQUILIBRIUM Crystal Clear

Solving this implies IONIC EQUILIBRIUM Crystal Clear

by Shashank Todwal -
Number of replies: 2

In all the books it is written that a buffer solution can be made with weak acid(e.g.CH3COOH) and strong base (e.g.NaOH).Suppose i have some CH3COOH solution.Then I have foll. equilibriums established:

CH3COOH----><----CH3COO- + H+      -----(1) ---Ka(small)                      AND

CH3COO-+H20----><----CH3COOH+OH----------(2)----Kh(small)

if i add some NaOH then reaction (2) is favoured backwards and almost all of my added NaOH gets converted to acetate ions as 1/Kh is very very large.But then b'coz of increase of acetate ions reaction (1) is favoured and almost all extra ions get converted to acid as 1/Ka is also very very large.

Then increase in acid again causes backward reaction of (2) and ions formed again and the whole process is repeated again and again.So at the end poor OH- is almost negligible.

This is just an example but if we speak in general then i can't maintain a pH of 10,11,etc as then in that case we are assuming [OH]=104etc. which i have proved above is impossible.

Plz give your suggestions on this!!!

In reply to Shashank Todwal

Re: Solving this implies IONIC EQUILIBRIUM Crystal Clear

by bharadwaj bharadwaj -
if OH- is added , reaction 2 goes in backward dirn giving more ch3coo- .

but since 1/k of reaction 1 is large doesnt mean 1 should go in backward reaction.
this is because the small value of H
+ ion and finite values of ch3coo- and ch3cooh
make Q=K for reaction 1 .


In reply to Shashank Todwal

Re: Solving this implies IONIC EQUILIBRIUM Crystal Clear

by srinath murthy -
The OH- will react with H+ , this proceeds instantenously , i.e.the moment you add OH- to the reaction mixture.