[arctic-cube]

Hi there,

By titrating a diprotic acid with OH^-, I can calculate the amount of substance of the acid by doing so:

1/2*n(Acid) = c(OH^-)*V(OH^-)

"1/2" because 2 parts OH^- are reacting with 1 part acid. Is this correct so far?

Now the problem: The acid has a pH-value of 4 at the beginning, which meant in other terms that the first equivalent point has already passed. Let's say the reaction looks like:

H₂A + OH^-  HA^- + H₂O

As I said that is the starting situation. Now I beginn the titration and note the volume of OH- till the "second" equivalent point.

HA^- + OH^- A^2- + H₂O

So at the beginning I already had something like HA^-, therefore I only need 1 part of OH- or do I still need 2 parts in the calculation?

[DocDingwall]

It might help to think about what's going on in the solution prior to the start of the titration.  At pH 4, is there really much OH- around? (No) So prior to the start of the titration you have:

AH₂ ::equil::AH^-+ H⁺

In other words, both protons are there.  The rest is easy, you titrate to the end point and # moles OH- = 1/2 # moles acid.