[born2dive00]

I have an acid base question for you all

I have 500ml of 2M of H3PO4, this gives me a H+ concentration of .1188 which has a pH of .9253 Easy enough.

I have a 100ml of 0.0143M of NaOH which gives me a OH- concentration of .01339 which has a pH of 12.1458 again easy enough.

To me I should just be able to take the .1188 and subtract the .01339 which would give us an H+ excess of .1048 which would equate to a pH of .97970 however find with several calculators it reports the pH at .927- .93 which is more acidic. even IF I take into consideration the amount of water formed in the reaction which would make the solution more base than the .97970.

What am I doing wrong? the challenge is here I do not want to use the moles value of the excess i.e. of the H3PO4 as there will be other multiproteic acids like EDTA added in to calculate the pH of multiple acids and bases.

Is there a way to do this?

Thanks everyone.

[Borek]

I have 500ml of 2M of H3PO4, this gives me a H+ concentration of .1188 which has a pH of .9253 Easy enough.

I have a 100ml of 0.0143M of NaOH which gives me a OH- concentration of .01339 which has a pH of 12.1458 again easy enough.

Looks more or less OK so far (that is, exact numbers depend on the values of Ka used, and in reality ionic strength of the solution will make the pH different, but these numbers are about OK as a first approximation). I get pH of 0.94 (or around 1, after getting ionic strength into consideration).

To me I should just be able to take the .1188 and subtract the .01339 which would give us an H+ excess of .1048 which would equate to a pH of .97970

It doesn't work like this. When you deal with a weak acid and you remove H⁺ from the solution, acid dissociation shifts to the right, producing more H⁺.

however find with several calculators it reports the pH at .927- .93 which is more acidic. even IF I take into consideration the amount of water formed in the reaction which would make the solution more base than the .97970.

These are not numbers I am getting.

Funnily enough, I do get 0.98 as a final pH, not because of the neutralization (amount of NaOH is two orders of magnitude lower that the amount of phosphoric acid), but because of the dilution.

To do calculation by hand in this specific case you can safely ignore Ka2 and Ka3. Then, use ICE table, assuming initial concentrations to be the ones given by the neutralization reaction.