Using approximation when working with polyprotic acids

[Nhi Pham 3D]

Hello! I understand that when the acidity constant (Ka1) of the polyprotic acid is significantly greater than the acidity constant when a proton is lost (Ka2), we approximate the pH by only taking into account the first deprotonation. If they are close together, we have to use an ICE table and take into account following deprotonations.

I was wondering if there was a specific number that the difference between Ka2 and Ka1 had to be greater than for us to make the approximation. In other words, how much of a difference between Ka2 and Ka1 is necessary for us to ignore the rest of the deprotonations.

Thank you!

[Nick_Graessle_2A]

Hello! I am not sure I completely understand your question; however, typically the cut off for approximation is 10^-3 and using that knowledge I have gotten through most problems I have worked through thus far. Hope that helps

[Yixin Wu 3B]

Hi, I agree with the previous answer that the typical cutoff is at 10^-3 of the Ka1. I think the reason is that when adding the concentration of dissociated H3O+ ions from the first deprotonation and second deprotonation, the added number is similar enough to the concentration of dissociated H3O+ only from the first deprotonation.

For example, if the concentration of H3O+ after the first deprotonation is 0.01, and 0.0000001 after second protonation, the concentration added 0.0100001 is similar enough to 0.01.