[Big-Daddy]

Have you tried to estimate pH using just first step of EDTA dissociation? For a weak acid it is relatively strong.

How can we estimate, we have no idea of EDTA's analytical concentration? Strong(er than most weak acids) as it may be, it might have been very dilute ...

[Borek]

The more diluted it is, the more to the right the dissociation is shifted, so concentration of H⁺ becomes more and more higher than the analytical concentration of the acid. In the limit for infinitely diluted H₄EDTA

[tex][H^+] = 4 \times C_{H_$EDTA}[/tex]

If the concentration of H⁺ is comparable with the acid concentration, solution can't be too diluted, which in turn means it must be acidic.

[Big-Daddy]

The more diluted it is, the more to the right the dissociation is shifted, so concentration of H⁺ becomes more and more higher than the analytical concentration of the acid. In the limit for infinitely diluted H₄EDTA

[tex][H^+] = 4 \times C_{H_$EDTA}[/tex]

If the concentration of H⁺ is comparable with the acid concentration, solution can't be too diluted, which in turn means it must be acidic.

Wouldn't we guess then that the concentration must be somewhat dilute for Ka1 to be going almost to completion? (Ka2 will only add a little on top of Ka1, water almost negligible) Is there maybe a mathematical way I can estimate concentration, given that [H⁺]=C_H4EDTA, from the Ka1 value alone?