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In class, the other day, we were discussing pH curves. Assume that I have HCl and NaOH of concentration 1 mol/mL. Now, consider that I have 50mL of HCl, and begin adding small quantities of NaOH. I observe that the pH increases, and once the volume added of NaOH reaches 50mL, the pH of the resulting solution becomes 7.

However, when mathematically calculating, if I added 49.99999 mL of NaOH, I find that the pH is way over 7 (the concentration of H+ becomes 50-49.99999 divided by 99.99999), which seems to be an anomaly. In essence, when finding the limit of - log (H+) as [H+] approaches 0, it needs to equal 7, but it doesn't. Does this mean that there should be a horizontal asymptote at pH = 7? Can somebody please explain this?

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  • $\begingroup$ You may be interested in this past answer of mine, where I derive the general equation of a titration curve and analyse its behaviour close to the equivalence point. $\endgroup$ – Nicolau Saker Neto Aug 6 '16 at 12:38
  • $\begingroup$ 1 mol/ml = 1000 M HCl solution! For every water molecule there would be ~18 HCl molecules. $\endgroup$ – Dan Burden Aug 6 '16 at 23:56
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You are neglecting the autoionization of water.

The autoionization of water has a $K_\text{eq}$ of $10^{-14}$ at room temperature and hence contributes $10^{-7}$M $\ce{H+}$ to solution, which is small (and thus negligible) when we deal with "regular" quantities of acid and base. Since we are not dealing with regular quantities of acid and base here, we must take into account the protons generated from autoionization; the addition of $10^{-7}$M $\ce{H+}$ will resolve any anomalies in pH calculations.

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