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I found somewhere in this site that question The reason behind the steep rise in pH in the acid base titration curve

when I tried to follow accepted answer, at some point I got stuck.

  • Answer says, that steep rise in PH must happen when d/dV([H+])( V is volume of added base in this case) is maximum. I disagree. It also depends on current [H+] concentration; maybe at some point d[H+]=1, and at another d[H+]=2, but if [H+]=0.1 in first case and [H+]=3 in second, then d/dV(pH) will be maximum in first case.
  • He differentiates this : enter image description here

    and gets this : enter image description here

I tried to solve, but got different results.So, if I'm correct, can you give me mathematical proof why d/dV(pH) is maximum at equivalence point ?

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Since $pH = -log(a_{[H^+]})$, $[H^+]_{initial}$ is irreverent when discussing the equivalence point. You can always show a full titration curve over all pH for a given acid-base titration regardless of initial proton concentrations. Therefore, equivalence point is independent of initial concentration. That being said, the fact that the equivalence point occurs when $\frac{pH}{dV}$ is at a maximum has to do with the shape of the pH vs Volume graph. Since titration curves look similar to $y = tan^{-1}(x)$, the maximum slope will occur at the equivalence point (that would be (0,0) for inverse tangent). This can be further confirmed by taking the second derivative and comparing where the graph crosses 0 and comparing to the maximum of $\frac{pH}{dV}$.

As for the derivative, using a derivative calculator to ensure the correct answer, I got:

$2[H^+]-(\frac{C_{A}V_{A}}{V_{A}+V_{B}})+(\frac{C_{B}V_{B}}{V_{A}+V_{B}})$ which simplifies to $2[H^+]+\frac{C_{B}V_{B}-C_{A}V_{A}}{V_{A}+V_{B}}$. A simple way to determine if these are the same is to use a CAS (computer algebra system) and divide the answer provided by one of these answers. If the result is 1, the difference is only in simplification.

I hope this answers the question adequately.

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