Let's say I have 1000 ml 1 M acetic acid ($\mathrm{p}K_\mathrm{a}$ = 4.76) and add 1000 ml 1M aniline ($\mathrm{p}K_\mathrm{aH}$ = 4.65). To what proportion is aniline protonated?

1M acetic acid has a pH of: $$\mathrm{pH} \approx 0.5*(4.76-\log 1)=2.38$$

Using Henderson Hasselbalch to determine fraction of protonated aniline: $$2.38=4.65-\log x$$ $$ x=\mathrm{[HA]/[A]}=186$$

so ~ 0.5 % would be aniline and 99.5 % would be protonated?

This can not be true. As aniline judging by its $\mathrm{p}K_\mathrm{a}$ does protonate less than the acetate anion. Does anilin get protonated after all?

Where is the mistake?

  • $\begingroup$ Looks like only thing you got right is pKa of acetic acid. $\endgroup$
    – Mithoron
    Commented Apr 21, 2023 at 22:36
  • $\begingroup$ Most pH related ready to use formulas are intended for specific scenarios AND specific simplifying conditions. When using them, you must know you can use them and why you can use them. Generally, you have to solve a set of nonlinear equations, based of equilibrium equations, substance amount inventory equations and charge balance equation. $\endgroup$
    – Poutnik
    Commented Apr 22, 2023 at 7:25
  • $\begingroup$ e.g. Famous pH=0.5(pKa - log c) is pH of weak acid solution IF there is no other pH active component AND IF c >> [H+] >> [OH-].(ignoring Pandora's box of activity coefficients.) $\endgroup$
    – Poutnik
    Commented Apr 22, 2023 at 7:28

2 Answers 2


Before you use any formula to calculate a pH, you first have to figure out what the major species are, and whether more than a small fraction of the species will react.

You calculated the pH of the acetic acid solution as 2.38. Try calculating the pH of the aniline solution. As you said, aniline is the deprotonated form, so the $\mathrm{p}K_\mathrm{a}$ given is that of the protonated form, and aniline is a weak base. So the pH is about 14 - 2.32 = 11.68.

This (and the fact that acetic acid and aniline are at the same concentration) tells you that there will be a reaction, and the pH will be somewhere in between.

As a first approximation, imagine that instead of acetic acid and aniline, you had acetic acid and acetate. In that case, you would have a 1:1 buffer with a pH of 4.76. If it were aniline and protonated aniline, you would have a 1:1 buffer with a pH of 4.65. The pH will be in that region, so the major species will be two weak acids and their two conjugate weak bases. This does not allow for using a weak acid formula or a buffer formula directly.

If acetic acid were to react completely with aniline, you would get an amphoteric salt of protonated aniline with acetate (at a concentration of 0.5 mol/L). It turns out there is a formula to estimate the pH of amphoteric salts and the pH of salts of weak acids and weak bases that you can try.

Once you have the pH, you can check if all equilibrium expressions (that for auto-dissociation of water, for the acetic acid and for the aniline) are satisfied simultaneously. Or you can follow the classical engineering approach with mass balance, charge balance and multiple equilibria (solve 3 equations simultaneously).

The $\mathrm{p}K_\mathrm{a}$ values are probably extrapolated to infinitely dilute solutions (ionic strength 0). At a concentration of 1 mol/L, the ionic strength is far from zero, so you can estimate the pH to one or two significant figures, say pH = 5. To get a better value, you could just make the solution and measure the pH, if the aniline is soluble at this concentration at a slightly acidic pH.

  • $\begingroup$ thanks. It may seem like a trivial question but non of my graduate chemistry colleagues could tell me $\endgroup$
    – Lukas4235
    Commented Apr 22, 2023 at 7:06
  • 1
    $\begingroup$ mass balance and multiple equilibria Charge balance is additional equation which is "free of charge". (Sum of charges of anions and cations is zero. [H+]+[BH+]=[OH-]+[A-]) $\endgroup$
    – Poutnik
    Commented Apr 22, 2023 at 11:53

One more thing that can be added maybe: You can predict the position of equilibrium between two moderately weak acid and base by comparing the pKa's and assume that oxonium and hydroxide is not the major species as explained here:


  • 2
    $\begingroup$ And then determine the pH from Henderson Hasselbalch for one or the other acid/base pair. $\endgroup$
    – Karsten
    Commented Apr 22, 2023 at 23:48
  • $\begingroup$ It is similar as pH of Na2HPO4 solution. One can approximate concentrations of H2PO4- and PO4^3- are the same. In our case, acetate and anilinium. $\endgroup$
    – Poutnik
    Commented Apr 23, 2023 at 8:00

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