From doing some research on the site I have found that many people have posted about solutions of $\ce{HCl}$ where $[\ce{HCl}] = \pu{1e-8 mol dm-3}$

Here they are able to deduce

$$[\ce{H+}] = \frac{[\ce{HCl}]_0}{2} + \sqrt{\frac{[\ce{HCl}]_0^2}{4} + K_\mathrm{w}}$$

where $[\ce{HCl}]_0$ is the initial concentration of $\ce{HCl}$ dissolved.

Can we make a generalisation of this and say if $\ce{H_xA}$ is a strong acid that will dissociate fully when dissolved in water? Then if $[\ce{H_xA}]_0 < \sqrt{K_\mathrm{w}}$ then the $\mathrm{pH}$ of solution can be expressed by

$$\mathrm{pH} = -\log_{10} \left(\frac{x[\ce{H_xA}]_0}{2} + \sqrt{\frac{(x[\ce{H_xA}]_0)^2}{4} + K_\mathrm{w}}\right)$$


1 Answer 1


Your formulation us correct for an acid with one dissociable proton ($x=1$) below pH 7, and if you assume ideal solution behavior it even works beyond your upper bound. Typically you should take $x=1$ because, mostly, even most strong acids dissociate one proton. However...

One limitation, however, is when you have an acid with two or more dissociable protons. If the second proton dissociates before you get to pH 7, then you need to add that into your calculation (and then it gets a lot more complicated). Among common acids sulfuric acid is the one that will do this; in the limit of extremely dilute solutions (0.001 molar or less, roughly), the pH of a sulfuric acid solution will be about 0.3 unit lower than your formula with $x=1$ because of that second dissociable proton.

  • $\begingroup$ What would the formula for sulphuric be then? $\endgroup$
    – H.Linkhorn
    Jan 12, 2019 at 12:52
  • 1
    $\begingroup$ You'd have to solve a cubic equation. I said it would be complicated ... . $\endgroup$ Jan 12, 2019 at 12:53
  • $\begingroup$ what is the second proton didn't dissociate before pH=7? $\endgroup$
    – H.Linkhorn
    Jan 12, 2019 at 13:05
  • $\begingroup$ Put $x=1$ then, that is all there is. The remaining undissociated protons are part of the "A". $\endgroup$ Jan 12, 2019 at 13:06
  • $\begingroup$ Could you elaborate on that? $\endgroup$
    – H.Linkhorn
    Jan 12, 2019 at 13:07

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