For eg: If we are calculating pH of the soln of a strong acid, we shall do this: pH= - log [H+]

Where H+ is the concentration coming from that acid.

Why shall we not add to this concentration, the H+ ions coming from water itself (By the autoionization of water) before calculating the pH?

Does water not ionize in presence of the acid/base?

Thank you.

  • 2
    $\begingroup$ Unless you're working with extremely dilute solutions ($\pu{\mu mol/L}$ or lower) of a strong acid, the amount of $\ce{H+}$ contributed by water is negligible for pH calculations. $\endgroup$
    – Sam202
    Dec 22, 2022 at 15:39
  • 2
    $\begingroup$ It is rather removed than not added, in cases where simplification error is much smaller than error due ignoring activity cofficients <> 1. Presence of acids or bases suppresses water auto-ionization. If there is 1e-5 mol/L of H+ or OH-, there is 1e-9 mol/L of H+ or OH- due water ionization, i.e. 10000 times less. $\endgroup$
    – Poutnik
    Dec 22, 2022 at 18:09

1 Answer 1


Pure water is very weakly dissociated, meaning that the amount of H+ and OH- is very small. If we look at the autoprotolysis equilibrium of water

$$\ce{H2O <=> H+ + OH-}$$

and its autoprotolysis constant at 25 °C

$$\ce{K_w = [H+][OH-] = 1.01\times10^{-14}}$$

According to the equation above we know that the H+ and OH- are the same, so we can easily calculate that the concentration of H+ is around $\ce{1.005\times10^{-7}}$ mol/L

Now let's assume you add a small amount of a strong acid, such as HCl, so that the concentration of this acid solution is 0.001 mol/L. If we assume a complete dissociation of HCl according to the following equation

$$\ce{HCl -> H+ + Cl-}$$

we know that the concentrations of H+ and Cl- ions are the same, and equal to 0.001 or $\ce{1.0\times10^{-3}}$ mol/L. Thus, several orders of magnitude higher than the amount of H+ ions coming from water.

So when calculating the pH we can use only the concentration of H+ from HCl dissociation, where $\ce{pH = -log_{10}(1.0\times10^{-3}) = 3}$. Or you can sum up the concentration of H+ coming from HCl and $\ce{H2O}$, where $\ce{pH = -log_{10}(1.0\times10^{-3} + 1.005\times10^{-7}}) \approx 3$ or more specifically 2.99996. But of course, this amount of decimal numbers makes no sense.

  • 2
    $\begingroup$ Be aware of supression of water auto-dissociation by H+ excess. With pH=3 is pOH=11 so there is 1e-11 mol/L of H+ and OH- from water dissociation. $\endgroup$
    – Poutnik
    Dec 22, 2022 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.