# Why should we not add the H+ ion conc coming from water before calculating pH of acidic/basic solution?

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.

• 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. Commented Dec 22, 2022 at 15:39
• 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. Commented Dec 22, 2022 at 18:09

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.

• 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. Commented Dec 22, 2022 at 18:50