What is the correct way to calculate the concentration $\ce{H3O+}$ in a solution with $\ce{pH}=6.99$?

Attempt 1.

pH<7, therefore there are only $\ce{H3O+}$ particles in the solution. $[\ce{H3O+}] = 10^{-\ce{pH}} = 10^{-6.99} = 1.02 \cdot 10^{-7}$

Attempt 2.

We have $[\ce{H3O+}] = 10^{-\ce{pH}} = 10^{-6.99} = 1.02 \cdot 10^{-7}$ and $[\ce{OH-}] = 10^{-\ce{pOH}} = 10^{-7.01} = 9.77 \cdot 10^{-8}$.

Because of $\ce{H3O+ + OH- -> 2 H2O}$ we are left with $[\ce{H3O+}] = 1.02 \cdot 10^{-7}- 9.77 \cdot 10^{-8} = 4.6 \cdot 10^{-9}$

When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking very large. So I wonder what the correct way is?

  • $\begingroup$ Let's put it bluntly. What is the concentration of $\ce{H3O+}$ in a solution with pH=7.00? Try calculating it using your first way. And your second way, too. Where is the truth now? $\endgroup$ Jan 24, 2016 at 12:27
  • $\begingroup$ @IvanNeretin I believe the second. So it should be always the second way. However, someone with a degree in chemistry claimed that chemists agreed one should use the first way, because using the second way would be superfluous work and the difference is small anyway. I did not believe it, hence my question. $\endgroup$
    – wythagoras
    Jan 24, 2016 at 12:31
  • $\begingroup$ The second attempt is wrong. There exists an equilibrium between the ions. The ions don't combine to form water molecules (they actually do but the rate at which they combine is equal to the rate at which water molecules dissociate to produce the ions at equilibrium, hence no net change). Your first attempt is correct. $\endgroup$
    – Yashas
    Jan 24, 2016 at 12:48
  • $\begingroup$ @wythagoras OK, let's try the other way around. At pH=7, using your second way (which is wrong, in case nobody told that before) you would get the concentration of $\ce{H3O+}$ as 0. But wait; what is pH? How it is defined? $\endgroup$ Jan 24, 2016 at 13:51
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    $\begingroup$ Here's what's wrong with the second method. When you subtract $[\ce{OH^{-}}]$ from $[\ce{H3O+}]$ to get the "excess" $[\ce{H3O+}]$, you are implicitly assigning an equilibrium constant of $+\infty$ to the neutralisation reaction. This is not true; the equilibrium constant is high ($\mathrm{k_{w}^{-1}=10^{14}}$) but it is not infinite. At these very low concentrations, you cannot perform such a subtraction, and must take the finite value of the equilibrium constant into account. $\endgroup$ Jan 24, 2016 at 21:35

5 Answers 5


If you take a sample of pure water, there will be few hydroxide and hydronium ions. Of course, they can combine to form water and yes they do combine but there will be few water molecules which break/combine to form the ions again. Hence, there exists a dynamic equilibrium between concentration of ions and water molecules.

$\textrm{pH}$ by definition is the negative logarithm of hydronium ion concentration.

$$\textrm{pH} = -\log [\ce{H^+}] = -\log [\ce{H3O^+}]$$

You can obtain the concentration of H+ ions by substituting the value of pH in the following formula,

$$[\ce{H3O^+}] = 10^{\mathrm{-pH}}.$$

Your attempt 2 is flawed because your assumption that all the ions combine to form water molecules is incorrect. There will always be some concentrations of the ions and all of them needn't combine to produce water molecules. Your attempt 1 is correct.

It appears like you are not aware of the concept of equilibrium and self ionization of water, I have picked few good materials which you might(should) want to refer to,

Chemical Equilibrium

Self Ionization of water

The concept of chemical equilibrium is very important and you will come across it frequently in chemistry, so you must learn it. Also, self-ionization of water along with chemical equilibrium are central concepts for learning acids and bases.


I think you are confusing two different concepts. If you want to know how much acid you need to add to get to a pH of 6.99, it is important to take account of the fact that water is slightly dissociated. But that was not the question. The question was simply

what is the concentration of H3O+

And that follows directly from the definition of p in pH:

$$\rm{pH = -\log_{10}([H_3O^+])}$$

A simple mathematical rearrangement gives you

$$\rm{[H_3O^+] = 10^{-6.99}}$$

Don't confuse yourself with random bits of science that don't belong in the answer... it just makes it harder than it needs to be.


Please discard earlier answer as there was a small misunderstanding.

Here self ionization of water will also take place which will increase H+ concentration and will reduce OH- concentration. Also [H+ ] from water will not be equal to 10-7 due to common ion effect. Net [H+]=10-pH

Also [H+ ] = [H3O+ ] because a single H+ combines with a single water molecule to give H3O+ without involving OH- as you did in attempt 2.

  • $\begingroup$ In earlier attempt by mistake I was considering that conc. Of HCl is given and pH is to be calculated $\endgroup$ Jan 24, 2016 at 12:55

The pH is close to 7. So the hydronium ion concentration of water can't be neglected. [H3O+fro water + H3O+ from acid][OH-]=10^-14

Please note that H2O dissociates partially to form H3O+ and OH- and that this process reaches equilibrium with finally the ionic product: [H+][OH-]=10^-14

If an acid is added to water. H+ increases and hence by the Law of Mass Action the equilibrium is pushed to the left and the concentration of OH- decreases. This is how concentration of H+ becomes greater than the concentration of OH-.

So you can in fact take H+ concentration as 10^(-ph) which gives the total concentration of H+ due to both acid and water. Your attempt 2 is conceptually erroneous as you have taken the difference of H+ and OH- and not found PH itself. I think the point that you have forgotten is that both H+ (rather H3O+) and OH- exist together in solution although one might be in excess of the other. So your first approach is more suitable. pH is by definition the negative of the common logarithm of total H+ concentration in the solution.

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    $\begingroup$ It is true that the $\ce{H+}$ from water shouldn't be neglected here. But it shouldn't be considered either. We know the pH already, so we don't care where those protons came from. $\endgroup$ Jan 24, 2016 at 12:30
  • $\begingroup$ What do you mean by "considered either"? We have to consider most of the H+ as coming from the water and not from the acid. Only a small excess amount is contributed by the acid and it is this which lowers the pH to 6.99. $\endgroup$ Jan 25, 2016 at 11:54
  • $\begingroup$ And it's true that now we know the pH we don't care about where the protons came from. $\endgroup$ Jan 25, 2016 at 11:55
  • $\begingroup$ Well, I mean just that: since we know the pH, we don't have to make any calculations involving the water self-ionization constant. $\endgroup$ Jan 25, 2016 at 12:02
  • $\begingroup$ Oh. In that sense. I simply explained about the ionization constant because the question had some confusion regarding that. $\endgroup$ Jan 25, 2016 at 14:08

PH = - log 10 [H3O+] [H3O+] = -antilog 10 (PH) [H3O+] = - 10^6.99 Because antilog b (x) = b^x Therefore , [H3O+] = 9772372.21

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    $\begingroup$ You've reached an impossibly high concentration because you've incorrectly used the properties of logarithms/exponentiation and misplaced the minus sign. $\endgroup$ Aug 10, 2019 at 13:16
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    $\begingroup$ This post is highly corrosive! Beware! $\endgroup$ Aug 10, 2019 at 13:28

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