A friend of mine was looking over the definition of pH and was wondering if it is possible to have a negative pH. From the equation below, it certainly seems mathematically possible—if you have a $1.1$ (or something $\gt 1$) molar solution of $\ce{H+}$ ions: $$\text{pH} = -\log([\ce{H+}])$$ (Where $[\ce{X}]$ denotes the concentration of $\ce{X}$ in $\frac{\text{mol}}{\text{L}}$.)

If $[\ce{H+}] = 1.1\ \frac{\text{mol}}{\text{L}}$, then $\mathrm{pH} = -\log(1.1) \approx -0.095 $

So, it is theoretically possible to create a substance with a negative pH. But, is it physically possible (e.g. can we create a 1.1 molar acid in the lab that actually still behaves consistently with that equation)?

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    $\begingroup$ The $0$ on the pH scale is an artefact of our system of units. Physically, there is nothing special about the concentration $1\textrm{ mol dm}^{-3}$ (pH $0$) any more than there is about the concentration $1 \textrm{ lb-mol ft}^{-3}$. $\endgroup$ Commented Aug 8, 2013 at 9:04
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    $\begingroup$ I won't go into technicalities as it has been abundantly discussed above but highest recorded $\mathrm{pH}$ is of fluoroantimonic acid with $\mathrm{pH}\ {-25}$, so yeah it's possible. $\endgroup$
    – Fakhruddin
    Commented Oct 19, 2014 at 16:19
  • $\begingroup$ Adding a little more background and maybe some boring technicalities may actually be the way to a good answer. In it's current state, this answer does not provide any more novel insight into the topic. $\endgroup$ Commented Oct 19, 2014 at 16:55
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    $\begingroup$ It's not a pH but pKa value $\endgroup$
    – Mithoron
    Commented Jan 11, 2015 at 17:35
  • $\begingroup$ This was featured in a Stack Overflow blog post (near "Chemistry isn’t collecting any rust"). $\endgroup$ Commented Jun 30, 2022 at 0:22

7 Answers 7


One publication for you: “Negative pH Does Exist”, K. F. Lim, J. Chem. Educ. 2006, 83, 1465. Quoting the abstract in full:

The misconception that pH lies between 0 and 14 has been perpetuated in popular-science books, textbooks, revision guides, and reference books.

The article text provides some counterexamples:

For example, commercially available concentrated HCl solution (37% by mass) has $\mathrm{pH} \approx -1.1$, while saturated NaOH solution has $\mathrm{pH} \approx 15.0$.


It's certainly possible theoretically. Solve for $\ce{pH < 0}$:

$\ce{-log[H+] < 0\\ log[H+] > 0\\ [H+] > 1}$

So, as you said, a solution in which the hydrogen ion concentration exceeds one should theoretically have a negative $\ce{pH}$. That said, at those extremes of concentration, the utility and accuracy of the $\ce{pH}$ scale breaks down for various reasons.

Even acids conventionally categorized as "strong" do not in fact dissociate 100%. In reality, their dissociation is also essentially an equilibrium process, though this only becomes apparent at surpassingly high concentrations. As the solution becomes more concentrated, any additional acid cannot be as thoroughly solvated, and the chemical equilibrium begins to favor dissociation progressively less and less. Hence, as the solution becomes increasingly saturated, the extent of dissociation begins to plateau and the hydrogen ion concentration approaches some practical upper limit. Furthermore, $\ce{pH}$ measured via molar concentration as a proxy for thermodynamic activity is inherently inaccurate at the extremes of concentration. Other phenomena, such as the formation of distinct chemical species by self-ionization in a concentration-dependent manner further complicate things (e.g., generation of $\ce{H3SO4+}$ in concentrated sulfuric acid, $\ce{H2F+}$ in concentrated hydrofluoric acid, etc.).

For highly concentrated solutions of strong acids, alternatives/extensions to $\ce{pH}$ exist that are functional beyond the limits of $\ce{pH}$ (see, for example, the Hammett acidity function).

As for whether solutions of negative $\ce{pH}$ have actually been experimentally prepared or observed, the answer is yes. Here's a link to one article describing the measurement of $\ce{pH}$ in acidic mine waters, which cites a figure of $-3.6$.

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    $\begingroup$ I completely agree with your answer. As I recall the pH is derived from the mass action law in aqueous solution (at $25^\circ{}C$), hence $\ce{pH = -\log_{10} c(H3O+)}$, with $\ce{K_{w} = [H3O+]\cdot[{}^{-}OH]\approx14}$. This is also one reason why those acidities are hard to measure. It is also worth mentioning that it is highly temperature depending. ($K=\exp\{-\frac{\Delta G}{RT}\}$). $\endgroup$ Commented Apr 10, 2014 at 3:46
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    $\begingroup$ @GregE. It's somewhat contradictory to discuss pH as -log[H+], and then say that pH = -3.6 is possible. There is no way you can cram over 1000 moles of hydronium ions into a liter! The -3.6 value only makes sense if you explain that pH = - log (H+ activity), and that it is the deviation of activity from concentration that makes the value -3.6 possible. -log[H+] instead would be about -1 for the mine water. $\endgroup$
    – DavePhD
    Commented Apr 15, 2015 at 11:19
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    $\begingroup$ @Dave I am not so sure of your comment. Thought the brackets technically meant activity. We just assume it's proportional to concentration in typical usage. $\endgroup$ Commented Aug 10, 2019 at 9:27

Any strong acid solution with concentration more than 1 mol/L has the negative pH. Think about any concentrate commonly used strong acid solution such as 3M $\ce{HCl}$, 6M $\ce{HNO3}$. Negative pH is actually very common.


It is very much possible.

Let’s say you put 3 moles of $\ce{HCl}$ into 1 mole of water. $\ce{HCl}$, being a strong acid dissociates completely into $\ce{H+}$ and $\ce{Cl-}$ ions as:

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

so after complete dissociation, $[\ce{H+}]=3~\mathrm{mol/L}$ (ignoring the very tiny contribution from water itself)

By definition, $$\mathrm{pH} = -\log[\ce{H+}]$$

therefore, $\mathrm{pH}= -\log 3= -0.48$

So it is very much possible to have solutions of strong acids whose $\ce{[H+]}$ is 1 molar or more, and thus whose pH is negative.

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    $\begingroup$ While I agree that a concentrated solution of a strong acid is an excellent counterexample, it should be mentioned that this treatment is bordering on being a bit simplistic. The pH is strictly defined as the negative logarithm of the activity of $\ce{H+}$, and the value of the activity deviates from the value of the concentration, especially at higher concentrations. Just as an example, the standard hydrogen electrode actually uses 1.18 M HCl to ensure that $a_{\ce{H+}} = 1$. $\endgroup$ Commented May 2, 2016 at 15:57
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    $\begingroup$ While negative pH values are most certainly possible (see all of the other answers), there is a practical element to the idea that the pH scale goes from 0-14. Most commercial pH meters, the ones used in educational settings (and many commercial settings) have a significant amount of error as you move to very high and very low pH. In addition, measuring pH values below 1 can also damage many pH meters, as the acid attacks the glass and other components of the probe. $\endgroup$ Commented Nov 8, 2017 at 13:31
  • $\begingroup$ What is the actual measured result? How much does it deviate? $\endgroup$ Commented Jun 30, 2022 at 0:26

It is possible to have $\mathrm{pH}<0$ and you don't need to create any substance. Take a concentrated solution of one of the strong inorganic acids (i.e. one with dissociation constant above 1000 like sulfuric acid) and here you are.


The pH scale is taken on our reference as 0 to 14 for concentration values from $1~\mathrm{M}$ to $\mathrm{10}^{-14} \, \mathrm{M}$. This range is such that our normal calculations in lab can be worked out easily. It is to be noted that this scale is at room temperature. If you increase the temperature then the limits changes. For example pH of pure water at $100 \, \mathrm{^\circ C}$ is $6.14$ and not $7$. Hence we can see that the scale has shifted with temperature.

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    $\begingroup$ This is downright incorrect. Yes the pH changes at different temperatures but that does not in any way mean that there are "limits" on the pH scale. "This range is such that our normal calculations in lab can be worked out easily" - no such thing. $\endgroup$ Commented May 2, 2016 at 15:58
  • $\begingroup$ If pH is about concentrations, then probably we have endless values for them. Depending them on temperature will increase those number of values just manifold. $\endgroup$ Commented May 27, 2023 at 12:27

$\mathrm{pH}$ is essentially a convention. It is defined as $$-\log_{10} [\ce{H+}]$$ since the concentrations of the solutions commonly used lie in the interval $$[10^{-14}\ \mathrm{mol/L},1\ \mathrm{mol/L}]$$ and thus the $\mathrm{pH}$ lies in $$[0,14]$$ But nothing constrains an aqueous solution from having a $\mathrm{pH}$ that does not lie in this interval. The only constraints are: $$[\ce{H+}]\lt[\ce{H2O}]_\text{liquid}$$ and $$[\ce{OH-}]\lt[\ce{H2O}]_\text{liquid}$$ The first limiting case is when you suppose all water has turned into $\ce{H+}$, which is not quite true, because there must be some water that has turned into $\ce{OH-}$ in order to $$K_\mathrm w=[\ce{H+}][\ce{OH-}]$$ But $$[\ce{H2O}]_\text{liquid}=\frac{1\ \mathrm{mol}}{18\ \mathrm g}\frac{1000\ \mathrm g}{1\ \mathrm L}=55.6\ \mathrm{mol/L}$$ And then we have $$[\ce{H+}]\lt55.6\ \mathrm{mol/L}$$ The last case implies $$[\ce{OH-}]\lt[\ce{H2O}]_\text{liquid}$$ which means (considering $K_\mathrm w=10^{-14}$) $$[\ce{H+}]\gt\frac{10^{-14}}{55.6}\ \mathrm{mol/L}$$ Then $$\frac{10^{-14}}{55.6}\ \mathrm{mol/L}\lt[\ce{H+}]\lt55.6\ \mathrm{mol/L}$$ $$-\log_{10}(55.6)\lt-\log_{10} [\ce{H+}]\lt-\log_{10}\left(\frac{10^{-14}}{55.6}\right)$$ $$-1.74\lt\mathrm{pH}\lt15.74$$

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    $\begingroup$ This is valid for aqueous solutions at room temperature, in which $K_w$ assumes that value and the density of water is approximately 1g/mL $\endgroup$
    – 1__
    Commented Aug 10, 2019 at 1:15
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    $\begingroup$ I'm afraid you started off with an approximate definition of pH, which makes the remainder of the derivation incorrect. pH is defined by the chemical activity of protons, not their concentration. The activity coefficient can differ enormously from 1 in concentrated solutions, allowing your proposed bounds to be exceeded. $\endgroup$ Commented Aug 10, 2019 at 2:26

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