What is the $\mathrm{pH}$ of a $\pu{5M}$ solution of hydrochloric acid?

So in order to solve this, apparently all you need to do is plug it into the equation:

$$\mathrm{pH} = -\log[\ce{H3O+}]$$

where the concentration of $\ce{H3O+}$ is $\pu{5M}$. But why can we assume this? Don’t we need to create an ice table from $\ce{HCl + H2O-> H3O+ + Cl-}$ and use the $K_\text{a}$ value of $\ce{HCl}$ to solve this?

  • 3
    $\begingroup$ It's because HCl is so strong in aqueous solution. In certain other solvents, you may need to actually calculate it. Also, note that in especially high concentrations you may need to consider the activity rather than just plugging in concentration $\endgroup$
    – Andy
    Apr 11, 2015 at 16:08
  • $\begingroup$ @Shadock if the dissociation constant were really only 1, at 5M, the HCl would only be about 36% dissociated, but because the dissociation constant is more like 1000000, HCl is almost 100% dissociated in 5M HCl $\endgroup$
    – DavePhD
    Nov 13, 2015 at 19:50
  • $\begingroup$ @DavePhD yeah you right i will erase my comment. But I'm not sure like you said that in HCl 5M, HCl is almost 100% dissociated. Because we are speaking about dissolution of HCl in a mixture of HCl and water which can't be approximate at just water here. My knowledge in dissolution is not really important but I have some doubts. If you know some good lectures please tell me. Friendly :) $\endgroup$
    – ParaH2
    Nov 13, 2015 at 21:20
  • $\begingroup$ @Shadock right, it can't just be approximated as water, so you need to use activity instead of concentration. See the references in my answer and also pubs.acs.org/doi/abs/10.1021/ac60302a030 for more information $\endgroup$
    – DavePhD
    Nov 13, 2015 at 21:40
  • $\begingroup$ @DavePhD Yes we use activities but I'm not sure we can speak about dissociation as easely as you said. And than you for the document. :) $\endgroup$
    – ParaH2
    Nov 13, 2015 at 22:04

4 Answers 4


As explained in pH Paradoxes: Demonstrating That It Is Not True That pH ≡ -log[H+] Journal of Chemical Education Vol. 83 pages 752-757.

pH = -log (hydrogen ion activity).

and "pH of 7.6 M HCl is about -1.85 (not -0.88)"

So how far off is -log (5) = -0.69 from the real answer?

From this table of activity coefficients 5m HCl has an activity coefficient of 2.38, so activity is 11.9, which yields pH = -1.1.

Unfortunately the table is in terms of molality rather than molarity. 5M HCl is about 5.5m, and activity coefficient is dramaticly increasing with concentration (see table I of the pH paradox reference) so the true pH deviates somewhat further from -0.69, but pH = -1.1 is a reasonable estimate.


Method to calculate the pH of a solution with a strong monoacid

$$ \begin{array}{|c|c|c|c|c|}\hline &\text{Before}&&\text{After}&\\\hline &\ce{AH}&\ce{H2O}&\ce{A-}&\ce{H3O+}\\\hline \text{Initial}&n0&\text{excess}&0&0\\\hline \text{Equilbrium}&\epsilon&\text{excess}&n0&n0\\\hline \end{array} $$

Then we have $\mathrm{pH}=-\log\left[\ce{H3O+}\right]$ Now we have to verify if our result is true.

We consider that $\ce{H3O+}$ in the solution has exactly the same concentration as that of the acid. However they can react with $\ce{OH-}$ because of the water autoprotolysis! $K_\text{a}=10^{14}$ total reaction.

$$\ce{H3O^+ + HO^- -> 2H2O}$$

So the concentration of $\ce{H3O+}$ is approximatively constant if $\left[\ce{H3O+}\right]\gg\left[\ce{HO-}\right]$

For example if $\frac{1}{10}\left[\ce{H3O+}\right]>\left[\ce{HO-}\right]$ then we can find the limit for which our reasonning is false :

$$ \left[\ce{HO-}\right]\times \left[\ce{H3O+}\right] < \frac{1}{10} \times \left[\ce{H3O+}\right] \times \left[\ce{H3O+}\right] \iff K_\text{e}<\frac{1}{10}\left[\ce{H3O+}\right]^2 $$

Then we have $\mathrm{pH}<6.5$

So if with a strong monoacid you find $\mathrm{pH}=6.8$ with this formula the result is not correct and you have to make an other approximation or reasonning!

Here we have $\mathrm{pH}=-0.69<6.5$ so you find a reasonable answer. And in water $\mathrm{pH}$ is between $0$ and $14$ so your solution is at $\mathrm{pH} \approx 0$

  • 1
    $\begingroup$ There is no non-metrological reason for pH to be restricted to $[0,14]$ in water. Perhaps adding a point about activity vs concentration would also be a good idea. $\endgroup$ Feb 23, 2018 at 14:58
  • $\begingroup$ @LinearChristmas I can't reach the server of doi.org from where I am; when I'll have some more time I'll try to add a point on that it should be great yes. $\endgroup$
    – ParaH2
    Feb 23, 2018 at 22:55
  • 1
    $\begingroup$ It seems the link is also missing a slash /, which is why it won't open (doi should be fine). That's my mistake. $\endgroup$ Feb 23, 2018 at 23:01

There exists seven strong acids that completely disassociate with hydrogen ( $\ce{HCl}$, $\ce{HBr}$, $\ce{H2SO4}$, $\ce{HClO3}$, $\ce{HClO4}$, $\ce{HI}$, and $\ce{HNO3}$), so the only calculation you need to do is the negative log of the concentration (in molarity).

Therefore $\mathrm{pH} = -\log5.0=-0.69$. Yes, $\mathrm{pH} = −0.69$ even though the $\mathrm{pH}$ scale "normally" ends at $\ce{pH} = 0$.

(in case of $\ce{HCl}~\mathrm{pH}$ will be negative for any solution more then $\pu{2M}$)


Just to give a quick and logical answer, if you look at the following acids:

  • $\ce{HCl}$
  • $\ce{HBr}$
  • $\ce{HI}$
  • $\ce{H2SO4}$ - first dissociation only
  • $\ce{HClO4}$
  • $\ce{HClO3}$
  • $\ce{HNO3}$

These acids above are Strong Acids, meaning that they completely dissociate, so for that purpose there is no need of a certain K value, since none of this is an equilibrium reaction.

Just a forward reaction. $\ce{->}$ not $\ce{<->}$

NOTE: This is usually for general purposes. Not to be super accurate.

  • 5
    $\begingroup$ This really isn't true in the concentration range of the question. For example quantitatively for nitric acid the degree of dissociation is 4M: 90%, 6M: 75%, 8M: 55%. See Ionization of Strong Electrolytes. III. Proton Magnetic Resonance in Nitric, Perchloric, and Hydrochloric Acids J. Chem. Phys. 22, 2067 (1954) scitation.aip.org/content/aip/journal/jcp/22/12/10.1063/… $\endgroup$
    – DavePhD
    Apr 21, 2015 at 13:30

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