Wikipedia gives this as the levelling effect of water

Any acid that is stronger than $\ce{H3O+}$ reacts with $\ce{H2O}$ to form $\ce{H3O+}$. Therefore, no acid stronger than $\ce{H3O+}$ exists in $\ce{H2O}$. For example, aqueous perchloric acid ($\ce{HClO4}$), aqueous hydrochloric acid ($\ce{HCl}$) and aqueous nitric acid ($\ce{HNO3}$) are all completely ionized and are all equally strong acids.

So from what I can understand, if I take an aqueous medium, then $\ce{H2SO4}$ and $\ce{HNO3}$ both should be equally strong.

But then comes the nitration reaction, where the nitronium cation acts as the real attacking electrophile, which means that in this reaction, $\ce{HNO3}$ acted as a base and $\ce{H2SO4}$ acted as an acid.

nitration mechanism

I am also aware concentrated acids are taken in the nitration reactions, but even the most concentrated acids have some amount of water.

  1. How are these two concepts contradicting, where am I going wrong?
  2. What are the limitations of the levelling effect and where all is it applicable? Is it not applicable to concentrated solutions, and if so, how concentrated?
  • $\begingroup$ Hi, AFAIK for comparing acidic strengths of strong acids we use other solvents (rather that water, for the same reason as you mentioned) like acetic acid. Therefore, since nitric acid is weaker than sulfuric acid it should get protonated in nitrating mixture. There is a full chance, that I didn't get the question... $\endgroup$
    – Zenix
    Jan 11, 2020 at 16:43
  • $\begingroup$ Yes I wholly agree that Nitric acid is weaker, but dude in FC we do have water content @Zenix $\endgroup$ Jan 11, 2020 at 16:45
  • $\begingroup$ As you already mentioned that nitrating mixture consists of concentrated sulfuric acid, just a thought, won't it absorb water, if it's still left? Since it's a powerful desiccant... $\endgroup$
    – Zenix
    Jan 11, 2020 at 16:50
  • $\begingroup$ Nah Nah doesnt work that way, it forms a constant boiling azeotrope and it cant be purified further (anyone correct me here if i am wrong) $\endgroup$ Jan 11, 2020 at 16:57
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    $\begingroup$ The stable "concentrated sulfuric acid" is about 98% H2SO4, so the water activity is essentially 0, and there are many more molecules of acid than there are of H2O. Thus, the behavior of aqueous solutions is irrelevant. NItric acid has higher water content, but typically only a small volume of it is added to the sulfuric to make the mixed acid. $\endgroup$
    – Andrew
    Jan 11, 2020 at 18:32

1 Answer 1


What you have gone wrong in this question is you completely disregard the fact that mentioned stronger acids than water are all in aqueous medium, in other words, dissolved in water. Thus, fast acid-base reaction ($\ce{H2O}$ acts as a base here) happens to give $\ce{H3O+}$ as the only acid in these mixtures (leveling effect of water).

On the other hand, concentration of sulfuric acid is $\approx 98\% (w/w)$. That means, approximately only $\pu{2 g}$ of $\ce{H2O}$ in $\pu{100 g}$ of concentrated solution. Therefore, mole ratio of $\ce{H2O : H2SO4}$ would be about $0.1 : 1$. Thus, ($\ce{H2SO4}$ is not leveling with water. That extra remaining $\ce{H2SO4}$ can be acted on $\ce{HNO3}$ in nitrating mixture as shown in your scheme (vide supra).

Note: As a rule of thumb, you may simply consider $1:1$ mole ratio of water and any stronger acid than water would be the limiting level for the leveling effect.

  • $\begingroup$ Okay, but what we have learnt in ionic equilibrium as that the strongest acid always reacts with the strongest base. So, $\ce{H2O}$ is definitely much much stronger a base than nitric acid. $\endgroup$ Jan 12, 2020 at 12:55
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    $\begingroup$ @HahaHahaha - of course the $\ce{H2O}$ reacts with the sulfuric acid, but there is so little of it that even if it were all protonated to $\ce{H3O+}$, there would still be plenty of protons from $\ce{H2SO4}$ available to react with the nitric acid. You need to think of $\ce{H2SO4}$ as the solvent. Every molecule of $\ce{H2O}$ or of $\ce{HNO3}$ is surrounded by many molecules of $\ce{H2SO4}$. $\endgroup$
    – Andrew
    Jan 12, 2020 at 14:26
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    $\begingroup$ Okay great that makes sense thanks both of y'all, I am accepting this answer $\endgroup$ Jan 12, 2020 at 14:29
  • $\begingroup$ I didn't see OP's question until now (was in my sleep). Thanks Andrew for answering. :-) $\endgroup$ Jan 12, 2020 at 16:27

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