Why is $\ce{H_2SO_4}$ more acidic than $\ce{HNO_3}$. Shouldn't it be the other way around because $\ce{N}$ is more electronegative than $\ce{S}$?

My reason: $\ce{O}$ can take electrons more easily from $\ce{S}$ than from $\ce{N}$. Hence, it will be more difficult to remove $\ce{H}$ from $\ce{H2SO4}$ than from $\ce{HNO3}$. Hence, $\ce{HNO3}$ should be more acidic.

  • 2
    $\begingroup$ Note that by the Chemistry SE site policy, it is preferred not using MathJax formatting in titles, for indexing/searching reason. $\endgroup$
    – Poutnik
    Feb 11 '20 at 7:54
  • $\begingroup$ N is not more electronegative than O !!! $\endgroup$
    – Maurice
    Feb 11 '20 at 9:29
  • $\begingroup$ I think the OP means S, as argued in the second paragraph. Should be a typo. $\endgroup$
    – mck
    Feb 11 '20 at 9:39
  • $\begingroup$ What about $\ce{H2SO3 and HNO2}$? If you are just arguing S vs N, they would have the same strength as $\ce{H2SO4 and HNO3}$, respectively. $\endgroup$ Feb 14 '20 at 4:44
  • 1
    $\begingroup$ @Karsten Theis By analysis of $\ce{^{17}O}$ Raman IR spectroscopy, there has been reportedly detected presence of dissolved SO2 and HSO4-, but no H2SO3. So the equilibrium is like $\ce{SO2 + H2O <=> HSO3- + H+}$. $\endgroup$
    – Poutnik
    Mar 25 '20 at 14:11

The important thing about acidity is the stability of the conjugate base, which is enhanced by resonance effect and inductive effect. The more stable the conjugate base, the easier deprotonation becomes, and thus the stronger the acid.

In a $\ce{NO3^-}$ ion, the negative charge can delocalise among three O atoms, but in a $\ce{HSO4^-}$ ion, in addition to delocalisation among three O atoms, there is an additional OH group, which further stabilises the anion by inductive effect. So $\ce{HSO4^-}$ is the more stable conjugate base, and $\ce{H2SO4}$ is a stronger acid than $\ce{HNO3}$.

  • $\begingroup$ According to my argument, since O can easily take electrons from S, O will have a partially -ve charge. Due to this, H+ will be difficult to release and hence $\ce{H2SO4}$ should be less acidic. But I do understand the first reason so that's fine. $\endgroup$ Feb 11 '20 at 10:21
  • $\begingroup$ No that's not correct. One should not consider the attraction between the partial -ve charge on O and H+, but instead should consider the O-H bond strength. The more -ve O becomes, the less electrons it needs from H, and the weaker the O-H bond, so H actually can escape more easily. $\endgroup$
    – mck
    Feb 13 '20 at 10:07
  • $\begingroup$ But in that case, H- will be released not H+... $\endgroup$ Feb 13 '20 at 15:05
  • $\begingroup$ Then maybe I'm unable to rationalize your original idea. Answer edited to remove that part. $\endgroup$
    – mck
    Feb 16 '20 at 14:40

Here's my method for looking at it:

For determining acidity, we need to look which one of $\ce{H2SO4}$ or $\ce{HNO3}$ form a more stable anion after the $H+$ is extracted.

Lets condsider $\ce{H2SO4}$ first- enter image description here • There are two sites to extract a proton.

• After a proton is removed, the $O-$ is in resonance with 2 oxygen atoms attached to the sulfur atom.

Now, in $\ce{HNO3}$ - enter image description here • There is one site to extract proton

• After removal of $H+$ , there are two resonance structures but the eletrons of the oxygen that lost a proton have delocalised only once and over a shorter region as compared to $\ce{H2SO4}$

Generally, a compound with more resonating structures is more stable due to more charge distribution. Amongst other factors like electronegativity, resonance usually dominates.

Fun fact: $\ce{H2SO4}$ is stronger acid than $\ce{HNO3}$ , this is shown by following reaction: $\ce{H2SO4} + \ce{HNO3} ---> \ce{NO2+} + \ce{HSO4-} + \ce{H2O}$ This reaction is used in the nitration of benzene to form the electrophile.


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