Why is $\ce{HClO4}$ more acidic than $\ce{HBrO4}$? But $\ce{HCl}$ is less acidic than $\ce{HBr}$? What determines the acidity? Is it the concentration of $\ce{H+}$? If so, how do these trends develop along the halogen group?

  • 1
    $\begingroup$ Would you mind to give the $pK_a$ values and the solvent in which they were measured for the acids you are asking about? $\endgroup$ – Klaus-Dieter Warzecha Mar 23 '14 at 17:42

In the determination of the $\mathrm{pK_a}$ values of really strong acids, water as a medium is irrelevant at a certain point.

Ivo Leito and coworkers have examined the acidity of various compounds in 1,2-dichloroethane (DCE).

In this medium the $\mathrm{pK_a}$ values for $\ce{HCl}$, $\ce{HNO3}$, $\ce{HBr}$, $\ce{HI}$, $\ce{HBF4}$, an $\ce{HClO4}$ have been determined as -0.4, -1.7, -4.9, -7.7, -10.3 and -13.0.


In this survey, 1,1,2,3,3-pentacyanopropene (1), was found to be even more acidic with a $\mathrm{pK_a}$ of -15.0!

EDIT It is important to note that these values are likely to be correct in their relative order. But the values are based on picric acid, for which the $\mathrm{pK_a}$ was arbitrarily set to zero. This initially escaped my attention!

Thanks to Martin for pointing out this important detail in his comment below!

Being an oddball, the rather unstable perbromic acid ($\ce{HBrO4}$) was not in the list!

It was synthesized by Evan Appelman (even on a larger scale) by bubbling fluorine gas through an alkaline solution of 1M $\ce{NaBrO3}$ and 5M $\ce{NaOH}$ until the solution becomes acid.

Amazingly, the author further states in a footnote:

The reaction is not smooth, and small explosions may take place in the vapor above the solution.

  • 1
    $\begingroup$ Could you clarify/ prove your first statement. From my understanding $\mathrm{pK_a}$ values always involve a solvent. For superacids they are measured relative to each other. Your cited paper uses 1,2-dichloroethane as solvent and the $\mathrm{pK_a}$ values have to be anchored to an acid with a known value. For the cited case this was not possible as exact measurements in DCE were not accessible. So your posted numbers are quite arbitrary. $\endgroup$ – Martin - マーチン Mar 24 '14 at 8:45
  • 1
    $\begingroup$ @Martin Good observation! One does indeed need a reference. In the paper cited, pikric acid was arbitrarily set to 0 an I forgot to mention that. This does however not invalidate the relative order. $\endgroup$ – Klaus-Dieter Warzecha Mar 24 '14 at 9:29

Assuming you are talking about acids in water, and taking into account that the acids you mention are all regarded as strong acids, I would focus on the relative stability of the corresponding conjugated bases.

$\ce{Br-}$ being larger than $\ce{Cl-}$ can better allocate the negative charge. Then, $\ce{HBr}$ ($pK_{a}$= −8.7) is stronger acid than $\ce{HCl}$ ($pK_{a}$=-6.3).

On the other hand $\ce{ClO4-}$ is more stable than $\ce{BrO4-}$. Here, another factor plays a role and should explain the difference in stability: the charge dispersal associated to resonance. Being the structure similar, we would expect the same effect: however, the resonance stabilization that characterizes such kind of ions has to be more effective for $\ce{ClO4-}$, due to a better overlap between $p$ orbitals ($2p$ of oxygen with $3p$ of chlorine, better than with $4p$ of bromine). As a consequence, $\ce{HClO4}$ ($pK_{a}$= −8) is stronger acid than $\ce{HBrO4}$ ($pK_{a}$= ? haven't found it, but here $\ce{HBrO4}$ is said to be an "almost strong acid", therefore not stronger than $\ce{H3O+}$).

These considerations tentatively explain the difference in $pK_{a}$ for such strong acids.

  • $\begingroup$ Perbromic acid: the Wikipedia article has been heavily edited. You sure it still says perbromic acid is almost strong? $\endgroup$ – Oscar Lanzi Aug 26 '18 at 20:59

There are two different kinds of acids mentioned. First, binary acids such as $\ce{HCl}$, $\ce{HBr}$, $\ce{HI}$, etc. The other acids mentioned were inorganic oxyacids or oxoacids.

The most important factor in oxoacid strength is the ability to donate a proton which is influenced by electronegativity and bond polarity.

Oxoacids are usually designated with X as the central atom, and thus, their structure can be written as:

$$ -\text{X}-\text{O}-\text{H} $$

X can be bonded to other things. i.e. hydrogens, more -OH groups, etc.

If X was an alkali or alkaline earth metal, and thus electropositive, the corresponding compound would be a base. This is not an acid. The reason for this is the $X-O$ bond is extremely covalent, almost to the point where the strength is equivalent to bond in ionic strength. The oxygen atom thus has a large partial negative charge $\delta^- $ and the electron density accumulates on the $-O-H$ while the hydrogen, $\ce{H^+}$, binds tightly with the oxygen, preventing deprotonation.

As the number of lone oxygens on $-X$ increases, the strength of the oxoacid increases as well.

Taken from Principles of Modern Chemistry by Oxtoby et. al 7th Edition:

If the formula of these acids is written as $XO_nOH_m$, the corresponding acid strengths fall into distinct classes according to the value of n, the number of lone oxygen atoms Each increase of 1 in n increases the acid ionization constant $K_a$ by a factor of about $10^5$.

Another way to examine the strength of acids is to look at the stability of their conjugate base which can be written as $XO_{n+1}(OH)^-_{m-1}$. As more oxygens are added on the central atom, the net negative charge is distributed more evenly around the base. As bases become more stable, they are less prone to accepting a proton. Remember that conjugate bases of strong acids are weak so intuitively, this makes sense.

Put briefly, as $-X$ becomes more electronegative, electron density is pulled towards the central atom. As the electron density is pulled from the $-H$ bond to $-X-O$, the $O-H$ bond breaks easily, and a proton is produced.

If we discuss organic acids and not binary nor inorganic acids, we not only have to talk about electronegativies, but steric hinderance, and resonance structures as well. If you would like, I can explain those as well.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.