Sulfuric acid because of its low volatility can be used to manufacture more volatile acids from their corresponding salts.

How does volatility affect the production of acids? Isn't it that sulfuric acid being stronger than the salt of weak acid replaces the anion part of salt so the weak acid is produced or does just the low volatility of sulfuric acid matter? Will it be feasible to produce a highly volatile acid even stronger than $\ce{H2SO4}$ by exploiting the low volatile nature of $\ce{H2SO4}$? What is the real process involved?

  • $\begingroup$ Can you please tell where did you quote that statement from? $\endgroup$
    – Kartik
    Aug 26, 2016 at 14:35
  • 1
    $\begingroup$ @Kartik From NCERT chemistry class 12 p block pg.no:190 under uses of sulfuric acid. $\endgroup$
    – JM97
    Aug 26, 2016 at 14:39
  • $\begingroup$ @Kartik No, I can't find it on page number 190 $\endgroup$
    – Ritil
    Jul 18, 2022 at 16:09
  • $\begingroup$ Can anyone please give a list of many important inorganic acids arranged in the order of volatility ? I can't find it on internet $\endgroup$
    – Ritil
    Jul 18, 2022 at 16:14

2 Answers 2


Yes, this is indeed the case. The reasoning behind it is using chemical equilibria to their fullest.

If you have a Brønsted acid and a Brønsted base in the same vessel, you will always have an equilibrium of the following kind:

$$\ce{HA + B- <=> A- + HB}$$

It depends on the nature of the acid and the base — i.e. their $\mathrm{p}K_\mathrm{a}/\mathrm{p}K_\mathrm{b}$ values — which side of the equation is favoured in a closed system (i.e. solution). For example, dissolving $\ce{NaHSO4}$ and $\ce{NaSH}$ in the same sample of water will result in the following set of equations:

$$\ce{H2SO4 + Na2S <=>> NaHSO4 + NaSH <=>> Na2SO4 + H2S}$$

Because the $\mathrm{p}K_\mathrm{a}$ values of both compare as follows:

$$\begin{array}{ccc}\hline \text{acid} & \mathrm{p}K_\mathrm{a1} & \mathrm{p}K_\mathrm{a2}\\ \hline \ce{H2SO4} & \approx -3 & 1.9\\ \ce{H2S} & 7.0 & 12.9\\ \hline\end{array}$$

We already have a way of producing $\ce{H2S}$ by means of protonating sulfide ions in solution here. However, $\mathrm{p}K_\mathrm{a1}\left (\ce{HCl}\right ) \approx -6$, which is more acidic than sulphuric acid, so adding $\ce{NaCl}$ to $\ce{H2SO4}$ should result in a net reaction of nearly nothing.

This is where vapour pressure comes in as a second factor. $\ce{HCl}$ is a gas under standard temperature and pressure while $\ce{H2SO4}$ is a liquid. ($\ce{H2S}$ is also a gas. If we added liquid sulphuric acid to sodium sulfide, $\ce{H2S}$ gas would evolve, but that’s what the $\mathrm{p}K_\mathrm{a}$ value predicts anyway.) If we have an open vessel — i.e. gases can diffuse away — then any $\ce{HCl}$ produced in equilibrium will dissociate away in the long run. This is due to a second equilibrium that we could write as follows:

$$\ce{HCl (g) <=> HCl (aq)}$$

The ‘equilibrium constant’ of this physical process is more or less what we call vapour pressure, and for hydrogen chloride both sides are significantly more balanced than for sulphuric acid. Therefore, we should describe the entire system in the following way:

$$\ce{NaCl + H2SO4 <=> NaHSO4 + HCl (diss) <=>> NaHSO4 + HCl (g) ^}$$

Where $\ce{(diss)}$ is a shorthand for dissolved in sulphuric acid, our reaction medium.

Drawing away the hydrogen chloride gas in the rightmost equation, for example by reducing pressure or inducing an air flow, will shift all equilibria to the right since that product is constantly being removed. And thus we can liberate $\ce{HCl}$ gas with sulphuric acid, even though the $\mathrm{p}K_\mathrm{a}$ of hydrogen chloride is lower than that of sulphuric acid.


An example of this would be the reaction between sulfuric acid and potassium nitrate, the salt of nitric acid. When this salt is mixed with sulfuric acid, nitric acid and potassium acid sulfate are the result, although the reaction is incomplete. If the resulting mixture is then heated the nitric acid evolves as a vapor because it is much more volatile than sulfuric acid. This vapor can then be condensed separately. In this way relatively pure nitric acid can be produced.

  • $\begingroup$ Great, that's exactly what I needed in solving a textbook problem right now. +1 $\endgroup$ May 4, 2016 at 18:38

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