I have read that sulfuric acid has low volatility. Then it states that more volatile acids can be made using sufuric acid and their corresponding salts, because of the low volatility.

Can you help me understand the chemistry behind the second statement?

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    $\begingroup$ The gist is that you could take concentrated sulfuric acid, dilute it 50:50 by volume with water and add sodium chloride. You could then boil off HCl. Not an efficient way to get HCl but it would work. $\endgroup$ – MaxW Sep 15 '16 at 6:28
  • $\begingroup$ What does volatility have to do here? $\endgroup$ – Hani Sep 15 '16 at 6:29
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    $\begingroup$ What is boiling off in the reactions I outlined? Sulfuric acid or hydrochloric acid? Isn't the first one to boil off "more volatile"? $\endgroup$ – MaxW Sep 15 '16 at 6:31

An acid according to the Brønsted-Lowry definition is any substance that can release $\ce{H+}$ ions. By having done so, it becomes a Brønsted-Lowry (conjugate) base which can accept $\ce{H+}$ ions. This is shown in equation $(1)$:

$$\begin{array}{ccccc}\ce{HA} & \ce{<=>} & \ce{H+} & + & \ce{A-}\\ \text{acid} & & \text{proton} & & \text{conjugate base}\end{array}\tag{1}$$

As already hinted, a Brønsted base is any substance that can accept such an incoming $\ce{H+}$ ion (reaction $(2)$)— as such, an acid will always react with a base and vice-versa; it is not possible for only an acidic reaction to happen.

$$\begin{array}{ccccc}\ce{B} & + & \ce{H+} & \ce{<=>} & \ce{HB+}\\ \text{base} & & \text{proton} & & \text{conjugate acid}\end{array}\tag{2}$$

Let me quickly quote part of the question:

[M]ore volatile acids can be made using sulphuric acid and their corresponding salts

The corresponding salt of a more volatile acid could also be termed its conjugate base. So the reaction we are talking about includes the conjugate base of one acid and sulphuric acid as a second acid. To exemplify things, I will use hydrochloric acid as the first acid whose conjugate base is chloride. If we thus mix chloride ions (e.g. in sodium chloride) with sulphuric acid, the following equilibrium will result:

$$\ce{Cl- + H2SO4 <=> HCl + HSO4-}\tag{3}\\[1.2em] K = \frac{[\ce{HCl}][\ce{HSO4-}]}{[\ce{Cl-}][\ce{H2SO4}]} = \frac{K_\mathrm{a} \left ( \ce{H2SO4} \right )}{K_\mathrm{a} \left ( \ce{HCl} \right )}$$

The equilibrium constant $K$ is the ratio of the acid constants of hydrochloric and sulphuric acid. Since hydrochloric acid ($\mathrm{p}K_\mathrm{a} \approx -8$) is more acidic than sulphuric acid ($\mathrm{p}K_\mathrm{a} \approx -3$), we expect the equilibrium to be shifted towards the reactant side — which would be true for a closed, pressurised system. Once we open the system, and allow for gases to be exchanged, the picture changes. Compare equation $(\text{3'})$, wherein the aggregate states of the reaction partners are mentioned.

$$\ce{Cl- (s) + H2SO4 (l) <=> HCl (g) ^ + HSO4- (s)}\tag{3'}$$

Hydrochloric acid is a gas and can diffuse away while sulphuric acid is a liquid and thus cannot diffuse away. (And cannot drip away since we are using a proper beaker of some kind.) If we constantly remove $\ce{HCl}$ from the reaction mixture by letting it diffuse away, we are exercising pressure onto the system, and per Le Châtelier’s principle the product side (one of whose species we are removing) will now be favoured.

This general principle applies to any acid, as long as it is more volatile than sulphuric acid, i.e. has a lower boiling point under the pressure used. If the second acid happens to be liquid rather than gaseous, a simple distillation will suffice to remove it from the reaction mixture.

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