# Why don't soluble gases that evolve in a reaction remain in solution?

Take for instance the following process:

$$\text{2NaCl}_{\ (s)} + \text{H}_2\text{SO}_{4 \ (aq)} \to \text{2HCL}_{\ (g)} + \text{Na}_2\text{SO}_{4 \ (aq)}$$

Why does the evolved gas leave the system instead of reprotonating water?

• There is no water in the system, to begin with. You normally use concentrated acid for this reaction. With diluted acid, gas would remain in the solution indeed. – Ivan Neretin Jul 21 '18 at 14:32

Given the reaction:

$$\ce{2NaCl(s) + H2SO4(aq) -> HCl(?) + Na2SO4(aq)}$$

It depends...

If the system is open, like a beaker sitting out on the bench, then HCl tends to escape from solution, esspecially if there is a large excess of $\ce{H2SO4}$.

If the system is closed, then Henry's Law relates the partial pressure of the compound in the gas phase to the concentration of the dissolved gas. The relative amounts of the compound in each phase of course will depend on the relative volumes of the gas and liquid.

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

$$H^{cp} = c/p$$ where:

• $H^{cp}$ = Henry's constant
• c = concentration of compound in aqueous phase
• p = partial pressure of compound in gas phase

There is another consideration for HCl since it readily ionizes in water too. So the HCl which is dissolved in solution is dependent on two different equilibria.

$$\ce{HCl(aq) <=> H^+ + Cl-}$$