# Chemistry - How to separate HCl and NaOH from salt and water solution?

## Solution 1:

Unless time travel is an option, you could

1. Electrolyze the $\ce{NaCl}$ solution to obtain a solution of $\ce{NaOH}$, and $\ce{H2}$ and $\ce{Cl2}$ as gases.

2. Collect the gases and photolyze them. The dissociation energy of $\ce{Cl2}$ is $243\, \mathrm{kJ \cdot mol^{-1}}$, irradiation at $\lambda$ < 490 nm will cleave $\ce{Cl2}$ to chlorine radicals and initiate the chain reaction to yield $\ce{HCl}$ gas.

$$\ce{Cl2 ->[h\nu] 2 Cl*}$$

$$\ce{Cl* +\ H2 -> HCl + H*}$$ $$\ce{H* +\ Cl2 -> HCl + Cl*}$$

EDIT 1

Unless you know exactly about the risk of both processes, don't do it! This video gives a nice impression of how violently hydrogen and chlorine react upon irradiation! Note that it has nothing to do with the laser involved. A flashlight would give the same dramatic effect!

EDIT 2

The answer above, giving practical advices on how to get $\ce{HCl}$ and $\ce{NaOH}$ back apparently suggests that you can not simply revert the neutralisation reaction.

The neutralisation is irreversible.

In order to find out why this is the case, you might want to have a look at thermodynamics of this exothermic reaction, particularly at the entropy changes.

## Solution 2:

The question is whether you want to get pure $\ce{HCl}$ and pure $\ce{NaOH}$ or whether you want a hydrochloric acid solution and a sodium hydroxide solution. If the former, I hereby direct you to Klaus’ answer. If the latter, read on.

The neutralisation of hydrochloric acid and sodium hydroxide is often simplistically given as in equation $(1)$.

$$\ce{HCl + NaOH -> NaCl + H2O}\tag{1}$$

However, that reaction is only true for gasous $\ce{HCl}$ reacting with solid $\ce{NaOH}$. When one neutralises the acidic, aquaeous solution and the basic, aquaeous solution, you need to first consider equation $(2)$ (for $\ce{HCl}$) and $(3)$ (for $\ce{NaOH}$).

$$\ce{HCl (g) + H2O -> HCl (aq) + H2O <=>> Cl- (aq) + H3O+ (aq)}\tag{2}$$

$$\ce{NaOH (s) + H2O -> Na+ (aq) + OH- (aq)}\tag{3}$$

In both solutions, the compounds can be considered completely dissociated. Once you mix these two together, a standard acid-base reaction happens in which only half of the ions react at all — see equation $(4)$.

$$\ce{Cl- (aq) + H3O+ (aq) + Na+ (aq) + OH- (aq) -> Cl- (aq) + Na+ (aq) + 2 H2O}\tag{4}$$

Removing the spectator ions gives us equation $(4')$

$$\ce{H3O+ (aq) + OH- (aq) -> 2 H2O}\tag{4'}$$

And this, when viewed closely, is the autoprotolysis of water (equation $(5)$) reversed.

$$\ce{2 H2O <<=> H3O+ + OH-}\\ K_w = [\ce{H3O+}][\ce{OH-}] = 10^{-14}\tag{5}$$

The ion product $K_w$ I added there shows the concentration of $\ce{H3O+}$ and $\ce{OH-}$ ions in a neutral solution. To be able to reverse that reaction you would somehow need to generate substantial quantities of $\ce{H3O+}$ and $\ce{OH-}$ — but they just aren’t present at equilibrium in the concentrations you would need, since their formation is strongly endothermic (and thus their destruction strongly exothermic).

And that is the true reason why an acid-base neutralisation is not reversible in many cases.