Why doesn't HCl form when you dissolve NaCl in water?

Question

People say it dissolves, but shouldn't the acid form?

$$ \ce{NaCl + H2O ->[?] HCl + ???} $$

Answer 1

If you dissolve NaCl in water you will get some HCl molecules but there's definitely not going to be a significant concentration of HCl formed.

The reaction that you propose -

$\ce{Cl- + H2O -> HCl + HO-}$

is highly thermodynamically unfavorable.

We can ascertain this fact through consultation of any pKa/pKb table. In the equation above, the product acid (HCl) is a much (as in almost a trillion trillion times) stronger acid than water.

Given that HCl is several trillion times stronger than water as a acid, then naturally, HCl will want to protonate hydroxide ion, a byproduct of HCl formation from chloride ion. This is ignoring the fact that hydroxide ion is also a strong base in water, so it has a high proton affinity in water.

So even if the products were formed - again, very unfavorable from a thermodynamic standpoint because the reactant base and reactant acid are both so weak - then the products would certainly react with each other and form the reactants again, resulting in no net change in solution contents and pH.

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As a result, the reaction that you propose is more like this (except that the bottom/reverse arrow should be a lot bigger).

$\ce{Cl- + H2O <<=> HCl + HO-}$

Answer 2

It does form HCl, the only question is how much? Dissociation constant of hydrochloric acid is (reference):

$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{Cl-}]}{[\ce{HCl}]} = \pu{1.3e6 M} \tag{1}$$

$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = \pu{1e-14 M^2} \tag{2}$$

In neutral water, $[\ce{H+}] = \pu{1e-7 M}$.

So in $\pu{1 M}$ $\ce{NaCl}$, assuming neutral pH and substituting $[\ce{H+}] = \pu{1e-7 M}$ in equation (1), we find $\pu{1e-13 M}$ $\ce{HCl}$ is present.

Not very much, but still billions of molecules in a 1 liter sample.

If you want to calculate rigorously, instead of assuming neutral pH solution, to the above two equations you would add two more:

• Electrical neutrality requirement: $$\ce{[Na+] + [H+] = [Cl-] + [OH-]}$$
• Number of chlorine atoms is conserved: $$\ce{[HCl] + [Cl-] = [Na+]}$$

And you would have 4 equations and 4 unknowns ($\ce{[HCl]}$, $\ce{[Cl-]}$, $\ce{[H+]}$, and $\ce{[OH-]}$), with $\ce{[Na+]}$ being known from the amount of $\ce{NaCl}$ added to the water.

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The above is a simplification of the real situation.

As explained in Essentials of Chemistry by Hessler and Smith at page 327, which even though it is over 100 years old I still think is quite accurate:

in an aqueous solution of sodium chloride [are] eight distinct things: the ions hydrogen, hydroxyl, sodium, and chlorine, and the undissociated substances: water, sodium chloride, sodium hydroxide and hydrochloric acid.