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

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

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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.


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

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

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  • $\begingroup$ Ka of water is 1.8*10^-16 ... Ka of HCl is >55 that's for sure but I'm not sure exactly what. Likely less than 10000 if I remember correctly. A trillion is what, 10^12? So perhaps not a trillion trillion. $\endgroup$ – Dissenter Jan 10 '15 at 0:45
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    $\begingroup$ pKa of water is ~16. pKa of HCl is ~-7 or so. pKeq then would be ~23. That's just about a trillion trillion. $\endgroup$ – jerepierre Jan 10 '15 at 0:51
  • $\begingroup$ @Dissenter Thanks for your insightful answer, so what really happens is that HCL will donate the H to HO and then water will form again and that just leaves us with Cl. And the reason Cl and Na won't bond again is because they are dissolved (broke apart) by the $\ce{H2O}$ $\endgroup$ – Asker123 Jan 10 '15 at 1:11
  • $\begingroup$ @AwesomeFlame123 what will really happen isn't that ... the process of Cl- reacting with water won't even happen much if at all because that process is like trying to climb Mt. Everest ... it's energetically VERY unfavorable. $\endgroup$ – Dissenter Jan 10 '15 at 2:00
  • $\begingroup$ Wouldn't the sodium also help to keep the solution relatively pH neutral? $\endgroup$ – CJ Dennis Jan 10 '15 at 2:19
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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.


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.

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  • $\begingroup$ Yes you are right, I like your view into this problem, the math makes sense. $\endgroup$ – Asker123 Jan 11 '15 at 17:35

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