$HCl(g) + H_2O(l) -> H_3O^{+}(aq) + OH^{-}(aq)$

Assuming $HCl$ completely ionizes in water, if we prepare a $HCl$ solution, will the $[HCl]=[H_3O]$ ($HCl$ concentration = $H_3O$ concentration)? That is, the $HCl$ before the reaction, and the $H_3O$ after reaction?

If so, then if we observe the following equilibrium reaction for the auto-ionization of water:

$2H_2O \rightleftharpoons H_3O^{+} + OH^{-}$

it can be concluded that $[H^{+}][OH^{-}]$ is a constant value, that is, the product of their equilibrium concentrations is a constant value. However, if the $H_3O$ concentration increases by the ionization of $HCl$, wouldn't the reaction shift to the left to counteract the change (by Le Chatelier's principle)? Consequently, this would decrease the concentration of $H_3O$ ions in solution, and contradict the equality of $HCl$ concentration = $H_3O$ concentration, so was the original proposition incorrect?


1 Answer 1


No, the autoprotolysis equation equilibrium makes the concentration of OH- to decrease appropriately. Say, the H3O+ is 1 mole/ltr, the OH- will be 10exp-14 moles per liter. If You calculate that precisely, You will find that the OH- and H3O+ will differ a tiny bit from that values. But compared to the deviations of the quilibrium due to temperature and some other reasons, that is way below any acuracy of chemical measurement. (and the calculus becomes somewhat ugly) Chemists as I live very happily with such simplifications. pH differences of 0.1 are can measured quite reliantly, 0.01 affords a lot of experience and some tricky preparation and apparatus (unknown to say, 95 % of chemistry PHDs), and that is the end of all we can do. Imagine, that that logarithmic scale (for equilibrium constants as well!) is not just some trick of chemists, nature behaves like that! Chemical potential depends linearly on log of concentration. Summa: that is why chemistry works so well on base of thumb rules.

  • $\begingroup$ Thanks for writing. :) The $OH^{-}$ concentration will decrease appropriately, but the only way for that to happen is by shifting the reaction to the left, so wouldn't the $H^{+}$ concentration decrease correspondingly? Also, what deviations are you referring to with temperature and other reasons? $\endgroup$
    – astiara
    Commented Mar 24, 2011 at 11:42
  • $\begingroup$ Any equlibrium constant depends on temperature. That reflects the nature of the reaction and the thermodynamics involved: exothermic, endothermic, and often reaction entropy causes rather surprising results. $\endgroup$
    – Georg
    Commented Mar 24, 2011 at 11:48
  • $\begingroup$ kind of unrelated but the system flagged you as having deleted a bunch of posts recently - any particular reason? (Same as last time?) $\endgroup$
    – David Z
    Commented Apr 4, 2011 at 22:00
  • $\begingroup$ No, just as usual, I delete older answers from time to time, when I have the impression they were not understood. $\endgroup$
    – Georg
    Commented Apr 5, 2011 at 8:43

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