# Auto-ionization of water

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

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?

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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? –  Jaydon Z Mar 24 '11 at 11:42