So I know that there's auto-ionization of water according to:

$$\ce{2H2O <=> H3O+ +OH-} \tag{1}$$

If I have pure water where the concentrations of both $\ce{H3O+}$ and $\ce{OH-}$ is $10^{-7}$

Say I add $\pu{10^{-5} M}$ $\ce{HCl}$, I'd have [$\ce{H3O+}$] = $ 10^{-5} + 10^{-7}$ (approx $10^{-5}$) and the [$\ce{OH-}$] = $10^{-7}$ (still) but since these are in equilibrium, according to reaction ($1$), the reaction will shift to the left so this would consume $\ce{H3O+}$ and $\ce{OH-}$ until their concentrations product is $10^{-14}$, how will this happen?

Will the $\ce{OH-}$ decrease much more less than what the $\ce{H3O+}$ will decrease, in a way that what the $\ce{H3O+}$ loses is negligible compared to what $\ce{OH-}$ lost? in a way that would keep [$\ce{H3O+}$] = $10^{-5}$ and [$\ce{OH-}$] = $10^{-9}$?

In brief, how is equilibrium brought back to the concentration of ions are concerned?

  • 1
    $\begingroup$ There is no "how" to it. The reaction just happens according to the equation above, as blunt as that. One H3O+ consumes one OH-. Can it consume two? No. There is nothing left to choose. $\endgroup$ – Ivan Neretin Jul 26 '20 at 10:21
  • $\begingroup$ But when their product is brought back to $1* 10^-14 $, will the [H3O+] remain still $ 10^-5 $ ? $\endgroup$ – Stephen Alexander Jul 26 '20 at 12:12
  • $\begingroup$ Pure stoichiometry. The decrease happens in the proportion specified in the reaction, so 1:1, until the equilibrium is re-established. $\endgroup$ – Zhe Jul 26 '20 at 12:38
  • $\begingroup$ If you add HCl in pure water, this will immediately reduce the amount of autoionization of water. $\endgroup$ – Maurice Jul 26 '20 at 12:41
  • $\begingroup$ I just don't get it how the equlibrium restoration affect the concentration of these 2 ions. Doesn't the equilibrium derived equation (product of concentration of these ions) = 10^-14 always apply? If so, when we add HCl of 10^-5 M equilbrium will be disrupted, and when concentrations are adjusted in a way to restore equilbrium and apply the product of concentrations of ions law again, what would the new concentrations be? $\endgroup$ – Stephen Alexander Jul 26 '20 at 16:37

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