# Effect of adding an acid or base to water with respect to concentration of ions

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?

• 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. – Ivan Neretin Jul 26 '20 at 10:21
• But when their product is brought back to $1* 10^-14$, will the [H3O+] remain still $10^-5$ ? – Stephen Alexander Jul 26 '20 at 12:12
• Pure stoichiometry. The decrease happens in the proportion specified in the reaction, so 1:1, until the equilibrium is re-established. – Zhe Jul 26 '20 at 12:38
• If you add HCl in pure water, this will immediately reduce the amount of autoionization of water. – Maurice Jul 26 '20 at 12:41
• 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? – Stephen Alexander Jul 26 '20 at 16:37