# Autoionization of water

Consider as solution of pure water at $$\pu{25 ^\circ{}C}$$ with a $$K_{\mathrm{w}} = 10^{-14}$$.

If we are to add an acid to the mix we would observe an increase in hydronium and an equal decrease in hydroxide content.

Yet, I fail to see how the hydroxide ion concentration can decrease with an increase in hydronium concentration.

Lets add $$\ce{HCl}$$ to our solution: $$\ce{HCl + H2O -> Cl- + H3O+}$$ Concentration $$[\ce{H3O+}]$$ increases, concentration $$[\ce{OH-}]$$ decreases in order to fulfill the $$K_{\mathrm{w}}$$ constant of $$10^{-14}$$.

Through what reaction exactly then will the hydroxide content decrease (by turning into water I assume)? Where does $$\ce{OH-}$$ gets an $$\ce{H+}$$ from? Not from $$\ce{H3O+}$$, I assume seeing that every molecule of a strong acid is said to turn into an hydronium ion 1:1.

• Note that the kinetic rate of $\ce{H3O+ + OH- -> 2 H2O}$ is $\frac{\mathrm{d}[\ce{OH-}]}{\mathrm{d}t}=k \cdot [\ce{H3O+}][\ce{OH-}]$ and it is one of the fastest ever chemical reactions, controlled by diffusion. Commented Feb 11, 2021 at 9:32
• To answer your question, yes the $\ce{H+}$ cations that react with the $\ce{OH-}$ anions come from the acid. Remember that in pure water the $\ce{OH-}$ anion concentration is only $1\times 10^{-7}$ so if you have a 0.1 molar acid solution then a minuscule fraction of the acid is consumed.
Maybe a numerical example will help you. Let's start from pure water, with the following concentrations :$$\ce{[H+] = [OH-] = 10^{-7} M}$$. Now we will suppose you add $$\ce{10^{-7}} mol$$ $$\ce{HCl}$$ in one liter of this water. Suddenly, the concentration of $$\ce{H+}$$ should double. This is not possible in the long run. A part $$a$$ of these supplementary $$\ce{H+}$$ ions must react with the same amount $$a$$ of the $$\ce{OH-}$$ ions to form water. The final molar concentrations of both ions are given by : $$\ce{[H+] = 2.000·10^{-7}} - a$$ $$\ce{[OH-] = 1.000·10^{-7} } - a$$ This amount $$a$$ may be obtained by solving the equation : $$\ce{ (2.000·10^{-7}} - a)\ce{(1.000 10^{-7}} - a)= 1.000·10^{-14}$$ The solution is : $$a = 0.382·10^{-7} mol/L$$. So about $$38.2$$% of the initial $$\ce{OH-}$$ ions have been destroyed by the addition of the new acidic ions. And the numerical values of the final concentrations are $$\ce{[H+] = 2.00·10^{-7}}- a = 1.618·10^{-7} mol/L$$$$\ce{[OH-] = 1.00 10^{-7} } - a = 0.618· 10^{-7} mol/L$$ And of course : $$\ce{[Cl-] = 1.000·10^{-7} mol/L}$$. If you want you may check the product of the two final concentrations $$\ce{[H+]}$$ and $$\ce{[OH-]}$$. It is equal to $${1.000·10^{-14} mol^2L^{-2}}$$.