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.

Please enlighten me.

  • $\begingroup$ H+ is in acidobasic reaction context just convenience for H3O+ ( which is exactly convenience for H3O+ . n H2O ). As H+ is nothing else but a proton/deuteron, which creates locally so strong electrostatic field which does not allow its free existence in condensed phases. $\endgroup$
    – Poutnik
    Commented Feb 11, 2021 at 9:03
  • $\begingroup$ @Lars To facilitate rendering chemical equations in the body of a question, answer, or comment, ChemSE offers you to use mathJax and mhchem syntax. Which I applied to the current form of your question. You may familiarize yourself with the underlying rules e.g., here. Because it is not understood by all internet browsers, you should not use this peculiar syntax in the title of a question. $\endgroup$
    – Buttonwood
    Commented Feb 11, 2021 at 9:03
  • $\begingroup$ @Buttonwood Thanks, message deleted. $\endgroup$
    – Buck Thorn
    Commented Feb 11, 2021 at 9:08
  • $\begingroup$ 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. $\endgroup$
    – Poutnik
    Commented Feb 11, 2021 at 9:32
  • 1
    $\begingroup$ 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. $\endgroup$
    – MaxW
    Commented Feb 11, 2021 at 11:26

1 Answer 1


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 ${1.000·10^{-14} mol^2L^{-2}}$.


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