Let's say you add $\ce{HCl}$ to water. The $\ce{H+}$ ion concentration increases and that causes a decrease in $\mathrm{pH}$. But why would the $\mathrm{pOH}$ increase?

I can't see why added $\ce{H+}$ ions will decrease the $\ce{OH-}$ concentration? Is this because of the water auto-ionization? Is it because, according to Le Chatelier principle, when you add an amount of a component, the reaction will go in the other direction to counteract the disturbance, thus decreasing the concentration of $\ce{OH-}$ to form water?

  • $\begingroup$ Well, yeah, it may be thought of like that. $\endgroup$ Commented Mar 14, 2018 at 12:41
  • 2
    $\begingroup$ Increasing the concentration of $\ce{H+}$ increases the collisions of $\ce{H+}$ and $\ce{OH-}$ which can result in the formation of water. $\endgroup$ Commented Mar 14, 2018 at 12:51

1 Answer 1


At a particular temperature, the $K_{\text{eq}}$ for the following reaction (yes, it's the auto-protolysis of water):

$$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$

will be constant. Note that $K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $([\ce{H2O}])$ is assumed to be constant.

So, you can directly observe from here, that since $\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$ at a particular temperature, if $\ce{[H+(aq)]}$ increases (i.e. $\mathrm{pH}$ decreases), $[\ce{OH-(aq)}]$ must decrease (i.e. $\mathrm{pOH}$ must increase).

Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $Q$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $\ce{H+}$ ions as much as it decreases the concentration of $\ce{OH-}$ ions. See explanation below:

Consider adding $\pu{0.01M}$ $\ce{HCl}$ to pure water (at $\pu{25^\circ C}$) at $t=0$. This is what happens till achieving equilibrium at $t=t_0$:

$$ \begin{array}{ccc} &\ce{H2O(l) &<=>& H+(aq)&+&OH-(aq)}\\\hline -&-&&\pu{10^-7M}&&\pu{10^-7M}\\ t=0&-&&\pu{(10^-7+10^-2)M}&&\pu{10^-7M}\\ t=t_0&-&&(10^{-7}+10^{-2}-x)~\pu{M}&&(10^{-7}-x)~\pu{M}\\ \end{array} $$

Now, if you actually solve for $x$ at $t=t_0$, you'll find $x$ to be of the order $10^{-8}$ or below. Notice that, for the $\ce{H+}$ ions, this means a decreases in concentration from $\approx\pu{0.01M}$ at $t=0$ to $\pu{0.0099999M}$ at $t=t_0$. This a hardly noticeable change. However, it is a catastrophe for the $\ce{OH-}$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $\pu{10^-7 M}$! The change in $\mathrm{pOH}$ is, hence, much more pronounced.


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