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When you dilute an acidic solution, pH increases towards 7 and pOH decreases. However, the decrease of pOH indicates that the concentration of hydroxide ions increases. This seems to contradict my understanding of how equilibrium systems respond to external disturbances, as it suggests the auto-ionisation equilibrium of water shifts to favour the ions (according to Le Chatelier's principle) to such an extent that the hydroxide ion concentration increases from prior to the dilution. However, I was under the impression that equilibrium systems can never fully undo the change due to the bi-directional nature of equilibria, meaning the reaction is using up ions as they are replaced. Please help to clear up my confusion! Thank you.

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Recall that $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$

Where $\ce{[OH-]}$ is the concentration of hydroxide ions.

  1. Suppose a solution has $10^{-7}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-7}]$ which is equal to $7$.

  2. Suppose a solution has $10^{-4}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-4}]$ which is equal to $4$.

So as you can see, lower the $\mathrm p \ce{OH}$, the higher the concentration of $\ce{OH-}$ ions. Le Chatelier's principle is not violated.

(How I like to remember this is - the strongest acids have the lowest $\mathrm pK_\mathrm a$. So that extends to $\mathrm p\ce{OH}$ too. This is all due to the fact that $\mathrm p\ce{OH}$ is the negative logarithm of the $\ce{[OH-]}$ concentration in solution.)

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    $\begingroup$ Note that the chemistry typography expects chemical formulas and p operator to be upright, in SE site context implemented via using \ce{} and \mathrm{} MathJax formatting. $\endgroup$
    – Poutnik
    Apr 14 at 12:51
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    $\begingroup$ $\mathrm p K_\mathrm a$ is denoted as $\mathrm p K_\mathrm a$ and it's $\ce{[OH-]}$ written as $\ce{[OH-]}$. Also don't leave spaces since it creates unnecessry white space $\endgroup$ Apr 14 at 13:00
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    $\begingroup$ Very well. Almost correct. But note the bracket [] are used to express molar concetration of compound inside. For literal numbers or variable , it better to use implicit or explicit paretheses, like $\log{\pu{e-4}}$ $\log{\pu{e-4}}$ or $\log{(\pu{e-4})}$ $\log{(\pu{e-4})}$. $10^{-4}$ can be used as well, but \pu{} has advantage to be used together with units ) $\endgroup$
    – Poutnik
    Apr 14 at 13:02
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    $\begingroup$ You can find useful Notation basics and How can I format math/chemistry expressions on Chemistry SE. $\endgroup$
    – Poutnik
    Apr 14 at 13:06
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    $\begingroup$ @latin333 - the water that is added to perform the dilution has 10^-7 M HO- because of water autoionization. If the solution being diluted is acidic, initial [HO-] is less than 10^-7, so "dilution" actually results in an initial increase of [HO-] (to the weighted average of the initial and 10^-7), and the concentration decreases back towards initial during equilibration. $\endgroup$
    – Andrew
    Apr 14 at 13:33
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The autoionization equilibrium of water is satisfied irrespectively of $\mathrm p\ce{H}$. That's why $\mathrm p\ce{H}$ and $\mathrm p\ce{OH}$ add up to a constant:

$$K_\mathrm{w}=a_{\ce{H3O+}}a_{\ce{OH-}}$$ $$\rightarrow \mathrm p K_\mathrm{w}=\mathrm p\ce{H} + \mathrm p\ce{OH}$$

This is a bit different from the typical scenario in which Le Chaterlier's principle is invoked because $\ce{H3O+}$ and $\ce{OH-}$ are on the same side in the reaction equilibrium, that is they are both products of the autoionization of water.

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