Auto-ionisation and Dilution 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.
Recall that $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$
Where $\ce{[OH-]}$ is the concentration of hydroxide ions.

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 $7$.

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.)