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I confronted a problem, stating that we have a solution of potassium hydroxide. By adding extra water, when the volume of solution reaches 630 mL, the pH decreases by 0.5 units. How much extra water have we added to the solution?

And I know this is a simple problem, but I wondered, when we add water to the strong basic solution, the volume increases, which means a decrease in $[OH^-]$. and a decrease in $[OH^-]$ means an increase in $[H^+]$, and therefore a decrease in pH.

Everything sounds good in a macroscopic view, but what really happens to $[H^+]$ when we add more water? Why doesn't it decrease just like $[OH^-]$ when we increase the volume of soultion? What happens to $[H^+][OH^-] = constant$ from a microscopic view?

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    $\begingroup$ The volume unit is liter or cubic meters, but not millimeters.. A volume of 630 mm has no meaning. Anyway ad data is missing : the initial volume. $\endgroup$ – Maurice Apr 21 at 14:37
  • $\begingroup$ You are right. I will edit it. Thanks. But the initial volume is not given. $\endgroup$ – DRLOVER Apr 21 at 14:44
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    $\begingroup$ OK, If the pH decreased by $0.5$, it means that the concentration changes by $\ce{10^{-0.5} = 0.3162}$. So the initial volume was $\ce{630·0.3162 = 199.2 = 200}$ mL . So you can subtract the initial volume (200 mL) from the final volume, to get the amount of extra water, $\endgroup$ – Maurice Apr 21 at 16:07
  • $\begingroup$ And if you add water to an solution of a strong acid, then the pH does rise, which means that $\ce{[H+]_{\mathrm{final}} < [H+]_{\mathrm{initial}}}$. Of course to maintain $K_w$ then $\ce{[OH-]_{\mathrm{final}} > [OH-]_{\mathrm{initial}}}$. $\endgroup$ – MaxW Apr 21 at 18:06
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    $\begingroup$ @MaxW I don’t think that’s what the OP meant. I think H+ refers to the minor species in the basic solution $\endgroup$ – Karsten Theis Apr 22 at 1:57
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Microscopic view often means looking at the kinetics (thermodynamics is easier for looking at bulk where concentration, temperature, and equilibrium make sense).

In the initial solution, the auto-dissociation of water is at equilibrium:

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

Forward and reverse reactions occur at the same rate, with a zero net reaction.

If you add water, the concentration of hydrogen ions and hydroxide ions both decrease by the same factor, while the concentration of water hardly changes. As a consequence, the rate of the forward reaction stays the same while the reverse reactions slows down.

Now, we have a net forward reaction making additional hydrogen ions and hydroxide ions. The hydroxide ion is the major species, so it's concentration does not change much (it stays lower than before the dilution). The hydrogen ion is a minor species. It was present at very low concentration (potassium hydroxide solutions are basic...). The net forward reaction increases its concentration by a lot (0.5 pH units is a factor three, approximately), to compensate for the decrease caused by dilution, and beyond, giving a concentration that is higher than the initial concentration.

Once you solve the quantitative question, you can make a table listing the effects of dilution and of the reaction attaining equilibrium again to get a better picture.

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