The question is: given a $10$ ml solution with pH = $12.3$, which is then diluted to a $100$ ml solution, find the pH of the new solution. My attempt was as follows:

[H$_3$O$^+$] = $10^{-12.3}$, and $c = \frac{n}{V}$ gives $n = 10^{-12.3} \frac{\text{mol}}{\text{dm}^3} \cdot 0.01 \text{dm}^3 = 10^{-14.3} \text{mol}$. After dilution, we have $V_{\text{new}} = 100 \text{ml} = 0.1 \text{dm}^3$ and hence $c_{\text{new}} = \frac{10^{-14.3}}{0.1} \implies \text{pH} = 13.3$. This answer, however, seems wrong since the answer in the book is $\text{pH} = 11.3$. What did I do wrong? When I however switch pH to pOH and to the same process (but doing $14.0 - \text{pOH}$ at the end to get pH) I get the correct answer $\text{pH} = 11.3$. Anybody knows what's going on?

  • $\begingroup$ Instead of writing $100$ ml ($100$ ml), write $\pu{100 ml}$ ($\pu{100 ml}$). $\endgroup$
    – Poutnik
    Nov 7, 2023 at 11:40
  • $\begingroup$ Dilution of water solutions causes convergence of pH values toward 7 at infinite dilution (at 25 deg C). // Remember Kw = [H+][OH-]=10^-14 // Dilution of the above solution causes generally raise of H+ concentration, being produces by water autodissociation. // The target pH can be in interval approximately 11.3(strong base solution) - 12.3 (pH buffer). The question as is does not have a single right numerical answer. $\endgroup$
    – Poutnik
    Nov 7, 2023 at 11:44

2 Answers 2


As Poutnik writes in the comments, the question does not have a single answer unless you specify the components of the solution.

Given that the pH is fairly basic at pH = 12.3, you could guess or assume that this is a strong base dissolved in pure water, and ignore ionic strength effects, and get the numerical answer given by the answer key.

However, you can exclude your first answer even without knowing more information than given:


If you add water to a solution, you are diluting it. If you dilute it infinitely, you approach pure water with a pH of 7. You don't expect the pH to get more extreme as you add water (maybe there is a case where this happens at very high concentrations, but not in a textbook scenario such as this).


If you started with a $\pu{10^{−12.3} M}$ solution of some inert material, then diluting it by a factor of 10 would indeed result in a $\pu{10^{−13.3} M}$ solution.

$\ce{H3O+}$ is not inert. It reacts with $\ce{OH-}$ and water (and of course with many other things too, if they are present). This means that the quantity of $\ce{H3O+}$ can, and does, change as the solution is diluted. Ignoring this reaction gives you the wrong answer.

The authors of the problem almost certainly intended you to solve it in the way you did (computing $\ce{[OH-]}$ and dividing by 10). But let's try to solve it from first principles to see what's actually happening; hopefully this makes things less mysterious.

Let's assume that the solution was formed by dissolving a strong base in water. This essential assumption is not stated in the problem, because the problems in elementary textbooks are often poorly written. Suppose the base is $\ce{NaOH}$. We will then have the following formulas:

  1. $\ce{[Na+] + [H3O+]} = \ce{[OH-]}$ (charge balance)
  2. $\ce{[H3O+][OH-]} = \pu{10^{-14} M}$ (equilibrium for the acid/base reaction)

In the system before dilution, $\ce{[H3O+]} = \pu{10^{-12.3} M}$ is given and so we can use the two equations to find the remaining two values. For all practical purposes, $\ce{[OH-]} = \ce{[Na+]} = \pu{10^{-1.7} M}$. Technically the two values are not exactly equal, but the difference is $\pu{10^{-12.3} M}$, about 10 parts per trillion difference, so it can be ignored. (If that makes you nervous, go ahead and do the calculation exactly and see how much it affects the answer!)

During the dilution, the quaitites of $\ce{H3O+}$ and $\ce{OH-}$ can change, because there is a reaction involving those species and water. So the post-dilution values of these are unknown But there is no reaction involving $\ce{Na+}$, so the quantity must remain unchanged. You can therefore deduce that the final concentration of $\ce{Na+}$ is $\pu{10^{-2.7} M}$. Thus we once again have two equations and two unknowns, and we can solve them for the final values of $\ce{[OH-]} = \pu{10^{-2.7} M}$ and $\ce{[H3O+]} = \pu{10^{-11.3} M}$


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