Consider an acidic solution with Hydrogen ion concentration, $\ce{[H+]}$ of $10^{-5}\:\mathrm{M}$. Since $\:\mathrm{pH} = -\log \ce{[H+]}$ the $\:\mathrm{pH}$ of solution is $5$. Suppose we dilute solution 10 times with water. Now, $\ce{[H+]}$ is $10^{-6}\:\mathrm{M}$ and $\:\mathrm{pH}$ is $6$. Further dilution should increase $\:\mathrm{pH}$ from $6$ to $7$ and then from $7$ to $8$ and so on. Can this go in for ever? Does this not imply that an acidic solution can be made basic/alkaline simply by adding water? But that doesn't happen? What prevents it?

Is there anyone already found answer for this problem or it is just an unsolved basic problem of chemistry?


4 Answers 4


Water undergoes autoionization, i.e., it reacts as follows:

$$ \ce{H2O + H2O <=> H3O+ + OH-} $$

The equilibrium constant for this reaction at standard conditions is $K_w = [\ce{H3O+}][\ce{OH-}] \approx 1.0 \cdot 10^{-14}$. In pure water, $[\ce{H3O+}] = [\ce{OH-}]$, hence $[\ce{H3O+}] = \sqrt{K_w} \approx 1.0 \cdot 10^{-7}\ \textrm{M}$.

Suppose we dilute a solution with some initial concentration of $[\ce{H3O+}]_i = \frac{n_i}{V_i}$, where $n_i$ is the initial number of moles of $\ce{H3O+}$ and $V_i$ is the initial volume. If I now add a volume $\Delta V$ of pure water (which would contain $(1.0 \cdot 10^{-7}) \cdot \Delta V$ moles of $\ce{H3O+}$) the resulting final concentration can be crudely approximated as:

$$ [\ce{H3O+}]_f \approx \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V} $$

We can see what value this expression approaches as we make the solution more and more dilute by taking the limit as $\Delta V \to \infty$:

$$ \lim_{\Delta V \to \infty} \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V} = 1.0 \cdot 10^{-7} $$

Therefore, $[\ce{H3O+}]$ tends towards $1.0 \cdot 10^{-7}$ with further dilution, so $pH$ will approach a value of $7.0$.

  • $\begingroup$ I add a question here....what will be change in ph when a buffer soln containing 0.1 MCh3COOH and 0.1 M CH3CHOONa. is diluted 10 times $\endgroup$
    – Pole_Star
    Commented Nov 29, 2017 at 16:20

You forget that pure water already contains $\ce{H_3O^{+}}$ ($\ce{Kw}=\ce{[H_3O+][OH^{-}]}=14$,). If you have an acidic solution, you will be able to reach the neutral point ($\ce{pH} = 7$) after enough dilution, but it will never be alkaline. You can see this as similar to a limit or infinite sum $\lim_{x\rightarrow \infty } \frac{x}{x+1}=1$, but when you look at finite $x$ it will never actually be 1.


pH of diluted acid

I've uploaded a figure, just for illustration (red line: calculated pH without considering autoionization, blue is the actual one). You can see the difference.


It can not be alkaline because the water is not really alkaline, more like neutral. But dilution in an alkaline solvent will easily make the solution alkaline.

EDIT: seems my answer was not considered an answer. Well since the question is composed out of more than 1 question...

Using water, you won't be able to get past aprox 7 PH. Using an alkaline non reactive solvent will allow you to go as far as the PH of the solvent.

Additional clarification: If no reaction happens, the PH limit of the solution is bound between the substance with the highest PH, and the one with the lowest.

No, an acidic solution can not be made alkaline by adding water.

  • $\begingroup$ I don't think you sufficiently answered the why, which is the main question. That makes this not an answer. $\endgroup$
    – M.A.R.
    Commented Aug 22, 2017 at 13:40
  • $\begingroup$ I answered the question "how much does PH change through dilution. $\endgroup$ Commented Aug 22, 2017 at 13:51
  • 1
    $\begingroup$ @GregE and Jori's answers both give an answer using straightforward math and their answers seem to be pretty well liked by the community. $\endgroup$
    – Tyberius
    Commented Aug 22, 2017 at 14:37
  • 2
    $\begingroup$ I agree with Tyberius - As the question already has a (three year old) answer, you should really only be adding another answer if you think there's important information that hasn't been covered by the existing answers, or if you think existing answers are too unclear to adequately answer the question. -- As it looks like you're new to the site, please take the tour and look at the help section if you haven't already. Stack Exchange is a Q&A site, not a forum, so we do things a bit differently here. $\endgroup$
    – R.M.
    Commented Aug 22, 2017 at 14:58

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