Consider the following problem:

Find the $\mathrm{pH}$ of the solution formed by the mixing of two solutions of $\mathrm{pH}$(s) $2$ and $3$ of equal volumes.

The normal way:

Since the resulting solution will have twice the volume, the new $\mathrm{pH}$ will be: $$-\log\bigg(\frac{10^{-2}+10^{-3}}{2}\bigg)=-\log\bigg(0.0055\bigg)=\boxed{2.26}$$

Interestingly, my book mentions a trick:

Find the average $\mathrm{pH}$ of the two solutions and subtract $0.24$ to get the actual $\mathrm{pH}$.

This trick surprisingly works for many problems with a remarkable accuracy. I'm curious to find why this method works.

  • 1
    $\begingroup$ Answered here in detail: chemistry.stackexchange.com/questions/7051/… $\endgroup$ Dec 31, 2020 at 5:16
  • $\begingroup$ @NicolauSakerNeto thank you. It solved my query. Unfortunately I was not able to find this question before. $\endgroup$
    – newbie105
    Dec 31, 2020 at 5:24
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    $\begingroup$ The hidden part of this trick is that the volumes of solution A and B must be equal. Just be careful when you apply it. $\endgroup$
    – AChem
    Dec 31, 2020 at 5:49
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    $\begingroup$ The question, in the way it is formulated, has infinite number of possible solutions in near the whole pH range 2-3. I guess it implies, what should be stated explicitly, that both solutions are diluted strong acids. $\endgroup$
    – Poutnik
    Dec 31, 2020 at 8:28
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    $\begingroup$ The "trick" is valid accidentally for particular pH difference equal to 1 for the same volumes of diluted strong acid solutions. It is mathematical consequence and has no direct relation to chemistry. It has no practical value. $\endgroup$
    – Poutnik
    Dec 31, 2020 at 12:24


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