A trick to find resultant pH [duplicate]

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.

• Answered here in detail: chemistry.stackexchange.com/questions/7051/… Dec 31, 2020 at 5:16
• @NicolauSakerNeto thank you. It solved my query. Unfortunately I was not able to find this question before. Dec 31, 2020 at 5:24
• 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. Dec 31, 2020 at 5:49
• 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. Dec 31, 2020 at 8:28
• 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. Dec 31, 2020 at 12:24