Why does this shortcut work for calculating pH (proof required)?

My teacher gave me a shortcut regarding calculation of resultant pH if the solutions are mixed and if the difference between the pH of individual solutions is 1.

1. If there are two solutions, the pH is their mean - 0.24.


2. If there are three solutions, mean + 0.56.

But what is the logic behind it?

Answer

Let's look at the first case, mixing equal volumes of two solutions (a and b) with a pH difference of 1.

Initially we have:

$$[H^+]_a=x,\ \ \ \ pH_a=-log\ x,\ \ \ \ n_{H^+_a}=xV$$ $$[H^+]_b=x/10,\ \ \ \ pH_b=-log\ x+1,\ \ \ \ n_{H^+_b}=xV/10$$ $$\overline{pH}=\frac{pH_a+pH_b}{2}=\frac{-log\ x-log\ x+1}{2}=-log\ x+0.5 $$

After the mixture, we obtain:

$$n_{H^+_{mix}}=n_{H^+_a}+n_{H^+_b}=11xV/10, \ \ \ \ [H^+]_{mix}=\frac{11xV/10}{2V}=11x/20$$

$$pH_{mix}=-log\ 11x/20=-log\ x - log\ 11/20=\overline{pH}-0.5+0.2596=\overline{pH}-0.2404$$

For the mixture of three solutions (a, b and c), we have a similar calculation. Before the mixture:

$$[H^+]_a=x,\ \ \ \ pH_a=-log\ x,\ \ \ \ n_{H^+_a}=xV$$ $$[H^+]_b=x/10,\ \ \ \ pH_b=-log\ x+1,\ \ \ \ n_{H^+_b}=xV/10$$ $$[H^+]_c=x/100,\ \ \ \ pH_c=-log\ x+2,\ \ \ \ n_{H^+_c}=xV/100$$ $$\overline{pH}=\frac{pH_a+pH_b+pH_c}{3}=\frac{-log\ x-log\ x+1-log\ x+2}{3}=-log\ x+1 $$

After the mixture, we have:

$$n_{H^+_{mix}}=n_{H^+_a}+n_{H^+_b}+n_{H^+_c}=111xV/100, \ \ \ \ [H^+]_{mix}=\frac{111xV/100}{3V}=111x/300$$

$$pH_{mix}=-log\ 111x/300=-log\ x - log\ 111/300=\overline{pH}-1+0.4318=\overline{pH}-0.5682$$

(It seems you got the sign wrong on the last equation you wrote down)

There are caveats, however. Technically these calculations only work well if we assume that the activity of the solvated proton is the same as its concentration, which is a poor approximation below pH 0 and above pH 14. You also get a slight deviation if you mix solutions that have pH too close to neutral, because the autodissociation of water affects the equilibrium values of $\ce{[H+]}$ and $\ce{[OH^{-}]}$. For example, we mix two equal amounts of liquid, one being pure water (pH=7) and the other being aqueous $\ce{HCl}$ solution with $\ce{[HCl]}=1\times 10^{-6}\ M$. Considering the autodissociation constant for water as $k_w=1\times 10^{-14}$, the pH for the last solution is actually approximately 5.9957 because water autodissociation already affects it slightly. Neglecting that, we mix the liquids creating a solution with $\ce{[HCl]}=5\times 10^{-7}\ M$. Directly using $\ce{[HCl]=5\times 10^{-7}}\ M\ce{=[H+]}$ yields a pH of 6.3010. Using the trick, we get the pH for $\ce{[H+]}=5.5\times 10^{-7}$, which is 6.2596. However, the real value for $\ce{[H+]}$, taking into account the shifting of the water autodissociation equilibrium, is $\ce{[H+]}=5.19258\times 10^{-7}\ M$, such that the real pH is 6.28461. A small deviation, to be sure, but it's there.