# Shortcut for calculating pH of mixtures

My teacher gave me a shortcut to calculate the resultant pH if 2 or 3 solutions are mixed, which are relevant only if the difference between the pH of individual solutions is 1.

1. If there are two solutions, the pH of the mixture $$\approx$$ (the mean pH − 0.24).
2. If there are three solutions, the pH of the mixture $$\approx$$ (the mean pH + 0.56).

What is the logic behind this approximation?

Case 1: Mixing equal volumes of two solutions with a pH difference of 1.

Initially we have:

$$\begin{array}{|c|c|c|} \hline &[\ce{H+}]& \mathrm{pH} &n_\ce{H+}\\ \hline a& x & \log x & xV\\ b&\frac{x}{10} & -\log x + 1 & \frac{xV}{10}\\ \hline \end{array}$$ \begin{align} \overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b}{2}=\frac{-\log x-\log x+1}{2}=-\log x+0.5 \end{align}

After complete mixing, we obtain:

$$({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}=xV+\frac{xV}{10}=\frac{11xV}{10}\\ [\ce{H+}]_\mathrm{mix}=\frac{1.1xV}{2V}=\frac{11x}{20}\\ \mathrm{pH}_\mathrm{mix}=-\log \frac{11x}{20}=-\log x - \log\frac{11}{20}=\overline{\mathrm{pH}}-0.5+0.2596=\overline{\mathrm{pH}}-0.2404$$

Case 2: Mixing equal volumes of three solutions with consecutive pH differences of 1.

Before mixing:

$$\begin{array}{|c|c|c|} \hline &[\ce{H+}]& \mathrm{pH} &n_\ce{H+}\\ \hline a& x & \log x & xV\\ b&\frac{x}{10} & -\log x + 1 & \frac{xV}{10}\\ c&\frac{x}{100} & -\log x + 2 & \frac{xV}{100}\\ \hline \end{array}$$ \begin{align} \overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b+\mathrm{pH}_c}{3}=\frac{-\log x-\log x+1-\log x+2}{3}=-\log x+1 \end{align}

After complete mixing, we obtain:

$$({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}+({n_{\ce{H+}}})_{c}=\frac{111xV}{100}\\ [\ce{H+}]_\mathrm{mix}=\frac{1.11xV}{3V}=\frac{111x}{300}\\ \mathrm{pH}_\mathrm{mix}=-\log \frac{111x}{300}=-\log x - \log\frac{111}{300}=\overline{\mathrm{pH}}-1+0.4318=\overline{\mathrm{pH}}-0.5682$$

(It seems you got the sign wrong on your last equation.)

Caveats:

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 $$\mathrm{pH}$$ of 0 and above $$\mathrm{pH}$$ of 14.

You also get a slight deviation if you mix solutions that have $$\mathrm{pH}$$ too close to neutral, because the auto-dissociation of water affects the equilibrium values of $$[\ce{H+}]$$ and $$[\ce{OH-}]$$.

For example, if we mix two equal amounts of liquid, one being pure water($$\mathrm{pH} = 7$$) and the other being aqueous $$\ce{HCl}$$ solution with $$[\ce{HCl}] = \pu{10^{-6}M}$$. Considering the auto-dissociation constant for water as $$K_\mathrm{w} = 10^{-14}$$, the $$\mathrm{pH}$$ for the last solution is actually approximately $$5.9957$$ because water auto-dissociation already affects it slightly.

Neglecting that, we mix the liquids creating a solution with $$[\ce{HCl}] = \pu{5\times 10^{-7} M}$$. Directly using $$[\ce{HCl}] = \pu{5\times 10^{-7}M} = [\ce{H+}]$$ yields a $$\mathrm{pH}$$ of $$6.3010$$.

Using the trick, we get the pH for $$[\ce{HCl}] = \pu{5.5\times 10^{-7}M}$$ as $$6.2596$$. However, the real value for $$[\ce{H+}]$$, taking into account the shifting of the water auto-dissociation equilibrium, is $$[\ce{H+}] = \pu{5.19258\times 10^{-7}M}$$, such that the real $$\mathrm{pH}$$ is $$6.28461$$. A small deviation surely, but it's present.