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?


1 Answer 1


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.)


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.


This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .