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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?

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  • $\begingroup$ This isn't a very good shortcut, as it demands that the volumes mixed be the same, significantly limiting its use. It's more of a trick that shows off how the logarithmic scale is a bit different to a linear scale. If you want to try investigating for yourself, try writing down what would happen as you mix 1 L of two or three different pH solutions apart by 1 pH unit, and keep track of pH, $\ce{[H+]}$ and possibly $n_{\ce{H+}}$ at all times. Why the trick works should become apparent as you manipulate the numbers. If you manage, feel free to answer your own question! $\endgroup$ – Nicolau Saker Neto Nov 23 '13 at 12:04
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    $\begingroup$ @NicolauSakerNeto - This is a good answer to this question. I encourage you to post it as such so that you can get more rep for it. $\endgroup$ – Ben Norris Nov 23 '13 at 14:08
  • $\begingroup$ @BenNorris I'll wait to see how the asker handles the comment. If needed I shall answer in full later. I just wanted to let them give it a shot, especially since they seem to have some affinity for mathematics. $\endgroup$ – Nicolau Saker Neto Nov 23 '13 at 14:19
  • $\begingroup$ I am a bit biased towards math, but that doesnt mean chem bores me. Its just that i do not find so much logic in chemistry, especially inorganic. But im adapting a bit $\endgroup$ – Rohinb97 Nov 25 '13 at 13:18
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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.

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