In the German Wikipedia article on $\mathrm{pH}$, I found the following formula for calculating the $\mathrm{pH}$ of weak acids (which are there defined of having $4.5 < \mathrm{p}K_\mathrm{a} < 9.5:$

$$c(\ce{H3O+}) = c^\circ\cdot\sqrt{K_\mathrm{a}\cdot c_0/c^\circ}\label{eqn:1}\tag{1}$$

I am a bit confused about this as in school, I learned that the $\mathrm{pH}$ (or more precisely, the concentration of $\ce{H3O+};$ one would have to take $-\log c(\ce{H3O+})$ to get $\mathrm{pH})$ for weak acids is

$$c(\ce{H3O+}) = \sqrt{K_\mathrm{a}\cdot c_0}\label{eqn:2}\tag{2}$$

I see that these formula are somewhat similar, but they aren't identical. The Wikipedia formula \eqref{eqn:1} includes $c^\circ,$ of which I could not find out what it is (it also appears in the equations for strong and very weak acids).

So I have two questions:

  1. What is $c^\circ$?
  2. Are these formulae different and if yes, how?
  • 1
    $\begingroup$ That's not how it works. Until you know the meaning of every term in a formula, there is no formula, just some random heap of letters and math symbols. So there is only one formula here. $\endgroup$ Commented Dec 4, 2020 at 19:35
  • $\begingroup$ @IvanNeretin agreed. Not understanding the second term was probably my main motivation to ask this question $\endgroup$
    – jng224
    Commented Dec 4, 2020 at 22:07
  • $\begingroup$ Then why didn't you read that article and find it out? True, they never say that quite explicitly, but one might deduce that $c^0$ is some kind of activity coefficient. If we disregard it altogether (that is, consider $c^0=1$, as people often do), then the formula (1) becomes the same as (2). $\endgroup$ Commented Dec 4, 2020 at 22:14
  • $\begingroup$ @Jonas Your confusion probably arises from improper notations (which I edited for clarity). It's not "$c$ to the power of zero" (c^0) which indeed mathematically leads to $c^0 = 1$ and makes little sense in this context. It should be either $c^\circ$ (c^\circ) or $c^⦵$ (c^⦵), which is standard concentration. $\endgroup$
    – andselisk
    Commented Dec 5, 2020 at 7:25

1 Answer 1


The purpose of this $c^\circ$ is to ensure dimensional consistency. Keep in mind that Wikipedia can be edited by anyone, although the content is often of very good quality, sometimes the volunteer writer misses some points and assumes that the reader might be aware of his/her symbols. In your link, the writer does not explicitly define $c^\circ$.

The German version is ensuring that the equilibrium constant is dimensionless and all the quantities which are mathematically operated upon are dimensionless numbers. Therefore $c^\circ$ must be 1 mol/L, if $c_0$ has molar units.

$$c(\ce{H3O+}) = c^\circ\cdot\sqrt{K_a\cdot c_0/c^\circ} \tag{1}$$

Note the quantity under the square has been made dimensionless. In order to attach a unit to the square root term, $c^\circ$ has been multiplied.

If you want to dig deeper, there is something called quantity calculus, which I quote again from my previous answer

Are the units of mole of oxygen molecules the same with the units of mole of nitrogen molecules?

Calculus here is not the integration / differentiation, but rather the Latin calculus implying a method of calculation.

There is a very nice article "Quantity Calculus: Unambiguous Designation of Units in Graphs and Tables" by Mary Anne White in the Journal of Chemical Education. Please read this if you are seriously interested. Search on Google Scholar and it is free to download from there.

In quantity calculus Each physical quantity as the product of a numerical value and a unit:

physical quantity = numerical value × unit

This approach was introduced by British scientists and many leading physicists used it. Now there is there is nothing less or nothing more. Therefore your ambiguity arises from introducing another factor such as "oxygen" or "nitrogen". The unit mol does not know whether it belongs to oxygen or nitrogen.

As explained in the comments, suppose we write L symbolizing the height of a tree, then I can only write, L = 10 m. For mathematical purposes, I will not introduce "tree" anywhere in this equation. The tree is already incorporated in L (in your mind) but not in the mathematical equation. One can also write L/m =10. Now you have a pure number on both sides.

  • 2
    $\begingroup$ Just a dimensional unit, then. So I was wrong. Well, good to know. $\endgroup$ Commented Dec 5, 2020 at 7:29

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