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.
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. $\endgroup$