How do we know that water has [H+] = 10^-7?

In the derivation of $K\_\mathrm{w}$, the ionic product of water, it is stated that 


$$[\ce{H+}][\ce{OH-}] = K\_\mathrm{w}$$


and substituting $[\ce{H+}] = [\ce{OH-}] = 10^{-7}$ we can find that $K\_\mathrm{w} = 10^{-14}$.


How do we know, though, that $[\ce{H+}] = 10^{-7}$? And why is $[\ce{H+}] = [\ce{OH-}]$?





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> Why is $[\ce{H+}] = [\ce{OH-}]$?
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Because the water is electrically neutral. 


In general, for all solutions, the sum of the concentration of each ion, multiplied by the ion's respective charge, must be zero. 



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> How do we know $[\ce{H+}] = 10^{-7}$?
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This is found by experimentally by determining $K\_\mathrm{w}$ and substituting, or by measuring the pH. 


For details see: Bandura, A. V.; Lvov, S. N. The Ionization Constant of Water over Wide Ranges of Temperature and Density. *J. Phys. Chem. Ref. Data* **2006,** *35* (1), 15–30. [DOI: 10.1063/1.1928231](https://doi.org/10.1063/1.1928231).


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