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 $[\ce{H+}] = [\ce{OH-}]$?


Why is $[\ce{H+}] = [\ce{OH-}]$?

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.

How do we know $[\ce{H+}] = 10^{-7}$?

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.

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  • $\begingroup$ Thanks ! But How do we know concentration of H+ is 10^-7 while deriving Kw ? $\endgroup$ – Dimenein Dec 9 '15 at 17:24
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    $\begingroup$ $K_w$ is measured. It does depend on temperature. pH of pure water is 7.47 at 0 degrees C and 6.14 at 100 degrees C. chemguide.co.uk/physical/acidbaseeqia/kw.html $\endgroup$ – MaxW Dec 9 '15 at 17:26

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