# Why don't we take into account the auto-protonation of water when measuring its concentration?

When we calculate the concentration of water at $25~^\circ\rm C$, why don't we assume dissociation of water at this temperature?

$$\ce{H2O <=> H+ + OH-}$$

• I didn't get it. What should we assume, and how? Commented May 6, 2016 at 18:43
• Because the number of moles of dissociated water is so slight compared to the number of moles of undissociated water, it makes no practical difference to our calculations to ignore the dissociated fraction. Commented May 6, 2016 at 18:46
• Compare the free energy to perform the dissociation with the thermal energy $kT$. Commented May 6, 2016 at 18:51
• If you take a look at the equilibrium constant for your reaction, you'll see why. @Ivan please don't forget about bad titles, for the sake of Mart-the-mod's heart. :) Commented May 6, 2016 at 19:18

$$\ce{H2O <=> H+ + OH-}$$ is 10^-14 and hence only 1 atom of water in 10^7 dissociates which really is negligible.