What is the concentration of OH- in pure water?

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Answer 1 · 2017-02-10T17:52:19

#[HO^-]# #=# #10^-7*mol*L^-1# at #298*K#

Water undergoes autoprotolysis according to the following equation:

#2H_2O rightleftharpoons H_3O^(+) + HO^-#

Thru very careful measurement at #298*K# the following value for the ion product has been found to be:

#[H_3O^+][HO^-]=10^-14#

We could simplify this by taking #log_10# of each side, but clearly if the solution is neutral, then #[HO^-]=[H_3O^+]=10^-7*mol*L^-1#.

If we take logarithms, then we get the useful expression:

#pH+pOH=14#.

See this old [answer for further details.](https://socratic.org/questions/how-do-you-find-poh-from-ph)

At higher temperatures than #298K#, how do you think the equilibrium would evolve? Remember that this is a bond-breaking reaction.