What is the concentration of OH- in pure water?

1 Answer
Feb 10, 2017

Answer:

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

Explanation:

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.

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