Question

What is the concentration of OH- in pure water?

Answer 1

`[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.