# What is the concentration of OH- in pure water?

Feb 10, 2017

$\left[H {O}^{-}\right]$ $=$ ${10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$ at $298 \cdot K$

#### Explanation:

Water undergoes autoprotolysis according to the following equation:

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

Thru very careful measurement at $298 \cdot K$ the following value for the ion product has been found to be:

$\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$

We could simplify this by taking ${\log}_{10}$ of each side, but clearly if the solution is neutral, then $\left[H {O}^{-}\right] = \left[{H}_{3} {O}^{+}\right] = {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$.

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

$p H + p O H = 14$.

See this old answer for further details.

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