# What are the molar concentrations of [H+] and [OH-] in pure water at 25°C?

Jun 10, 2017

[H_3O^+]=[""^(-)OH]=10^-7*mol*L^-1 under the given conditions........

#### Explanation:

WE know from classic experiments that water undergoes autoprotolysis, i.e. self-ionization.

We could represent this reaction by $\left(i\right)$:

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

OR by $\left(i i\right)$:

${H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}^{+} + H {O}^{-}$

Note that (i) and (ii) ARE EQUIVALENT REPRESENTATIONS, and it really is a matter of preference which equation you decide to use. As far as anyone knows, the actual acidium ion in solution is
${H}_{5} {O}_{2}^{+}$ or ${H}_{7} {O}_{3}^{+}$, i.e. a cluster of 2 or 3 or 4 water molecules with an EXTRA ${H}^{+}$ tacked on.

We can use ${H}^{+}$, $\text{protium ion}$, or ${H}_{3} {O}^{+}$, $\text{hydronium ion}$, equivalently to represent this species.

The equilibrium constant for the reaction, under standard conditions, is..........K_w=[H_3O^+][""^(-)OH]=10^-14.

And so ${K}_{w} = {\left[{H}_{3} {O}^{+}\right]}^{2}$ because $\left[H {O}^{-}\right] = \left[{H}_{3} {O}^{+}\right]$ at neutrality, and thus..........

$\left[{H}_{3} {O}^{+}\right] = \left[H {O}^{-}\right] = \sqrt{{10}^{-} 14 \cdot m o {l}^{2} \cdot {L}^{-} 2} = {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$

And to make the arithmetic a bit easier we can use the $p H$ function, where $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$

And thus in aqueous solution under the given standard conditions, $p H + p O H = 14$.

How do you think $p H$ and $p O H$ would evolve under non-standard conditions, i.e. at a temperature of say $373 \cdot K$? Would ${K}_{w}$ decrease, remain constant, increase? Remember that the reaction as written is A BOND BREAKING reaction. How would this inform your choice?