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

1 Answer
anor277
Jun 10, 2017

[H3O+]=[−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 (i):

2H2O(l)⇌H3O++HO−

OR by (ii):

H2O(l)⇌H++HO−

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
H5O+2 or H7O+3, i.e. a cluster of 2 or 3 or 4 water molecules with an EXTRA H+ tacked on.

We can use H+, protium ion, or H3O+, hydronium ion, equivalently to represent this species.

The equilibrium constant for the reaction, under standard conditions, is..........Kw=[H3O+][−OH]=10−14.

And so Kw=[H3O+]2 because [HO−]=[H3O+] at neutrality, and thus..........

[H3O+]=[HO−]=√10−14⋅mol2⋅L−2=10−7⋅mol⋅L−1

And to make the arithmetic a bit easier we can use the pH function, where pH=−log10[H3O+], and pOH=−log10[HO−]

And thus in aqueous solution under the given standard conditions, pH+pOH=14.

How do you think pH and pOH would evolve under non-standard conditions, i.e. at a temperature of say 373⋅K? Would Kw decrease, remain constant, increase? Remember that the reaction as written is A BOND BREAKING reaction. How would this inform your choice?

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