Question

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

Answer 1

`[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 `(i)`:

`2H_2O(l) rightleftharpoonsH_3O^+ +HO^-`

OR by `(ii)`:

`H_2O(l) rightleftharpoonsH^+ +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
`H_5O_2^+` or `H_7O_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 `H_3O^+`, `"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=[H_3O^+]^2` because `[HO^-]=[H_3O^+]` at neutrality, and thus..........

`[H_3O^+]=[HO^-]=sqrt(10^-14*mol^2*L^-2)=10^-7*mol*L^-1`

And to make the arithmetic a bit easier we can use the `pH` function, where `pH=-log_10[H_3O^+]`, and `pOH=-log_10[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 `K_w` decrease, remain constant, increase? Remember that the reaction as written is A BOND BREAKING reaction. How would this inform your choice?