A sample of pure water is found to have a pH of 6.5; which of the following is a possible explanation? (Hint: The reaction of two water molecules to form hydronium ion and hydroxide ion is an endothermic process.)?

The `K_w` of water is a function of temperature, and not pressure. And since the process is endothermic, it can only be that the temperature is warmer than `25^@ "C"`, favoring the products' side so that

`"pH" = -log["H"_3"O"^(+)]`

for a higher `["H"_3"O"^(+)]` is lower than `7.0`. What is `"pH"` `+` `"pOH"` at this temperature if `"pH" = 6.5`?

The autoionization of water is endothermic:

`2"H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "OH"^(-)(aq)`

`K_w = 10^(-14)` at `25^@ "C"`

We can write `K_w = ["H"_3"O"^(+)]["OH"^(-)]` for pure water, which has `["H"_3"O"^(+)] = ["OH"^(-)]` by definition of pH-neutral water.

• An endothermic reaction favors the products at higher temperature, so an endothermic reaction has higher `[H_3O^(+)]`.

Therefore, neutral `"pH"` is lower than `7` at warmer temperatures, NOT colder temperatures, eliminating `(3)` that it could be at a lower temperature than `25^@ "C"`.

• Since no species in the reaction is gaseous, `K_w` is NOT a function of pressure. Liquids and aqueous solutions are highly incompressible...

That eliminates `(1)` that it could be several miles above sea level, `(4)` that the sample is at sea level, and `(5)` that the sample is pressurized to `"10 atm"`.

Under standard conditions which your syllabus will detail, we represent the autoprotolysis of water as....

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

And this is an EQUILBRIUM RXN for which...

`K_w=[H_3O^+][HO^-]=10^-14`

And clearly, this is a bond-breaking reaction....we would expect at HIGHER temperatures than `298*K` that the given equilibrium should shift to the RIGHT.....and so since `pK_a=-log_10[H_3O^+]`, should not `pH` DECREASE.....?

At `100` `""^@C` I seem to remember that the `pH` of water is `6.14`. Is this consistent with the analysis presented here?