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?