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.)?

  1. The sample is several miles above sea
    level.
  2. The sample’s temperature is above 25◦C.
  3. The sample’s temperature is below 25◦C.
  4. The sample is exactly at sea level.
  5. The sample is in a container pressurized
    to 10 atm.

2 Answers
May 14, 2018

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"#.

May 14, 2018

Would it not be #"option 2...."#

Explanation:

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?