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 Kw 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 25C, favoring the products' side so that

pH=log[H3O+]

for a higher [H3O+] is lower than 7.0. What is pH + pOH at this temperature if pH=6.5?


The autoionization of water is endothermic:

2H2O(l)H3O+(aq)+OH(aq)

Kw=1014 at 25C

We can write Kw=[H3O+][OH] for pure water, which has [H3O+]=[OH] by definition of pH-neutral water.

  • An endothermic reaction favors the products at higher temperature, so an endothermic reaction has higher [H3O+].

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

  • Since no species in the reaction is gaseous, Kw 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....

2H2O(l)H3O++HO

And this is an EQUILBRIUM RXN for which...

Kw=[H3O+][HO]=1014

And clearly, this is a bond-breaking reaction....we would expect at HIGHER temperatures than 298K that the given equilibrium should shift to the RIGHT.....and so since pKa=log10[H3O+], 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?