The ion-product of pure water, Kw, is
2.51 * 10(-14) at the body temp of 37 degrees Celcius. What are the molar H2O and OH concentrations in the pure water temperature?

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asked by Leroy
  1. I expect you meant H3O^+ and not H2O.
    (H3O^+)(OH^-) = Kw
    (H3O^+)=(OH^-) = sqrt Kw.

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  2. Yes you are correct for the change to H3O^

    But do I need the plug the 37 degrees in somewhere or is the response above the final answer?

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    posted by Gina
  3. I saw only this one extra question whereas your post said "additional" questions. If there are more than this please just post everything again at the top of the page.
    You don't do anything with the 37C temperature. That's the temperature of the water at body temperature and Kw changes with T. At this T it is 2.51E-14 (at room T it is 1E-14). So (H3O^+) = (OH^-) = sqrt 2.51E-14.
    So I would write (H3O^+) = 1.58E-7
    and (OH^-) = 1.58E-7

    By the way it makes it hard to follow the thread when you change screen names.

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  4. I didn't put units on my answers and I should have done so. They are
    (H3O^+) = 1.58E-7 M
    (OH^-) = 1.58E-7 M.

    And just to make things a little more interesting let me point out that this is a NEUTRAL solution with a pH = 6.80. We always talk about pH = 7.0 as being neutral and solutions less than 7.0 as being acid BUT all of that is based on the ion product for H2O being 1E-14. :-)

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