At 40 degrees Celsius, the value of Kw is 2.92 X 10^-14

a.) calculate the [H+] and [OH-] in pure water at 40 degrees celsius

b.)what is the pH in pure water at 40 degrees celsius

c.)if [OH-] is .18M , what is the pH

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  1. H2O ==> H^+ + OH^-
    You know Kw = (H^+)(OH^-) = 2.92 x 10^-14.
    AND you know (H^+) = (OH^-)in pure water.
    Then pH = - log (H^+)
    Post your work if you get stuck.

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  2. I already worked out (c), but the rest is still just mush. I'm sorry for my stupidity, but this just doesn't make sense. I can find [H+] or [OH-] as long as I have the concentration of 1 of them, so, ???

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    posted by Rebekah
  3. By the way, how do you find H+ and OH- concentrations from the pH? I have scoured my text book, but found nothing.

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    posted by Rebekah
  4. (a)
    H2O ==> H^+ + OH^-
    for every x mols H2O that dissociate, there are x mols H^+ and x mols OH^- so
    since Kw = (H^+)(OH^-) = 2.92 x 10^-14
    then x*x= 2.92 x 10^-14
    so x = sqrt 2.92 x 10^-14

    (b)I assume you can do pH = -log(H^+) now that you know (H^+).
    The answer you should obtain is 6.76731 which rounds to 6.77

    Here is how you find pH from (H^+).
    Suppose pH = 5.32
    pH = - log(H^+)
    5.32 = -log(H^+)
    -5.32 = log(H^+)
    So you take the antilog of -5.32. To do that, enter 5.32 on your calculator, change the sign to - (or enter -5.32 at the beginning), then punch the 10x key on your calculator. If you have done it right, you should get 4.7863 x 10^-6. Of course that's too many significant figures; however, I copied ALL of the digits so you can check your calculator ability. Now just to make sure you get it, use pH = 6.76731 and see if you get the same answer as you have for (a).
    Let me know if this isn't clear.

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