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PH problem

  1. Mar 26, 2009 #1
    The pH of pure water at 37°C is 6.80. Calculate Kw,
    pOH, and [OH-] at this temperature.

    pH: 6.8
    pOH = 14 -6.8 = 7.2
    [OH]: 10-7.2 = 6.3 x 10-8 M
    Kw= [H3O] [OH] = 1.5 x 10-7 x 6.3 x 10-8 = 1x10-14

    Real answer:
    [OH]: 1.6x10-7
    Kw: 2.5 x 10-14

    What am I doing wrong? Any help would be greatly appreciated.
  2. jcsd
  3. Mar 26, 2009 #2


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    Beware my response here may be fundamentally flawed: but the question should give you both pH and pOH if you want to find Kw for water at 37 celslius.

    In your "Work" section you bluntly assumed 1x10-14 for water, but this is NOT the value for water at 37 celsius.
  4. Mar 26, 2009 #3
    Thanks for the response.

    Well Kw= [H3O] x [OH]
    So I got the value for pOH from the equation pk = 14 = pOH + pH.
    I then found the concentrations for H3O and OH by using the formulas for pH (pH = -log [H3O]) and pOH ( pOH= -log[OH]). I then plugged those values into the orginal equation and got Kw = 1x10-14. I realize that is the value for water at 25o C but apart from the fact that K increases with temperature I don't know how else it ties in.
  5. Mar 27, 2009 #4


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    Where does the OH- comes from? Think about stoichiometry of the reaction. How is [OH-] related to [H+] if there are no other sources of both but water autodissociation?
  6. Mar 27, 2009 #5


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    Now this is clear (Borek, #4). The [H+] concentration is equal to the [OH-] concentration. All the values needed to determine the dissociation constant are present.
  7. Mar 29, 2009 #6
    Edit- Never mind, I understand. The water dissociates into equal parts so the pH and pOH will both be 6.8. I think at least.

    Thanks for the help!
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