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2.0e-8 M NaOH in water

  1. Mar 2, 2005 #1
    there are 2.0e-8 M NaOH in water. I have to calculate the pH of the solution considering the OH from the water. The answear to this is pH=7.04 I know this is calculated with the quadratic equation. what I dont get is how to set this up. I alway get a wrong answear. Can somone help me with this?
     
  2. jcsd
  3. Mar 2, 2005 #2

    Borek

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    Staff: Mentor

    You have two equations:

    Kw = [OH-][H+]

    and charge balance:

    [OH-] = [H+] + [Na+]

    This is enough to solve the problem, but in case you don't know what to do next:

    You have to get rid of [OH-], so rearrange first equation

    [OH-] = Kw / [H+]

    and use [OH-] in the second equation:

    Kw/[H+] = [H+] + [Na+]

    That's your quadratic equation.

    Try BATE for such (and much more complicated) pH and acid - base equilibrium calculations.
    --

    Chemical calculators for labs and education
    BATE - pH calculations, titration curves, hydrolisis
     
  4. Mar 2, 2005 #3
    Worked out fine. Thanks a lot :smile:
     
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