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PH scale

  1. Oct 28, 2008 #1
    1. The problem statement, all variables and given/known data
    How would you prepare 5.00 L if an aqueous solution having a pH = 10.70, if you had a supply of 0.250 M NaOH?


    2. Relevant equations
    I don't even know anymore....



    3. The attempt at a solution
    I've tried a million different ways but I never come up with the answer provided in the book, which is 10 mL.

    Here are two ways I've tried:
    pH = 10.7; so pOH = 14.0-10.7 = 3.3
    antilog (-3.3) = 5.0x10^-4 M

    m1v1 = m2v2
    v2 = m1v1 / m2 = (0.250 M x 5000 mL) / (5.0x10^-4) = 2500000 mL
    That's DEFINITELY not right....

    Second attempt:
    5000 mL has x moles NaOH
    5.0x10^-4 M / 5 L = 1x10^-4 moles NaOH
    1x10^-4 mol NaOH x (40.0 g/mol) = 0.004 g NaOH
    density of NaOH = 2.13 g/mL; so 0.004 g / 2.13 g/mL = 0.0018 mL
    And that's not right either...

    I'm missing SOMETHING and I've been working on this problem forever. Please help?
     
  2. jcsd
  3. Oct 28, 2008 #2
    Ah... you know what, never mind. I flipped the equation for M1V1 = M2V2 incorrectly. I've got the right answer now.

    Sorry for wasting space!
     
  4. Oct 28, 2008 #3

    Borek

    User Avatar

    Staff: Mentor

    OK

    Good, approach, but lousy realisation. Write which solution - volume & concentration - is 1, and which is 2.

    That's completely off - 2.13 is a denstisty of solid NaOH, not solution.


    Edit: my uplink went down for 10 minutes so you have answered your own question before I was able to post.
     
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