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Acid Base Equilibrium

  1. Dec 20, 2016 #1
    1. The problem statement, all variables and given/known data
    To unclog a drain you add 26g of sodium hydroxide to 150mL of water. Calculate the pH and pOH for the solution you prepared. Can someone tell me if I'm doing it right.

    2. Relevant equations


    3. The attempt at a solution
    n = m/M
    n = 26/40
    n = 0.65mol for 0.15L

    Does this mean that I have to convert into mol/L?

    So that would be 4.33 mol/L and this would also be my H+ value/ And then would I plug it into my equation

    pH = -log[H+] and find my pH value.

    Then plug my pH value into my pH + pOH = 14 equation and solve for pOH?

    Is this the correct method?
     
  2. jcsd
  3. Dec 20, 2016 #2

    Borek

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    Staff: Mentor

    No.

    Trivial mistake, but you got it reversed. NaOH doesn't produce H+ when dissociating. Other than that, you are on the right track.

    Actually the final concentration is a bit lower than 4.33 M as the volume is not 150 mL, it gets a bit higher after NaOH dissolution. But you can safely ignore it.
     
  4. Dec 20, 2016 #3
    Sorry I meant to put OH-

    okay here's what I've gotten

    NaOH = 26/40 = 0.65mol

    And 0.65 mol would be my 0.65 mol/L?
     
  5. Dec 20, 2016 #4
    I put that value into my pOH equation and ended up with 0.187

    I then put that into the next equation
    pH = 14 - pOH and got
    pH = 13.81

    Is this correct or way off?
     
  6. Dec 20, 2016 #5

    Borek

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    Staff: Mentor

    No, you have to take the volume into account, Actually you did it - and correctly - in your first post.
     
  7. Dec 20, 2016 #6
    Okay so basically I'm using the formula c= n/v?
     
  8. Dec 21, 2016 #7

    Borek

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    Staff: Mentor

    Yes, just like you did initially.
     
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