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Homework Help: Calculate equilibrium dissolved oxygen at 19.8 Celsius

  1. Oct 11, 2014 #1
    1. The problem statement, all variables and given/known data
    1. Calculate the equilibrium dissolved oxygen concentration for each temperature if PO2 = 21% (this means that if atmospheric pressure is 1 atm then O2 pressure is 21% of that)

    2. Relevant equations
    K19.8/RT = Kp

    3. The attempt at a solution
    K19.8/RT = Kp

    K19.8 = 1.475x10-3

    (1.457x10-3)/(8.314x10-3)(273.15+19.8) = Kp

    5.982x10-4 = Kp

    Kp = (O2 (aq)) / (O2 (g))

    5.982x10-4 = (O2 (aq)) / (0.21)

    1.256x104 atm = O2 (aq)

    ICE table:

    __________________O2 (g)_______O2 (aq)


    Change___________0.21 -x_______+x

    Equilibrium_______0.21 -x_______+x

    5.982x10-4 = x/(0.21 – x)

    5.982x10-4 (0.21 – x) = x

    1.256x10-4 - 5.982x10-4 x = x

    1.256x10-4 = x + 5.982x10-4 x

    1.256x10-4 = 1.000598x

    1.255x10-4 = x

    PV = nRT

    n/V = P/RT

    M = 1.255x10-4 atm/(0.0821atm L/mol K) (273.15 + 19.8)

    M = 5.218x10-6

    This doesn't look right to me but I'm not sure what else to try. Thanks for the help!
  2. jcsd
  3. Oct 11, 2014 #2


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    Staff: Mentor

    No idea what is what of what, can you please post whole question and all information given?
  4. Oct 11, 2014 #3
    Ok I'll simplify it a bit. All I need is how to calculate the equilibrium dissolved oxygen concentration for 7 degrees celsius if PO2 = 21% (this means that if atmospheric pressure is 1 atm then O2 pressure is 21% of that). K was previously found to be 1.813x10-3 at 7 degrees celsius. My professor is asking this so there must be a way! My head hurts. The chemical equation is O2 (g) = O2 (aq), thats it.
  5. Oct 11, 2014 #4


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    Staff: Mentor

    What is K specifically? How is it defined?

    When it comes to solubility of gases, we typically describe it with Henry's law, not with a typical equilibrium constant. These are perfectly equivalent, but you need to be sure how K is defined and what units it has.
  6. Oct 11, 2014 #5
    K is for concentration. So if I use Henry's law, Henry's coefficient for O2 at 25 degrees celsius is 1.3×10−3 mol L/atm. How can I adjust Henry's coefficient to work at 7 degrees celsius? Thanks for the help!
  7. Oct 12, 2014 #6


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    Staff: Mentor

    Huh? No idea what you just said.

    Van 't Hoff equation comes to mind. Yes, it requires more data. But that's a general case - every approach will require some additional information that you have not listed so far.
  8. Oct 12, 2014 #7
    Got it. Henry's law worked!

    temp 7°C

    ln(kHT/kH25) = (ΔH°/R)(1/298 – 1/T)


    ln(kHT/1.26x10-3mol/Latm) = 3.04x10-1

    e ln(kHT/1.26x10-3mol/Latm) = e3.04x10-1

    kHT/1.26x10-3mol/L atm = 1.36

    kHT = 1.71x10-3 mol/L atm

    C = (1.71x10-3 mol/L atm) 0.21atm

    C = 3.59x10-4 M
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