1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    PV=nRT

    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)

    Initial______________0.21_________0

    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

    Borek

    User Avatar

    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

    Borek

    User Avatar

    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

    Borek

    User Avatar

    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)=(-11.71kJ/mol/8.314x10-3kJ/molK)(1/298–1/280)

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted