# Calculate equilibrium dissolved oxygen at 19.8 Celsius

1. Oct 11, 2014

### spooky01

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. Oct 11, 2014

### Staff: Mentor

No idea what is what of what, can you please post whole question and all information given?

3. Oct 11, 2014

### spooky01

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.

4. Oct 11, 2014

### 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.

5. Oct 11, 2014

### spooky01

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!

6. Oct 12, 2014

### 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.

7. Oct 12, 2014

### spooky01

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