[TaggerFeld]

Hi everyone. I need help on this. I am trying out this problem and trying to see if I am doing this right, but I can't quite figure it out. The original question is: "A water sample , at 25 C, Eh= 0.2 and pH=5.40. Assuming this system is at equilibrium, calculate the amt. of dissolved oxygen in ppm in the water sample.

What I know is that given KH = 0.0012589:

We know that O2 makes up about 21% of the atmosphere molecules, so its partial pressure is 0.21 atm (210,000 ppm)....Solving using Henrys Law

O2(aq) = PO2 x KH = 0.21 atm x 0.0012589 (M/atm) = 2.6 x 10-4 M...

given this M, we can convert from molarity to ppm...but it just doesn't seem to work... or seem right. Anyone help me on what I'm doing wrong?

[enahs]

2.6 x 10^-4 is correct.

What/how are you converting to PPM? You just say it does not seem right without telling us how and what you are getting!

[TaggerFeld]

I wasn't sure if that answer is right or wrong. The teach gave me a hint that the molar conversion I need was 8.5 mg/L. I wasn't sure what he meant. I know that 1 ppm equals 1 mg/L. But I don't know how molarity (moles per liter) equals this. I also had tried took the moles using the atomic weight of O2 and used that to change it into grams, took the grams changed it into mg. We tried using an online converter which yielded the same results.

[enahs]

Umm, your hint is the answer, actually!


You know you have 2.6 x 10^-4 M


2.6 x 10^-4 M = (2.6 x 10^-4 mol)   *   32 g  = 0.0085 g/L  = 8.5 mg/L
                    (                L )         mol

mg/L ~= ppm.

Can you reason reason out why that is true? (I put almost equal to, because that uses the assumption  that 1ml water = 1g, or 1kg water = 1L).

[TaggerFeld]

Yes, that is exactly what I tried as well. His original hint was: "This molar concentration is equivalent to 8.5 mg/L so you can make the desired unit conversion which you need." Based on what you did, is exactly what I had tried... 8.5 was my answer as well. But he told me this was wrong. He didn't tell me anything else. I am still confused... it seems based on the Eh & pH given...maybe I was suppose to use those values or something? I dunno. I'm still confused.

[Borek]

What is Eh? And what chapter you are doing right now?

[TaggerFeld]

Eh is oxidation potential. The chapter I am on is Chemistry of Water + Pollution, specifically either ground water contamination/acid-base chemistry in water.

[Borek]

If you know oxidation potential and pH - shouldn't you use Nernst equation?

O₂ + H₂O + e^- <-> OH^-

[TaggerFeld]

So not using Henrys Law and use the Nernst equation, but how? I'm a bit confused.

[Borek]

Write Nernst equation for the reaction I have shown (note it is not balanced) - look up oxygen reduction potential in tables, solve for dissolved oxygen concentration... Everything else is given. You will still have conversion to ppm problem, but at least you will know real concentration.

[TaggerFeld]

Well the original hint to the problem was: "[O2-aq]/[O2-g]=0.0012589" ... given this, this should be the KH in using Henry's Law though, right? If not, how do I use the Nernst equation? I didn't think originally we had to understand it...