Bunsen Coefficient Solubility Calculation Units?

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SUMMARY

The discussion centers on the calculation of dissolved gas concentrations in water using the Bunsen coefficient equation: ln(β) = exp(A1 + A2*(T/100) + A3*ln(T/100) + A4*(T/100) + S*(B1 + B2*(T/100) + B3*(T/100)^2), where β is expressed in mL/L. Participants clarify that the equation assumes constant pressure, likely at 760 mmHg, and that it is primarily applicable for single gas solubility calculations. The relationship between mL/L and mg/L or mol/L is explored, emphasizing that partial pressures are not needed for pure gas calculations, but may be relevant in mixtures. The discussion concludes that while the equation can be used for pure gases, it may not yield accurate results for gas mixtures due to interdependencies in solubility.

PREREQUISITES
  • Understanding of the Bunsen coefficient and its application in gas solubility calculations.
  • Familiarity with the ideal gas law and its components (P, V, T).
  • Knowledge of Henry's Law and its implications for gas solubility.
  • Basic grasp of logarithmic and exponential functions as they relate to scientific equations.
NEXT STEPS
  • Research the application of Henry's Law in calculating gas solubility under varying pressures.
  • Explore the derivation and use of the Bunsen coefficient in different temperature and salinity conditions.
  • Investigate the effects of gas mixtures on solubility and the limitations of using the Bunsen coefficient.
  • Learn about the relationship between partial pressures and gas solubility in multi-component systems.
USEFUL FOR

Researchers, environmental scientists, and chemists involved in aquatic chemistry, particularly those studying gas solubility dynamics in water under varying conditions.

jeanius
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I've been reading about calculating dissolved gas concentrations in water at varying temperature and salinities using an equation of the form:

ln(β)=exp(A1 + A2*(T/100) + A3*ln(T/100) + A4*(T/100) + S*(B1 + B2*(T/100) + B3*(T/100)^2

where β is in units of mL/L, and A# and B# are empirically derived least square fit constants, different for specifics gases. My question is how to relate mL/L solubility to mg/L or mol/L of the specific gas in solution, say of oxygen in water. Any help would be appreciated! Thanks!
 
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Coefficients are given for a specific pressure - you know P, T & V, use ideal gas equation.
 
So I would determine the volume of gas present by multiplying my sample volume by beta to get mL of gas, convert to m^3 for V, and use the partial pressure of oxygen for P? Or rather the pressure of the environment?
 
Why partial pressure? You are calculating volume of pure gas dissolved.

Whether you will convert volume to L or m3 is a secondary thing, you can always choose such an R value that will incorporate any volume units you need.
 
But say it was a gas mixture or oxygen, nitrogen, argon, hydrogen, etc. all at different partial pressures, would I use partial pressures then?
 
Equation gives you answer in terms of pure gas, so no, no partial pressures.
 
Would I use dalton's law at that point? Or would I calculate the average molar mass for R_specific, then determine each gas' moles from their respective partial pressures?
 
Last edited:
Actually I am no longer sure.

Equation doesn't contain any pressure related expressions, so it most likely assumes constant pressure - and what pressure is assumed should be given together with coefficients, otherwise they are useless. If the answer calculated is given as mL per L, it would be logical to assume it is mL of the PURE gas dissolved in L of the water. That in turn means you don't need Dalton's law, partial pressures, averaged molar masses - you do calculations for each gas separately, assuming its pressure is that for which coefficients are given.

Could be coefficients are given assuming solution in equilibrium with atmosphere - in such a case you will have to use partial pressures of oxygen, nitrogen and argon (assuming some total pressure - again, probably atmospheric). But that's like scratching right ear with the left hand - clumsy and inconvenient, so it sounds highly unlikely.

Note: I doubt you will get correct results while trying to calculate concentrations for a mixture using equation you have listed - my bet is that it works OK for a solution containing one gas only, when there is more than one gas, their solubility are not independent.
 
I believe the equation should be valid for 760mmHg. If you go by Henry's Law, you'd be able to determine concentration of gases at various partial pressures no problem, but you'd be out a salinity term. Regardless, an actual pressure compensation isn't present in the equation, and I'm wondering how I'd go about accounting for that. If I'm given solubility of a gas at 760mmHg, regardless of how many gases are present, shouldn't I be able to scale that according to the partial pressure present in the environment? A mL of gas at 760 should scale linearly to a mL of that same game at a different pressure right?
 
  • #10
jeanius said:
If I'm given solubility of a gas at 760mmHg, regardless of how many gases are present, shouldn't I be able to scale that according to the partial pressure present in the environment? A mL of gas at 760 should scale linearly to a mL of that same game at a different pressure right?

Linear scaling with pressure sounds logical - that would be my approach.
 
  • #11
Borek said:
Linear scaling with pressure sounds logical - that would be my approach.

This does sound right, though I can't back it up very solidly. Is there a way you know of that this couples to Henry's Law?
 
  • #12
Linear scaling with pressure IS Henry's law. Equation you listed is not, as it doesn't say a word about pressure. The way the Bunsen coefficient is defined suggests it is an equivalent of proportionality constant from the Henry's law. This coefficient is always a function of temperature and solution composition (that is, presence of other solutes) and your formula takes it partially into account.
 

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