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Topic: Urgent: How to calculate the concentration of a volatile compound in a solution?  (Read 3822 times)

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Offline kcy

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Hello!

I want to calculate the concentration of a volatile compound such as toluene or naphthalene in a solution.
From the references, I know that adding 250uM toluene in a 2 ml liquid would actually be adding of 91 uM toluene based on Henry's law constant of 0.27.

To a 15 ml sealed container, 2 mL of water and 250 uM of toluene (kH of toluene is 6,64 atm-
L/mol; thus K(AW) is 0.27) were added. So, according to the reference; the final aqueous concentration of toluene in the container is 91 uM at 25C.

So; how do I get this?

Thanks in advance!
« Last Edit: November 07, 2015, 02:58:58 PM by kcy »

Offline Borek

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Please elaborate, when you just mix liquids Henry's law doesn't apply.
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Offline kcy

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Thanks for notice. I just edited it!

Offline Borek

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From the references, I know that adding 250uM toluene in a 2 ml liquid would actually be adding of 91 uM toluene based on Henry's law constant of 0.27.

I think you are misreading what they wrote. Adding 2 mL of 250 μM means adding 2 mL of 250 μM, nothing else.

I have not checked numbers, but chances are they take into account fact some of the toluene leaves the solution, so you have toluene in both pases - liquid and gaseous. Then finding the concentration is just a matter of solving the equilibrium - you are given volumes of both phases, you are given Henry's constant, you know total amount of toluene, solving it is trivial.
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Offline kcy

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The thing is; this is a protocol which I'm going to use.
But first, I want to understand the logic behind that.
Of course, some of the compound is in gaseuos phase and some in the liquid phase, I know that. And I used this equilibrium but couldn't find the same answer as they put in the journal reference.
mol(total) = mol(air) + mol(water)
mol(total) = CaVa + CwVw = (KAW)CwVa + CwVw = Cw(KAWVa + Vw)
a:air
w:water
at 25C
KAW:0.27 (Henry's law, dimensionless)

In a 15 ml sealed vial, if all toluene was in the 2 ml solution, it would be 250uM toluene as total. But because of its volatility, only 91 uM is in the 2 ml liquid. I know volatility of toluene is the reason; but I want to calculate this. If the equilibrium on top is wrong, would you mind to write me the exact equation which I can get this result?

Thanks in advance.

Offline Borek

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if all toluene was in the 2 ml solution, it would be 250uM toluene as total

First things first - no, 250 μM is not total, it is a concentration, and not an amount of the substance.

mol(total) = mol(air) + mol(water)
mol(total) = CaVa + CwVw = (KAW)CwVa + CwVw = Cw(KAWVa + Vw)

Logic behind looks OK to me.

Ho did you convert between kH of 6.64 atm×L/mol and 0.27?
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Offline kcy

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I got this info from a document in the attachment:
It says:
Henry's Law constant is often used in the form of a dimensionless
air/water partition coefficient
KAW = KH/RT
R = gas constant (0.0821 L-atm/mol-K)
T = 298K

That's how I found 0.27 from 6.64 atm×L/mol.

Offline kcy

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I got this info from a document in the attachment:
It says:
Henry's Law constant is often used in the form of a dimensionless
air/water partition coefficient
KAW = KH/RT
R = gas constant (0.0821 L-atm/mol-K)
T = 298K

That's how I found 0.27 from 6.64 atm×L/mol.

And also here is the reference I've found:

Two milliliters of concentrated cell suspensions in Tris-HNO3 buffer (50 mM, pH 7.0) were contacted with 250uM toluene (from a 100-mM stock solution in ethanol) in 15-ml serum vials sealed with Teflon-coated septa and aluminum crimp seals.
Toluene liquid phase concentration was 90 uM based on Henry's law constant of 0.27; 250 uM was added if all toluene in the liquid phase.

Offline kcy

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I got this info from a document in the attachment:
It says:
Henry's Law constant is often used in the form of a dimensionless
air/water partition coefficient
KAW = KH/RT
R = gas constant (0.0821 L-atm/mol-K)
T = 298K

That's how I found 0.27 from 6.64 atm×L/mol.

And also here is the reference I've found:

Two milliliters of concentrated cell suspensions in Tris-HNO3 buffer (50 mM, pH 7.0) were contacted with 250uM toluene (from a 100-mM stock solution in ethanol) in 15-ml serum vials sealed with Teflon-coated septa and aluminum crimp seals.
Toluene liquid phase concentration was 90 uM based on Henry's law constant of 0.27; 250 uM was added if all toluene in the liquid phase.


Okay Borek! I finally understood! I've made a mistake while putting all the knowns into the equation.

I got 2 ml solution and inside 250uM toluene so; I need to calculate the mole first.
Finally the result is 90,7uM.

Thanks for the brainstorming!



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