Using Henry's law to calculate ammonia concentration?

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Discussion Overview

The discussion revolves around calculating the concentration of ammonia in the air above a 5% ammonia solution in a closed chamber of 20 liters at 40°C, with a focus on applying Henry's law and understanding vapor pressures.

Discussion Character

  • Exploratory, Technical explanation, Homework-related

Main Points Raised

  • One participant seeks help in calculating the ammonia concentration in the chamber air using Henry's law, expressing uncertainty about the process.
  • Another participant suggests that understanding vapor pressures is essential for the calculation.
  • A later reply indicates that the original poster may need to educate themselves on vapor pressures, providing external resources for further learning.
  • Further contributions mention specific resources and links related to ammonia solution vapor pressure, emphasizing the importance of checking units.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there is a mix of guidance and requests for clarification, with no definitive method or solution provided for the calculation.

Contextual Notes

Limitations include the original poster's lack of knowledge about vapor pressures and the need for external resources to fully understand the concepts involved in the calculation.

Hopper295
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TL;DR
How do I calculate the concentration of ammonia in air above a ammina solution of a known concentration
We have a closed chamber of 20 liters. The bottom of this chamber is filled with ammonia solution of 5% (=2,7M/liter). The solution temperature is controlled at 40°C

I would like to calculate the concentration of ammmonia in the chamber air, above the solution. It should be something with henry's law but I don't knwo how to do this...
An approximation would already be super...

Hope someone can help me, thanks!
 
Science news on Phys.org
You'll want to learn about the concept of vapour pressures first.
 
Hi dr. nate,

To be honest I don't know much about this... :rolleyes:

can you explain some more what you mean?
 
Ok thanks. I was already trying to do it myself using googel, but this doesn't go so well.
I was actually hoping someone would find this easy and show me how to do it.
 

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