Quantitative equilibrium calculations

Fundamentals
Problems
Excercises for improved understanding
     5 ml water instead of 1 ml
     2 ml air volume instead of 1 ml
     25 ml air volume instead of 5 ml
     Volumes are doubled
     Total amount of the substance is doubled
     Initial situation changes
Questions for recapitulation
Good to know
Minesweeper-problems

Excercises for an improved intuitive understanding

After working through these problems you should have developed a good overview of how partition constants can be applied to solve quantitative partition problems. But do you also have a good ´intuitive´ understanding of how partition systems respond to changes of one or several variables? If you change the volume of a phase or a partition constant then some of the system responses (like changes in equilibrium concentrations, mass fractions) are proportional to the changed variable while others are not. An example for a non-linear response is the fact that you will need 10 times more solvent to extract 99% of a chemical out of a water sample than if you wanted to extract only 90% (see problem Organic pollutants in water). The following exercise is intended to train your intuitive understanding of partition processes (corresponds to Box 3 in the script).

 

Exercise for a better understanding of the system response of a simple partition system:

Use sheet 1 of the Mehr-Phasensystem.xls to calculate the equilibrium state of 1 ng of a compound i in the following partition system: 1 ml air, 1 ml water and a Ki aw of 0.1 [Lwater / Lair ]. Try to estimate (not calculate yet) how the concentrations of i and the mass fraction of i in both phases change if you change the system as follows:

2 ml air volume instead of 1 ml

Answer: The mass distribution of i will shift towards the air. In any case the mass fraction in air under the new equilibrium conditions must be between 9 and 18%. This implies that the mass fraction in water will be between 82 and 91%. In turn this means that the concentration in water will not change much. If the mass fraction in air can double at best while the air volume is doubled, then the new concentration in air must be between 1/2 and 2/2 of the original value. Compare this with the exact solution to the problem (calculated with the spread sheet).

 

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