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Distribution Coeff


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#1 Guest_dopey_*

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Posted 10 March 2005 - 11:43 AM

The distribution coefficient in this case is the concentration in the continuous phase/ concentration in the dispersed phase.
The dispersed phase is the NaOH.
What concentration value is used for the continuous phase when there is a mixture of substances in it? And how do I figure the conc. values out in order to get the coeff.?

Dont worry, you're not doing my homework.

#2 gvdlans

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Posted 10 March 2005 - 12:29 PM

Please do not start new threads every time you have something to clarify, but continue the existing topic. You only start a new thread if you have a new topic.

About the distribution coefficient, all I can say is that I would expect the H+, Fe+ and Cl- ions to move from the (a-polar) benzene phase to the (polar) water phase. In the water phase the H+ ions react with the OH- ions to form H2O. If you have sufficient OH-, all H+ ions will eventually be removed from the benzene phase.
This process will go faster when your water droplets are smaller, so it is important that the two phases are well mixed.

#3 Guest_dopey_*

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Posted 10 March 2005 - 01:12 PM

But I still dont know how to get the distribution coefficient.

#4 mbeychok

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Posted 10 March 2005 - 04:21 PM

Dopey:

The distribution coefficient (or partition coefficient) defines how a specific component will partition itself between two insoluble liquid phases. The distribution coefficient is primarily influenced by the solubility of the distributed component in each of the two insoluble liquid phases.

In your case, the two insoluble liquid phases are water and liquid hydrocarbon. Thus, the coefficent (K) that you seek is:

K = (NaOH wt. % in hydrocarbon liquid) / (NaOH wt. % in water)

The best way to determine the coefficient is measure it experimentally in the laboratory. I doubt very seriously that you will find any published data on your specific case.

The presence of HCl very much complicates using relative solubilities because it will react with the NaOH and therefore reduce the "apparent" solubility of NaOH in the hydrocarbon liquid while increasing the "apparent" solubility of NaOH in the water phase. In effect, this is what Gvdlans also told you.

I am curious as to why you need to know the distribution coefficient. Can you tell us?

#5 MrShorty

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Posted 10 March 2005 - 05:24 PM

A couple of things I will add:

1) What component do you want the distribution coefficient of? mbeychok assumed you were looking for K(NaOH). Is that the case? I wouldn't call myself an expert, but to solve the problem rigorously, you probably need to determine the distribution coefficient of all components.

2) You have a reactive system (H+ + OH- => H2O, FeCl3 + H2O => FeO + HCl). Do you want to know the state of the system at some point before the reaction reaches equilibrium, or after the acid/base have been neutralized, at which point it could be argued that there is no "NaOH" or "HCl" in the system. Rather, you would have NaCl (which I would expect to be essentially all in the aqueous phase).

Your main question seems to be about calculating distribution coefficients. I expect you would need a robust thermodynamic model to calculate distribution coefficients rigorously. I'm fairly familiar with non electrolyte calculations using an activity coefficient model (like the NRTL equation), but I really don't know much about calculating electrolyte activity coefficients (other than it can be done). In the absence of a rigorous thermodynamic model, you'll need to make some assumptions (such as assuming all of the electrolytes will be in the aqueous phase at equilibrium).




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