Distributions within saturated solutions

[JeffEvarts]

There are some really really good treatises out there on fractional crystallization, and I'm ploughing through them one at a time.

One very basic thing has me confused, though: If you have 1 liter of water at 100C, it will dissolve 455g of NaCO3 or 1150 of KCO3, or pretty much any linear combination of those two. (The saturation curve is definitely nonlinear at lower temperatures, but that's not important for this question.)

Suppose I
1) put 1.5kg of each in a single flask,
2) intermix them carefully
3) add a liter of water
4) raise the combined temp to 100 degrees
5) let the system stabilize
6) draw off 100ml of clear fluid

Am I going to get
1) A saturated solution containing ONLY the most soluble salt
2) A combination of salts based on available amounts (50/50)
3) a mixture based on something else (temperature? atomic weight?)

Thanks for your time,
-Jeff Evarts

 

[Chestermiller]

Have you learned about the concept of solubility product yet?

Chet

 

[JeffEvarts]

ChesterMiller: Yes, at least the basics.

The solubility product is an equilibrium expression (like SO2/O2 vs SO3 in gasses) that tells us how much of a single salt will be solvated when a system of that salt and water stabilizes.

It is my current (limited) understanding that there is cross-inhibition between salts, that is: If a solution is saturated with (say) NaOH, then less CaOH will enter solution than if there was no NaOH. It is about this that I am asking: If there's enough to saturate either way, how do I determine what the solution will contain?

-Jeff

 

[Chestermiller]

Hi Jeff,

If you know how much NaCO3 and how much KCO3 dissolve in isolation, then you know the solubility product for each. When you dissolve both at the same time, you must satisfy the solubility products for both of them simultaneously. Let x be the amount of KCO3 that dissolves, and y be the amount of NaCO3 that dissolves. CO3 is common to both so there will be x + y CO3 formed. This gives you two equations and two unknowns (the two solubility product equations) to solve for x and y. (If you divide one equation by the other, you immediately have the ratio of x to y).

Chet

 

[JeffEvarts]

LOL. I tend to oversimplify things, but in this case I overcomplicated them.

Thank you Chet

 

[Borek]

Just remember in such concentrated solutions calculations are quite difficult, because of the very high ionic strength.

 

[Chestermiller]

Just remember in such concentrated solutions calculations are quite difficult, because of the very high ionic strength.

Yes. You need to use an appropriate activity coefficient model, often an extended version of Debye Huckel. See Handbook of Aqueous Electrolyte Thermodynamics: Theory & Application by Joseph F. Zemaitis Jr., Diane M. Clark, Marshall Rafal and Noel C. Scrivner

Chet

 

[JeffEvarts]

So back in April, I thanked you and went away, missing the next post by borek.

You bastards! :)

First I get a really simple answer that I can understand, then you tell me go read a book with the title "Handbook of Aqueous Electrolyte Thermodynamics: Theory and Application"?

That squishy noise you just heard was thousands of brain cells beating themselves to death against the inside of my skull in an effort to avoid parsing the title of that book.

Perhaps I shall just ask the same question again:

If I have equal amounts of solid K2CO3 and Na2CO3, thoroughly intermixed, and "wet" the mixture, such that the water is exposed to both products in vast excess, then collect that water and evaporate it, what will I find?

-Jeff

 

[Chestermiller]

First I get a really simple answer that I can understand,

This sounds very interesting Jeff. Please fill us in on the details of your calculation.

If I had to make a quick and dirty guess on this problem, I would say that neither material would dissolve as much as when they are each in isolation. And more KCO3 would dissolve than Na2CO3.

I have asked a friend of mine to look over this thread, and to provide his expert judgement. He is one of the authors of the Electrolyte Solution book that I mentioned, and he owns an engineering software company that features a product that can provide an accurate quantitative answer to your question.

Chet

 

[Chestermiller]

Good news guys. I have received a reply from my friend Marshall Rafal, and he has graciously offered to solve this problem using the OLI Systems concentrated electrolyte solution software. Here is an excerpt from Marshall's response:

There are two ways that I know of to get the answer to this question. The first is to find a specific experimentally measured point on this ternary. The other is to use the OLI software. I have asked Jim Berthold of OLI to analyze this problem. Jim will go ahead and run the calculation and post the answer on the Physics Forums site. (Hopefully in our thread)

Jeff, you will finally have your answer. I am looking forward to seeing what they come up with.

Chet