# Distributions within saturated solutions

by JeffEvarts
Tags: distributions, saturated, solutions
 P: 34 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
 Sci Advisor HW Helper Thanks PF Gold P: 5,067 Have you learned about the concept of solubility product yet? Chet
 P: 34 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
 Sci Advisor HW Helper Thanks PF Gold P: 5,067 Distributions within saturated solutions 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
 P: 34 LOL. I tend to oversimplify things, but in this case I overcomplicated them. Thank you Chet
 Admin P: 23,400 Just remember in such concentrated solutions calculations are quite difficult, because of the very high ionic strength.