Questions about the common-ion effect

[ProjectFringe]

I have two questions about the common-ion effect. Sorry if my terminology is not correct, I’ll try to be as clear as possible. My goal is to be able to understand what will happen to a solution of two compounds with a common ion, when subject to evaporation and temperature change. So, I really need to first completely understand the common-ion effect, so that I can predict what changes will occur in the solution when evaporation occurs or the temperature of the solution changes.

1. Firstly, is my fundamental understanding of the common-ion effect correct?

I’m sure it is not, but I will try my best to express how I understand it and maybe someone can tell me where I go wrong. My guess is that I’m thinking about the solution as two compounds and not as one solution with its own solubility and concentration, but I’m not sure.

-As I understand, according to the common-ion effect:

Combining two compounds with a common ion in solution will DECREASE the SOLUBILITY (by this I mean the maximum amount that a compound will dissolve in a solution, expressed in mol/L) of the solution, including BOTH compounds EQUALLY (or by similar amounts).

And therefore, if ONE or BOTH of the two compounds CONCENTRATION (actual molarity of a compound in the solution) is ABOVE their newly reduced SOLUBILITY, then the compound with the LOWER CONCENTRATION will precipitate.

The precipitation of this compound will cause a DECREASE in the CONCENTRATIONS of BOTH compounds in the solution (in a way because the common ion in the solution is being reduced), and it will continue to precipitate until the CONCENTRATIONS of BOTH compounds are below their SOLUBILITY.2. Secondly, what happens when three compounds with the same common ion are combined, does ONLY the compound with the lowest concentration form as a precipitate, or is it possible for the compound with the SECOND LOWEST CONCENTRATION to also precipitate?Thanks in advance for your time!

 

[Borek]

Common ion effect is basically just a rule of thumb, so while your predictions are more or less correct (although not always, if one thing precipitates, concentration of the other in general doesn't change - but there are some nitpicking details).

But: there is no need to use rule of thumb when there is a solid chemistry behind the solubility.

Every salt has its solubility product, Ksp, which defines when the salt will precipitate out. For the simplest case of dissolving AgCl

AgCl ↔ Ag⁺ + Cl^-

solubility product is

Ksp = [Ag⁺][Cl^-]

(as it is common in equilibrium formulas, concentrations should be raised to powers being stoichiometry coefficients, here 1 for both ions).

Here comes the most important part: AgCl won't precipitate till product of the concentrations of ions present in the solution gets higher than Ksp. And if the product of the concentrations is higher than Ksp, salt will precipitate (removing ions from the solution and decreasing their concentrations) till it is not.

That, plus stoichiometry of the precipitation, plus mass conservation, let's you calculate what precipitates out of every solution and how the concentrations of all ions present in the solution change (assuming you know Ksp for every salt involved, these are tabularized, for AgCl Ksp happens to be almost exactly 10^-10).

 

[ProjectFringe]

Thank you for your response, and sorry for my late reply. I have been trying to figure out how to apply the math to what I want to figure out and am still struggling. I believe I know how to calculate the Ksp correctly, but I don’t know what to do after that.

For example, I can calculate the amount of grams/L, mol/L, and Ksp values for NaCl and KCl, at saturation in solution, at a given temperature. But I don’t know what formula or how to calculate how those values change when the solutions of the two compounds are combined.

For example:
NaCl solubility @ 25C: 359g/L, 6.14M, Ksp 37.7
KCl solubility @ 25C: 360g/L, 4.83M, Ksp 23.3

I believe that due to the common-ion effect, when combined, the solubility of both NaCl and KCl will change. I want to calculate what the new solubility values are when a solution of 6.14M NaCl is combined with a solution of 4.83M KCl.

 

[Borek]

I am afraid you are out of luck here.

That is: for weakly soluble salts which produce diluted solutions this is a trivial exercise. Let's say you have a solution that is saturated in regard to both AgCl (Ksp 10^-10) and PbCl₂ (Ksp around 1.7×10^-5). There are three unknowns - concentrations of Ag⁺, Pb²⁺ and Cl^-. There are three equations - solutions is saturated, so two equations for Ksp, and it must be electrically neutral, so there is a third equation, for charge balance:

[Ag⁺] + 2[Pb²⁺] = [Cl^-]

Three equations, three unknowns, just solve.

However, what is trivial for diluted solutions, becomes difficult for concentrated ones. That is, you can do naïve calculations exactly the same way as the one I described, but it won't produce result that is even remotely accurate. That's because equilibrium equations actually don't work with concentrations, but with so called activities of ions. For diluted solutions activity equals concentration, but the more concentrated solution and the more ions it contains the higher the difference is. There is an easy to calculate parameter called "ionic strength", which helps predicting these differences and can be even used to calculate activity coefficients, which tell us what is the difference between the concentration and the observed activity of the ion - but it helps for solutions that are in well below 0.1M range.

In short: the more concentrated the solution, the better it is to experimentally measure its properties instead of trying to calculate them.

 

[ProjectFringe]

Okay, got it. Thanks again for your detailed and thoughtful explanations.