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Solubility Equilibria
As you know, ionic solids dissolves in a polar solvent (Like Dissolves Like). As shown in the animation from SolutionBookNotes1 handout ([click here to go to website (by mhhe) that animates this process]), and ([Notes on Which Substance Dissolve Miscible in Another Substance] webpage and You Tube video on Dissolve and Miscible, the ionic solid will dissolve into its ions until such time that the given solvent volume can not fit any more solute in it.
Therefore, the solution is said to be saturated or it is at its saturation point. The rate of the ionic solid dissolving is equal to rate of ions going back into solid phase. There is a dynamic EQUILIBRIUM established. This is called Solubility Equilibria which is treated like any other equilibrium problems.
Please remember that adding less than the amount of solute to get to the saturation point, the solution is called unsaturated and adding more is called a saturated solution with solid (stuff) in the bottom of the bucket. It is not a supersaturated solution. Supersaturation requires a lot of work (increase temp then saturate the solution and let cool down, etc).
At the saturation point, the concentration (amount) of the solid that has dissolved into its ions is called the solubility (sometimes molar solubility) of that ionic solid. Of course, the equilibrium concentration of ions that have formed in the solution is determined using the stoichiometry of the reaction (it is really not a chemical reaction but a physical process however it behavior the same with respect to the calculations).
MATH FOR SOLUBILITY EQUILIBRIUM (MOST OF TIME 1st YEAR CHEM STUDENTS DO NOT DO)
Solving Solubility Equilibria problems will be accomplished the same as all other equilibrium problems. Therefore, we will follow [same procedures as previous equilibrium problems (click here to get procedures)].
The following are a few changes/comments specific for solubility equilibria:
(These notes should supplement your video (Solubility Equilibria by TMW)notes and reading Section 18.9 (Pg533) and not replace them)
Step 1: There are no actual chemical reactions here. It is a physical process. You are simply dissolving an ionic solid into its ion (forward reaction)and the ions are reforming back into a solid (backward reaction). So the reaction would look like the following for sodium chloride: NaCls <==> Na1+ + Cl1-
Step 2: You determine the equilibrium constant expression the same way. The new concept here is with the solid. Molarity [ ], is mole per volume. Well for a solid as you take or gain moles you also take or gain volume so we consider the molarity change of any solid equal to 1 so the solid term can be removed from the equilibrium constant expression. There is a special name for this equilibrium constant, Solubility product, Ksp. So the Ksp expression will simply be the products side (concentration of ions raised to their coefficent as powers). For the above reaction, Ksp = [Na1+][Cl1-].
Step 3 through end: Done the same as any equilibrium problem. You as first year students will not need to create an ICE Box table for these problems.
In general there are usually only two types of questions first year students will be asked.
a. Given the Ksp of the solid, calculating the molar solubility of the solid (and/or the equilibrium concentrations of the ions.)
The trick here is emphasized in the video (so look at your notes) where he uses "x" as solubility of solid and determines the equilibrium concentration of the ions by the stoichiometry of the reaction.
b. Given the molar solubility of the solid (and/or equilibrium concentration of ions), determine the Ksp of the solid.
There is no trick here, you just use stoichiometry to determine the equilibrium concentration of the ions and plug into the Ksp expression. I would like you to show how you used the molar ratio to determine the equilibrium concentration of the ions, though.
Homework Problems
Make sure that you follow the above directions and video notes and show all work including writing the equilibrium chemical reaction and solubility product expression for each problem.
Please do problems on a separate sheet of paper.
1. The concentration of lead (II) ion, Pb2+, in a saturated solution of PbI2 is 1.30E-3M (1.30X10-3M). What is the Ksp(solubility product constant)?
2. What is the molar solubility of silver chromate (Ag2CrO4) and the equilibrium concentration of the ions if the Ksp = 1.10E-12.
The answers are 8.79E-9 (unit of M3) and 6.50E-5M, 6.50E-5M and 1.30E-4M, respectively.