Chem 30: Solubility The Common Ion Effect

The Common Ion Effect: Remember that when a salt dissolves in water, some of it dissolves and some may stay as a solid. There is a point at which the salt dissolving reaches equilibrium. This means that for each formula unit of salt that dissolves, a pair of ions come together to create a formula unit. When we dissolve ions in water we know that the product of the concentrations of the ions, to the power of their coefficients in a balanced equation, is equal to the Ksp. As long as the product of the concentration of the ions does not exceed the Ksp value, no precipitate forms.

The Common Ion Effect: Let’s look at the following example: Copper (I) iodide is not very soluble in water (Ksp = 4 × 10−19): CuI(s) ⇌Cu+(aq) + I−(aq) Since Ksp = [Cu+][I-], if the product of the concentrations of Cu+ and I- stays below 4 × 10−19, no precipitate will form, however, if we add extra iodine or copper ions (the common ion) things will change. Let`s say we added some iodide ions; this will increase the concentration of iodide and the product of the copper ions and iodide ions would be higher than the Ksp value. This is not possible so the concentrations would need to be lowered. In order to lower the concentration of iodide ions, some copper (I) iodide would need to precipitate out. This follows Le Chatelier`s Principle because we increased the concentration of the products (iodide ions) so the reaction shifts to the reactants.

The Common Ion Effect: After adding iodide ions to the solution, copper (I) iodide will precipitate out, but once the reaction reaches equilibrium again, the concentration of copper ions and iodide will no longer be equal (there will be more iodide ions in solution than copper ions). This also means the concentration of Cu+ will be lower than at the original equilibrium and the concentration of I- will be higher than at the original equilibrium. Overview: If a slightly soluble ionic compound is dissolved in water, you can force precipitation of that salt by adding a readily soluble ionic compound that has an ion in common with the slightly soluble salt. This shift is known as the common ion effect.

The Common Ion Effect: Ex. The Ksp of silver iodide is . What is the iodide ion concentration of a 1.00 L saturated solution of AgI to which 0.020 mol of AgNO3 is added? AgI(aq) ⇌ Ag+(aq) + I-(aq) Ksp = [Ag+][I-]=8.3x10-17

Selective Precipitation Selective precipitation is when you separate ions from two solutes to create a precipitate. To calculate selective precipitation we use initial concentrations (Q), instead of the equilibrium concentrations. In selective precipitation there are restraints in forming a precipitate; a precipitate doesn't always form in a solution. It depends on the relationship between K and Q: If Q>K  a precipitate forms and the reaction proceeds to the left If Q<K  no precipitate forms and the reaction proceeds to the right If Q=K  the reaction is at equilibrium

Example: If 40L of 0. 0050M KCl is mixed with 60L of 0. 0030M Pb(NO3)2 Example: If 40L of 0.0050M KCl is mixed with 60L of 0.0030M Pb(NO3)2. Will the precipitate PbCl2 (Ksp= 1.6 X 10-5) form? First we calculate the initial concentrations of Pb2+ and Cl-: Calculate value of Q:

Example: If 100L of 0. 03M Pb(NO3)2 is mixed with 200L of 0. 09M KCl Example: If 100L of 0.03M Pb(NO3)2 is mixed with 200L of 0.09M KCl. Does a precipitate (PbCl2) form? (Ksp = 1.6 X 10-5). What are the equilibrium concentrations of Pb2+  and Cl-?

By using selective precipitation, we can calculate the amount of a reagent necessary to react with the salt. Example: If a solution contains  1.0 x 10-5 M  of  Pb2+  and 2.0 x 10-5 M  Ag+ and Cl-  is slowly added to the  solution, will AgCl (Ksp=1.6 x 10-10)  or PbCl2 (Ksp=1.6 x 10-5) precipitate first? Which will be in excess?

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