Precipitation and Dissolution

Some ionic compounds do not dissolve in water.
In reality, even “insoluble” compounds dissolve to a small extent. They are “sparingly soluble.”
- An equilibrium exists between the precipitate and its ions.
- Example: AgCl(s) ⇌ Ag+(aq) + Cl–(aq) K << 1
The low value of K indicates that the AgCl(s) dissolves only slightly to form those two ions.
Writing the equilibrium expression for the process gives us
[Ag+][Cl-] = K '
- Since the undissolved salt is a SOLID, it does not appear in the expression. Therefore, there is no denominator!!
Ksp: solubility product constant

At equilibrium, the solution is saturated with dissolved ions
In ordinary equilibria, we have a reaction quotient, Q, the products over the reactants (to the correct exponents).
Q can have any value at all. But when Q = K, we have equilibrium. (When Q < K, reactions must go forward)
Since the solubility expression has no denominator, it is a product, not a quotient.
- We call it the ION PRODUCT.
- When the ion product = the Ksp, the solution is saturated!
Like all equilibrium constants, Ksp has a fixed value at a given temperature.
The product of the two ion concentrations at equilibrium must have this value regardless of the direction from which equilibrium is approached.
We can use the Ksp to find the solubility.

Add a sparingly soluble solid to water and stir.
Eventually the solution becomes saturated with ions.
In other words, the compound reaches its solubility.
In a saturated solution, an equilibrium exists between the dissolved ions and undissolved solid.
We can calculate the solubility of a compound from its Ksp value.
Molar solubility: when solubility has units of moles per liter (M)

Instead of calculating solubility from Ksp, it is possible to calculate Ksp of a compound from its solubility.
- There are many sources listing the solubility of sparingly soluble compounds.
- Depending on the source, these solubility values are expressed in many different units.
- We must always make sure that the solubility is the molar solubility (units of M) when using the Ksp expression.

Common ion effect: the presence of an ion common to the salt REDUCES solubility
Equilibrium can be approached from either direction.
Starting with just reactants—Add solid compound to water.
Starting with just products—Add ions to water.

Ksp values can be used to predict whether a precipitate will form when ions (from solutions) are added to water.
- Q is of the same form as Ksp, only difference is the concentrations at any particular moment can be used with Q and only equilibrium concentrations can be used with Ksp.
If Q > Ksp: a precipitate will form, decreasing the ion concentrations until equilibrium is established.
If Q < Ksp: the solution is unsaturated, no precipitate will form, and if desired more compound can be dissolved.
If Q = Ksp: the solution is saturated, just at the point of precipitation.

Consider a solution of two cations.
- One way to separate the cations is to add an anion that will form a precipitate with only one of them.
- The precipitate can then be removed by filtration.
- The other cation will remain in the filtrate.
- This approach is called selective precipitation, and is often used as a method of purification.
When can Ksp values alone be used to predict which compound will precipitate first?
1. Both ions to be separated must be at the same initial concentration.
2. Both complexes must have the same cation : anion ratio.
3. Both of these criteria must be met. If not then calculations must be done to determine which compound will precipitate first.