Solubility

Understanding solubility requires us to understand how things (especially ionic compounds) dissolve, reversible reactions, and the mathematics of equilibrium - especially “little x” problems.

Let’s look at solubility with respect to one particular ionic compound - lead II chloride (\(PbCl_2\)). All of the ideas that are developed here, will then be applicable to other compounds. We can write the process of lead II chloride dissolving as a simple reaction:

\(PbCl_{2~(s)} \rightleftharpoons PbCl_{2~(aq)}\)

Although that reaction is correct, it isn’t very useful. It is much more interesting to write it as an ionic equation:

\(PbCl_{2~(s)} \rightleftharpoons Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

Remembering that solids are not included in the equilibrium expression, we can write the following:

\(K=[Pb^{+2}] \cdot [Cl^{-1}]^2\)

Since this equilibrium expression has an “odd” look (nothing on the bottom), we give it a special name and add a subscript to the K:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2\)

where \(K_{sp}\) stands for the Solubility Product.

There are several important ideas embedded in this:

1. \(K_{sp}\) is always for an ionic solid dissolved in water breaking up into its constituent ions
2. Any problem that mentions \(K_{sp}\) does not need to give the reaction, because it is implied.

There is also some vocabulary that flows from this:

• Saturated solutions are solutions for which \(Q_{sp} = K_{sp}\). In other words, a saturated solution is at equilibrium. Remember that this does NOT mean that the process of dissolving has stopped. It means that the rate at which ions are entering solution is matched by the rate at which ions are coming together and dropping out of solution. Therefore the concentrations of the ions will not be changing, nor will the amount of solid present. In addition, no solid must be visibly present. Saturation is based on the concentrations of dissolved ions only.
• Unsaturated solutions are those that can hold more solute than is currently dissolved. That is \(K_{sp} > Q_{sp}\). If solid is present, then the rate of solvation is faster than the rate of precipitation and the concentrations will be rising. If no solid is present, then the solution will remain as it is. 
• Supersaturated solutions are those where “too much” solute is dissolved. That is, solutions where \(K_{sp} < Q_{sp}\). Generally, we would expect such a situation to result in ions in solution coming together and precipitating, thus decreasing the concentrations of the ions and increasing the amount of solid present in the container. In rare occasions, when the crystalline structure of the solid is complex, it is possible for a solution to temporarily remain in a state of supersaturation until random chance allows the beginning of precipitation or until crystalization is seeded.
• Molar Solublity is the amount of a solid that will dissolve in water in units of \(\frac{moles~solid}{liters}\)

The only way to really understand all of this is, of course, to look at some problems. We’ll look at five different problems: 

1. Finding the concentrations of ions in a saturated solution 
2. Finding the concentrations of ions in a saturated solution in which one ion was already present (this is called the common ion effect) 
3. Finding the molar solubility of an ionic compound 
4. Finding the solubility of an ionic compound in \(\frac{g}{100~mL}\) 
5. Determining whether a precipitate will form when solutions are mixed