Solution Equilibria

What is $K_{sp}$?

The product of a reaction's product's concentrations at equilibrium.

What is the equilibrium expression for $K_{sp}$?

Take for example $AgCl_{s} \rightleftharpoons Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$, at equilibrium, the solubility constant is equivalent to:

$$K_{sp} = [Ag^{+}][Cl^{-}]$$ (36.1)

Now, for the dissociation constant of $PbCl_{2 \ (s)} \rightleftharpoons Pb^{+2}_{(aq)} + 2Cl^{-}_{(aq)}$, since there are two moles of chloride ions for every mole of lead, this needs to be taken into account - raise each ion to the number of moles that get dissociated. In the AgCl example, you would raise each to one. Since this adds no added value to the calculation (because when there is an absence of a power its assumed to equal one), none are noted above. But for lead chloride, one would write the solubility constant in the following manner:

$$K_{sp} = {[Pb^{+}][Cl^{-}]^{2}}$$ (36.2)

Note: If you are given a $K_{sp}$ and balance the equation, you can calculate the concentrations of the ions.

How is the $K_{sp}$ related to the amount of dissociation that occurs for a given molecule?

They are directly related: The higher the $K_{sp}$, the more dissociation will occur at equilibrium.

This takes it back to $K_{acid}$ and $K_{base}$ - no? Remember, a higher $K_{a}$ or $K_{b}$ reflects an acid or base which actively is dissociating to create H$^{+}$ or OH$^{-}$.

This makes sense though when you look at how the $K_{sp}$ is calculated: Because the numerator represents the product of the concentrations of each of the products, if more dissociation occurs, then the concentration of the products will be higher and multiplying them in turn will create a larger number.