What is Ksp?
Ksp, or the solubility product constant, is a unitless measure of how well a solid dissolves in an aqueous solution. To be precise, Ksp is an equilibrium constant, which tells how much a solid dissociates when the dissolution has reached equilibrium. Solubility is a specific case of acid or base dissociation, so solubility is calculated similarly to Ka or Kb.
The more soluble a salt is, the higher the Ksp value will be. This is because the salt will dissolve into more product ions, making the concentration of each product higher. According to the solubility constant expression, greater product = greater Ksp. A slightly soluble salt, as a result, will have a relatively low Ksp value.
How to Calculate Ksp
As mentioned above, solubility is a specific case of acid or base dissociation, in which the original reactant is a solid. Since Keq = [product][product]/[reactant][reactant], Ka = [product][product]/[acid].
In the case of acid dissociation, acid → H3O+ + conj. base
Thus, Ka = [H3O+][conj. base]/[acid]
However, only aqueous substances are included in equilibrium expressions. Solids, liquids, and gases are not included. In the case of solubility, where the reactant is a solid, the equilibrium product should only include the aqueous products.
Thus, Ksp = [product][product]
How to Calculate Molar Solubility
Using the Ksp equation, you can calculate solubility given the balanced dissolution equation and the Ksp.
In the dissolution equation: aA(s) → bB(aq) + cC(aq), where each of the lowercase letters symbolizes the coefficients of the corresponding substance, A is the solid reactant, and B and C are aqueous products.
The solubility constant expression for this reaction would be Ksp = [B]b[C]c. Since all products come from the same reactant, they will always be produced in proportion to each other. Thus, you can replace B and C with x and add the coefficients into the equation.
Ksp = [bx]b[cx]c = bc(x)b + c
If the value of Ksp is given, the molar solubility of A can be found by solving for x.
For any questions, you will likely not need to memorize the Ksp or molar solubility values for any compounds – they should be given in the problem!