19.4: Slightly Soluble Salts
Lesson Objectives
The student will:
 define solubility product constants.
 write solubility product constant expressions.
 calculate solubility product constants.
Vocabulary
 solubility product constant
Introduction
Since equilibrium constant expressions omit solids, the equilibrium constant expressions for the dissolution of a salt will have a denominator of 1, so no denominator will be shown. Here are some examples with their equilibrium constant expressions.
In this final section of the chapter, we will look at solubility equilibria.
Salts in an Equilibrium System
Consider the dissociation reaction for copper(II) hydroxide:
The equilibrium constant for this reaction is very low with a value of . A very small value for means that the reactant (the solid salt) is heavily favored, so very few and ions will actually be formed when is allowed to dissolve in water. In other words, the solution will become saturated very quickly.
This type of equilibrium will be established for any saturated solution that also contains some of the solid salt. When writing the equilibrium expression for a dissociation reaction, the equilibrium constant is often given the subscript “sp,” which stands for solubility product. The solubility product constant works the same as any other equilibrium constant, but because calculations for this type of reaction are often quite similar, they are frequently differentiated in this way.
Example:
Tooth enamel is composed of the compound hydroxyapatite, , which has a value of . Write the dissociation equation and comment on the value of the equilibrium constant.
Solution:
The equilibrium constant is very small, so the reactants are heavily favored. This means that tooth enamel will not dissolve readily in water (definitely a good thing).
Example: Write the dissociation reaction and the solubility product constant expression for each of the following solids.
Solution:
Calculating K_{sp} From Solubility
The of a slightly soluble salt can be calculated from its solubility. Solubilities are usually given in grams/liter, or occasionally moles/liter. If the solubility is given in moles/liter, you can follow the same process as demonstrated below, except you would skip the first step.
Given the solubility of copper(I) bromide (), the molarity of a saturated solution can be determined. From that value, we can find the concentrations of the individual ions.
In this case, one ion and one ion is formed from each formula unit of , so when of dissolves, the concentration of the ions in solution will be and .
The expression for is . Substituting the calculated values into the expression will give:


 .

The process is only slightly more complicated for salts that dissociate into more than two ions. Consider the salt calcium phosphate. The solubility of calcium phosphate is . First, we convert the solubility in to .
Then, using the dissolving equation, we determine the molarity of the ions in solution.
For every of calcium phosphate that dissolves, there will be 3 times as many calcium ions and 2 times as many phosphate ions in solution. Therefore,
We finish the calculation by writing the expression and substituting the molarities of the ions into the expression and calculate.
Calculating Solubility From K_{sp}
Not only can be calculated from the solubility, but the solubility can be calculated from the value. Given the for (), we can use the dissociation equation and some algebra to calculate the concentrations of each ion in solution.
If we allow to represent the moles of that dissolves in one liter of water, and , since only one of each ion is produced per formula unit of . We can then write the expression for , set it equal to the given value, substitute the assigned variables into the equation, and solve.
Because of how we defined , we can see that the molarity of will be the same so the solubility of is . To find the solubility of in , we simply need to multiply the molarity by the molar mass.
Solubility of
This problem becomes slightly more difficult when the salt generates more than two ions. Let's try lead(II) fluoride. Here is the dissolving equation, the expression, and the value for :
Once again, we let represent the solubility of . Therefore, the concentration of lead ions in solution will be and the concentration of fluoride ions in solution will be . Substitute these variables into the expression and solve for .
When we assigned the variables, represented the molarity of the lead ions in solution, but also represented the molarity of lead(II) fluoride that had dissolved, so this is the solubility in moles/liter. To get grams/liter, we multiply by the molar mass.


 Solubility of

You should look back at the assignment of variables and determine the molarity of fluoride ions in solution. There will be questions where you are asked to find the concentrations of the ions in solution from . You would use this same process, but the desired answer may not always be equal to .
Lesson Summary
 Equilibrium constants for slightly soluble salts are called solubility product constants.
Review Questions
 What is the solubility product constant? Give an example.
 Why is solubility considered a special case for chemical equilibria?
 Nickel hydroxide is a slightly soluble salt. Its dissociation reaction is represented as: . Which of the following represents the solubility product constant expression, ?
 The for is . What is at equilibrium?
 not enough information is given
 The for is . What is at equilibrium?
 Magnesium hydroxide is the active component in milk of magnesia, a suspension used to cure indigestion. It has an equilibrium constant of . Write the dissociation equation and comment on the value of the equilibrium constant.
 Write the dissociation reactions for the following salts as well as the expressions.
 calcium fluoride
 chromium(II) carbonate
 arsenic(III) sulfide