In this explainer, we will learn how to explain and calculate the solubility product.
There are many ionic substances that we would classify as being soluble, such as sodium chloride. However, there are also many substances that we would consider insoluble, such as barium sulfate or calcium carbonate. There are even some ionic substances that we may consider to be slightly soluble, such as calcium hydroxide.
When something is described as insoluble, it is often taken to mean that the substance does not dissolve at all in water. However, for insoluble substances such as barium sulfate or copper(II) hydroxide, the reality is that these substances can dissolve in water but only in very small amounts.
If we were to express this dissolution as an equilibrium for a general metal ion, , and a general anion, , we could write the dissolution equation as follows:
Here, we can imagine that the equilibrium lies very far to the left-hand side with only a very small number of ions actually being in solution.
This equilibrium can be expressed in the form of an equilibrium constant known as the solubility product, . for this general equation can be written as
Example 1: Constructing the Equation for the Solubility Product of a Generic Inorganic Compound
What is the equation for the solubility product of a generic inorganic compound with the formula ?
The solubility product of a compound involves the product of the concentration of the ions raised to the power of their respective stoichiometric coefficients. This generic compound, , dissociates into a single positive ion of and a single negative ion of . Square brackets are used to indicate that this is the concentration, and it is their product that is equal to .
The equation is, therefore, .
The square brackets here indicate concentration, measured in mol⋅dm−3 or mol/L, and as such, the units to this particular solubility product equation would be mol2⋅dm−6, as calculated below:
This particular type of equilibrium constant is different from the others where some value of the products is usually divided by some value of the reactants. For example, the equilibrium constant for concentration is calculated by multiplying together the concentrations of the products and then dividing that value by the product of the concentrations of the reactants. In the case of solubility products, the reactants in this heterogeneous equilibrium are the insoluble solid, and the amount that changes between the initial start of the dissolution and the equilibrium is so small that it can be considered constant and is therefore incorporated into .
Definition: Solubility Product
The solubility product of a compound is the product of the concentrations of the ions in a saturated solution raised to the power of their respective stoichiometric coefficients.
The following is an example:
When equals the product of the ion concentration, the solution is said to be saturated:
However, when the product of the ion concentration becomes greater than the value of , then a precipitate starts to form:
Finally, when is greater than the product of the ion concentration, then more of the solid will dissolve before reaching equilibrium:
Furthermore, we all know that it is easier to dissolve more sugar in hot rather than cold tea, and as such, it should be no surprise that values of are temperature dependent and vary according to the temperature at which they were measured.
Before looking at how to calculate the solubility product, it is important to correctly write the equation for for ionic substances with more than one of the same ion:
Nickel phosphate has the formula , and so can be written as follows:
The units can be calculated as follows:
Notice in the examples above how the different powers each of the concentrations were raised to affect the units of .
Example 2: Constructing the Equation for the Solubility Product of Vanadium(III) Carbonate
What is the equation for the solubility product of vanadium(III) carbonate ?
The solubility product can be defined as the product of the concentrations of the ions in a saturated solution raised to the power of their respective stoichiometric coefficients. In this instance, the two ions in solution will be vanadium and carbonate . When vanadium carbonate dissolves in water, two vanadium ions and three carbonate ions are produced:
Therefore, we must raise the concentration of vanadium ions to the power of two and raise the concentration of carbonate ions to the power of three, giving us our final equation:
Questions related to either rely on calculating values of from given concentrations or on calculating concentrations and related pieces of information from values of . We know that, for insoluble ionic substances, the equilibrium lies very much toward the reactants, and so, unsurprisingly, the values of are very small.
For example, the value of for silver bromide is mol2⋅dm−6. Using this value, we can determine how much silver bromide, in grams, will actually dissolve in 1 dm3 (1 000 mL) of water.
The equilibrium for the dissolution of silver bromide can be written as follows:
This gives us the equation for :
We know that when the ionic substance dissolves, equal quantities of silver and bromide ions are produced. This allows us to state the following:
We can then substitute into the equilibrium equation:
And we can solve for by taking the square root:
We can therefore see that moles of silver bromide has dissolved in 1 dm−3 (1 000 mL) of water. Finally, we can use the molar mass of silver bromide (188 g/mol) and the number of moles to determine the mass that has dissolved:
The final value tells us that 0.000137 g of silver bromide is dissolved in 1 dm3 (1 000 mL) of water at 298 K.
We have just seen in the example above that the solubility product for silver bromide is mol2⋅dm−6. However, we cannot really appreciate yet what this value means in comparison to the solubility product of other ionic solids. In the table below, we can see more examples of different ionic solids, the solubility product of these solids, and the amount in grams we are able to dissolve in 1 L of water at 298 K.
|Ionic Solid||Chemical Formula||Relative Formula Mass, (g/mol)||Decreasing (mol2⋅L−6)||Solubility in Water (g/L, 298 K)|
As we can see from the table, the trend shows that as the solubility product decreases, the amount of the ionic substance that we can dissolve in a fixed volume of water also decreases. We should, however, acknowledge that the relative formula mass and the number of moles of ions in solution both play a role in the amount of the substance that can be dissolved. There may be exceptions to this general trend in the solubility product of two different chemicals of similar magnitudes.
Example 3: Calculating the Mass of Zinc Carbonate That Will Dissolve in 1000 mL of Water
Taking the solubility product of zinc carbonate to be mol2⋅L−2 at 298 K, how many grams of zinc carbonate with molar mass 125.38 g/mol will dissolve in 1 000 mL of water? Give your answer in scientific notation to 2 decimal places.
We can begin by writing out the equation for the dissolution of zinc carbonate at equilibrium:
From this, we can write out the equation for :
From this equation, we can also determine the units of to be mol2⋅L−2.
We know that at equilibrium the concentrations of zinc ions and carbonate ions are equal, and we can symbolize this as :
We can now substitute in the value given to us in the question for :
We can solve for by taking the square root:
We can therefore see that moles of zinc carbonate has dissolved in 1 000 mL of water.
Finally, we use the molar mass to calculate the grams of zinc carbonate that will dissolve:
We can also do this approach in reverse and calculate the value of from the amount of substance that has dissolved in water to create a saturated solution.
Example 4: Calculating the Value of 𝐾𝐬𝐩 for a Saturated Solution of Copper(II) Hydroxide
A saturated solution of copper(II) hydroxide contains g of in every 1 dm3 (1 L) of water. Taking the molar mass of to be 97.56 g/mol, answer the following questions.
- What is the value of without units, to two decimal places in scientific notation?
- What is the unit of
for this hydroxide?
Firstly, we can write the equation for :
In order to calculate , we firstly need to calculate the concentration of the ions at equilibrium using where is the number of moles and is the volume.
We can calculate the number of moles using the molar mass and mass of copper(II) hydroxide:
As the volume is equal to 1 dm3, then the concentration is simply equal to . Therefore, the concentration of ions at equilibrium is mol.
Next, we set the concentration of the ions to equal :
We know that, at equilibrium, the concentration of hydroxide ions is double the concentration of copper ions:
We equivalate this to and simplify it as follows:
We then solve for using the concentration we calculated earlier for :
If we look at the equation for , we can see that there are three instances of concentration: . Each set of square brackets indicates a concentration in mol⋅dm−3. As such, we have , which, when simplified, gives us mol3⋅dm−9.
For silver bromide, it was necessary for us to use the square root operation to resolve and determine the concentration. However, for some ionic substances that contain more than one anion or cation, more complicated roots may be necessary.
Consider an aqueous solution of aluminum hydroxide. If we know that the value of , how can we calculate the concentration of in a saturated solution of aluminum hydroxide?
Initially, we would write out the equation for the equilibrium and :
We can then substitute in :
We can then substitute in our value for and solve for :
We are now able to solve for to give the concentration of ions in the saturated solution:
- The solubility product is a heterogeneous equilibrium constant.
- The solubility product can be defined as the product of the concentrations of the ions in a saturated solution raised to the power of their respective stoichiometric coefficients.
- When the solubility product equals the product of the ion concentration, the solution is saturated.
- Solubility products are temperature dependent with values usually quoted at 298 K.
- Certain ionic substances may necessitate the use of cubed roots or fourth roots if calculating the concentration from .