Solubility Product

Solubility product: Defining solubility

Some ionic compounds, such as common table salt, are definitely soluble.

Other ionic species, however, seem to be insoluble. Take the calcium phosphate in your teeth - it's another example of an ionic compound, and luckily for us, it doesn’t tend to dissolve in water. But in fact, no ionic species is completely insoluble. All ionic species dissolve in solution to some extent. For example, calcium phosphate becomes much more soluble in acidic solutions, which is why dentists advise you to limit your intake of fizzy drinks - they're full of phosphoric acid.

We can measure how soluble or insoluble an ionic compound is in a more quantifiable way using solubility.

Solubility tends to be measured in grams of solute per 100 grams (or sometimes 1 kilogram) of water. However, we can also measure it in moles per 100 grams (or 1 kilogram) of water.

It is hard to assign exact solubility values to the terms soluble and insoluble. However, as a general rule:

• If more than 1g of a species dissolves in 100g of water, we say it is soluble.

• If between 0.1g and 1g of a species dissolves in 100g of water, we say it is slightly soluble.

• If less than 0.1g of a species dissolves in 100g of water, we say it is insoluble.

Saturated solutions

In the definition of solubility, we mentioned saturating a solution.

Saturation relates to solubility in a simple way:

• If the amount of a solute in a solution is less than the solute’s solubility, the solution is unsaturated. The solute dissolves into aqueous ions.

• If the amount of solute in a solution is equal to the solute’s solubility, the solution is saturated. If you add more solute, it remains as a solid and doesn’t dissolve into aqueous ions.

• If you add a large amount of a fully soluble salt to the water, you might not form a saturated solution - soluble salts have a high solubility and so typically all of the solid dissolves. However, if you add the same amount of a slightly soluble salt to water, you quickly saturate the solution. Slightly soluble salts have a low solubility, so some of the salt doesn't dissolve. It instead remains as a solid.

Fig. 1 - Soluble and slightly soluble species. Soluble compounds typically dissolve entirely in solution, whilst slightly soluble compounds quickly form saturated solutions

But that final sentence is perhaps an oversimplification. Once you reach saturation, there is no net dissolution of solid ionic species. However, that doesn’t mean that saturated solutions are static. Instead, they are examples of dynamic equilibria, in which the processes of dissolving and precipitation continuously take place. We call these equilibria solubiilty equilbria. Let’s consider them next.

Solubility product: Solubility equilibria

If you add more of an ionic compound to a saturated solution, overall, there is no net dissolution. However, at an atomic level, both dissolution and precipitation are ongoing. The solid ionic compound dissolves into aqueous ions in solution as aqueous ions simultaneously precipitate to reform the solid ionic compound. The rate of dissolution equals the rate of precipitation, and so the overall amount of solute dissolved in the solution remains constant.

Sound familiar? What we are describing here is a type of dynamic equilibrium. We call it a solubility equilibrium.

In solubility equilibria, the forward reaction is dissolution whilst the backward reaction is precipitation. Hence, dissolution and precipitation are two sides of the same reversible reaction.

We can represent the solubility equilibrium for the dissolution of the general ionic compound A_aB_b in contact with its own saturated solution with the following equation:

$$A_aB_b(s)\rightleftharpoons aA^{b+}(aq)+bB^{a-}(aq)$$

Solubility equilibria follow all the rules of standard equilibria. For example:

• At equilibrium, the rate of the forward reaction (dissolution) equals the rate of the backward reaction (precipitation).
• At equilibrium, the concentrations of products and reactants don't change.
• The equilibrium concentrations of products and reactants can be represented using an equilibrium constant, which has a fixed value at a certain temperature.
• The equilibrium always tends towards a certain position. This position can be changed by changing external factors.

Let's now explore some of the ideas mentioned above in more detail. We'll start with the equilbrium constant used for solubility equilibria: the solubility product.

Solubilty product: Definition and principle

All dynamic equilibria are characterised by a certain proportion of products to reactants in the system at equilibrium. We measure this ratio using an equilibrium constant.

Well, solubility equilibria are no different. Their particular equilibrium constant is known as the solubility product.

Note the following:

• The solubility product is based on equilibrium concentrations.
• Like with other equilibrium constants, we don't include the concentrations of pure solids or liquids in the solubility product expression. This is because their concentrations don't change.
• Again, like with other equilibrium constants, the solubility product is constant for a certain reaction at a specific temperature. However, if you change the temperature, you change the value of the solubility product.

Ready to try writing an expression for the solubility product? Let's look at that now.

Solubility product: Formula

Try writing a solubility product expression for the dissolution reaction equation we gave you above. It is much like the expression for the more common equilibrium constant K_c.

With any luck, you’ll end up with the following:

$$K_{sp}={[A^{b+}]_{eqm}}^a\space {[B^{a-}]_{eqm}}^b$$

Remember that here, square brackets represent concentration, whilst the small letters eqm show that we take the measurement at equilibrium.

Confused as to how we got to that formula? Simply put, to find an expression for the solubility product:

1. Write a balanced solubility equilibrium equation.
2. Take the equilibrium concentration of each of the aqueous ion products.
3. Raise each equilibrium concentration to the power of its molar coefficient in the balanced equilibrium equation.
4. Multiply these terms together.

We'll look at an example of writing solubility product expressions in just a second. But before we do that, we need to learn about the solubility product's units.

Solubility product units

The solubility product's units differ depending on the reaction equation itself. You find them in much the same way as you find the units for any other equilibrium constant:

1. Substitute the units you used for the equilibrium concentrations of aqueous ions into each term in the solubility product expression. If not given, the typical units of concentration are mol dm^-3.
2. Expand each term according to its exponential, then simplify the expression to find your final units.

Let's put everything we just learned together and have a go at writing solubility product expressions for particular ionic species.

Here's another example. In this one, you need to write an equilibrium equation yourself.

Uses of the solubility product

It is all well and good knowing what the solubility product is, but how is it helpful?

Well, the solubility product has a few uses. In particular:

• The solubility product gives us an indication of solubility. In general, the higher the value of the solubility product, the greater the solubility of the ionic species.
• We can use the solubility product to quantitatively calculate a slightly soluble species' solubility.
  • We typically count species with a solubility product less than 1.0 as slightly soluble.
  • Species with a solubility product greater than 1.0 are instead considered to be fully soluble. Solubility product values aren't really relevant for these species and so we don't tend to use them in solubility calculations.
• We can also use the solubility product to predict whether a salt will precipitate or not if we change the equilibrium conditions. We'll look at that next.

Solubility product and the common ion effect

The solubility product is an equilibrium constant. The clue is in the name - it is constant for a certain reaction at a specific temperature. We can change other factors, such as pH and the concentration of ions, but the solubility product doesn't change. Instead, the position of the solubility equilibrium changes in order to keep the solubility product the same. This changes the ionic species' solubility.

Let's explore the idea of changing solubility by changing the concentration of a particular ion in solution. Adding a common ion to a saturated solution of a slightly soluble salt causes more of the salt to precipitate out of solution, thus decreasing its solubility. This is known as the common ion effect.

For example, consider what happens when you add aqueous MgSO₄ to a saturated solution of MgCO₃.

• MgSO₄ is fully soluble. This means that the first solution, which we'll call A, contains a high concentration of Mg²⁺ and SO₄^2- ions.
• However, MgF₂ is only slightly soluble. This means that the second solution, which we'll call B, contains a low concentration of Mg²⁺ and CO₃^2- ions. Because the solution B is saturated, you'll find the aqueous ions alongside undissolved solid MgCO₃. We can represent the solubility equilibrium with the equation \(MgCO_3(s)\rightleftharpoons Mg^{2+}(aq)+CO_3^{2-}(aq)\)
• Notice that Mg²⁺ is a common ion - it is found in both solutions.
• If we add solution A to solution B, we increase the concentration of Mg²⁺ ions in solution. This disturbs the equilibrium. According to Le Chatelier's principle, the position of the solubility equilibrium shifts to the left, causing additional MgCO₃ to precipitate out of solution in order to offset the disturbance and restore equilibrium. As a result, the solubility of MgCO₃ decreases.

We can also look at the common ion effect in terms of the solubility product. Adding A to B increases the concentration of Mg²⁺. If we were to substitute this new concentration into the expression for the solubility product, we would find that the solubility product increases. However, we know that the solubility product is constant for a certain reaction at a specific temperature - it can't change unless we change the temperature. Therefore, the system reacts by decreasing the overall concentration of the ions involved in the solubility equilibrium (Mg²⁺ and CO₃^2-) by turning them into solid MgCO₃. In this way, the solubility product can be used to predict whether a salt will precipitate out of solution or not.

Fig. 2 - a) A system is at equilibrium.b) Adding a common ion increases the concentration of dissolved solute in solution and disturbs the equilibrium.c) The position of the equilibrium shifts to the left, causing more of the dissolved solute to precipitate out of solution in order to restore equilibrium. Thus, the solute's solubility decreases

I hope you are comfortable with solubility products and their use now. If you want to take a look at here are some Solubility "Product Calculations" for you to study.