You’re probably all familiar with crystallisation experiments - they’re simple and visually exciting examples of chemical reactions. Mix an acid with a base, evaporate off most of the water, leave it to dry in the open air, and then you can watch a unique and colourful salt crystal grow before your eyes.
There is actually some important chemistry behind crystal formation. By evaporating the water, you increase the concentration of dissolved aqueous salt ions, and this causes the salt to precipitate into solid crystals. The relationship between aqueous ion concentration and precipitation is explained by something called the solubility product.
- This article is about the solubility product in physical chemistry.
- We'll start by looking at solubility as a whole. This will include defining soluble and saturated solution.
- We'll then explore how slightly soluble ionic species form solubility equilibria when in contact with their own saturated solution.
- After that, we'll learn about the solubility product, which is the equilibrium constant for solubility equilibria.
- You'll find out how to write a formula for the solubility product and work out its units.
- Finally, we'll see how the solubility product relates to changing equilibrium conditions, such as dictated by the common ion effect.
Solubility product: Defining solubility
Some ionic compounds, such as common table salt, are definitely soluble.
A soluble substance is one that is able to dissolve in water. We call the substance in question a solute.
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 is the maximum amount of a species that can dissolve in a given volume of solvent. In other words, it is the amount of a compound needed to completely saturate a solution.
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.
100g of water is equivalent to 1 decilitre. Likewise, 1 kilogram of water equals 1 litre, which again is equivalent to 1 decimetre cubed. Therefore, solubility often takes the units g dL-1, g L-1, mol dL-1, mol L-1., or mol dm-3. In solubility calculations at A level, you'll probably use mol dm-3.
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.
Another term for dissolving is dissolution. The opposite of dissolving is precipitation.
Saturated solutions
In the definition of solubility, we mentioned saturating a solution.
A solution is saturated when no more of a solute can dissolve in it.
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.
Solubility equilibria are equilibria formed between all ionic salts that are in contact with their own saturated solution.
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 AaBb 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.
Never encountered equilibria before? Or perhaps you simply want a quick reminder of their properties. Either way, head over to Chemical Equilibrium for a detailed look at these chemical systems.
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.
The solubility product is the equilibrium constant for the dissolution of an ionic species in a solubility equilibrium. It tells you about the relative concentration of aqueous ions in a dissolution reaction at equilibrium.
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.
You can find out all about equilibrium constants, including the more common equilibrium constant Kc, over at Equilibrium Constants. You might want to have a quick look at that article anyway to refresh your knowledge on equilibrium constant expressions before we move on to the next section.
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 Kc.
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:
- Write a balanced solubility equilibrium equation.
- Take the equilibrium concentration of each of the aqueous ion products.
- Raise each equilibrium concentration to the power of its molar coefficient in the balanced equilibrium equation.
- 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.
You might remember from
Equilibrium Constants that the expression for K
c is a fraction - it features the
equilibrium concentrations of any reactants in the bottom half. However, in
solubility equilibria, the reactant is always an
ionic solid. We never include any pure solids or liquids in equilibrium constant expressions because
their concentration always remains constant. Therefore, we leave the ionic solid reactant out of the expression for K
sp.
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:
- 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.
- 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.
Find the solubility product expression for the following solubility equilibrium, then work out its units:
$$CuCl(s)\rightleftharpoons Cu^+(aq)+Cl^-(aq)$$
To find an expression for the solubility product (Ksp), we first need an equilibrium equation for the reaction. Luckily, this has been given to us. We then take the equilibrium concentrations of the aqueous ion species, raise them to the power of their molar coefficients in the balanced equation, and multiply the two terms together. Here, the ion products are Cu+ and Cl-. They both have a molar coefficient of 1, and so we raise their equilibrium concentrations to the power of 1. Raising something to the power of 1 doesn't change its value. Therefore, we end up with the following expression:
$$K_{sp}=[Cu^+]_{eqm}\space [Cl^-]_{eqm}$$
To calculate the necessary units, substitute the units of concentration used into the solubility product expression and expand out the brackets:
$$units=(mol\space dm^{-3})\space (mol\space dm^{-3})$$ $$units=mol^2\space dm^{-6}$$
Here's another example. In this one, you need to write an equilibrium equation yourself.
Find the solubility product expression and its units for the solubility equilibrium formed between solid calcium fluoride and its own saturated solution.
First, we need to write a balanced equilibrium equation. Calcium fluoride has the formula CaF2. One mole of CaF2 dissolves into one mole of positive Ca2+ ions and two moles of negative F- ions:
$$CaF_2(s)\rightleftharpoons Ca^{2+}(aq)+2F^-(aq)$$
That means that in the solubility product expression, we raise [Ca2+] to the power of 1 and [F-] to the power of 2:
$$K_{sp}=[Ca^{2+}]_{eqm}\space {[F^-]_{eqm}}^2$$
If we substitute the units of concentration into each term of the expression and expand all of the brackets, we arrive at our final units for the solubility product:
$$units=(mol\space dm^{-3})\space (mol\space dm^{-3})^2$$ $$units=mol^3\space dm^{-9}$$
Of course, these are just expressions for the solublity product. To find the value of the solubility product itself, you need to substitute the equilibrium concentrations of dissolved ions into the solubility product expression you just worked out, then expand all of the brackets to reach your final answer. We'll show you how to do this (alongside other mathematical problems involving solubility equilibria) in Solubility Product Calculations.
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.
Working out solubility from the solubility product is another example of solubility equilibrium calculations, which are worthy of their own article - head over to Solubility Product Calculations for plenty of worked examples and handy tips.
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.
Shifting the position of an equilibrium - you should recognise this phrase from Le Chatelier's Principle. Now would be a great time to look over that article again if you feel a little rusty on the subject, as a thorough understanding of Le Chatelier's principle will help you make sense of the following ideas.
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.
A common ion is an ion found in two different compounds. The common ion effect states that ionic species are less soluble in solutions containing a common ion.
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 SO42- 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 CO32- 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 CO32-) 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
The same principle can be applied to explain the process of crystallisation. Evaporating the water from a saturated solution increases the concentration of aqueous solute ions until it exceeds the value of the solubility product, bringing the system out of equilibrium. In order to keep the solubility product the same, some of the aqueous solute ions precipitate into a solid, thus decreasing their concentration and bringing the system back to equilibrium.
Fig. 3 - a) A system is at equilibrium.b) Evaporating water increases the concentration of dissolved solute ions, bringing the system out of equilibrium. c) Le Chatelier's principle tells us that the position of the equilibrium shifts to the left to oppose the disturbance. This causes more of the solute to precipitate and restores equilibrium
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.
Solubility Product - Key takeaways
- A soluble species is one that is able to dissolve in solvent.
- All systems containing a solid ionic salt in contact with its own saturated solution form a dynamic equilibrium known as a solubility equilibrium.
- We can use an equilibrium constant known as the solubility product (Ksp)to show the relative concentration of aqueous ions in a solubility equilibrium.
- For the equilibrium \(A_aB_b(s)\rightleftharpoons aA^{b+}(aq)+bB^{a-}(aq)\), the solubilty product has the expression \(K_{sp}={[A^{b+}]_{eqm}}^a\space {[B^{a-}]_{eqm}}^b\)
- The units of the solubility constant vary depending on the reaction equation.
- The solubility product is constant for a certain reaction at a specific temperature.
- The solubility product helps explain the common ion effect: Ionic species are less soluble in solutions containing a common ion.
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