7.12 The Common Ion Effect

In the last section, you learned about Ksp, the solubility product for a dissolution. However, we only discussed solubility equilibria in water. What would happen if a dissolution took place in another solution? Specifically, what if a solute was added to a solution that included an ion that the solute dissolves into? This concept brings into discussion the common-ion effect.

The common-ion effect describes how a common ion can suppress the solubility of a substance. For example, if we try dissolving AgBr in a solution of NaBr, there will already be some concentration of the Bromite ion (Br-) present in the system. Therefore we know that our solubility equilibrium will adjust by producing more reactants. This is explained by thinking about Le Chatelier’s Principle. The Common-Ion effect, like other situations where a system is given a stress, is just another example of Le Chatelier’s Principle that can be justified with Q. Let’s take a look:

We know that for a reaction AB (s) ⇌ A+(aq) + B-(aq), Ksp = [A][B]. If we have a solution in which we have some initial concentration of say, B, we can find Qsp at those conditions as being Qsp = [A][B+X] where X is the initial concentration. We know from here that Q > K and therefore the dissolution of AB is inhibited by the existence of the B- solution. This will make a bit more sense when we take a look at a real example in the next section.

Let’s start by first finding the molar solubility in water like we did in 7.11:

We know the following about AgBr: AgBr (s) ⇌ Ag+ (aq) + Br- (aq).

Next, let’s set up an ICE Box and solve to find our molar solubilities:

Ksp = [Ag+][Br-] = x^2 = 7.7*10^(-13) ⇒ x = 8.8 * 10^(-7) M

Let’s compare this molar solubility to the molar solubility in a 0.0010M solution of NaBr. We can simplify this by saying we have a 0.0010M concentration of Br- at the beginning of the reaction. Adjusting our ICE Box we find:

It follows then that: Ksp = [Ag+][Br-] = [x][0.0010+x] ≈ [x][0.0010] = 7.7*10^(-13)

Therefore, x is 7.7 * 10^(-10) M. Notice that this is a smaller value than the one found in just a solution of pure water. 

We dug a little bit into why the common ion effect is the way it is and how to calculate molar solubilities when common ions come into play, but why exactly does it work? The simple answer is Le Chatelier’s Principle! Like most other rules involving non-standard conditions in equilibrium, Le Chatelier’s Principle can help us understand why the common ion effect impacts molar solubility. As a brief review, Le Chatelier’s Principle tells us that whenever there is an external stress placed upon a system, the system will respond by either producing more reactants or producing more products in order to return to equilibrium. For example, increasing the concentration of a reactant will push equilibrium towards the products and induce a reaction to produce more products. 

Let’s apply this to the common ion effect. As we’ve established, the common ion effect kicks in when there is a presence of an ion in a solution in which the same ion is being dissolved into. For example, the common ion effect would take effect if CaSO4 (Ksp = 2.4 * 10^-5) was dissolved in a solution of CuSO4, the common ion being the sulfate ion. Applying Le Chatelier’s Principle, let’s write out the equilibrium for the dissolution of CaSO4:

CaSO4 (s) ⇌ Ca2+ (aq) + SO42- (aq)

The common ion effect can be thought of as an external stress on our system, that stress being an already existing concentration of the sulfate ion. Therefore, Le Chatelier’s Principle tells us that this will drive the reaction towards the reactants, decreasing the molar solubility. Therefore, Le Chatelier’s Principle can be used to qualitatively justify the Common Ion Effect. As we saw before, this can be qualitatively seen either through calculating Q or simply calculating the new molar solubility and observing it to be lower. However, the approach shows this in general.