Common-Ion Effect

Common Ion Effect Definition

Let's begin by looking at the definition of Le Chatelier's principle, and at the definition of common ion effect.

Le Chatelier’s principle and the common ion effect are directly related. To understand the common ion effect, we must first understand Le Chatelier’s principle. Above, we’ve written the basic importance and definition of Le Chatelier’s principle for this very reason.

To facilitate an even better understanding of this article, we will go over the basics of Le Chatelier below:

Disturbance in this case means a change in the system’s conditions. Which include:

1. change in temperature by cooling or heating

2. change in concentration by adding or removing a reactant/product

3. change in pressure by adding a gas or decreasing the volume of a reaction

The chemical equilibrium we refer to is the overall net reaction. When a reaction is in equilibrium, the reaction rate of the reactants and products are equivalent.

If a reaction is not in equilibrium, then Le Chatelier states that the reaction will shift back to equilibrium by adjusting the concentrations of reactants or products in order to re-establish the equilibrium ratio.

1. When we increase the concentration of the reactants, the equilibrium shifts to the right and the forward reaction is favored, resulting in more product being made until equilibrium is reached again.
2. When we increase the concentration of the products, the equilibrium shifts to the left and the backward reaction is favored, resulting in more reactants being made until equilibrium is reached again.

In the illustration above, increasing the reactants causes a shift in the equilibrium to the other side until the balance is re-established by the production of more products. Keep in mind that if we were to add more products instead of reactants the equilibrium would shift towards the products until more reactants are made to re-establish balance.

Now that we’ve had a brief look at Le Chatelier’s principle, we can move on to defining our common ion effect.

Common Ion Effect on Solubility

Since we’ve defined the common ion effect as a shift in equilibrium, the next question to ask would be how does this shift in equilibrium affect our reaction?

The simple answer to that would be that the shift in equilibrium caused by adding a common ion impacts solubility.

This is because when we add a common ion, the reaction shifts towards the reactants (left) to balance out the excess product, resulting in more precipitation for a saturated solution.

Figure 2: The Common-ion effect illustrated. Daniela Lin, Study Smarter Originals.

In the graphic above, KCl’s solubility in water is greater than its solubility in an aqueous NaCl solution, as KCl and NaCl contain common ions (Cl).

Common Ion Effect on Solubility of Ionic Salts

After several experiments involving ionic salts, scientists have obtained patterns in data that they have coined as solubility rules.

It’s important for us to be aware of these general trends in order to, later on, calculate the molar solubility of common ions.

Figure 3: Solubility table for different salts. The red asterisk indicates exceptions, which are shown in the image below. Daniela Lin, Study Smarter Originals.

Figure 4: Exceptions for solubility of ionic salts shown. Daniela Lin, Study Smarter Originals.

In particular, when an excess of a slightly soluble salt dissolves in water, this creates an equilibrium between the solid salt and salt ions in the saturated solution. Another important thing to know about solubility is the K_sp.

At a given temperature, the solubility product constant, \(K_{sp}\) , maintains a constant numerical value even when the concentration of the salt changes. Thus, the higher the \(K_{sp}\) the more soluble the compound is. It isn’t necessary to memorize the solubility tables given above. Instead, familiarize yourself with them and understand that those that form insoluble salts will have lower \(K_{sp}\) ’s and vice-versa for more soluble salts.

This also implies that adding a salt that has a common ion with a given slightly soluble salt (solute) would lead to lower solute solubility and the formation of precipitate sooner than that predicted by the solubility product constant, \(K_{sp}\), for the pure slightly soluble salt in water (solvent) alone. Recall that the \(K_{sp}\) , is calculated for a solute dissolved into a pure solvent.

Molar Solubility and the Common Ion Effect

Now, let's talk about molar solubility and the common ion effect.

We noted that, once the solution becomes saturated, the solute will no longer dissolve. As mentioned above we usually use the \(K_{sp}\), the product solubility constant, to describe the dissolution of a slightly soluble salt in an aqueous solution. Also, solids are usually not included because their change in concentrations doesn't affect the \(K_{sp}\) since they don't dissolve.

On the other hand, molar solubility describes the dissolution of a substance in a liter of solution. The unit for molar solubility is mol/L or M.

This means that the product solubility constant is not the same thing as the molar solubility.

Consider the equation:

$$X_2Y\left(s \right) \rightarrow X^+\left(aq \right)+Y^-\left(aq \right)$$

where, X₂Y (s), stands for a salt of interest, the cation of the salt is, X ⁺, and the anion of the salt is, Y ^-. Let's say the final concentration of the cation is, [X ⁺] = 0.5 M, then the concentration of the anion would then be, [Y ^- ] = 0.25 M (because, in this salt, X₂Y, there are 2 moles of, X ⁺, for every one mole of, Y ^- ). As such, the product solubility constant would be given by:

$$K_{sp}=\left[X^+ \right]^2\left[Y^- \right]=\left[0.5 \right]^2\left[0.25 \right]=0.0625$$

While the molar solubility of the hypothetical salt, X₂Y, that we added to the solution would have just been, [X₂Y] = 0.25 M.

Common Ion Effect Examples

Now that we have read about what \(K_{sp}\) is, how the common ion effect affects the solubility of ionic salts and some general solubility rules, we can move on to studying some common ion effect examples.

You’ve reached the end of this article. We have reviewed the basics of Le Chatelier’s principle and related it to the common ion effect. Furthermore, we have learned about the molar solubility of ions, solubility rules, and gone over some examples to promote more understanding.

For more exercises, please head over to the flashcards!