The common-ion effect | Equilibrium | AP Chemistry | Khan Academy

The presence of a common ion can affect a solubility equilibrium. For example, let's say we have a saturated solution of lead(II) chloride. Lead(II) chloride is a white solid. So, here's the white solid on the bottom of the beaker, and the solid is at equilibrium with the ions in solution, which would be Pb²⁺ and Cl⁻. Notice how the mole ratio is 1:2 of Pb²⁺ to 2 Cl⁻. Therefore, if we have two Pb²⁺ ions in our diagram, there should be twice as many chloride anions at equilibrium. The rate of dissolution is equal to the rate of precipitation; therefore, the concentration of ions in solution remains constant. So, our system is at equilibrium.

Now, let's add some solid potassium chloride. Potassium chloride is a soluble salt, so it will dissociate and turn into K⁺ and Cl⁻ in solution. Adding a source of chloride anion means the system is no longer at equilibrium. So, let me write in here: not at equilibrium at that moment in time. The system was at equilibrium, and a stress was added to the system. In this case, the stress was increased chloride anion.

So, there's an increase in the concentration of Cl⁻. According to Le Chatelier's principle, the system will move in the direction that decreases the stress. If the stress is increased concentration of chloride anion, the system will move to the left to get rid of some of that extra chloride anion. When the system moves to the left, Pb²⁺ ions will combine with chloride anions to form PbCl₂, and we can see that down here in the diagram.

So, imagine, say, this Pb²⁺ ion combined with these two chloride anions to form some more of the white solid. Looking at the third diagram, the amount of white solid has increased from the second diagram, and we've lost this Pb²⁺ ion and these two chloride anions. The amount of our precipitate PbCl₂ will keep forming until equilibrium is reached.

Let's just say this third diagram does represent the system at equilibrium. So, I'll write in here: at equilibrium. This is an example of the common ion effect. For this problem, the common ion is the chloride anion because there were two sources of it. One was from the dissolution of PbCl₂. If we had dissolved some solid to make a saturated solution, the source of these chloride anions would be from PbCl₂, and the second source is from the added KCl, which of course dissolves to form chloride anion.

So, the chloride anion is the common ion, and we use Le Chatelier's principle to predict the system will move to the left to get rid of the extra chloride anion. When the system moved to the left, we formed more of the solid PbCl₂, and that's why this amount got bigger over here. If we compare the first diagram with the third diagram, the first diagram has more of the lead(II) chloride in solution, and the third diagram has less of it. Therefore, the addition of the common ion of chloride anion decreased the solubility of lead(II) chloride.

So, the common ion effect says that the solubility of a slightly soluble salt like lead(II) chloride is decreased by the presence of a common ion. Another way to think about this is using the reaction quotient Q. For the diagram on the left, we're at equilibrium; therefore, the reaction quotient Qsp is equal to the Ksp value for lead(II) chloride, which means the system is at equilibrium. Adding chloride anion increases the value for Qsp. So now Qsp is greater than Ksp, and the system is not at equilibrium.

To decrease the value for Q, the system needs to move to the left. The system will continue to move to the left until Qsp is equal to Ksp again, and the system is at equilibrium. A shift to the left means an increase in the amount of PbCl₂, which therefore decreases the solubility of PbCl₂, but it doesn't change the value for Ksp. Ksp for PbCl₂ stays the same at the same temperature.

Next, let's see how the presence of a common ion affects the molar solubility of lead(II) chloride. To do that, let's calculate the molar solubility of lead(II) chloride at 25 degrees Celsius in a solution that is 0.10 M in potassium chloride. The Ksp value for lead(II) chloride at 25 degrees Celsius is 1.7 times 10 to the negative fifth.

To help us calculate the molar solubility, we're going to use an ICE table where I stands for the initial concentration, C is the change in concentration, and E is the equilibrium concentration. First, let's say that none of the lead(II) chloride has dissolved yet. If that's true, the concentration of lead(II) ions would be zero, and the concentration of chloride anions from the lead(II) chloride would also be zero. However, there's another source of chloride anions, and that's because our solution is 0.10 M in KCl.

KCl is a soluble salt, so KCl dissociates completely to turn into K⁺ and Cl⁻. Therefore, if the concentration of KCl is 0.10 M, that's also the concentration of Cl⁻ from the KCl. So we can add here: plus 0.10 M, thinking about that as being from our KCl. There are going to be two sources of chloride anions here, and so the Cl⁻ anion is our common ion. The other source of chloride anion is PbCl₂ when it dissolves.

So, some of the PbCl₂ will dissolve; we don't know how much, so I like to write minus X in here. If some of that dissolves, the mole ratio of PbCl₂ to Pb²⁺ is a 1:1 mole ratio. So for losing X for PbCl₂, we're gaining X for Pb²⁺. Looking at our mole ratios, if we're gaining X for Pb²⁺ and it's a 1:2 mole ratio, we would write in here plus 2X for the chloride anion.

So for the equilibrium concentration of Pb²⁺, it would be 0 plus X or just X. For the equilibrium concentration of the chloride anion, it would be 0.10 plus 2X. The 0.10 came from the potassium chloride, and the 2X came from the dissolution of lead(II) chloride. Next, we need to write a Ksp expression, which we can get from the dissolution equation.

So, Ksp is equal to the concentration of lead(II) ions raised to the first power times the concentration of chloride anions. Since there's a 2 as a coefficient in the balanced equation, we need to raise that concentration to the second power. Pure solids are left out of equilibrium constant expressions, so we don't write anything for PbCl₂.

Next, we plug in our equilibrium concentrations. For lead(II) ions, the equilibrium concentration is X, and for the chloride anion, the equilibrium concentration is 0.10 plus 2X. We're also going to plug in the Ksp value for lead(II) chloride here. We have the Ksp value plugged in, X, and 0.10 plus 2X.

Now, let's think about 0.10 plus 2X for a second here. With a very low value for Ksp, 1.7 times 10 to the negative fifth, that means that not very much of the PbCl₂ will dissolve. If that's true, X is a pretty small number, and if X is a small number, 2X is also pretty small. So, we're going to make an approximation and say that 0.10 plus a pretty small number is approximately equal to just 0.10, and that's going to make the math easier on us.

So, instead of writing 0.10 plus 2X squared, we just have 0.10 squared. Solving for X, we find that X is equal to 0.0017, which we could just write as 1.7 times 10 to the negative third molar. It's okay to write molar here because this X value represents the equilibrium concentration of Pb²⁺, and if that's the equilibrium concentration of Pb²⁺, that's also the concentration of lead(II) chloride that dissolved.

So, this number, this concentration, is the molar solubility of lead(II) chloride in a solution at 25 degrees where the solution is 0.10 M in KCl. Most textbooks leave this minus X out of their ICE tables because the concentration of a solid doesn't change. I like to just leave it in here, though, to remind me that X represents the molar solubility of the slightly soluble salt.

Finally, if we calculate the molar solubility of lead(II) chloride without the presence of a common ion, this 0.10 would have been gone from everything. Doing the math that way, we would have found that the molar solubility at 25 degrees Celsius, using this value for the Ksp, the molar solubility comes out to 0.016 M.

So, comparing these two molar solubilities, 0.016 M versus 0.0017 M, that's approximately a factor of ten. Therefore, the addition of a common ion decreased the solubility by approximately a factor of 10. So, doing the common ion effect in a quantitative way also shows a decrease in the solubility of a slightly soluble salt because of the presence of a common ion.