15.3 Common Ion Effect

In the last unit, we were exploring the solubilities of different salts in pure water. However, when salts are added to a solution that has a common ion, solubilities can change drastically. Remember if Qsp > Ksp a precipitate will form in the solution. Take an example of NaCl solution.

NaCl will dissociate and produce Na⁺ and Cl^– ions.

NaCl                  Na⁺ (aq)  +  Cl^– (aq)

If we try dissolving PbCl₂ in this NaCl solution, an equilibrium for dissolution of PbCl₂ will establish like this.

PbCl₂ (s)                           Pb⁺² (aq) + 2Cl^–(aq)

Since NaCl is highly soluble in water, it will dissociate completely and produce lot of Cl- ions in the solution. So, if you apply Le Chatelier’s principle here, a high concentration of chloride ion coming from dissociation of NaCl will shift the equilibrium for dissolution of PbCl₂ to the left side. In other words, you can say less PbCl₂ will be able to dissolve. We can calculate the solubility of a salt in a solution with the help of the Ksp value.

Example 6: How much CaF₂ can be dissolved in 0.15 M NaF solution. Given the Ksp for CaF₂ is 4.0 x 10^-11.

Since NaF is highly soluble in water, we will begin with the dissolution of NaF in the solution.

NaF (s)                 Na⁺ (aq)  +  F^– (aq)

Based on the mole ratio from the reaction above, we can say the concentration of Na⁺  and F^– ions in the solution will be 0.15 M.

[Na⁺] = [F^–] = 0.15 M

Next, CaF₂ will dissociate in the solution as per the reaction below.

CaF₂ (s)              Ca⁺² (aq)  + 2 F^–  (aq)

Here some fluoride ions are already present in the solution, due to the dissolution of NaF. However, Calcium ions will come solely from the dissolution of CaF₂.

Since the mole ratio of Ca⁺² ions and CaF₂ is 1: 1, we can assume that the concentration of Ca⁺² ions in solution will be the same as the solubility of CaF₂ in solution. However, F^– ions’ concentration coming from dissociation of CaF₂ in solution will be twice the solubility of CaF₂, as the mole ratio of F^–: CaF₂ is 2:1. We can set up an ICE chart and solve the equilibrium concentration of Ca⁺² ions and F^– ions.

Reaction	CaF2 (s)	Ca+2 (aq)	2 F–  (aq)
Initial		0	0.15 M
Change		x	2x
Equilibrium		x	0.15 + 2x

Ksp = [Ca⁺²] [F^–]²

4.0 x 10^-11 = (x) (0.15 + 2x)²

Assuming 5% rule will apply in this case, we can assume 2x is much smaller than 0.15, hence can be ignored.

4.0 x 10^-11 = (x) (0.15 )²

x =

x = 1.8 x 10^-9 M                     

(Comparing the value of x with 0.15, we can say our assumption was correct and 5% rule was applicable).

This value of x = 1.8 x 10^-9 M represents the solubility of CaF₂ in 0.15 M NaF solution.

Look at the video below to see more examples of the common ion effect.

https://www.youtube.com/watch?v=u3GL0faEcxc

Learning Check:

1. Find the solubility of BaF₂ in 0.15 M NaF solution. Ksp for BaF₂ = 1.6 x 10^-6
2. Find the maximum concentration of AgNO₃ in 0.15 M NaCl solution without having any precipitation. Ksp for AgCl = 1.7 x 10^-10

Answer: 9) 7.11 x 10^-5 mol/L             10) 1.13 x 10^-9 mol/L