Chemistry » Equilibria of Other Reaction Classes » Precipitation and Dissolution

# Common Ion Effect

## Common Ion Effect

As we saw when we discussed buffer solutions, the hydronium ion concentration of an aqueous solution of acetic acid decreases when the strong electrolyte sodium acetate, NaCH3CO2, is added. We can explain this effect using Le Châtelier’s principle. The addition of acetate ions causes the equilibrium to shift to the left, decreasing the concentration of $${\text{H}}_{3}{\text{O}}^{\text{+}}$$ to compensate for the increased acetate ion concentration. This increases the concentration of CH3CO2H:

$${\text{CH}}_{3}{\text{CO}}_{2}\text{H}+{\text{H}}_{2}\text{O}⇌{\text{H}}_{3}{\text{O}}^{\text{+}}+{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}$$

Because sodium acetate and acetic acid have the acetate ion in common, the influence on the equilibrium is called the common ion effect.

The common ion effect can also have a direct effect on solubility equilibria. Suppose we are looking at the reaction where silver iodide is dissolved:

$$\text{AgI}(s)⇌{\text{Ag}}^{\text{+}}(aq)+{\text{I}}^{\text{−}}(aq)$$

If we were to add potassium iodide (KI) to this solution, we would be adding a substance that shares a common ion with silver iodide. Le Châtelier’s principle tells us that when a change is made to a system at equilibrium, the reaction will shift to counteract that change. In this example, there would be an excess of iodide ions, so the reaction would shift toward the left, causing more silver iodide to precipitate out of solution.

### Resource:

View this simulation to see how the common ion effect works with different concentrations of salts.

## Example

### Common Ion Effect

Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr2). The Ksp of CdS is 1.0 $$×$$ 10–28.

### Solution

The first thing you should notice is that the cadmium sulfide is dissolved in a solution that contains cadmium ions. We need to use an ICE table to set up this problem and include the CdBr2 concentration as a contributor of cadmium ions:

$$\text{CdS}(s)⇌{\text{Cd}}^{\text{2+}}(aq)+{\text{S}}^{\text{2−}}(aq)$$

$${K}_{\text{sp}}=[{\text{Cd}}^{\text{2+}}][{\text{S}}^{\text{2−}}]=1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-28}$$

$$(0.010+x)(x)=\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-28}$$

$${x}^{2}+\text{0.010}x-\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-28}=0$$

We can solve this equation using the quadratic formula, but we can also make an assumption to make this calculation much simpler. Since the Ksp value is so small compared with the cadmium concentration, we can assume that the change between the initial concentration and the equilibrium concentration is negligible, so that 0.010 + x ~ 0.010. Going back to our Ksp expression, we would now get:

$${K}_{\text{sp}}=[{\text{Cd}}^{\text{2+}}]{[\text{S}}^{\text{2−}}]=\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-28}$$

$$(0.010)(x)=\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-28}$$

$$x=\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-26}$$

Therefore, the molar solubility of CdS in this solution is 1.0 $$×$$ 10–26M.