# 19.15: Common Ion Effect

Lithium carbonate is an essential component of lithium batteries, which tend to be longer-lasting than regular alkaline batteries. The material is obtained from lithium ores by adding $$\ce{CO_2}$$ under high pressure to form the more soluble $$\ce{LiHCO_3}$$. The mixture is then depressurized to remove the carbon dioxide and the lithium carbonate precipitates out of solution.

## Common Ion Effect

In a saturated solution of calcium sulfate, an equilibrium exists between the solid calcium sulfate and its ions in solution.

$\ce{CaSO_4} \left( s \right) \rightleftharpoons \ce{Ca^{2+}} \left( aq \right) + \ce{SO_4^{2-}} \left( aq \right) \: \: \: K_\text{sp} = 2.4 \times 10^{-5}$

Suppose that some calcium nitrate was added to this saturated solution. Immediately, the concentration of the calcium ion in the solution would increase. As a result, the ion product of the $$\left[ \ce{Ca^{2+}} \right]$$ times the $$\left[ \ce{SO_4^{2-}} \right]$$ would increase and now be greater than the $$K_\text{sp}$$. According to Le Châtelier's principle, the equilibrium above would shift to the left in order to relieve the stress of the added calcium ion. Additional calcium sulfate would precipitate out of the solution until the ion product once again becomes equal to the $$K_\text{sp}$$. Note that in the new equilibrium the concentrations of the calcium ion and the sulfate ion would no longer be equal to each other. The calcium ion concentration would be larger than the sulfate ion concentration.

This situation describes the common ion effect. A common ion is an ion that is in common to both salts in a solution. In the above example, the common ion is $$\ce{Ca^{2+}}$$. The common ion effect is a decrease in the solubility of an ionic compound as a result of the addition of a common ion. Adding calcium ion to the saturated solution of calcium sulfate causes additional $$\ce{CaSO_4}$$ to precipitate from the solution, lowering its solubility. The addition of a solution containing sulfate ion, such as potassium sulfate, would result in the same common ion effect.

Example 19.15.1

What is the concentration of zinc ion in $$1.00 \: \text{L}$$ of a saturated solution of zinc hydroxide to which $$0.040 \: \text{mol}$$ of $$\ce{NaOH}$$ has been added?

Solution:

Step 1: List the known quantities and plan the problem.

Known

• $$K_\text{sp} = 3.0 \times 10^{-16}$$ (from table in "Conversion of $$K_\text{sp}$$ to Solubility")
• Moles of added $$\ce{NaOH} = 0.040 \: \text{mol}$$
• Volume of solution $$= 1.00 \: \text{L}$$

Unknown

• $$\left[ \ce{Zn^{2+}} \right] = ? \: \text{M}$$

Express the concentrations of the two ions relative to the variable $$s$$. The concentration of the zinc ion will be equal to $$s$$, while the concentration of the hydroxide ion will be equal to $$0.040 + 2s$$.

Step 2: Solve.

The $$K_\text{sp}$$ expression can be written in terms of the variable $$s$$.

$K_\text{sp} = \left[ \ce{Zn^{2+}} \right] \left[ \ce{OH^-} \right]^2 = \left( s \right) \left( 0.040 + 2s \right)^2$

Because the value of the $$K_\text{sp}$$ is so small, we can make the assumption that the value of $$s$$ will be very small compared to 0.040. This simplifies the mathematics involved in solving for $$s$$.

\begin{align} K_\text{sp} &= \left( s \right) \left( 0.040 \right)^2 = 0.0016s = 3.0 \times 10^{-16} \\ s &= \frac{K_\text{sp}}{\left[ \ce{OH^-} \right]^2} = \frac{3.0 \times 10^{-16}}{0.0016} = 1.9 \times 10^{-13} \: \text{M} \end{align}

The concentration of the zinc ion is equal to $$s$$ and so $$\left[ \ce{Zn^{2+}} \right] = 1.9 \times 10^{-13} \: \text{M}$$.

The relatively high concentration of the common ion, $$\ce{OH^-}$$, results in a very low concentration of zinc ion. The molar solubility of the zinc hydroxide is less in the presence of the common ion than it would be in water.