KEY2CHEM

Effect of Common Ions and pH on Solubility

External factors such as the presence of a common ion or changes in pH can alter the solubility of some salts.

For an ionic compound $$\require{mhchem}\ce{MX2(s)}$$, the $$K_{sp}$$ value describes the equilibrium $$\require{mhchem}\ce{MX2(s) <=> M^2+(aq) + 2 X^{-}(aq)}$$.

Adding either $$\require{mhchem}\ce{M^2+}$$ or $$\require{mhchem}\ce{X-}$$ to the equilibrium solution disturbs the equilibrium position, shifting it to the left (toward undissolved solid), as predicted by Le Chatelier’s Principle. This decreases the solubility of $$\require{mhchem}\ce{MX2}$$ since now less ionic compound is dissolved. The ions $$\require{mhchem}\ce{M^2+}$$ and $$\require{mhchem}\ce{X-}$$ are referred to as common ions, since they are present in the equilibrium equation.

Similarly, lowering the pH of the solution, by increasing [$$\require{mhchem}\ce{H3O+}$$] shifts the equilibrium position for ionic compounds containing anions that are conjugate bases of weak acids. For example, adding acid to a saturated solution of $$\require{mhchem}\ce{Mg(OH)2}$$ increases the solubility of the ionic compound. Added hydronium will react with the hydroxide ion dissolved in the solution from the equilibrium: $$\require{mhchem}\ce{Mg(OH)2(s) <=> Mg^2+(aq) + 2 OH^{-}(aq)}$$. Decreasing [$$\require{mhchem}\ce{OH-}$$] will shift the equilibrium position to the right, meaning more ionic compound dissolves.

Example 1.

Which compound will be more soluble when dissolved in $$0.1 \text{ M } \require{mhchem}\ce{HCl}$$ than when dissolved in pure water?

A. $$\require{mhchem}\ce{AgCl}$$

B. $$\require{mhchem}\ce{CaCO3}$$

C. $$\require{mhchem}\ce{CuBr}$$

Solution

B. $$\require{mhchem}\ce{CaCO3}$$

$$\require{mhchem}\ce{CO3^{2-}}$$ is a weak base that will react with $$\require{mhchem}\ce{H3O+}$$ in the acidic solution. Consuming the carbonate ion will shift the equilibrium position to the right, meaning more of the ionic compound dissolves. $$\require{mhchem}\ce{CaCO3(s) <=> Ca^2+(aq) + CO3^{2-}(aq)}$$

Example 2.

What will happen to the molar solubility of $$\require{mhchem}\ce{Mg(OH)2}$$ when $$0.1 \text{ M } \require{mhchem}\ce{Mg(NO3)2}$$ is added to a saturated solution of $$\require{mhchem}\ce{Mg(OH)2}$$?

A. Solubility will increase.

B. Solubility will remain unchanged.

C. Solubility will decrease.

Solution

C. Solubility will decrease.

The $$\require{mhchem}\ce{Mg(NO3)2}$$ added contains a common ion ($$\require{mhchem}\ce{Mg^2+}$$). Adding a common ion decreases the solubility of the ionic compound, as predicted by Le Chatelier’s Principle.  $$\require{mhchem}\ce{Mg(OH)2(s) <=> Mg^2+(aq) + 2 OH^{-}(aq)}$$

Example 3.

What will happen to the molar solubility of $$\require{mhchem}\ce{AgCl}$$ when $$0.1 \text{ M } \require{mhchem}\ce{HNO3}$$ is added to a saturated solution of $$\require{mhchem}\ce{AgCl}$$?

A. Solubility will increase.

B. Solubility will remain unchanged.

C. Solubility will decrease.

Solution

B. Solubility will remain unchanged.

$$\require{mhchem}\ce{AgCl}$$ will not have its solubility impacted by addition of strong acid, since the chloride ion is a spectator ion (negligible base) that will not react with added acid. As such, the equilibrium position is unchanged.