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.