KEY2CHEM

Solubility and $$K_{sp}$$ Values

The equilibrium constant to describe the solubility of an ionic compound is the solubility product constant, $$K_{sp}$$. For the ionic compound $$\require{mhchem}\ce{MX2(s)}$$, the $$K_{sp}$$ value describes the equation $$\require{mhchem}\ce{MX2(s) <=> M^2+(aq) + 2 X^{-}(aq)}$$. For this equation, $$K_{sp} = [\require{mhchem}\ce{M^2+}][\require{mhchem}\ce{X^{-}}]^2$$

The more soluble the ionic compound, the more ionic compound will dissolve and the larger the $$K_{sp}$$ value.

Example 1.

Silver chloride ($$\require{mhchem}\ce{AgCl}$$) is less soluble in pure water than is sodium chloride ($$\require{mhchem}\ce{NaCl}$$). What is true about the $$K_{sp}$$ values of these two ionic compounds?

A. $$K_{sp }\; \require{mhchem}\ce{AgCl} = K_{sp} \;\require{mhchem}\ce{NaCl}$$

B. $$K_{sp} \;\require{mhchem}\ce{AgCl} < K_{sp} \;\require{mhchem}\ce{NaCl}$$

C. $$K_{sp} \;\require{mhchem}\ce{AgCl} > K_{sp} \;\require{mhchem}\ce{NaCl}$$

Solution

B. $$K_{sp} \; \require{mhchem}\ce{AgCl} < K_{sp} \;\require{mhchem}\ce{NaCl}$$

A less soluble compound will have a smaller $$K_{sp}$$ value, since less of the compound is dissociated into its ions at equilibrium.

Example 2.

The following pictures represent three different ionic compounds (MX, MY, and MZ) at equilibrium. The open spheres are the cation and the shaded spheres are the respective anion. Which compound has the largest $$K_{sp}$$?

A. MX

B. MY

C. MZ

Solution

C. MZ

Ionic compound MZ has the largest number of dissolved ions at equilibrium, so it will have the largest $$K_{sp}$$ value.

Example 3.

The molar solubility of $$\require{mhchem}\ce{Mg(OH)2}$$ in pure water at $$25^\circ C$$ is $$5.4 \times 10 ^{-4}$$. What is the $$K_{sp}$$ of $$\require{mhchem}\ce{Mg(OH)2}$$ at this temperature?

A. $$6.3 \times 10 ^{-10}$$

B. $$5.4 \times 10^{-4}$$

C. $$1.6 \times 10^{-3}$$

Solution

A. $$6.3 \times 10 ^{-10}$$

Let $$S =$$ molar solubility of $$\require{mhchem}\ce{Mg(OH)2}$$, so $$[\require{mhchem}\ce{Mg^2+}] = S$$ and $$[\require{mhchem}\ce{OH-}] = 2S$$

$$\require{mhchem}\ce{Mg(OH)2(s) <=> Mg^2+(aq) + 2 OH^{-}(aq)}$$

$$K_{sp} = [\require{mhchem}\ce{Mg^2+}][\require{mhchem}\ce{OH-}]^2 = S\times (2S)^2 = 4S^3 = 4(5.4 \times 10^{-4})^3 = 6.3 \times 10 ^{-10}$$