Solubility describes the maximum amount of a material that can dissolve in a specific amount of liquid at a given temperature. While solubility can be expressed in various units, the most useful measure for chemical calculations is molar solubility. This metric expresses the concentration of the dissolved substance in moles per liter of solution (molarity).
Defining Molar Solubility and the \(K_{sp}\) Constant
Molar solubility (\(s\)) quantifies the amount of solute, in moles, that dissolves to form exactly one liter of a saturated solution. A saturated solution exists at dynamic equilibrium, where the rate of dissolution equals the rate of precipitation. The value of \(s\) represents the concentration of the dissolved compound at this equilibrium point, typically expressed in \(\text{mol/L}\).
This equilibrium condition for a sparingly soluble ionic compound is described by the Solubility Product Constant (\(K_{sp}\)). The \(K_{sp}\) is an equilibrium constant that applies specifically to the dissolution of ionic solids in water. Like all equilibrium constants, the \(K_{sp}\) value remains constant at a specific temperature, providing a fixed reference point for maximum solubility.
The relationship between \(s\) and \(K_{sp}\) is derived directly from the balanced chemical equation for the compound’s dissociation. For a simple 1:1 salt, \(K_{sp}\) is the product of the ion concentrations. For compounds with complex stoichiometry, the concentration of each ion must be raised to the power of its stoichiometric coefficient. The stoichiometric ratio connects the molar solubility (\(s\)) to the numerical value of the \(K_{sp}\) constant.
Calculating Molar Solubility from \(K_{sp}\)
Calculating molar solubility (\(s\)) from a known \(K_{sp}\) value requires careful attention to the compound’s dissociation stoichiometry. The process begins by writing the balanced equation for the solid dissolving into its constituent ions. For example, calcium fluoride, \(\text{CaF}_2\), dissociates into one \(\text{Ca}^{2+}\) ion and two \(\text{F}^-\) ions.
Next, set up the \(K_{sp}\) expression, which for \(\text{CaF}_2\) is \(K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2\). Introduce the variable \(s\) to represent the molar solubility of \(\text{CaF}_2\). Since one mole of \(\text{CaF}_2\) yields one mole of \(\text{Ca}^{2+}\) and two moles of \(\text{F}^{-}\), the equilibrium concentrations are \([\text{Ca}^{2+}] = s\) and \([\text{F}^{-}] = 2s\).
Substituting these expressions into the \(K_{sp}\) equation gives the algebraic relationship: \(K_{sp} = (s)(2s)^2\), which simplifies to \(K_{sp} = 4s^3\). This formula links the constant to the molar solubility for any compound with a 1:2 stoichiometry. To solve for \(s\), divide the known \(K_{sp}\) value by four and take the cube root. For instance, if the \(K_{sp}\) for \(\text{CaF}_2\) is \(3.9 \times 10^{-11}\), solving the equation yields \(s = 2.1 \times 10^{-4} \text{ mol/L}\).
Calculating \(K_{sp}\) from Molar Solubility
The inverse calculation involves starting with a known molar solubility (\(s\)) to determine the corresponding \(K_{sp}\) constant. This calculation begins with the balanced dissociation equation and the \(K_{sp}\) expression. In this case, the value of \(s\) is already a numerical quantity, typically provided in \(\text{mol/L}\).
The known molar solubility (\(s\)) is used along with the reaction stoichiometry to find the specific equilibrium concentration of each ion in the solution. For example, if the molar solubility of silver sulfate, \(\text{Ag}_2\text{SO}_4\), is \(1.4 \times 10^{-2} \text{ mol/L}\), the dissociation (\(\text{Ag}_2\text{SO}_4(s) \rightleftharpoons 2\text{Ag}^{+} + \text{SO}_4^{2-}\)) dictates the ion concentrations. The sulfate ion concentration is \(s\), but the silver ion concentration is twice that amount due to the \(2:1\) ratio (\(2.8 \times 10^{-2} \text{ mol/L}\)).
These numerical concentrations are substituted into the \(K_{sp}\) expression, which for silver sulfate is \(K_{sp} = [\text{Ag}^{+}]^2[\text{SO}_4^{2-}]\). The calculation becomes \(K_{sp} = (2.8 \times 10^{-2})^2(1.4 \times 10^{-2})\), yielding the final \(K_{sp}\) value.
The Common Ion Effect on Solubility
Molar solubility calculations assume the compound dissolves in pure water, but other substances can alter the result. The common ion effect describes the reduction in solubility when an ion common to the equilibrium is added from a different source. This effect is a direct application of Le Châtelier’s Principle, where a system at equilibrium shifts to counteract applied stress.
If a soluble salt containing a product ion is introduced, the ion concentration increases, stressing the system. To re-establish equilibrium, the system shifts the dissolution reaction back toward the solid reactants, causing more of the sparingly soluble salt to precipitate. This results in a lower overall molar solubility (\(s\)) compared to dissolution in pure water. The mathematical calculation of this new solubility requires incorporating the initial concentration of the common ion into the \(K_{sp}\) expression, making the algebraic solution more involved.