Solubility is the ability of an ionic compound to dissolve in a solvent, typically water, forming a solution where the substance separates into its constituent ions. For sparingly soluble compounds, the saturation point is reached quickly, establishing a dynamic equilibrium between the undissolved solid and the dissolved ions. The Solubility Product Constant (\(K_{sp}\)) is a quantitative measure that describes this equilibrium for sparingly soluble ionic compounds. Determining \(K_{sp}\) requires precise experimental measurement of the dissolved ions combined with chemical theory.
Understanding the Solubility Product Constant
The Solubility Product Constant is an equilibrium constant for the dissolution of a solid ionic compound in an aqueous solution at saturation. The general dissolution process involves a solid compound, \(\text{A}_x \text{B}_y\), in equilibrium with its aqueous ions, \(x\text{A}^{y+}\) and \(y\text{B}^{x-}\). The \(K_{sp}\) expression is the product of the ion concentrations, each raised to the power of its stoichiometric coefficient: \(K_{sp} = [\text{A}^{y+}]^x [\text{B}^{x-}]^y\). The solid reactant is omitted because its concentration is constant. A smaller \(K_{sp}\) value indicates lower solubility for the compound.
Molar solubility (\(s\)) is the concentration of the dissolved compound in moles per liter of solution (\(\text{mol}/\text{L}\)) at saturation. This molar solubility is directly related to the concentration of the individual ions in the \(K_{sp}\) expression through the stoichiometry of the dissolution reaction. For example, if \(s\) moles of the solid dissolve per liter, the resulting ion concentrations will be \(x \cdot s\) and \(y \cdot s\) for the cation and anion, respectively.
Determining Molar Solubility Through Measurement
To calculate \(K_{sp}\), the molar solubility (\(s\)) must first be determined experimentally. This requires creating a saturated solution where undissolved solid remains present to ensure equilibrium is reached. Maintaining a constant temperature is necessary because \(K_{sp}\) is temperature-dependent.
Titration
Titration is a common analytical technique used to find the concentration of dissolved ions. A solution of known concentration (the titrant) is incrementally added to react completely with one of the ions in the saturated solution. For example, the concentration of the hydroxide ion in a saturated calcium hydroxide solution can be determined by titrating it with a standardized acid solution. The volume and concentration of the titrant used at the equivalence point allow for the calculation of the ion’s molarity in the saturated solution.
Conductivity Measurement
Conductivity measurement relies on the fact that dissolved ions conduct electricity in a solution. A conductivity meter measures the electrical conductance of the saturated solution, which is directly proportional to the total concentration of the dissolved ions. This measured conductivity is then used in conjunction with the limiting molar conductivity of the ions to calculate the molar concentration, \(s\).
Gravimetric Analysis
Gravimetric analysis involves carefully evaporating a precise volume of the saturated solution to dryness. The mass of the recovered solid residue is then measured and converted into moles. Dividing the moles by the original solution volume yields the molar solubility, \(s\). Regardless of the method used, the goal is to obtain the exact concentration of the dissolved ionic species in \(\text{mol}/\text{L}\) at equilibrium.
Calculating the Solubility Constant from Experimental Data
Once the molar solubility (\(s\)) is experimentally determined, it is used to calculate the numerical value of the Solubility Product Constant. This calculation requires correctly applying the stoichiometric relationship between the dissolved solid and its constituent ions. The first step is writing the balanced dissolution equation and the corresponding \(K_{sp}\) expression.
1:1 Stoichiometry
For an ionic compound with 1:1 stoichiometry, such as silver bromide (\(\text{AgBr}\)), the dissolution is \(\text{AgBr}(s) \rightleftharpoons \text{Ag}^{+}(aq) + \text{Br}^{-}(aq)\). Since \(s\) equals the concentration of both ions, the \(K_{sp}\) expression simplifies to \(K_{sp} = [\text{Ag}^{+}][\text{Br}^{-}] = s^2\). Substituting the measured value of \(s\) yields the final \(K_{sp}\).
1:2 or 2:1 Stoichiometry
For a compound like calcium fluoride (\(\text{CaF}_2\)), which has 1:2 stoichiometry, the reaction is \(\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^{-}(aq)\). If \(s\) moles dissolve, the ion concentrations are \(s\) for \(\text{Ca}^{2+}\) and \(2s\) for \(\text{F}^{-}\). The \(K_{sp}\) expression is \(K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2\), which calculates to \(K_{sp} = (s) \cdot (2s)^2 = 4s^3\). This \(4s^3\) relationship also applies to 2:1 stoichiometry, such as silver sulfate (\(\text{Ag}_2\text{SO}_4\)). The final \(K_{sp}\) value is a unitless constant.
External Factors Affecting Solubility
The \(K_{sp}\) value is a true constant only under specific conditions, and external factors can significantly influence the measured solubility of an ionic compound.
Temperature
Temperature is the most significant factor, as \(K_{sp}\) is an equilibrium constant that changes with temperature. For most ionic solids, the dissolution process is endothermic, meaning increased temperature increases solubility and results in a larger \(K_{sp}\) value.
Common Ion Effect
The presence of a common ion decreases the solubility of the salt, an effect predicted by Le Châtelier’s Principle. If a salt containing one of the constituent ions is added to the saturated solution, the equilibrium shifts back toward the solid. This causes precipitation and lowers the maximum amount of the original salt that can dissolve.
pH
The pH of the solution affects the solubility of salts containing an ion that is the conjugate base of a weak acid. For example, carbonate salts (\(\text{CO}_3^{2-}\)), which are the conjugate bases of the weak acid \(\text{HCO}_3^{-}\), become more soluble in acidic solutions. The added \(\text{H}^{+}\) ions react with the \(\text{CO}_3^{2-}\) ions, effectively removing them from the equilibrium and causing more of the solid to dissolve.