Solubility defines the maximum amount of a substance (the solute) that can dissolve in a given amount of a solvent at a specific temperature. This property is measured in a saturated solution, which is the point where the solvent can hold no more dissolved material. Calculating solubility is fundamental to understanding chemical reactions, especially in fields like pharmaceutical development and environmental monitoring. Determining this limit allows scientists to predict how much of a compound will remain dissolved and how much will precipitate out of a solution.
Expressing Solubility as Concentration
Solubility is most practically expressed as a concentration, quantifying the amount of dissolved solute relative to the volume of the solution. The simplest measure involves dissolving a known mass of a substance in a measured volume of solvent, often expressed in grams of solute per liter of solution (g/L).
A more chemically useful expression is molarity, which is the number of moles of solute dissolved per liter of solution (mol/L). To convert the g/L value to molarity, one must use the solute’s molar mass. Dividing the mass in grams by the molar mass yields the number of moles, which is then divided by the solution volume in liters. Molar solubility is the standard unit for complex equilibrium calculations, as it directly relates to the number of individual particles in the solution.
Calculating Molar Solubility Using the Solubility Product Constant
For sparingly soluble salts, solubility is determined using the solubility product constant (\(K_{sp}\)). This constant is an equilibrium expression describing the balance between the undissolved solid and its dissociated ions in a saturated solution. The first step is writing the balanced dissolution equation, showing the solid breaking apart into its constituent aqueous ions.
For a generic salt, such as \(AB_2\), the dissolution equation is \(AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^{-}(aq)\). The \(K_{sp}\) expression is the product of the ion concentrations, each raised to the power of its stoichiometric coefficient: \(K_{sp} = [A^{2+}][B^{-}]^2\). Molar solubility, represented by ‘\(x\)‘, is defined as the concentration of the dissolved salt in moles per liter.
Substituting ‘\(x\)‘ into the \(K_{sp}\) expression accounts for the stoichiometry. In the \(AB_2\) example, if ‘\(x\)‘ moles dissolve, the concentration of \(A^{2+}\) is ‘\(x\)‘ and \(B^{-}\) is ‘\(2x\)‘. The expression becomes \(K_{sp} = (x)(2x)^2\), simplifying to \(K_{sp} = 4x^3\). Solving this algebraic equation for ‘\(x\)‘ yields the molar solubility of the salt in pure water.
This method allows for the prediction of solubility provided the \(K_{sp}\) value is known. For compounds with different stoichiometries, the \(K_{sp}\) expression and the resulting algebraic solution for ‘\(x\)‘ will change accordingly.
Adjusting Solubility Calculations for the Common Ion Effect
Solubility changes significantly when the solvent is not pure water but a solution already containing one of the salt’s constituent ions. This is known as the common ion effect, an application of Le Chatelier’s principle. Adding an existing ion shifts the dissolution equilibrium to the left, favoring solid formation and reducing the salt’s overall solubility.
To calculate the new, lower solubility, the initial concentration of the common ion must be incorporated into the \(K_{sp}\) expression. Consider the salt \(AB_2\) dissolving in a solution containing ion \(B^{-}\) at an initial concentration, \(C\). If molar solubility is ‘\(x\)‘, the concentration of \(A^{2+}\) remains ‘\(x\)‘, but the concentration of \(B^{-}\) is now \((C + 2x)\).
The \(K_{sp}\) expression is written as \(K_{sp} = (x)(C + 2x)^2\). Since \(K_{sp}\) values for sparingly soluble salts are very small, the amount of common ion added by the dissolving salt (‘\(2x\)‘) is negligible compared to the initial concentration, \(C\). This allows for the simplification \(K_{sp} \approx (x)(C)^2\). Solving for ‘\(x\)‘ in this simplified equation provides the new molar solubility, which confirms the reduction caused by the common ion effect.