Molar solubility (\(S\)) is the maximum amount of solute, expressed in moles per liter (mol/L), that can dissolve in a solvent to form a saturated solution. Determining \(S\) relies directly on the compound’s solubility product constant (\(K_{sp}\)). The \(K_{sp}\) links the solid compound and its dissolved ions at equilibrium, serving as the starting point for calculating molar solubility. The calculations are algebraic and vary based on how the solid compound dissociates into its constituent ions.
Defining the Solubility Product Constant (\(K_{sp}\))
The solubility product constant (\(K_{sp}\)) is an equilibrium constant that specifically describes the dissolution of a sparingly soluble ionic solid in an aqueous solution. When an ionic compound dissolves, it establishes a dynamic equilibrium between the undissolved solid and its dissociated ions in the solution. This constant is a measure of the extent to which a solid dissolves, where a smaller \(K_{sp}\) value indicates lower solubility.
To establish the \(K_{sp}\) expression, first write a balanced chemical equation for the dissolution process. For a generic ionic solid, \(AB\), the equilibrium reaction is \(AB(s) \rightleftharpoons A^+(aq) + B^-(aq)\). The \(K_{sp}\) is defined as the product of the dissolved ion concentrations, each raised to the power of its stoichiometric coefficient. For the \(AB\) example, the expression is \(K_{sp} = [A^+][B^-]\). The concentration of the pure solid reactant, \(AB(s)\), is omitted from the expression.
The Step-by-Step Calculation Procedure
The process for finding molar solubility from a known \(K_{sp}\) value can be illustrated using a simple salt with a 1:1 ion ratio, such as silver chloride (\(AgCl\)). The calculation begins by writing the balanced dissolution equation: \(AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)\). Next, the corresponding \(K_{sp}\) expression is formulated as \(K_{sp} = [Ag^+][Cl^-]\).
In a saturated solution of \(AgCl\), every mole of \(AgCl\) that dissolves yields one mole of \(Ag^+\) and one mole of \(Cl^-\) ions. If \(S\) represents the molar solubility, the equilibrium concentration of both ions is equal to \(S\): \([Ag^+] = S\) and \([Cl^-] = S\).
Substituting these \(S\) expressions into the \(K_{sp}\) equation results in the algebraic relationship \(K_{sp} = (S)(S)\), or \(K_{sp} = S^2\). The final step involves solving for \(S\) by taking the square root of the \(K_{sp}\) value: \(S = \sqrt{K_{sp}}\).
Applying the Calculation to Varied Stoichiometries
When the ionic solid does not dissociate in a simple 1:1 ratio, the calculation requires an algebraic adjustment to account for the varied stoichiometry. Consider a salt like calcium fluoride, \(CaF_2\), which dissociates according to the equation \(CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)\). The balanced equation shows that for every mole of \(CaF_2\) that dissolves, one mole of \(Ca^{2+}\) ions and two moles of \(F^-\) ions are produced.
If \(S\) is the molar solubility of \(CaF_2\), the equilibrium concentrations of the ions must reflect the \(1:2\) ratio: \([Ca^{2+}] = S\) and \([F^-] = 2S\). The \(K_{sp}\) expression remains the product of the ion concentrations raised to their stoichiometric powers: \(K_{sp} = [Ca^{2+}][F^-]^2\). Substituting the \(S\) terms into this expression yields \(K_{sp} = (S)(2S)^2\).
The expression must be simplified algebraically before solving for \(S\). Since \((2S)^2\) becomes \(4S^2\), the final equation is \(K_{sp} = 4S^3\). To find \(S\), the \(K_{sp}\) value is first divided by four, and then the cube root of the result is calculated. This adjustment for stoichiometric coefficients is necessary for any non-1:1 salt.