The Solubility Product Constant (\(K_{sp}\)) measures the degree to which an ionic solid dissolves in water. It represents a specific type of chemical equilibrium that occurs when a solid compound is only slightly soluble in a solvent. The value of \(K_{sp}\) is an indicator of solubility; a larger value means the compound dissolves more readily, while a smaller value suggests it is less soluble. This constant is temperature-dependent. Solubility equilibrium centers on a saturated solution, which holds the maximum possible amount of dissolved solute at a given temperature. In this state, the solid dissolves into ions at the same rate that ions recombine to form the solid. For a generic ionic solid, \(A_x B_y\), the equilibrium is \(A_x B_y (s) \rightleftharpoons xA^{y+} (aq) + yB^{x-} (aq)\). The \(K_{sp}\) expression is \(K_{sp} = [A^{y+}]^x [B^{x-}]^y\), where ion concentrations are raised to the power of their stoichiometric coefficients.
Understanding the Solubility Equilibrium
A saturated solution exists when the system achieves a state of dynamic equilibrium between the undissolved solid and the dissociated ions in the solution. This equilibrium is the foundation for defining the Solubility Product Constant. Because the concentration of a pure solid is considered constant, it is conventionally excluded from the mathematical expression for \(K_{sp}\).
The resulting \(K_{sp}\) equation is the product of the molar concentrations of the ions, with each concentration term raised to the power of its stoichiometric coefficient. For example, in the dissolution of a salt like \(\text{AgCl}\), the equilibrium is \(\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)\), and the expression is \(K_{sp} = [\text{Ag}^+][\text{Cl}^-]\).
Calculating Ksp from Solubility Data
To calculate the \(K_{sp}\) of a compound, first write the balanced chemical equation for its dissolution in water, paying close attention to the stoichiometric coefficients. Next, the molar solubility, ‘\(s\)‘, must be determined, which represents the moles of the solid that dissolve per liter of solution. If the solubility is given in mass per volume, it must first be converted to molar solubility using the compound’s molar mass.
For a simple 1:1 salt like silver chloride (\(\text{AgCl}\)), the equilibrium concentrations of \(\text{Ag}^+\) and \(\text{Cl}^-\) are both equal to the molar solubility ‘\(s\)‘. Substituting these into the \(K_{sp}\) expression yields \(K_{sp} = (s)(s) = s^2\). If the measured molar solubility (\(s\)) of \(\text{AgCl}\) is \(1.3 \times 10^{-5}\) M, the \(K_{sp}\) is \(1.7 \times 10^{-10}\).
The calculation becomes more complex with salts that have non-unity coefficients, such as lead(II) chloride, \(\text{PbCl}_2\). This salt dissociates as \(\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq)\). If the molar solubility is ‘\(s\)‘, the concentration of \(\text{Pb}^{2+}\) is ‘\(s\)‘, but the concentration of \(\text{Cl}^-\) is \(2s\) due to the 1:2 stoichiometry. The \(K_{sp}\) expression is \(K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2\), which becomes \(K_{sp} = (s)(2s)^2 = 4s^3\).
For a salt like silver sulfide, \(\text{Ag}_2\text{S}\), the dissociation is \(\text{Ag}_2\text{S}(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{S}^{2-}(aq)\). The concentrations are \(2s\) for \(\text{Ag}^+\) and \(s\) for \(\text{S}^{2-}\). The \(K_{sp}\) expression is \(K_{sp} = [\text{Ag}^+]^2[\text{S}^{2-}]\), resulting in the relationship \(K_{sp} = (2s)^2(s) = 4s^3\). The correct use of stoichiometric coefficients to relate the ion concentrations to the molar solubility ‘\(s\)‘ is the most significant factor in accurately determining the \(K_{sp}\) value.
Determining Molar Solubility from Ksp
Determining molar solubility from a known \(K_{sp}\) value is the reverse of the previous calculation and is commonly used to predict how much of a compound will dissolve. This calculation requires rearranging the established \(K_{sp}\) expression to solve for the molar solubility, ‘\(s\)‘. This unknown ‘\(s\)‘ represents the concentration of the dissolved compound in moles per liter (M).
For a 1:2 salt where \(K_{sp} = 4s^3\), the algebraic rearrangement involves dividing the \(K_{sp}\) value by four and then taking the cube root of the result to find ‘\(s\)‘. For example, if a compound has a \(K_{sp}\) of \(3.9 \times 10^{-11}\), the molar solubility ‘\(s\)‘ would be the cube root of \((3.9 \times 10^{-11})/4\). The molar solubility represents the maximum concentration of the compound that can be dissolved before precipitation begins.
Using Ksp to Predict Solution Behavior
The calculated \(K_{sp}\) value serves as a reference point for predicting whether an ionic solid will precipitate out of a solution when two solutions are mixed. This prediction is made by calculating the Ion Reaction Quotient, or \(Q\). \(Q\) has the exact same mathematical form as the \(K_{sp}\) expression but uses the momentary, non-equilibrium concentrations of the ions present. The comparison between \(Q\) and \(K_{sp}\) yields three possible scenarios for the solution’s behavior.
Unsaturated Solution
If \(Q\) is less than \(K_{sp}\) (\(Q < K_{sp}[/latex]), the solution is unsaturated. This means more solute could dissolve, and no precipitate will form. The system will shift forward toward the dissolved ions to reach equilibrium.
Saturated Solution
If [latex]Q\) is equal to \(K_{sp}\) (\(Q = K_{sp}\)), the solution is saturated and is precisely at equilibrium. The rates of dissolution and precipitation are equal, and no net change occurs.
Supersaturated Solution
If \(Q\) is greater than \(K_{sp}\) (\(Q > K_{sp}\)), the solution is supersaturated, and a precipitate will form. The system must shift backward toward the solid reactants to lower the ion concentrations until \(Q\) equals \(K_{sp}\).