When an ionic compound is placed in water, it dissolves until the solution becomes saturated. This means the rate of dissolution equals the rate at which the dissolved ions recombine to form the solid. This dynamic state is known as solubility equilibrium, which primarily concerns ionic compounds that are only slightly soluble in water. Understanding this equilibrium allows chemists to quantify how much of a substance can dissolve and predict whether a solid will form. These calculations rely on two related quantities: solubility (a measurable concentration) and the solubility product (a fixed constant for that compound at a specific temperature).
Defining Solubility and the Solubility Product
Molar solubility (\(S\)) refers to the maximum amount of a solute that can dissolve in a solvent, measured in moles per liter (M). It is the concentration of the dissolved compound in a saturated solution. It directly tells us the concentration of the ions released into the solution, based on the compound’s chemical formula.
The solubility product constant (\(K_{sp}\)) is the equilibrium constant for the dissolution of a solid ionic compound into its constituent ions in an aqueous solution. For a generic compound \(A_x B_y\), the equilibrium is \(A_x B_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)\). The \(K_{sp}\) expression is defined as the product of the concentrations of the dissolved ions, each raised to the power of its stoichiometric coefficient.
The solid reactant \(A_x B_y(s)\) is never included in the \(K_{sp}\) expression because the concentration of a pure solid is considered constant. For the generic salt \(A_x B_y\), the expression is \(K_{sp} = [A^{y+}]^x [B^{x-}]^y\). The exponents \(x\) and \(y\) are the coefficients from the balanced dissolution equation.
If the molar solubility of \(A_x B_y\) is \(S\), then the concentration of the cation \(A^{y+}\) is \(xS\), and the concentration of the anion \(B^{x-}\) is \(yS\). Substituting these terms into the \(K_{sp}\) expression yields \(K_{sp} = (xS)^x (yS)^y\). This algebraic relationship is used to convert between the two values.
Calculating the Solubility Product from Solubility
Calculating \(K_{sp}\) requires knowing the molar solubility (\(S\)) and the compound’s stoichiometry. This process involves writing the balanced dissolution equation, expressing ion concentrations in terms of \(S\), and substituting these into the \(K_{sp}\) equation.
Consider silver chloride (\(\text{AgCl}\)), which has a simple 1:1 stoichiometry. The dissolution equilibrium is \(\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)\). If the molar solubility is \(S\), then \([\text{Ag}^+]\) is \(S\) and \([\text{Cl}^-]\) is \(S\). The \(K_{sp}\) expression is \(K_{sp} = [\text{Ag}^+][\text{Cl}^-]\), which simplifies to \(K_{sp} = S^2\).
If \(S\) for \(\text{AgCl}\) is \(1.34 \times 10^{-5}\) M at \(25^\circ\text{C}\), the calculation is \(K_{sp} = (1.34 \times 10^{-5})^2\). This results in a \(K_{sp}\) value of \(1.80 \times 10^{-10}\). This method applies universally, but the algebraic relationship changes with the compound’s formula.
For a compound with 1:2 stoichiometry, such as calcium fluoride (\(\text{CaF}_2\)), the dissolution is \(\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)\). If the molar solubility is \(S\), then \([\text{Ca}^{2+}]\) is \(S\), but \([\text{F}^-]\) is \(2S\) due to the coefficient of two. The \(K_{sp}\) expression is \(K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2\).
Substituting the solubility terms yields \(K_{sp} = (S)(2S)^2\), which simplifies to \(K_{sp} = 4S^3\). If \(S\) for \(\text{CaF}_2\) is measured as \(3.3 \times 10^{-4}\) M, we calculate \(K_{sp} = 4(3.3 \times 10^{-4})^3\). This results in a \(K_{sp}\) of \(1.4 \times 10^{-10}\), demonstrating how stoichiometry alters the mathematical relationship between \(S\) and \(K_{sp}\).
Calculating Solubility from the Solubility Product
Determining the molar solubility (\(S\)) when the \(K_{sp}\) is known is often the more common problem. This calculation involves setting up the same algebraic relationship between \(K_{sp}\) and \(S\) and then solving for \(S\). The steps rely heavily on the stoichiometry of the salt.
Consider lead(II) iodide, \(\text{PbI}_2\), which has a known \(K_{sp}\) of \(8.7 \times 10^{-9}\) at \(25^\circ\text{C}\). The dissolution equilibrium is \(\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)\). Based on the 1:2 stoichiometry, \([\text{Pb}^{2+}]\) is \(S\), and \([\text{I}^-]\) is \(2S\).
The solubility product expression \(K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2\) leads to the algebraic equation: \(K_{sp} = (S)(2S)^2\), which simplifies to \(K_{sp} = 4S^3\). To solve for \(S\), we rearrange the equation: \(S^3 = K_{sp} / 4\).
Using the given \(K_{sp}\) value, \(S^3 = (8.7 \times 10^{-9}) / 4\), which equals \(2.175 \times 10^{-9}\). Taking the cube root of this result gives a molar solubility \(S\) of \(1.29 \times 10^{-3}\) M.
This calculated value means that a maximum of \(1.29 \times 10^{-3}\) moles of \(\text{PbI}_2\) can dissolve in one liter of water at \(25^\circ\text{C}\). For salts with higher coefficients, such as aluminum hydroxide, \(\text{Al}(\text{OH})_3\), the relationship is \(K_{sp} = 27S^4\), requiring taking the fourth root to find \(S\).
Using the Solubility Product to Predict Precipitation
The solubility product constant, \(K_{sp}\), applies only when the solution is saturated and at equilibrium. To determine whether a solid ionic compound will form (precipitate) when two solutions are mixed, we use the Ion Product (\(Q\)). \(Q\) is calculated using the exact same mathematical expression as \(K_{sp}\), but it uses the initial concentrations of the ions present in the solution.
Comparing \(Q\) to the known \(K_{sp}\) predicts the solution’s behavior. This comparison reveals three possible conditions:
- If \(Q < K_{sp}[/latex], the solution is unsaturated, and more solid can dissolve.
- If [latex]Q = K_{sp}\), the solution is saturated, and the system is at equilibrium with no net change.
- If \(Q > K_{sp}\), the solution is supersaturated, meaning ion concentrations exceed the limit defined by \(K_{sp}\). The system will spontaneously shift to form a solid precipitate until \(Q\) equals \(K_{sp}\).