The dissolution of a slightly soluble ionic compound in water establishes a state of solubility equilibrium. This dynamic process involves the solid dissolving into its constituent ions and reforming from the ions back into the solid state. A solution is considered saturated when these two opposing processes occur at equal rates, meaning the maximum amount of solute is dissolved at a specific temperature. The solubility product constant, \(K_{sp}\), is an equilibrium constant used to quantify this solubility for sparingly soluble salts.
Defining the Solubility Product Expression
To define the solubility of a compound, first write the balanced chemical equation for its dissolution. For a general ionic solid \(M_x A_y\), it dissociates as: \(M_x A_y (s) \rightleftharpoons x M^{y+} (aq) + y A^{x-} (aq)\). The solid reactant is not included in the equilibrium expression because the concentration of a pure solid remains constant. The \(K_{sp}\) expression is defined only by the concentrations of the dissolved ions, each raised to the power of its stoichiometric coefficient.
For a simple 1:1 salt like silver chloride, \(\text{AgCl} (s) \rightleftharpoons \text{Ag}^{+} (aq) + \text{Cl}^{-} (aq)\), the expression is \(K_{sp} = [\text{Ag}^{+}][\text{Cl}^{-}]\). When the stoichiometry is different, the coefficients become exponents in the expression. For example, calcium fluoride, \(\text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^{-} (aq)\), yields \(K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2\). Similarly, for silver phosphate, \(\text{Ag}_3 \text{PO}_4 (s) \rightleftharpoons 3\text{Ag}^{+} (aq) + \text{PO}_4^{3-} (aq)\), the expression is \(K_{sp} = [\text{Ag}^{+}]^3[\text{PO}_4^{3-}]\).
Calculating \(K_{sp}\) from Molar Solubility
The molar solubility (\(s\)) is the concentration of the dissolved solid in a saturated solution, typically measured in moles per liter (\(\text{mol/L}\)). If \(s\) is known, it can be used to determine the \(K_{sp}\) value for the compound. This process requires a Reaction, Initial, Change, Equilibrium (RICE) table to relate the solubility of the solid to the equilibrium concentrations of its ions.
Consider the example of lead(II) chloride, \(\text{PbCl}_2\), with a molar solubility of \(s\). The dissociation reaction is \(\text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Cl}^{-} (aq)\). If \(s\) moles dissolve per liter, the equilibrium concentration of \(\text{Pb}^{2+}\) is \(s\) and the concentration of \(\text{Cl}^{-}\) is \(2s\) due to the \(1:2\) stoichiometric ratio.
Substituting these equilibrium concentrations into the \(K_{sp}\) expression, \(K_{sp} = [\text{Pb}^{2+}][\text{Cl}^{-}]^2\), yields \(K_{sp} = (s)(2s)^2\). This simplifies mathematically to \(K_{sp} = 4s^3\). If the molar solubility, \(s\), is \(1.59 \times 10^{-2} \text{ M}\), the \(K_{sp}\) is calculated as \(4 \times (1.59 \times 10^{-2})^3\), giving a value of \(1.61 \times 10^{-5}\).
Calculating Molar Solubility from \(K_{sp}\)
Determining the molar solubility (\(s\)) when the \(K_{sp}\) value is known is the reverse calculation. This process involves setting up the RICE table and the \(K_{sp}\) expression in terms of the unknown \(s\), and then solving the resulting algebraic equation. For a salt like silver chromate, \(\text{Ag}_2 \text{CrO}_4\), the dissolution is \(\text{Ag}_2 \text{CrO}_4 (s) \rightleftharpoons 2\text{Ag}^{+} (aq) + \text{CrO}_4^{2-} (aq)\).
If \(s\) is the molar solubility, the equilibrium concentrations are \(2s\) for \(\text{Ag}^{+}\) and \(s\) for \(\text{CrO}_4^{2-}\). The \(K_{sp}\) expression is \(K_{sp} = [\text{Ag}^{+}]^2[\text{CrO}_4^{2-}]\), which simplifies to \(K_{sp} = 4s^3\). If the \(K_{sp}\) for silver chromate is \(1.1 \times 10^{-12}\), the calculation requires solving the equation \(1.1 \times 10^{-12} = 4s^3\) for \(s\).
First, divide the \(K_{sp}\) value by four, resulting in \(s^3 = 2.75 \times 10^{-13}\). To find \(s\), the cube root of this value must be calculated. The molar solubility, \(s\), is thus found to be \(6.50 \times 10^{-5} \text{ M}\).
Accounting for the Common Ion Effect
The solubility of a sparingly soluble salt is reduced when it is dissolved in a solution that already contains one of its constituent ions, a phenomenon known as the common ion effect. This effect is a direct application of Le Châtelier’s principle: adding a product (the common ion) shifts the equilibrium toward the reactants, decreasing the solid’s solubility. For example, the solubility of \(\text{PbCl}_2\) is lower in a sodium chloride (\(\text{NaCl}\)) solution, which provides \(\text{Cl}^{-}\) ions, than it is in pure water.
To calculate the solubility (\(s\)) under the common ion effect, the initial concentration of the common ion must be incorporated into the RICE table. If calculating the solubility of \(\text{PbCl}_2\) in a \(0.050 \text{ M}\) solution of \(\text{NaCl}\), the initial concentration of \(\text{Cl}^{-}\) is \(0.050 \text{ M}\). The equilibrium concentration of \(\text{Cl}^{-}\) then becomes \((0.050 + 2s)\), and the \(K_{sp}\) expression is \(K_{sp} = (s)(0.050 + 2s)^2\).
Because the \(K_{sp}\) for sparingly soluble salts is very small, the amount of additional ion contributed by the dissolving solid (\(2s\)) is often negligible compared to the initial concentration of the common ion (\(0.050 \text{ M}\)). This allows for the mathematical approximation that \(0.050 + 2s \approx 0.050\), simplifying the calculation. Using this approximation with the \(\text{PbCl}_2\) \(K_{sp}\) of \(1.61 \times 10^{-5}\) gives \(1.61 \times 10^{-5} \approx s(0.050)^2\), which yields a much smaller solubility value for \(s\) than in pure water.
Determining if Precipitation Will Occur
When two solutions containing different ions are mixed, it is necessary to determine if the resulting concentrations will exceed the solubility limit, leading to the formation of a solid precipitate. This determination relies on calculating the reaction quotient, \(Q_{sp}\). \(Q_{sp}\) is mathematically equivalent to the \(K_{sp}\) expression but uses the momentary, non-equilibrium concentrations of the ions immediately after mixing.
The \(Q_{sp}\) value is then compared to the known \(K_{sp}\) value for the potential precipitate. If \(Q_{sp} < K_{sp}[/latex], the solution is unsaturated, and no precipitation will occur because more solid could still dissolve. If [latex]Q_{sp} = K_{sp}[/latex], the solution is exactly saturated and at equilibrium, meaning no net change is currently taking place. The condition for precipitation is that [latex]Q_{sp}[/latex] must be greater than [latex]K_{sp}[/latex], indicating the solution is supersaturated. The excess ions will spontaneously combine to form the solid precipitate until the ion concentrations decrease enough for [latex]Q_{sp}[/latex] to equal [latex]K_{sp}[/latex], establishing a new equilibrium. For instance, if lead nitrate is mixed with sodium chloride, the concentrations of [latex]\text{Pb}^{2+}[/latex] and [latex]\text{Cl}^{-}[/latex] are used to calculate [latex]Q_{sp}[/latex] for [latex]\text{PbCl}_2[/latex]; if [latex]Q_{sp} > K_{sp}\), precipitation of lead(II) chloride will occur.