The relationship between a solid ionic compound and its dissolved components in water is a core concept in chemistry. Chemists use molar solubility (\(S\)) and the solubility product constant (\(K_{sp}\)) to understand the extent of dissolution. Molar solubility (\(S\)) is the number of moles of solute that can dissolve in one liter of solution (mol/L). \(K_{sp}\) is an equilibrium constant that describes the balance between a solid and its dissolved ions in a saturated solution. Knowing the \(K_{sp}\) allows one to calculate the molar solubility, revealing how much of the substance will dissolve.
Deriving the Solubility Product Expression
The first step in connecting \(K_{sp}\) to molar solubility is establishing the correct chemical equilibrium expression. This begins with writing a balanced equation for the dissolution of the ionic solid in water. For any sparingly soluble ionic compound \(A_x B_y\), the solid is in equilibrium with its constituent ions: \(A_x B_y (s) \rightleftharpoons xA^{y+} (aq) + yB^{x-} (aq)\).
The solubility product constant, \(K_{sp}\), is derived from the law of mass action. The concentration of the pure solid reactant is considered constant and is excluded from the final expression. The resulting \(K_{sp}\) expression is the product of the dissolved ion concentrations, with each concentration raised to a power equal to its stoichiometric coefficient. The expression takes the form \(K_{sp} = [A^{y+}]^x [B^{x-}]^y\).
To relate \(K_{sp}\) to the molar solubility (\(S\)), the ion concentrations must be expressed in terms of \(S\). Since the solid dissolves to produce \(x\) moles of \(A^{y+}\) and \(y\) moles of \(B^{x-}\) per mole of \(A_x B_y\), the equilibrium concentrations are \([A^{y+}] = xS\) and \([B^{x-}] = yS\). Substituting these terms yields the general formula \(K_{sp} = (xS)^x (yS)^y\). This algebraic setup is used to solve for \(S\) when \(K_{sp}\) is known.
Calculation Method 1: Simple 1:1 Stoichiometry
The simplest calculation involves compounds where the cation and anion combine in a one-to-one (1:1) ratio, such as silver chloride (\(\text{AgCl}\)). When solid silver chloride dissolves, it establishes an equilibrium with its ions: \(\text{AgCl} (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq)\).
Because the stoichiometric coefficients for both ions are one, the solubility product expression is straightforward: \(K_{sp} = [\text{Ag}^+][\text{Cl}^-]\). If \(S\) represents the molar solubility of \(\text{AgCl}\), then both \([\text{Ag}^+]\) and \([\text{Cl}^-]\) are equal to \(S\). Substituting these values simplifies the relationship to \(K_{sp} = S^2\).
To find the molar solubility, the equation is rearranged to \(S = \sqrt{K_{sp}}\). For example, the \(K_{sp}\) for \(\text{AgCl}\) is approximately \(1.8 \times 10^{-10}\) at \(25^\circ\text{C}\). Plugging this value in, \(S = \sqrt{1.8 \times 10^{-10}}\), which results in a molar solubility of \(1.34 \times 10^{-5}\) mol/L. This method applies to all ionic compounds that dissociate into two ions in a 1:1 molar ratio, such as barium sulfate (\(\text{BaSO}_4\)) or lead(II) sulfate (\(\text{PbSO}_4\)).
Calculation Method 2: Complex Stoichiometry
Calculations become more involved when the ionic compound has a stoichiometry other than 1:1. A common example is calcium fluoride (\(\text{CaF}_2\)), which has a 1:2 ratio of cation to anion. The dissolution equilibrium is \(\text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq)\), meaning two fluoride ions are produced for every one calcium ion.
The \(K_{sp}\) expression for this reaction is \(K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2\). If \(S\) is the molar solubility of \(\text{CaF}_2\), the equilibrium concentration of calcium ions, \([\text{Ca}^{2+}]\), is \(S\). Because of the 2:1 stoichiometric ratio, the concentration of fluoride ions, \([\text{F}^-]\), must be \(2S\). Substituting these terms gives \(K_{sp} = (S)(2S)^2\).
Algebraic simplification shows that the \((2S)^2\) term expands to \(4S^2\). Multiplying this by \(S\) results in the final algebraic relationship: \(K_{sp} = 4S^3\). To calculate \(S\) from a known \(K_{sp}\), the equation is rearranged to solve for \(S^3\), giving \(S^3 = K_{sp}/4\). The molar solubility \(S\) is then found by taking the cube root: \(S = \sqrt[3]{K_{sp}/4}\).
Using a typical \(K_{sp}\) value for \(\text{CaF}_2\), such as \(1.7 \times 10^{-10}\), the calculation proceeds as \(S^3 = (1.7 \times 10^{-10})/4\), which equals \(4.25 \times 10^{-11}\). Taking the cube root yields a molar solubility of \(S = 3.49 \times 10^{-4}\) mol/L. This systematic approach is necessary for accurately determining the molar solubility of all non-1:1 ionic compounds.