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The solubility product constant (Ksp) is the equilibrium expression for a dissolution process of an ionic solid. For a sparingly soluble salt, Ksp = [ions]^stoichiometric coefficients at saturation. A small Ksp indicates low solubility; calculating Ksp from solubility data and using Ksp to predict precipitation are core skills.
Write Ksp expressions for various ionic compounds, calculate Ksp from solubility data, and use Ksp to find solubility. Verify with experimental data.
You already know that dissolving an ionic solid in water is a reversible process — at some point the rate of dissolution equals the rate of precipitation, and the solution is saturated. You also know how to write equilibrium expressions for reversible reactions. The solubility product constant, Ksp, is simply the equilibrium expression applied to that dissolution process. For a generic salt A_mB_n dissolving as A_mB_n(s) ⇌ mAn+(aq) + nBm−(aq), the Ksp expression is Ksp = [An+]m · [Bm−]n. The solid does not appear in the expression because its activity is constant — exactly the same rule you learned when writing equilibrium expressions for heterogeneous equilibria.
The numerical value of Ksp tells you how far the dissolution proceeds before equilibrium is reached. A very small Ksp (like 1.8 × 10⁻¹⁰ for AgCl) means the ions barely accumulate before the solution is saturated. A larger Ksp means more solid can dissolve. To calculate molar solubility from Ksp, define a variable *s* for the moles of salt that dissolve per liter, express each ion concentration in terms of *s* and its stoichiometric coefficient, substitute into the Ksp expression, and solve. For example, if PbCl₂ dissolves as PbCl₂ → Pb²⁺ + 2Cl⁻, then [Pb²⁺] = s and [Cl⁻] = 2s, giving Ksp = (s)(2s)² = 4s³. Solving for *s* yields the molar solubility.
The reverse calculation — finding Ksp from experimental solubility data — follows the same algebra in the opposite direction. If you know that 0.0015 mol of CaF₂ dissolves per liter, you can compute [Ca²⁺] = 0.0015 M and [F⁻] = 0.0030 M, then multiply: Ksp = (0.0015)(0.0030)² = 1.35 × 10⁻⁸. The stoichiometric coefficients matter enormously here — forgetting to double the fluoride concentration (or to square it in the Ksp expression) is the most common arithmetic error.
The real power of Ksp emerges when you compare the ion product Q to Ksp. The ion product is calculated identically to Ksp but uses the actual ion concentrations in solution rather than equilibrium values. If Q < Ksp, the solution is unsaturated and more solid can dissolve. If Q > Ksp, the solution is supersaturated and precipitation will occur until Q falls back to Ksp. This comparison is the foundation for predicting whether a precipitate forms when two solutions are mixed — a skill you will use extensively when you move on to precipitation equilibria and the common-ion effect.