# Solubity Generic (Don't Use)

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Chemical Equilibrium – Solubility The solubility of a substance is dependent on the forces holding the crystal together (the lattice energy) and the solvent acting on these forces. For now, we will consider only water as the solvent. As the solid dissolves, water molecules surround the ions in the solution by a process called hydration. During hydration, energy is released. The extent to which the energy of hydration is greater than the lattice energy determines the solubility. The equilibrium involved in the solubility of a substance is: MaXb (s) ( a Mn+ (aq) + b Xn- (aq) ( a general equation) where M is the cation of change n+ and X is the anion of charge n-. The subscripts of the salt become the coefficients for the ions. The equilibrium mass action expression would be: Ksp = [Mn+]a [Xn-]b Notice that only the aqueous ions are part of the mass action expression. Solids are never included. Also notice that the coefficient for each ion becomes the exponent in the mass action expression. There are four common types of solubility equilibria problems: solubility in pure water (will a precipitate form or finding the molar solubility), solubility in the presence of a common ion and selective precipitation. We will look at each of these in turn. Solubility in pure water Numerical problems involving solubility of a compound in pure water may fall into two categories: (1) determining whether a precipitate will form when two aqueous solutions are mixed or (2) determining the actual concentration of ions in a saturated solution. Let’s examine the mixing of two solutions. The problem might read as follows: Suppose 100.0 mL of a 0.0100 M CaCl2 solution and 200.0 mL of a 0.0200 M Na3PO4 solution were mixed. Would a precipitate form? To tackle this problem, you must do a stoichiometry problem first. Start