Solubility Equilibria

Sodium chloride is considered soluble because large quantities of it will dissolve in water, but when lead chloride is added to water, only a small amount dissolves, while the rest remains insoluble.

The undissolved solid coexists with the lead and chloride ions that are in solution. Some of the solid lead chloride continues to dissolve, while some of the ions in solution recombine to form a precipitate.

When the rate of dissolution equals the rate of precipitation, a solubility equilibrium is established.

The equilibrium constant can be calculated from the equilibrium concentrations of the ions according to the dissolution reaction—where lead chloride dissociates into one lead and two chloride ions.

So, the equilibrium constant is given by the molar concentration of lead ions multiplied by the square of the molar concentration of chloride ions. Because the concentration of the solid lead chloride remains constant, it is excluded from the calculation.

This equilibrium constant is called the solubility product, denoted by K_sp. At 25 °C, the K_sp of lead chloride is 1.17 × 10⁻⁵.

The value of K_sp represents the extent to which a compound can dissolve to form a saturated aqueous solution. At a given temperature, the K_sp of a compound is constant.

The solubility of a compound in moles per liter, known as the molar solubility, is often used to express the concentration of the dissolved solid in a saturated solution. The solubility of a compound can vary depending on factors, such as the pH of the solution and if there are other ions present.

The molar solubility of a compound, x, can be calculated from its K_sp using an ICE table.

The initial concentrations of lead ions and chloride ions in the solution are zero.

At equilibrium, the molar concentration of lead ions is represented by x, while that of chloride ions is 2x.

Substituting into the equilibrium expression, the solubility product for lead chloride is equal to x times 2x², which equals 4x³.

As the K_sp for lead chloride is 1.17 × 10⁻⁵, x is solved to be 1.43 × 10⁻² molar.

For compounds that have the same dissociation stoichiometry, such as lead chloride and calcium fluoride, where 1 mole of each compound produces 3 moles of dissolved ions, the respective K_sp values can be used directly to compare their relative solubilities.