Example. The solubility of AgCl in water is 0.000013 mole liter^-1 at 25C. What is the solubility-product constant, K_sp?

Solution. Since one mole of AgCI produces one mole each of Ag⁺ and Cl^-, the concentration of each ion in a saturated solution is

Hence,

Example. Would AgCl be more, or less, soluble in 0.01-molar sodium chloride than in pure water? What would its solubility be?

Solution. The solubility-constant expression tells us that the product of concentrations of Ag⁺ and Cl^- is fixed, no matter what the source of the ions. Hence, if we let x be the solubility of AgCl under these conditions, there will be 0.01 mole of Cl^- ion per liter from NaCI, and x moles per liter from AgCl. The total ion concentrations will be

The solubility-product constant will be the same as in the previous example; thus,

We can guess in advance that adding more chloride ion will make AgCl less soluble, because the product of [Ag⁺] and [Cl^-] remains unchanged, or that x will turn out to be less than 1.3 x 10^-5 mole liter^-1. We then can neglect x in comparison with 0.01 in the sum, and thereby avoid having to solve a quadratic equation. With this simplification,

If you compare this result with that of the preceding example, you will find that the solubility of AgCl in 0.01-molar NaCI is only one 76th its solubility in pure water.

This decrease in solubility of a slightly soluble salt in the presence of another salt that shares an ion is known as the common ion effect. We could have predicted it from Le Chatelier's principle: When a system at equilibrium is subjected to a stress, the equilibrium shifts in such a way as to partially relieve that stress.