THE COMMON-ION EFFECT
The solubility of a salt is considerably reduced when it is dissolved in a solution that already contains
one of its ions, rather than in a pure solvent. For example, if a salt such as CaF 2 is dissolved in a
solution already containing Ca2+ ions, the dissociation equilibrium will shift toward the production
of the solid salt. This reduction in solubility, called the common-ion effect, is another example of Le
Châtelier’s principle.
Example: The Ksp of AgI in aqueous solution is 1 × 10−16 mol/L. If a 1 × 10−5 M solution of AgNO 3 is
saturated with AgI, what will be the final concentration of the iodide ion?
Solution: The concentration of Ag+ in the original AgNO 3 solution will be 1 × 10−5 mol/L. After AgI
is added to saturation, the iodide concentration can be found by the formula:
where x = molar solubility of AgI. Since the Ksp of AgI is so small compared with the
initial concentration of Ag+ ions, we can assume that dissociation of AgI will not
contribute much to the final concentration of Ag+ ions; that is, that (1 × 10−5 + x) ≈ 1 ×
10 −5. Solving for x:If the AgI had been dissolved in pure water, the concentration of both Ag+ and I− would have been 1 ×
10 −8 mol/L. The presence of the common ion, silver, at a concentration one thousand times higher
than what it would normally be in a silver iodide solution has reduced the iodide concentration to
one thousandth of what it would have been otherwise. An additional 1 × 10−11 mol/L of silver will, of
course, dissolve in solution along with the iodide ion, but this will not significantly affect the final
silver concentration, which is much higher, thus validating the approximation we made.