We can write the solubility expression in this way:
PbCl_2(s) rightleftharpoons Pb^(2+) + 2Cl^-, is the reaction, AND
K_(sp) = [Pb^(2+)][Cl^-]^2 is the solubility expression; sp stands for solubility product, and K_(sp) has been measured for a variety of salts at various temperatures.
The solubility expression depends on just the concentration of the lead and chloride ions; from where the ions come makes no difference. So if [Cl^-] is artificially raised (by adding sodium chloride), it logically follows that [Pb^(2+)] must be correspondingly reduced in order for the solubility expression to be obeyed. The only way for [Pb^(2+)] to be reduced is for more PbCl_2 to precipitate. The same effect would pertain if I added lead nitrate, a soluble lead salt. [Pb^(2+)] would increase, and how would [Cl^-] evolve?
Such a phenomenon is known as "salting out". It pushes the solubility equilibrium to the left hand side as WRITTEN. IF you were trying to isolate a precious metal, i.e. gold or platinum or rhodium, you want to salt out the precious metal salt and leave little of its ions in solution.
"MORE ADVANCED TREATMENT,"
"2nd year analytical chemistry:"
Given what I have said above, it might seem that I could reduce [Pb^(2+)] to any desired level, simply by ramping up the concentration of chloride/halide ion. At very high concentrations of halide, however, soluble complex ions of lead may occur, i.e. [PbX_4]^(2-). For simple solubility equilibria, however, the formation of these complex ions may be ignored.
