Why does adding sodium chloride to a saturated solution of lead chloride reduce the solubility of #PbCl_2#?

1 Answer
Mar 19, 2016

Answer:

This is a standard problem in solubility equilibria; the solubility of #PbCl_2# depends on the concentration of ions FROM ALL SOURCES.

Explanation:

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.