How do we use #K_"sp"# values for solubility calculations....?

1 Answer
Nov 7, 2016

Answer:

Well, for a start write the solubility expression, and then do a bit of work...

Explanation:

Let's take a sparingly soluble salt, #PbCl_2#, for which #K_(sp)=5.89xx10^(−5)# at #25# #""^@C#.

We can write the solubility expression as follows:

#PbCl_2(s) rightleftharpoons Pb^(2+) + 2Cl^-#

#K_(sp)=5.89xx10^(−5)=[Pb^(2+)][Cl^-]^2#, and if we represent the #"solubility of lead chloride"# under these conditions as #S#, then #K_(sp)=5.89xx10^(−5)=Sxx2S^2=4S^3#. We could usually solve this for #S# fairly easily.

Now that's the solubility product. However, there are scenarios when the ion product is greater than #K_(sp)#, i.e. if we did the reaction in a salt solution, where #[Cl^-]# was artificially high. Because this ion product #># #K_"sp"#, #"lead chloride"# would precipitate from solution until the ion product #-=K_"sp"#. Such a process is normally called #"salting out"#, and if the metal were precious, say a salt of gold, or rhodium, or iridium, we would want to #"salt out"# the metal species so as maximize recovery of the metal.

See https://socratic.org/questions/what-is-ksp-in-chemistry for another treatment of the problem.