Question

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

Answer 1

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.