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

Nov 7, 2016

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

#### Explanation:

Let's take a sparingly soluble salt, $P b C {l}_{2}$, for which K_(sp)=5.89xx10^(−5) at $25$ ""^@C.

We can write the solubility expression as follows:

$P b C {l}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s P {b}^{2 +} + 2 C {l}^{-}$

K_(sp)=5.89xx10^(−5)=[Pb^(2+)][Cl^-]^2, and if we represent the $\text{solubility of lead chloride}$ 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 ${K}_{s p}$, i.e. if we did the reaction in a salt solution, where $\left[C {l}^{-}\right]$ was artificially high. Because this ion product $>$ ${K}_{\text{sp}}$, $\text{lead chloride}$ would precipitate from solution until the ion product $\equiv {K}_{\text{sp}}$. Such a process is normally called $\text{salting out}$, and if the metal were precious, say a salt of gold, or rhodium, or iridium, we would want to $\text{salt out}$ 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.