Professor Dave again, I wanna tell you about selective precipitation.
We’ve learned quite a bit about ionic compounds that are water insoluble, and we know that
certain ions, when in saturated solution, will combine to form a solid precipitate.
But there are situations when multiple different precipitates could potentially form within
a solution.
In such situations, we will often observe a phenomenon called selective precipitation.
Let’s learn what this means by looking at an example.
The silver halides, like silver chloride, silver bromide, and silver iodide, are compounds
that are quite water insoluble, and from these diminishing Ksp values, we can see that as
the size of the ion increases from chloride to bromide to iodide, the solubility of the
silver halide decreases.
There are technological applications for mixtures of these silver halides, such as photochromic
eyeglass lenses, but it is interesting to note that if silver ions are added to a mixture
of chloride, bromide, and iodide ions, the three silver halides will not precipitate
in equal amounts.
In fact, the precipitate that forms will be almost exclusively silver iodide.
This is a phenomenon known as selective precipitation.
The reason this occurs, is because silver iodide is much less soluble in water than
the other silver halides.
If combining ions in solution in such a way that more than one precipitate is possible,
and if one of those precipitates has a Ksp value that is more than two orders of magnitude
smaller than the others, almost all of that less soluble compound will precipitate before
any of the more soluble compounds can begin to precipitate.
For this reason, in order to get a homogeneous mixture of the three silver halides, it is
the case that a mixture of the halide ions must be slowly added to a solution of silver ions.
This way, the few halide ions that are in solution will precipitate without competition.
We can see selective precipitation in action even when the halide ion concentrations are unequal.
Let’s say we have a solution that is 0.001 molar in potassium iodide and 0.1 molar in
potassium chloride, and silver nitrate is slowly added, introducing silver ions to the solution.
We know that a precipitate, whether silver chloride or silver iodide, will form when
Q equals Ksp, so let’s take the solubility product expressions for each possible precipitate,
silver chloride and silver iodide, and find the concentration of silver ion upon which
each precipitate will form.
All we have to do is divide the Ksp value by the concentration of the respective halide ion.
Silver chloride can begin to precipitate when the silver ion concentration is 1.8 x 10^-9
molar, while silver iodide can begin to precipitate when the silver ion concentration is only
1.5 x 10^-13 molar.
So we can clearly see that silver iodide will precipitate at a much lower silver concentration
than silver chloride, which means the silver iodide will certainly precipitate first, and
will probably precipitate completely before any silver chloride will form.
Selective precipitation is an interesting phenomenon, and represents yet another application
of the solubility product.