It’s Professor Dave, let’s learn about the common ion effect.
As we continue to learn about solubility and precipitates, we are beginning to understand
ways that we can predict or even control precipitation.
One such approach we must understand involves something called the common ion effect, so
let’s learn precisely what this means.
The common ion effect is an application of Le Chatelier’s principle, which we learned
all about in an earlier tutorial, so if you need to review, check that out now.
Otherwise, we know that under this principle, which describes the behavior of equilibria,
if we were to add an additional amount of one of the components in an equilibrium, the
equilibrium would shift towards the other side, using up some of that additional component
and creating more of the substances on the other side.
Let’s now look at an equilibrium between, say, silver iodide and the aqueous silver
and iodide ions.
If we are interested in forming a silver iodide precipitate, one way we can do this is to
add some potassium iodide.
Potassium iodide shares a common ion with silver iodide, that being the iodide ion.
If there is more iodide ion in solution, that will affect the equilibrium, and it will shift
left to use up the excess iodide ions.
If the equilibrium shifts left, that means more silver iodide precipitate will form,
so this is a method of promoting precipitation.
Let’s look at this more quantitatively.
Cadmium sulfide has a Ksp value of 1 x 10^-28.
If we place some cadmium sulfide in a 0.01 molar cadmium bromide solution, what will
be the molar solubility of the cadmium sulfide?
So first we must recognize that the common ion in this situation is cadmium, as it is
present in both compounds.
Let’s then set up an icebox for the cadmium sulfide equilibrium.
We know the initial concentration of cadmium ion is 0.01, from the cadmium bromide, and
there is no sulfide at first.
Then, a tiny amount of the cadmium sulfide will also dissolve, so we must add a change
of X to each initial concentration, to get our equilibrium concentrations, which will
be 0.01 + X, and X.
We can then plug these terms into the Ksp expression, like so.
Now, since the Ksp is so incredibly tiny when compared with the initial cadmium ion concentration,
we can consider X, or the additional concentration of cadmium that will come from cadmium sulfide,
to be negligible.
In other words, the equilibrium concentration of cadmium will certainly round down to 0.01,
so let’s just call it 0.01, which simplifies the expression considerably.
Solving for X, since the sulfide concentration will be equal to the molar solubility of cadmium
sulfide, we get a molar solubility of 1.0 x 10^-26.
Without the cadium bromide, the molar solubility would simply have been the square root of
the solubility product, which would have been significantly larger than the value we just
calculated, so the presence of the cadmium ions in solution has reduced the molar solubility
and will thus promote precipitation.
And that wraps up an introduction to the common ion effect.