- [Instructor] The presence of a common ion
can affect a solubility equilibrium.
For example,
let's say we have a saturated solution of lead II chloride.
Lead II chloride is a white solid,
so here's the white solid on the bottom of the beaker.
And the solid's at equilibrium with the ions in solution.
So that would be Pb2+ and Cl-.
Notice how the mole ratio is one-to-two of Pb2+ to 2Cl-.
So if we have two Pb2+ ions in our diagram,
there should be twice as many chloride anions.
At equilibrium, the rate of dissolution is equal to
the rate of precipitation.
Therefore, the concentration of ions in solution
remains constant.
So our system is at equilibrium.
And let's add some solid potassium chloride.
Potassium chloride is a soluble salt.
So it will dissociate and turn into K+ and Cl- in solution.
Adding a source of chloride anion means
the system is no longer at equilibrium.
So let me write in here,
not at equilibrium at that moment in time.
So the system was at equilibrium
and a stress was added to the system.
In this case, the stress was increased chloride anion.
So there's an increase in the concentration of Cl-.
According to Le Chatelier's principle,
the system will move in the direction
that decreases the stress.
So if the stress is
increased concentration of chloride anion,
the system will move to the left
to get rid of some of that extra chloride anion.
When the system moves to the left,
Pb2+ ions will combine with chloride anions to form PbCl2.
And we can see that down here in the diagram.
So imagine,
say this Pb2+ ion combined with these two chloride anions
to form some more of the white solid.
Looking at the third diagram,
the amount of white solid has increased
from the second diagram,
and we've lost this Pb2+ ion and these two chloride anions.
And the amount of our precipitate PbCl2
will keep forming until equilibrium is reached.
Let's just say this third diagram,
it does represent the system at equilibrium.
So I'll write on here, at equilibrium.
This is an example of the common ion effect.
For this problem, the common ion is the chloride anion,
because there were two sources of it.
One was from the disillusion of PbCl2.
If we had dissolved some solid to make a saturated solution,
the source of these chloride anions would be from PbCl2.
And the second source is from the added KCl,
which of course dissolved to form chloride anion.
So the chloride anion is the common ion.
And we use Le Chatelier's principle
to predict the system will move to the left
to get rid of the extra chloride anion.
When the system moved to the left,
we formed more of the solid PbCl2.
And that's why this amount got bigger over here.
So if we compare the first diagram with the third diagram,
the first diagram has more of
the lead II chloride in solution.
And the third diagram has less of it.
Therefore, the addition of the common ion of chloride anion,
that decreased the solubility of lead II chloride.
So the common ion effect says that
the solubility of a slightly soluble salt,
like lead II chloride,
is decreased by the presence of a common ion.
Another way to think about this is using
the reaction quotient, Q.
For the diagram on the left, we're at equilibrium.
Therefore the reaction quotient Qsp is equal to
the Ksp value for lead II chloride,
which means the system is at equilibrium.
Adding chloride anion increases the value for Qsp.
So now Qsp is greater than Ksp
and the system is not at equilibrium.
In order to decrease the value for Q,
the system needs to move to the left.
And the system will continue to move to the left
until Qsp is equal to Ksp again
and the system is at equilibrium.
A shift to the left means an increase in the amount of PbCl2
which therefore decreases the solubility of PbCl2.
But it doesn't change the value for Ksp.
Ksp for PbCl2 stays the same at the same temperature.
Next, let's see how the presence of a common ion affects
the molar solubility of lead II chloride.
And to do that,
let's calculate the molar solubility of lead II chloride
at 25 degrees Celsius
in a solution that is 0.10 molar in potassium chloride.
The Ksp value for lead II chloride at 25 degrees Celsius is
1.7 times 10 to the negative fifth.
To help us calculate the molar solubility,
we're going to use an ICE table,
where I stands for the initial concentration,
C is the change in concentration,
and E is the equilibrium concentration.
First, let's say that
none of the lead II chloride has dissolved yet.
And if that's true,
the concentration of lead II plus ions would be zero,
and the concentration of chloride anions
from the lead II chloride would also be zero.
However, there's another source of chloride anions,
and that's because our solution is 0.10 molar in KCl.
KCl is a soluble salt.
So KCl associates completely to turn to K+ and Cl-.
Therefore, if the concentration of KCl is 0.10 molar,
that's also the concentration of Cl- from the KCl.
So we can add here plus 0.10 molar.
And think about that as being from our KCl.
So there are two sources.
There's going to be two sources of chloride anions here.
And so the Cl anion and is our common ion.
The other source of chloride anion is PbCl2
when it dissolves.
So some of the PbCl2 will dissolve,
we don't know how much, so I like to write -X in here.
And if some of that dissolves,
the mole ratio of PbCl2 to Pb2+ is a one-to-one mole ratio.
So if we're losing X for PbCl2, we're gaining X for Pb2+.
And looking at our mole ratios,
if we're gaining X for Pb2+
and it's a one-to-two mole ratio,
we would write in here +2X for the chloride anion.
So for the equilibrium concentration of Pb2+,
it would be zero plus X, or just X.
And for the equilibrium concentration of the chloride anion,
it would be 0.10 plus 2X.
So the 0.10 came from the potassium chloride,
and the 2X came from the dissolution of lead II chloride.
Next, we need to write a Ksp expression,
which we can get from the disillusion equation.
So, Ksp is equal to the concentration of lead II plus ions
raised to the first power
times the concentration of chloride anions.
And since there's a two as a coefficient
in the balanced equation,
we need to raise that concentration to the second power.
Pure solids are left out of
equilibrium constant expressions.
So we don't write anything for PbCl2.
Next, we plug in our equilibrium concentrations.
So for lead II plus, the equilibrium concentration is X.
And for the chloride anion,
the equilibrium concentration is 0.10 plus 2X.
We also need to plug in the Ksp value for lead II chloride.
Here we have the Ksp value plugged in.
X and 0.10 plus 2X.
And let's think about 0.10 plus 2X for a second here.
With a very low value for Ksp,
1.7 times 10 to the negative fifth,
that means that not very much of the PbCl2 will dissolve.
And if that's true, X is a pretty small number.
And if X is a small number, 2X is also pretty small.
So we're going to make an approximation
and say that 0.10 plus a pretty small number
is approximately equal to just 0.10.
And that's going to make the math easier on us.
So instead of writing 0.10 plus 2X squared,
we just have 0.10 squared.
Solving for X, we find that X is equal to 0.0017,
which we could just write as
1.7 times 10 to the negative third molar.
It's okay to write molar here
because this X value represents
the equilibrium concentration of Pb2+.
And if that's the equilibrium concentration of Pb2+,
that's also the concentration of lead to chloride
that dissolved.
So this number, this concentration,
is the molar solubility of lead II chloride
in a solution at 25 degrees
where the solution is 0.10 molar in KCl.
Most textbooks leave this -X out of their ICE tables
because the concentration of a solid doesn't change.
I like to just leave it in here though
to remind me that X represents
the molar solubility of the slightly soluble salt.
Finally, if we'd calculate
the molar solubility of lead II chloride
without the presence of a common ion,
this 0.10 would have been gone from everything.
And doing the math that way, we would have found that
the molar solubility at 25 degrees Celsius
and using this value for the Ksp,
the molar solubility comes out to 0.016 molar.
So comparing these two smaller solubilities,
0.016 molar versus 0.0017,
that's approximately a factor of 10.
Therefore the addition of a common ion
decreased the solubility by approximately a factor of 10.
So doing the common ion effect in a quantitative way
also shows a decrease in
the solubility of a slightly soluble salt
because of the presence of a common ion.