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Current time:0:00Total duration:10:08

AP.Chem:

SPQ‑5 (EU)

, SPQ‑5.B (LO)

, SPQ‑5.B.1 (EK)

, TRA‑7 (EU)

, TRA‑7.F (LO)

, TRA‑7.F.1 (EK)

- [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.