Solubility Equilibrium
Chapter 10
What happens to an insoluble salt when it
is added to some water?
What would you see?
What does it mean to be “insoluble”?
“Insoluble” is a lie. All solids dissolve, but
some dissolve a very, very tiny amount.
What would this look like if you could see
it?
Why are the arrows wrong?
Pb2+(aq)
I-(aq)
PbI2(s)
I-aq
Because we have no change in the amount
of solid sitting at the bottom of the water,
yet ions are still forming, it has to be
concluded that ions are also precipitating
out at an equal rate. What do we call this?
Pb2+(aq)
I-(aq)
PbI2(s)
I-aq
Equilibrium of an insoluble, or not very
soluble, salt is called Solubility Product, so
the constant is the Solubility Product
Equilibrium Constant, or Ksp.
Pb2+(aq)
I-(aq)
PbI2(s)
I-aq
For the reaction PbI2(s)  Pb2+(aq) + 2I-(aq)
Ksp = [Pb2+][I-]2
Why is [I-] squared?
What is missing from this K? Why?
Where there’s a K, there’s a Q.
When would you use Q?
What would happen if Q is less than K?
What would happen if Q is greater than K?
What would happen if Q is equal to K?
For the reaction PbI2(s)  Pb2+(aq) + 2I-(aq)
According to LeChatelier’s Principle, what
would happen if you added NaI?
What is the name for this situation?
What would happen to the value of K?
What would happen to the solubility of
PbI2?
Let’s Prove it:
For the reaction PbI2(s)  Pb2+(aq) + 2I-(aq)
Ksp = 9.8 x 10-9
What is the [Pb2+]?
Let’s Prove it:
For the reaction PbI2(s)  Pb2+(aq) + 2I-(aq)
Ksp = 9.8 x 10-9
If 0.20 M NaI is added, what is the new [Pb2+]?
Ksp will not only tell us how much of an
insoluble salt would dissolve (ionize), but
how much of an ion would need to be
added to get a precipitate. This could
even be used to precipitate out one ion
from a mixture of many for isolation of that
ion - often called selective precipitate.
Quick example of selective precipitation:
A beaker has mixture of Na+ ions and Ag+
ions. How can I- ions be used to separate
the Na+ from the Ag+?
Not quick example of selective
precipitation:
A beaker has mixture of Cl- ions, Br- ions,
and I- ions. If [Ag+] was added, in what
order would the ions precipitate?
Ksp AgCl = 1.8 x 10-10
Ksp AgBr = 3.3 x 10-13
Ksp AgI = 1.5 x 10-16
Not quick example of selective
precipitation:
A beaker has mixture of Cl- ions, Br- ions,
and I- ions, each present at 0.0010 M. How
much [Ag+] would be needed to precipitate
each ion?
Ksp AgCl = 1.8 x 10-10
Ksp AgBr = 3.3 x 10-13
Ksp AgI = 1.5 x 10-16
Ksp AgCl = 1.8 x 10-10
Ksp AgBr = 3.3 x 10-13
Ksp AgI = 1.5 x 10-16