Reading Notes
for Chapter 15
These are Dr. Bodwin's reading notes for Chapter 15 of "Chemistry
2e" from OpenStax.
I am using a local .pdf copy that was downloaded in May 2020.
Chapter
Summary:
Equilibrium shows up in every type of chemical interaction, although in
some cases the reaction appears to go to completion (large K) or there
appears to be no reaction (small K). As we continue to re-visit some of
the earlier topics with this new perspective, we can appreciate how
equilibrium influences many of the observations we may have made or
"rules" we may have learned. Again, the most important thing to
remember is that even though we may look at equilibria that apply to
specific types of reactions or specific conditions, they are all still
equilibria that exhibit the same behaviours and follow the same rules.
Solubility:
In Gen Chem I (Chapter 4), we learned some solubility rules that were
very binary; we had a list of "soluble" things and a list of
"insoluble" things. But real
solubility isn't quite so cleanly defined.
Consider barium fluoride. According to the rules, barium fluoride is
"insoluble". But what happens when some barium fluoride solid is put in
water?
BaF2(s)
<=>
Ba+2(aq) + 2 F-1(aq)
This equilibrium has an
equilibrium constant value of 2.4E-5; reactant-favored, but not
overwhelmingly so. The equilibrium constant expression for this
reactions is:
Ksp = [Ba+2] [F-1]2
{NOTE: pure solids do not appear in equilibrium constant expressions.}
This is called a solubility product constant
because it describes a solubility
reaction, and because the reactants do not appear in the equilibrium
constant expression, it is just the product
of the ion concentrations. This equilibrium behaves just like every
other equilibrium, although it is a little odd because there is no
denominator. What are the concentrations of barium and fluoride ions in
solution when solid barium fluoride is added to water? It's an
equilibrium problem, so let's put together a table...
|
BaF2(s) |
Ba+2(aq) |
F-1(aq) |
[ ]initial
|
XXXXXXX
|
0
M
|
0
M
|
[
]change
|
XXXXXXX
|
+x M
|
+2x M
|
[
]eq
|
XXXXXXX
|
x M
|
2x M
|
So we can plug in to the equilibrium constant expression:0
Ksp = 2.4E-5 = [x] [2x]2
= [Ba+2] [F-1]2
2.4E-5 = 4x3
x = 0.039
So when solid barium fluoride reaches equilibrium with its ions in
solution, the concentration of barium ions will be 0.039M and the
concentration of fluoride ions will be 0.078M.
What if we turn that reaction around?
Ba+2(aq) + 2 F-1(aq)
<=> BaF2(s)
That looks like a precipitation
reaction, with an equilibrium constant of (1 / 2.4E-5) = 42,000. We could turn invert the equilibrium
constant expression and use this new equilibrium constant value, OR we
could just use the Ksp expression and constant because in an
equilibrium is doesn't matter if we start with "reactants" or if we
start with "products", we always reach the same equilibrium condition.
If I combine 50.0mL of 0.100M barium nitrate solution with 50.0mL of
0.150M sodium fluoride solution, what will the concentrations of barium
ions and fluoride ions be after precipitation has occurred. It's an
equilibrium problem so... yep, set up a table:
|
BaF2(s) |
Ba+2(aq) |
F-1(aq) |
[ ]initial
|
XXXXXXX
|
0.0500
M
(dilute to the "initial" concentration after mixing)
|
0.0750
M
(dilute to the "initial" concentration after mixing)
|
[
]change
|
XXXXXXX
|
-x M
(now we're losing some of
this)
|
-2x M
(and losing twice as much of
this)
|
[
]eq
|
XXXXXXX
|
(0.0500 - x)
M
|
(0.0750 -
2x) M
|
So we can plug in to the equilibrium constant expression:
Ksp = 2.4E-5 = [0.0500 - x]
[0.0750 - 2x]2 = [Ba+2] [F-1]2
2.4E-5 = [0.0500 - x] (0.005625 - 0.3000x + 4x2)
2.4E-5 = -4x3 + 0.5000x2 + 0.020625x +
0.00028125
Hmm, that one's getting a little thick... solving 3rd-order polynomials
is a little "beyond the scope of this course". Assumptions won't help a
whole lot... there are certainly ways to solve this problem, but I
don't know that it's important to dive into that much math in this
course at this time. If you'd like to work through an example that's a
little less of a math mountain, try a 1:1 salt like barium carbonate or
silver bromide. Solubility products are found in Appendix J of your
textbook.
Lewis
Acid-Base Reactions and Formation Constants:
Most of our discussion of acids and bases focussed on Bronsted-Lowry
acids and bases. We can expand the acid-base concept by using the Lewis
definitions (electron acceptors and donors) and use the Lewis acid-base
concept to describe formation of coordination complexes.
Coordination complexes are important in variety of systems, including
biological systems that contain metals. The Lewis base (electron donor)
is called a ligand
(rhymes with "BIG end"). Formation of these complexes are described by
a formation reaction, with a formation constant, Kf. Again, it's
an equilibrium constant that does everything other equilibrium
constants do, it's just applied to a specific type of reaction, as
shown on page 840 of your textboook.
The reverse reaction is a dissociation or decomposition, and has a
corresponding Kd equilibrium. Same idea, just the reverse direction.
Manipulating and Combining Equilibrium Constants:
Equilibria are all around us and often combine in interesting ways. To
understand these manipulated and combined equilibria, we need to
remember a few key relationships. These are all logical relationships
that are usually easier to understand if we remember that an
equilibrium constant is "products over reactants.
- Reversing an equilibrium - We've already seen this a few times.
If we reverse the direction of an equilibrium, the reactants become
products and the products become reactants. If we reverse the identity
of products and reactants, we have to invert the equilibrium constant
expression. This makes sense; if an equilibrium is product-favored and
we reverse it, the "new" reaction will be reactant-favored.
- Multiplying an equilibrium by a constant - This can be helpful in
working with reaction mechanism. If we multiply a reaction by a
constant, the equilibrium constant is raised to that power. Doube a
reaction? Square the equilibrium constant. That's also why it is always
important to include a balanced chemical equation with your analysis;
there can be more than 1 "right" answer depending upon exactly how you
balance the chemical equation.
- Adding equilibria together - This is another mechanism tool. If equilibria are added together, their equilibrium constants are multiplied together.
These manipulations are important for studying mechanisms, but they
also belp when we're trying to find an equilibrium constant. If I need
the equilibrium constant for the reaction "A <=> B" but I can
only find "B <=> A", I can just invert the equilibrium.
Similarly, if I can find "2A <=> 2B" or "0.5A <=> 0.5B", I
can just raise the equilibrium constant to an appropriate power.
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