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.
  1. 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.
  2. 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.
  3. 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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